From 2cf4e26075ee4599c708c896f366b0ace06c2107 Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Mon, 9 Mar 2026 10:25:30 +0100 Subject: [PATCH 1/8] P4 intro: minor beauty fixes --- .../solved_notebooks/P4_probmod/probmod_1-intro.jl | 11 +++++------ 1 file changed, 5 insertions(+), 6 deletions(-) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl index 5f0b0b0..7f6d394 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl @@ -78,9 +78,9 @@ x(t) &= \mathrm{cos}(α) \, v_0 \, t \, , ``` where the parameters are: -- $v_0$ is the initial velocity. -- $\alpha$ is the launching angle. -- $g = 9.8$ is the gravitational acceleration. +- $v_0$: the initial velocity. +- $\alpha$: the launching angle. +- $g = 9.8$: the gravitational acceleration. """ # ╔═╡ ee9cd05c-7fa5-4412-b3b4-32b0a1774c87 @@ -143,7 +143,7 @@ plot( ) # ╔═╡ 314ba0ec-015f-488d-ac63-780b1dbf5bc2 -md"The latter is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." +md"The third is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." # ╔═╡ c070f3c9-dd80-49f9-b279-c77f46316116 md"## Defining the model" @@ -358,7 +358,6 @@ md""" # ╠═╡ show_logs = false let # model definition (comment out for speed) - g = 9.8 @model function splash(x_f) v0 ~ Normal(5, 1) α ~ TriangularDist(0, pi/2) @@ -379,12 +378,12 @@ let # calculate interesting things prob_face_hit = mean([1.5 <= y_s <= 1.7 for y_s in y_s_samples]) + trajectories = [ x -> sin(α_samples[i]) / cos(α_samples[i]) * x - g / 2 * ( x / (cos(α_samples[i])*v0_samples[i]) )^2 for i in 1:length(v0_samples) ] - plot(trajectories, xlims = (0, 5), ylims = (0, 2), legend = false, color = :blue, alpha = 0.3, title = "Probability of getting hit ≈ $(round(100*prob_face_hit, digits = 3))%" From b84b24f6ba3940b271499b39568724e684fff8c7 Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Mon, 9 Mar 2026 10:37:24 +0100 Subject: [PATCH 2/8] P4 intro: copy changes to student notebook --- .../student_notebooks/P4_probmod/probmod_1-intro.jl | 11 +++++------ 1 file changed, 5 insertions(+), 6 deletions(-) diff --git a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl index 5f0b0b0..7f6d394 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl @@ -78,9 +78,9 @@ x(t) &= \mathrm{cos}(α) \, v_0 \, t \, , ``` where the parameters are: -- $v_0$ is the initial velocity. -- $\alpha$ is the launching angle. -- $g = 9.8$ is the gravitational acceleration. +- $v_0$: the initial velocity. +- $\alpha$: the launching angle. +- $g = 9.8$: the gravitational acceleration. """ # ╔═╡ ee9cd05c-7fa5-4412-b3b4-32b0a1774c87 @@ -143,7 +143,7 @@ plot( ) # ╔═╡ 314ba0ec-015f-488d-ac63-780b1dbf5bc2 -md"The latter is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." +md"The third is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." # ╔═╡ c070f3c9-dd80-49f9-b279-c77f46316116 md"## Defining the model" @@ -358,7 +358,6 @@ md""" # ╠═╡ show_logs = false let # model definition (comment out for speed) - g = 9.8 @model function splash(x_f) v0 ~ Normal(5, 1) α ~ TriangularDist(0, pi/2) @@ -379,12 +378,12 @@ let # calculate interesting things prob_face_hit = mean([1.5 <= y_s <= 1.7 for y_s in y_s_samples]) + trajectories = [ x -> sin(α_samples[i]) / cos(α_samples[i]) * x - g / 2 * ( x / (cos(α_samples[i])*v0_samples[i]) )^2 for i in 1:length(v0_samples) ] - plot(trajectories, xlims = (0, 5), ylims = (0, 2), legend = false, color = :blue, alpha = 0.3, title = "Probability of getting hit ≈ $(round(100*prob_face_hit, digits = 3))%" From 3ef3ee190ec9022f63d43a2d41f5696c8b219d5b Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Mon, 9 Mar 2026 11:21:50 +0100 Subject: [PATCH 3/8] minor notebook updates 2: basics - beauty fixes - minor chicken question update 3: advanced - beauty fixes - update petrigrowth question: return `splittable` in model, define logistic functions outside of model & add time as input 4: review - beauty fixes --- .../P4_probmod/probmod_2-basics.jl | 14 +++--- .../P4_probmod/probmod_3-advanced.jl | 49 +++++++++---------- .../P4_probmod/probmod_4-review.jl | 8 +-- .../P4_probmod/probmod_2-basics.jl | 8 +-- .../P4_probmod/probmod_3-advanced.jl | 40 ++++++--------- .../P4_probmod/probmod_4-review.jl | 8 +-- 6 files changed, 57 insertions(+), 70 deletions(-) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl index 5551d01..4a5cd63 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -256,13 +256,13 @@ md"## 4: Super eggs" # ╔═╡ 372436c4-262f-49b8-b1cf-626b043542bf md""" -When a chicken lays an egg, there's a small chance it contains two egg yolks. This chance, as well as the number of eggs a chicken lays per year, go down as the chicken gets older. +When a chicken lays an egg, there's a small chance it contains two egg yolks. This chance, as well as the number of eggs a chicken lays per year, goes down as the chicken gets older. """ # ╔═╡ 20111742-008a-44c3-8c27-62791cce3e1e md""" You can make the following assumptions -- The age $A$ of a random chicken (in years) is discrete and Uniformly distributed between 0 and 12. +- The age $A$ of a random chicken (in years) is discrete and Uniformly distributed between 0 and 10. - The number of eggs $N$ an $A$-year old chicken lays in a year is Poisson distributed with mean $300 - 20 \, A$. - The probability $P$ of an $A$-year old chicken's egg having a double yolk is distributed as a `Beta(1, 800 + 100*A)`. """ @@ -271,7 +271,7 @@ You can make the following assumptions md""" !!! questions 1. If someone hands you a random chicken, what is the probability it will lay 2 or more double eggs in a year? - 1. Compare the distributions of double eggs for 1-year old and 3-year old chickens. + 1. Compare the distributions of double eggs for 1-year old and 5-year old chickens. """ # ╔═╡ 6ebe676b-aee6-459b-bcda-de0fe14418a3 @@ -287,7 +287,7 @@ md""" # ╔═╡ 2b3d930f-53d9-4869-9e4a-86a1a681b9d8 @model function eggs() - A ~ DiscreteUniform(0, 12) + A ~ DiscreteUniform(0, 10) N ~ Poisson(300 - 20*A) P ~ Beta(1, 800 + 100*A) @@ -318,7 +318,7 @@ chicken_ages = chain_egg[:A]; histogram(sp_egg[chicken_ages .== 1], normalize = :probability) # ╔═╡ 9d9b4091-cf44-46e6-b81b-74663c9c8dfe -histogram(sp_egg[chicken_ages .== 3], normalize = :probability) +histogram(sp_egg[chicken_ages .== 5], normalize = :probability) # ╔═╡ ff06c070-50a2-43d0-9729-1c47e728ff52 md"## 5: Birthdays" @@ -331,7 +331,7 @@ Sometimes, people are born on the same day of the year. # ╔═╡ 01648616-bf50-4f66-82fc-eaae3de22a38 md""" !!! question - What is the probability that, in a class of 150 students, 3 or more share a birthday? + What is the probability that, in a class of 150 students, 3 or more share a birthday? Assume the probability for a person to be born is equal on every day of the year. """ # ╔═╡ 29438eef-a725-4873-9372-194f2f899f12 diff --git a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl index 874e41d..62a3ba3 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -7,7 +7,7 @@ using InteractiveUtils # This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error). macro bind(def, element) #! format: off - quote + return quote local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end local el = $(esc(element)) global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el) @@ -52,7 +52,7 @@ Bacteria follow **logistic growth**, and you can use the following assumptions: md""" !!! questions 1. Plot the prior distribution of P0. - 2. What is the probability your bacteria are in a splittable state 8 hours after inoculation? + 2. What is the probability that your bacteria are in a splittable state 8 hours after inoculation? 3. Plot 100 of the sampled logistic growth curves from 0 to 12 hours. """ @@ -84,41 +84,31 @@ rand(dropletdist, 10000) |> histogram # ╔═╡ 107533fc-b300-4b8d-bea2-a3aa6a37938d md"### 2: Probability" -# ╔═╡ 38c5b9cd-0baf-4a62-912e-4993614ddbf3 -md""" -!!! tip - You can `return` the logistic function estimated within the model and retrieve it using `generated_quantities` to make plotting easier later on. - - Remember: anonymous functions can be defined using `myfun = x -> ...` -""" - # ╔═╡ bde57599-1dca-41a4-94aa-498da72c2012 logistic(t, P0, r, K) = K / (1 + (K - P0)/P0 * exp(-r*t)) # ╔═╡ 976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 -@model function petrigrowth() - P0 ~ dropletdist +@model function petrigrowth(t) + P0 ~ MixtureModel([Poisson(10), Poisson(30)], [0.75, 0.25]) r ~ LogNormal(0.0, 0.3) K ~ Normal(1e5, 1e4) - logfun = t -> logistic(t, P0, r, K) - return logfun + num_bacteria = logistic(t, P0, r, K) + splittable = 1e4 < num_bacteria < 1e5 + return splittable end # ╔═╡ 4345b3dd-0731-4dd5-a319-627b4f91306e -petri_model = petrigrowth(); +petri_model = petrigrowth(8); # ╔═╡ 9d747eef-a883-49b3-acb7-f0d077a2b902 chain_petri = sample(petri_model, Prior(), 2000); # ╔═╡ ce8b53c3-0af1-4e9a-ad80-6fd7a2bf020b -logfuns = generated_quantities(petri_model, chain_petri); - -# ╔═╡ e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 -sp_petri = [logfun(8.0) for logfun in logfuns]; +sp_splittable = generated_quantities(petri_model, chain_petri); # ╔═╡ e3092915-a083-425e-8fa1-b7bf370abc8a -prob_splittable = mean((sp_petri .>= 1e4) .&& (sp_petri .<= 1e5)) +prob_splittable = mean(sp_splittable) # ╔═╡ e4f42f16-5cce-4fc2-aa01-8971f37c710e md"### 3: Plot" @@ -126,11 +116,17 @@ md"### 3: Plot" # ╔═╡ 6253f255-e92d-4b5a-9d89-d4fe384fc438 md""" !!! tip - Plotting a function `myfun` is as simple as entering `plot(myfun)`. The same syntax applies if `myfun` is a vector of functions. However, don't forget it was asked to plot only **100** functions. + Remember: anonymous functions can be defined using `myfun = x -> ...`, and can be visualized using `plot(myfun)`. The same syntax applies if `myfun` is a vector of functions. However, don't forget it was asked to plot only **100** functions. """ +# ╔═╡ 551e2c0c-faaf-476b-9021-7a10a75e92cf +logfuns = [ + t -> logistic(t, chain_petri[:P0][i], chain_petri[:r][i], chain_petri[:K][i]) + for i in 1:length(chain_petri) +] + # ╔═╡ 14a8bc88-e10e-42d1-a4c0-7f3ebc5512a9 -plot(logfuns[1:100], color = :violet, alpha = 0.3, label = false, xlims = (0, 12)) +plot(logfuns[1:100], color = :violet, alpha = 0.5, label = false, xlims = (0, 12)) # extra arguments are for making the plot prettier but are not required # ╔═╡ c8941726-9e81-47fc-9b7e-cb3b5c0c61ca @@ -138,7 +134,7 @@ md"# 2: Attraction" # ╔═╡ 431023df-3724-4325-b0ac-96dbf5e4fd20 md""" -Following a course on electromagnetism will teach one that computing the net force between 2 arbitrary shapes can be a terrifying task. Tragedy has it then, that this is a very general problem with application from making fusion reactors to space travel. We can ease the pain by turning it into a sampling problem. +Following a course on electromagnetism will teach one that computing the net force between 2 arbitrary shapes can be a terrifying task. Tragedy has it then, that this is a very general problem with applications from making fusion reactors to space travel. We can ease the pain by turning it into a sampling problem. We'll start in a humble manner and simulate **the gravitational force between 2 cubes**. Both cubes are size 1. The first cube is in [0, 1] x [0, 1] x [0, 1], and the second cube in [1.1, 2.1] x [0, 1] x [0, 1], as shown in the figure below. """ @@ -149,7 +145,7 @@ begin ye = [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1] ze = [0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1] - xe2 = [0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0] .+ 1.1 + xe2 = xe .+ 1.1 plot(xlims = (-0.5, 2.5), ylims = (-1, 2), zlims = (0, 3)) plot!(xe, ye, ze; color = :blue, linewidth = 0.5, label = "cube 1") @@ -241,16 +237,15 @@ histogram(estimator_sp) # ╠═6517d682-b524-42e3-8a13-78ed5c1ce0dc # ╠═60eabb68-7d65-4f78-97d9-b1e20c2dbd0a # ╟─107533fc-b300-4b8d-bea2-a3aa6a37938d -# ╟─38c5b9cd-0baf-4a62-912e-4993614ddbf3 # ╠═bde57599-1dca-41a4-94aa-498da72c2012 # ╠═976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 # ╠═4345b3dd-0731-4dd5-a319-627b4f91306e # ╠═9d747eef-a883-49b3-acb7-f0d077a2b902 # ╠═ce8b53c3-0af1-4e9a-ad80-6fd7a2bf020b -# ╠═e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 # ╠═e3092915-a083-425e-8fa1-b7bf370abc8a # ╟─e4f42f16-5cce-4fc2-aa01-8971f37c710e # ╟─6253f255-e92d-4b5a-9d89-d4fe384fc438 +# ╠═551e2c0c-faaf-476b-9021-7a10a75e92cf # ╠═14a8bc88-e10e-42d1-a4c0-7f3ebc5512a9 # ╟─c8941726-9e81-47fc-9b7e-cb3b5c0c61ca # ╟─431023df-3724-4325-b0ac-96dbf5e4fd20 diff --git a/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl b/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl index e23730f..44851cc 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -15,7 +15,7 @@ md"# Review exercise: Buffon's needles" # ╔═╡ a6435b94-f1af-4609-abfc-93d88730d023 md""" -A wise man once said: ["there is no greater joy than estimating π"](https://en.wikipedia.org/wiki/Approximations_of_%CF%80). Next to throwing darts at the unit square, another method to accomplish this is using [Buffon's needle problem](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem). +A wise man once said: ["there is no greater joy than estimating π"](https://en.wikipedia.org/wiki/Approximations_of_%CF%80). One method to accomplish this is using [Buffon's needle problem](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem). The experiment is as follows: consider a floor with parallel lines all a distance of 1 away from eachother. Now drop a needle of length 1 (and width ~0) on the floor with a **random position and angle**. What is the probability $P_{cross}$ that the needle will cross one of the lines? """ @@ -24,8 +24,8 @@ The experiment is as follows: consider a floor with parallel lines all a distanc md"The following image illustrates the problem (imagine $l$ = $t$ = 1) for two needles, where `a` crosses a line and `b` does not." # ╔═╡ c2ba2183-957f-4b94-a22e-0c2f3dd957ad -md""" -![Buffon's needles](https://upload.wikimedia.org/wikipedia/commons/thumb/5/58/Buffon_needle.svg/1920px-Buffon_needle.svg.png) +html""" + """ # ╔═╡ d71cf8dc-99e5-48e0-9abe-2242a6ccc30b diff --git a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl index 27f2d54..d96a552 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl @@ -218,13 +218,13 @@ md"## 4: Super eggs" # ╔═╡ 372436c4-262f-49b8-b1cf-626b043542bf md""" -When a chicken lays an egg, there's a small chance it contains two egg yolks. This chance, as well as the number of eggs a chicken lays per year, go down as the chicken gets older. +When a chicken lays an egg, there's a small chance it contains two egg yolks. This chance, as well as the number of eggs a chicken lays per year, goes down as the chicken gets older. """ # ╔═╡ 20111742-008a-44c3-8c27-62791cce3e1e md""" You can make the following assumptions -- The age $A$ of a random chicken (in years) is discrete and Uniformly distributed between 0 and 12. +- The age $A$ of a random chicken (in years) is discrete and Uniformly distributed between 0 and 10. - The number of eggs $N$ an $A$-year old chicken lays in a year is Poisson distributed with mean $300 - 20 \, A$. - The probability $P$ of an $A$-year old chicken's egg having a double yolk is distributed as a `Beta(1, 800 + 100*A)`. """ @@ -233,7 +233,7 @@ You can make the following assumptions md""" !!! questions 1. If someone hands you a random chicken, what is the probability it will lay 2 or more double eggs in a year? - 1. Compare the distributions of double eggs for 1-year old and 3-year old chickens. + 1. Compare the distributions of double eggs for 1-year old and 5-year old chickens. """ # ╔═╡ 98fcfcef-bf63-4eae-a325-ed4cef6d4fdd @@ -275,7 +275,7 @@ Sometimes, people are born on the same day of the year. # ╔═╡ 01648616-bf50-4f66-82fc-eaae3de22a38 md""" !!! question - What is the probability that, in a class of 150 students, 3 or more share a birthday? + What is the probability that, in a class of 150 students, 3 or more share a birthday? Assume the probability for a person to be born is equal on every day of the year. """ # ╔═╡ da44d18c-8be3-446e-a5c2-905af545d2c6 diff --git a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl index 8dac90d..0fee41b 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -7,7 +7,7 @@ using InteractiveUtils # This Pluto notebook uses @bind for interactivity. When running this notebook outside of Pluto, the following 'mock version' of @bind gives bound variables a default value (instead of an error). macro bind(def, element) #! format: off - quote + return quote local iv = try Base.loaded_modules[Base.PkgId(Base.UUID("6e696c72-6542-2067-7265-42206c756150"), "AbstractPlutoDingetjes")].Bonds.initial_value catch; b -> missing; end local el = $(esc(element)) global $(esc(def)) = Core.applicable(Base.get, el) ? Base.get(el) : iv(el) @@ -52,7 +52,7 @@ Bacteria follow **logistic growth**, and you can use the following assumptions: md""" !!! questions 1. Plot the prior distribution of P0. - 2. What is the probability your bacteria are in a splittable state 8 hours after inoculation? + 2. What is the probability that your bacteria are in a splittable state 8 hours after inoculation? 3. Plot 100 of the sampled logistic growth curves from 0 to 12 hours. """ @@ -77,25 +77,18 @@ dropletdist = missing; # ╔═╡ 107533fc-b300-4b8d-bea2-a3aa6a37938d md"### 2: Probability" -# ╔═╡ 38c5b9cd-0baf-4a62-912e-4993614ddbf3 -md""" -!!! tip - You can `return` the logistic function estimated within the model and retrieve it using `generated_quantities` to make plotting easier later on. - - Remember: anonymous functions can be defined using `myfun = x -> ...` -""" - # ╔═╡ bde57599-1dca-41a4-94aa-498da72c2012 logistic(t, P0, r, K) = K / (1 + (K - P0)/P0 * exp(-r*t)) # ╔═╡ 976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 -@model function petrigrowth() - P0 ~ dropletdist +@model function petrigrowth(t) + P0 ~ missing r ~ missing K ~ missing - logfun = missing - return logfun + num_bacteria = missing + splittable = missing + return splittable end # ╔═╡ 4345b3dd-0731-4dd5-a319-627b4f91306e @@ -104,11 +97,8 @@ petri_model = missing # ╔═╡ 9d747eef-a883-49b3-acb7-f0d077a2b902 chain_petri = missing -# ╔═╡ ce8b53c3-0af1-4e9a-ad80-6fd7a2bf020b -logfuns = missing - # ╔═╡ e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 -sp_petri = missing +sp_splittable = missing # ╔═╡ e3092915-a083-425e-8fa1-b7bf370abc8a prob_splittable = missing @@ -119,9 +109,12 @@ md"### 3: Plot" # ╔═╡ 2b39e0bd-91cd-456e-9053-7b8fe5a395fb md""" !!! tip - Plotting a function `myfun` is as simple as entering `plot(myfun)`. The same syntax applies if `myfun` is a vector of functions. However, don't forget it was asked to plot only **100** growth curves. + Remember: anonymous functions can be defined using `myfun = x -> ...`, and can be visualized using `plot(myfun)`. The same syntax applies if `myfun` is a vector of functions. However, don't forget it was asked to plot only **100** functions. """ +# ╔═╡ ac37dc49-ff1d-41c6-a2e3-74be2aaabd4c +logfuns = missing + # ╔═╡ d6c193d7-9fe2-413b-801f-ebc33c772ee9 missing # plot @@ -130,7 +123,7 @@ md"# 2: Attraction" # ╔═╡ 431023df-3724-4325-b0ac-96dbf5e4fd20 md""" -Following a course on electromagnetism will teach one that computing the net force between 2 arbitrary shapes can be a terrifying task. Tragedy has it then, that this is a very general problem with application from making fusion reactors to space travel. We can ease the pain by turning it into a sampling problem. +Following a course on electromagnetism will teach one that computing the net force between 2 arbitrary shapes can be a terrifying task. Tragedy has it then, that this is a very general problem with applications from making fusion reactors to space travel. We can ease the pain by turning it into a sampling problem. We'll start in a humble manner and simulate **the gravitational force between 2 cubes**. Both cubes are size 1. The first cube is in [0, 1] x [0, 1] x [0, 1], and the second cube in [1.1, 2.1] x [0, 1] x [0, 1], as shown in the figure below. """ @@ -141,7 +134,7 @@ begin ye = [0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1] ze = [0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1] - xe2 = [0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0] .+ 1.1 + xe2 = xe .+ 1.1 plot(xlims = (-0.5, 2.5), ylims = (-1, 2), zlims = (0, 3)) plot!(xe, ye, ze; color = :blue, linewidth = 0.5, label = "cube 1") @@ -228,16 +221,15 @@ missing # histogram # ╟─bebe6f6b-9603-4062-8685-363ed7ff3dcb # ╠═eb6dc4e7-e779-4bfc-b865-3defa3894181 # ╟─107533fc-b300-4b8d-bea2-a3aa6a37938d -# ╟─38c5b9cd-0baf-4a62-912e-4993614ddbf3 # ╠═bde57599-1dca-41a4-94aa-498da72c2012 # ╠═976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 # ╠═4345b3dd-0731-4dd5-a319-627b4f91306e # ╠═9d747eef-a883-49b3-acb7-f0d077a2b902 -# ╠═ce8b53c3-0af1-4e9a-ad80-6fd7a2bf020b # ╠═e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 # ╠═e3092915-a083-425e-8fa1-b7bf370abc8a # ╟─e4f42f16-5cce-4fc2-aa01-8971f37c710e # ╟─2b39e0bd-91cd-456e-9053-7b8fe5a395fb +# ╠═ac37dc49-ff1d-41c6-a2e3-74be2aaabd4c # ╠═d6c193d7-9fe2-413b-801f-ebc33c772ee9 # ╟─c8941726-9e81-47fc-9b7e-cb3b5c0c61ca # ╟─431023df-3724-4325-b0ac-96dbf5e4fd20 diff --git a/exercises/student_notebooks/P4_probmod/probmod_4-review.jl b/exercises/student_notebooks/P4_probmod/probmod_4-review.jl index f2b9423..c12d084 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_4-review.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_4-review.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -15,7 +15,7 @@ md"# Review exercise: Buffon's needles" # ╔═╡ a6435b94-f1af-4609-abfc-93d88730d023 md""" -A wise man once said: ["there is no greater joy than estimating π"](https://en.wikipedia.org/wiki/Approximations_of_%CF%80). Next to throwing darts at the unit square, another method to accomplish this is using [Buffon's needle problem](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem). +A wise man once said: ["there is no greater joy than estimating π"](https://en.wikipedia.org/wiki/Approximations_of_%CF%80). One method to accomplish this is using [Buffon's needle problem](https://en.wikipedia.org/wiki/Buffon%27s_needle_problem). The experiment is as follows: consider a floor with parallel lines all a distance of 1 away from eachother. Now drop a needle of length 1 (and width ~0) on the floor with a **random position and angle**. What is the probability $P_{cross}$ that the needle will cross one of the lines? """ @@ -24,8 +24,8 @@ The experiment is as follows: consider a floor with parallel lines all a distanc md"The following image illustrates the problem (imagine $l$ = $t$ = 1) for two needles, where `a` crosses a line and `b` does not." # ╔═╡ c2ba2183-957f-4b94-a22e-0c2f3dd957ad -md""" -![Buffon's needles](https://upload.wikimedia.org/wikipedia/commons/thumb/5/58/Buffon_needle.svg/1920px-Buffon_needle.svg.png) +html""" + """ # ╔═╡ d71cf8dc-99e5-48e0-9abe-2242a6ccc30b From 37b57432135de438918f1beab213c4403c68818e Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Mon, 9 Mar 2026 11:29:04 +0100 Subject: [PATCH 4/8] advanced notebook: less help in student notebooks --- .../P4_probmod/probmod_3-advanced.jl | 34 ++----------------- 1 file changed, 2 insertions(+), 32 deletions(-) diff --git a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl index 0fee41b..0e35844 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl @@ -72,7 +72,7 @@ md""" """ # ╔═╡ eb6dc4e7-e779-4bfc-b865-3defa3894181 -dropletdist = missing; +missing # plot # ╔═╡ 107533fc-b300-4b8d-bea2-a3aa6a37938d md"### 2: Probability" @@ -82,24 +82,10 @@ logistic(t, P0, r, K) = K / (1 + (K - P0)/P0 * exp(-r*t)) # ╔═╡ 976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 @model function petrigrowth(t) - P0 ~ missing - r ~ missing - K ~ missing - - num_bacteria = missing - splittable = missing + missing return splittable end -# ╔═╡ 4345b3dd-0731-4dd5-a319-627b4f91306e -petri_model = missing - -# ╔═╡ 9d747eef-a883-49b3-acb7-f0d077a2b902 -chain_petri = missing - -# ╔═╡ e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 -sp_splittable = missing - # ╔═╡ e3092915-a083-425e-8fa1-b7bf370abc8a prob_splittable = missing @@ -162,25 +148,13 @@ md"### 1: Net Force" # ╔═╡ 06d0e92f-1f07-41fd-b6ee-e94eb539627d @model function cubeforce() - x1 ~ missing - y1 ~ missing - z1 ~ missing - - x2 ~ missing - y2 ~ missing - z2 ~ missing - F = missing - return F end # ╔═╡ 43119060-bd35-4c4c-8831-d5c5df1d8dd5 cubemodel = cubeforce(); -# ╔═╡ d26ad946-2bb4-4383-860f-d601903ce1be -force_sp = missing - # ╔═╡ 7b117453-0875-4b41-a228-866c6c0a8208 force_average = missing @@ -223,9 +197,6 @@ missing # histogram # ╟─107533fc-b300-4b8d-bea2-a3aa6a37938d # ╠═bde57599-1dca-41a4-94aa-498da72c2012 # ╠═976cfb96-3196-4a61-bd5f-e4f5d24ba1e9 -# ╠═4345b3dd-0731-4dd5-a319-627b4f91306e -# ╠═9d747eef-a883-49b3-acb7-f0d077a2b902 -# ╠═e1d0c5d5-2b7a-4af4-a7ac-e1ca46e665c8 # ╠═e3092915-a083-425e-8fa1-b7bf370abc8a # ╟─e4f42f16-5cce-4fc2-aa01-8971f37c710e # ╟─2b39e0bd-91cd-456e-9053-7b8fe5a395fb @@ -239,7 +210,6 @@ missing # histogram # ╟─9be28327-d87f-4d50-bbcc-91e799f14dbf # ╠═06d0e92f-1f07-41fd-b6ee-e94eb539627d # ╠═43119060-bd35-4c4c-8831-d5c5df1d8dd5 -# ╠═d26ad946-2bb4-4383-860f-d601903ce1be # ╠═7b117453-0875-4b41-a228-866c6c0a8208 # ╠═b28cfbae-2fac-4b38-b234-53f71e381bcd # ╟─304d6052-6d3d-487c-8c01-dab259276d6f From 7fc61c4359e39f048f2fc92656cb46806e9677ac Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Thu, 12 Mar 2026 13:27:30 +0100 Subject: [PATCH 5/8] introduce alternative method for getting model output --- .../P4_probmod/probmod_1-intro.jl | 17 +++++++++++++++++ .../P4_probmod/probmod_1-intro.jl | 17 +++++++++++++++++ 2 files changed, 34 insertions(+) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl index 7f6d394..062b891 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl @@ -335,6 +335,19 @@ md""" We set a lot of keyword arguments of our plot to make the figure pretty here, but don't panic: this is still no programming course, so we will always clearly provide which one(s) you need if you need to plot something. If you ever do want to look for one yourself, you can find them on the function's help page (`?plot` or the `🔍 Live docs` in the bottom right corner). """ +# ╔═╡ 1de58e6a-4ee2-4d67-b97d-e63bbb9ed057 +md"## Sampling alternative" + +# ╔═╡ 83a8b9b6-72da-4aff-8364-37846b527a2f +md"It's also possible to skip the chain construction entirely and simply call your model as a function to get a sample of the model output. Note that it's not possible to get the corresponding values of the stochastic variables using this approach, so only use this method if you **only** care about the model output." + +# ╔═╡ 771b9cb1-d809-46de-9380-2db73ccd014d +# get a sample of the output 500 times +y_s_samples_alternative = [splash_model() for _ in 1:500] + +# ╔═╡ 0f1cccd8-937d-4fc2-8657-c9967db483b8 +histogram(y_s_samples_alternative, bins = 15) # yup, that's the same distribution + # ╔═╡ befb8f05-6ab7-4594-9cf1-71df47c1df03 md"## Essentials" @@ -456,6 +469,10 @@ end # ╟─00572e1a-d1e6-4c84-b1b2-b53404087dd8 # ╠═f550ed7a-1bba-40b5-9338-cdeb0520aae9 # ╟─3b465fc0-3a9d-4e67-8d4a-a14b3cd61f4e +# ╟─1de58e6a-4ee2-4d67-b97d-e63bbb9ed057 +# ╟─83a8b9b6-72da-4aff-8364-37846b527a2f +# ╠═771b9cb1-d809-46de-9380-2db73ccd014d +# ╠═0f1cccd8-937d-4fc2-8657-c9967db483b8 # ╟─befb8f05-6ab7-4594-9cf1-71df47c1df03 # ╟─52af5de0-b516-4947-84f7-dde63e4c513e # ╟─2ceede7e-8f0c-420c-bde2-f16b726cd204 diff --git a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl index 7f6d394..062b891 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl @@ -335,6 +335,19 @@ md""" We set a lot of keyword arguments of our plot to make the figure pretty here, but don't panic: this is still no programming course, so we will always clearly provide which one(s) you need if you need to plot something. If you ever do want to look for one yourself, you can find them on the function's help page (`?plot` or the `🔍 Live docs` in the bottom right corner). """ +# ╔═╡ 1de58e6a-4ee2-4d67-b97d-e63bbb9ed057 +md"## Sampling alternative" + +# ╔═╡ 83a8b9b6-72da-4aff-8364-37846b527a2f +md"It's also possible to skip the chain construction entirely and simply call your model as a function to get a sample of the model output. Note that it's not possible to get the corresponding values of the stochastic variables using this approach, so only use this method if you **only** care about the model output." + +# ╔═╡ 771b9cb1-d809-46de-9380-2db73ccd014d +# get a sample of the output 500 times +y_s_samples_alternative = [splash_model() for _ in 1:500] + +# ╔═╡ 0f1cccd8-937d-4fc2-8657-c9967db483b8 +histogram(y_s_samples_alternative, bins = 15) # yup, that's the same distribution + # ╔═╡ befb8f05-6ab7-4594-9cf1-71df47c1df03 md"## Essentials" @@ -456,6 +469,10 @@ end # ╟─00572e1a-d1e6-4c84-b1b2-b53404087dd8 # ╠═f550ed7a-1bba-40b5-9338-cdeb0520aae9 # ╟─3b465fc0-3a9d-4e67-8d4a-a14b3cd61f4e +# ╟─1de58e6a-4ee2-4d67-b97d-e63bbb9ed057 +# ╟─83a8b9b6-72da-4aff-8364-37846b527a2f +# ╠═771b9cb1-d809-46de-9380-2db73ccd014d +# ╠═0f1cccd8-937d-4fc2-8657-c9967db483b8 # ╟─befb8f05-6ab7-4594-9cf1-71df47c1df03 # ╟─52af5de0-b516-4947-84f7-dde63e4c513e # ╟─2ceede7e-8f0c-420c-bde2-f16b726cd204 From bb865e115f34e268f1bb75a1291ff8d85f130a29 Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Thu, 12 Mar 2026 13:30:19 +0100 Subject: [PATCH 6/8] added more explanation to gravity question after harsh comments by the assistent formerly known as jarjarbinks02 --- .../P4_probmod/probmod_3-advanced.jl | 71 +++++++++++++++---- .../P4_probmod/probmod_3-advanced.jl | 38 +++++----- 2 files changed, 75 insertions(+), 34 deletions(-) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl index 62a3ba3..80536cf 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl @@ -123,7 +123,7 @@ md""" logfuns = [ t -> logistic(t, chain_petri[:P0][i], chain_petri[:r][i], chain_petri[:K][i]) for i in 1:length(chain_petri) -] +]; # ╔═╡ 14a8bc88-e10e-42d1-a4c0-7f3ebc5512a9 plot(logfuns[1:100], color = :violet, alpha = 0.5, label = false, xlims = (0, 12)) @@ -154,18 +154,19 @@ end # ╔═╡ bb8e4cd6-6106-4d3f-9e35-0dfd8b2c45f5 md""" -The gravitational force can be estimated by **randomly sampling a point from both cubes** and using the formula for gravitational force between those points, ignoring all constants: +You can sample the gravitational force between both cubes by **randomly sampling a point from both cubes** and using the formula for gravitational force between those points, ignoring all constants: ```math -F = \frac{1}{r^2} +F = \frac{1}{r^2} \, . ``` """ # ╔═╡ 058d2a08-7d40-46c5-8c5b-2b5ce9d2eb18 md""" !!! questions - 1. What is the estimated net force between the two cubes? Is this the same as if you had treated the cubes as point masses? - 1. How many samples do you need to estimate this force reliably? Define a reliable estimator as one having a standard deviation of 0.1. Visualise the distribution of the estimator. + 1. What is the estimated total force between the two cubes? Is this the same as if you had treated the cubes as point masses? + 1. To estimate the net force, you take the average of $n$ samples. Of course, the result will vary every time you take a new sample: if you take the average of only 10 samples, your estimated total force will vary wildly! We can quantify how much the estimated total force varies by taking a **sample of sample averages** and then calculating the variance. Assuming you want your sample average to have a variance of no more than $0.01$, how many samples do you need? Visualise the distribution of the sample average. + - EXTRA: How does the [central limit theorem](https://en.wikipedia.org/wiki/Central_limit_theorem) apply to this question? Can you use it to answer the question with less trial and error? """ # ╔═╡ 3cd12379-5e7f-4f5d-a7bb-8b3a9d2e8497 @@ -195,7 +196,7 @@ cubemodel = cubeforce(); force_sp = [cubemodel() for _ in 1:2000]; # ╔═╡ 1ce8f808-b917-4831-89b4-4c318f05724d -force_average = mean(force_sp) +estimated_force = mean(force_sp) # ╔═╡ 420120c3-1c8c-46e4-a7c0-0f2405dd159a pointmass_force = 1 / 1.1^2 @@ -207,20 +208,58 @@ histogram(force_sp) md"### 2: Variance of Estimator" # ╔═╡ e38e8f51-90db-4136-84e2-f06cd03d502a -@bind required_samples Slider(10:10:200, default = 140, show_value = true) - # manually change until standard deviation is about 0.1 +@bind n Slider(10:10:200, default = 20, show_value = true) + # manually change until variance is about 0.01 # ╔═╡ 788f60fc-b1a1-4635-a36a-f954738bcffc -estimator = mean([cubemodel() for _ in 1:required_samples]) +mean([cubemodel() for _ in 1:n]) # one sample of the estimated force # ╔═╡ a7e975a9-beb5-4169-9d22-3e970be2838b -estimator_sp = [mean([cubemodel() for _ in 1:required_samples]) for _ in 1:2000]; +estimated_force_sp = [mean([cubemodel() for _ in 1:n]) for _ in 1:2000]; # "n" samples of the estimated force # ╔═╡ 7ff5cb25-47bd-432f-a404-359abaf44e3a -estimator_sp_σ = std(estimator_sp) +estimated_force_sp_var = var(estimated_force_sp) # ╔═╡ 99c7f6d6-eada-491b-b966-fdb1195fc111 -histogram(estimator_sp) +histogram(estimated_force_sp) + +# ╔═╡ 5c8ba3b4-fee0-49bd-a5e4-4bef400807e0 +md"#### Extra: The central limit theorem" + +# ╔═╡ d695bdfb-1557-4923-9638-1f57689085ca +md""" +According to the central limit theorem, for samples $X_i$ following some distribution with mean $μ$ and variance $σ^2$, the distribution of the sample average $\bar{X}_n$ will converge to a normal distribution with mean $μ$ and variance $σ^2/n$. + +If we know the variance $σ_1^2$ for some sample size $n_1$, we can then find the required amount of samples $n_r$ for $σ_r^2 = 0.01$ using the following simple derivation: + +Given that +```math +\begin{align} +σ^2_1 \approx σ^2 / n_1 +\\ σ^2_r \approx σ^2 / n_r +\end{align} +``` +it follows +```math +\begin{align} +σ^2 &= σ^2 +\\ \Rightarrow n_1 σ^2_1 &\approx n_r σ^2_r +\\ \Rightarrow n_r &\approx \frac{σ_1^2}{σ_r^2} n_1 +\end{align} +``` +""" + +# ╔═╡ 9799aead-e2f5-418a-9079-34f985286fcb +n_1 = 20 + +# ╔═╡ f5769c0e-b452-49b6-9553-324eeeaad229 +var_1 = var([mean([cubemodel() for _ in 1:n_1]) for _ in 1:2000]) + +# ╔═╡ 1a916042-ab26-4e52-9462-6880ca18072c +var_r = 0.01 + +# ╔═╡ 07ec86cc-84f1-497e-8cb8-5eaf3a004221 +n_r = var_1 / var_r * n_1 # we need at least ~ 130 samples # ╔═╡ Cell order: # ╟─52a38b60-178b-4a1d-ac32-e73fafd339f9 @@ -263,5 +302,11 @@ histogram(estimator_sp) # ╠═e38e8f51-90db-4136-84e2-f06cd03d502a # ╠═788f60fc-b1a1-4635-a36a-f954738bcffc # ╠═a7e975a9-beb5-4169-9d22-3e970be2838b -# ╠═7ff5cb25-47bd-432f-a404-359abaf44e3a # ╠═99c7f6d6-eada-491b-b966-fdb1195fc111 +# ╠═7ff5cb25-47bd-432f-a404-359abaf44e3a +# ╟─5c8ba3b4-fee0-49bd-a5e4-4bef400807e0 +# ╟─d695bdfb-1557-4923-9638-1f57689085ca +# ╠═9799aead-e2f5-418a-9079-34f985286fcb +# ╠═f5769c0e-b452-49b6-9553-324eeeaad229 +# ╠═1a916042-ab26-4e52-9462-6880ca18072c +# ╠═07ec86cc-84f1-497e-8cb8-5eaf3a004221 diff --git a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl index 0e35844..e685e50 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl @@ -127,20 +127,21 @@ begin plot!(xe2, ye, ze; color = :orange, lw = 0.5, label = "cube 2") end -# ╔═╡ bb8e4cd6-6106-4d3f-9e35-0dfd8b2c45f5 +# ╔═╡ 4cb55330-8a8a-4622-af4a-ab6f0b643123 md""" -The gravitational force can be estimated by **randomly sampling a point from both cubes** and using the formula for gravitational force between those points, ignoring all constants: +You can sample the gravitational force between both cubes by **randomly sampling a point from both cubes** and using the formula for gravitational force between those points, ignoring all constants: ```math -F = \frac{1}{r^2} +F = \frac{1}{r^2} \, . ``` """ # ╔═╡ ceeeee76-e31c-4429-8ed0-e1c503433dbf md""" !!! questions - 1. What is the estimated net force between the two cubes? Is this the same as if you had treated the cubes as point masses? - 1. How many samples do you need to estimate this force reliably? Define a reliable estimator as one having a standard deviation of 0.1. Visualise the distribution of the estimator. + 1. What is the estimated total force between the two cubes? Is this the same as if you had treated the cubes as point masses? + 1. To estimate the net force, you take the average of $n$ samples. Of course, the result will vary every time you take a new sample: if you take the average of only 10 samples, your estimated total force will vary wildly! We can quantify how much the estimated total force varies by taking a **sample of sample averages** and then calculating the variance. Assuming you want your sample average to have a variance of no more than $0.01$, how many samples do you need? Visualise the distribution of the sample average. + - EXTRA: How does the [central limit theorem](https://en.wikipedia.org/wiki/Central_limit_theorem) apply to this question? Can you use it to answer the question with less trial and error? """ # ╔═╡ 9be28327-d87f-4d50-bbcc-91e799f14dbf @@ -156,7 +157,7 @@ end cubemodel = cubeforce(); # ╔═╡ 7b117453-0875-4b41-a228-866c6c0a8208 -force_average = missing +estimated_force = missing # ╔═╡ b28cfbae-2fac-4b38-b234-53f71e381bcd pointmass_force = missing # (doesn't require Turing, only maths) @@ -165,22 +166,18 @@ pointmass_force = missing # (doesn't require Turing, only maths) md"### 2: Variance of Estimator" # ╔═╡ e38e8f51-90db-4136-84e2-f06cd03d502a -@bind required_samples Slider(10:10:200, show_value = true) - -# ╔═╡ b753fa82-28ab-46d1-9085-4c65301db046 -estimator = mean([cubemodel() for _ in 1:required_samples]) - # a single estimation of the force given `required_samples` samples +@bind n Slider(10:10:200, show_value = true) # ╔═╡ 55cf127f-5534-4926-955a-487ea9553b70 -estimator_sp = missing - # a sample of estimations given `required_samples` samples - -# ╔═╡ f6a67c1e-3677-4e51-b6dd-5d11a5146ea7 -estimator_sp_σ = missing - # standard deviation of the force estimator +estimated_force_sp = missing + # multiple samples of your estimated force, using `n` samples # ╔═╡ 45f0813b-c4c9-4f13-8d66-1e58293c4422 -missing # histogram +missing # histogram of estimated force samples + +# ╔═╡ f6a67c1e-3677-4e51-b6dd-5d11a5146ea7 +estimated_force_sp_var = missing + # variance of the estimated force # ╔═╡ Cell order: # ╟─52a38b60-178b-4a1d-ac32-e73fafd339f9 @@ -205,7 +202,7 @@ missing # histogram # ╟─c8941726-9e81-47fc-9b7e-cb3b5c0c61ca # ╟─431023df-3724-4325-b0ac-96dbf5e4fd20 # ╟─599ac984-ef1d-4c7a-8e87-9d4ddb1aa710 -# ╟─bb8e4cd6-6106-4d3f-9e35-0dfd8b2c45f5 +# ╟─4cb55330-8a8a-4622-af4a-ab6f0b643123 # ╟─ceeeee76-e31c-4429-8ed0-e1c503433dbf # ╟─9be28327-d87f-4d50-bbcc-91e799f14dbf # ╠═06d0e92f-1f07-41fd-b6ee-e94eb539627d @@ -214,7 +211,6 @@ missing # histogram # ╠═b28cfbae-2fac-4b38-b234-53f71e381bcd # ╟─304d6052-6d3d-487c-8c01-dab259276d6f # ╠═e38e8f51-90db-4136-84e2-f06cd03d502a -# ╠═b753fa82-28ab-46d1-9085-4c65301db046 # ╠═55cf127f-5534-4926-955a-487ea9553b70 -# ╠═f6a67c1e-3677-4e51-b6dd-5d11a5146ea7 # ╠═45f0813b-c4c9-4f13-8d66-1e58293c4422 +# ╠═f6a67c1e-3677-4e51-b6dd-5d11a5146ea7 From c58d1cd45bd508de6b3b619f93ec642a1433f2a3 Mon Sep 17 00:00:00 2001 From: Bram Spanoghe <70662421+bspanoghe@users.noreply.github.com> Date: Thu, 12 Mar 2026 14:43:57 +0100 Subject: [PATCH 7/8] notebook 2: remove combinations exercise --- .../P4_probmod/probmod_2-basics.jl | 88 +------------------ .../P4_probmod/probmod_2-basics.jl | 62 +------------ 2 files changed, 7 insertions(+), 143 deletions(-) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl index 4a5cd63..9eefb34 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl @@ -101,74 +101,9 @@ var(Poisson(15)) # analytical answer of var(Y|X=15) # Given X, Y follows a Poisson distribution with mean X -# ╔═╡ ce7d57ed-4f31-4dcf-af3b-b37a2e2a9393 -md"## 2: Combinations" - -# ╔═╡ 087994ce-a26d-40c4-87eb-ef9f0ce7f1fb -md""" -Let `U ~ Uniform(0, 4)`, `V ∼ Normal(U, 1)` and `W ~ TriangularDist(0, 4, U)`. -1. Use sampling (n = 10_000) to make a histogram of `|V − W|`. -2. Estimate `P(V > W)` and `P(V * W >= 10)` -3. Are `V` and `W` independent? -""" - -# ╔═╡ 9e5cc347-c74f-46a3-9534-c5ad812844bf -md"### 1: Histogram" - -# ╔═╡ 47a43282-3892-4a9a-94b7-c359fa74e12b -@model function combinations() - U ~ Uniform(0, 4) - V ~ Normal(U, 1.0) - W ~ TriangularDist(0, 4, U) -end - -# ╔═╡ 890eaeb5-c63f-470c-85eb-d77cec68b740 -comb_model = combinations() - -# ╔═╡ 0d0297a3-60b6-47e5-b42b-bde99c09a16e -comb_chain = sample(comb_model, Prior(), 10_000); - -# ╔═╡ c3cec28a-f472-46f5-9082-ddfdeb01ddf0 -spV = comb_chain[:V]; - -# ╔═╡ 1911f031-d668-47df-995a-ca339ad4ae47 -spW = comb_chain[:W]; - -# ╔═╡ 5c13b506-c05b-4258-95e7-93b648defa17 -spVW = abs.(spV - spW) - -# ╔═╡ d8ae7b0a-afee-41c0-af0c-f8b1d30086f9 -histogram(spVW) - -# ╔═╡ faa105b6-0700-4f4d-92fe-2bb72a4d6e44 -md"### 2: Probabilities" - -# ╔═╡ b4f156df-0b9c-495a-b95b-000c7f166243 -probVW1 = mean(spV .> spW) - -# ╔═╡ f7eb6faf-8884-4fea-9974-330d7fd555aa -probVW2 = mean(spV .* spW .>= 10) - -# ╔═╡ 4decd959-aeb9-47d4-a381-14bbf4dbc5ab -md"### 3: Independence" - -# ╔═╡ a465722f-49f9-4d37-9a2b-00d1120f5c06 -md""" -!!! hint - One way to prove dependence is showing that $E[V] \neq E[V \mid W \leq w]$ for at least one value $w$. - - If the expected value of $V$ can change based on some information about $W$, they can't be independent! -""" - -# ╔═╡ b2999f6e-2dbd-44ab-a303-97234f665d33 -mean(spV) - -# ╔═╡ 5ea80b00-941a-4226-87cb-66da7ee7d76d -mean(spV[spW .<= 1]) # E[V] != E[V | W <= 1] => they are dependent - # ╔═╡ 9740ea64-cd4f-46b1-a741-02e392280601 md""" -## 3: Dice +## 2: Dice """ # ╔═╡ 187854bb-9e30-454d-9e03-cccf77aebb6b @@ -252,7 +187,7 @@ md"### 3: Comparison" p_watercube_is_better = mean(sp_w .> sp_d) # ╔═╡ 34f3014f-f4d4-43d1-b46f-bdca73aee33f -md"## 4: Super eggs" +md"## 3: Super eggs" # ╔═╡ 372436c4-262f-49b8-b1cf-626b043542bf md""" @@ -321,7 +256,7 @@ histogram(sp_egg[chicken_ages .== 1], normalize = :probability) histogram(sp_egg[chicken_ages .== 5], normalize = :probability) # ╔═╡ ff06c070-50a2-43d0-9729-1c47e728ff52 -md"## 5: Birthdays" +md"## 4: Birthdays" # ╔═╡ 6ac2238a-16fd-4a8d-b779-8627d87367ed md""" @@ -414,23 +349,6 @@ E_triplebday = p_triplebday * 365 # The expected value of the amount of triple b # ╠═597d97fb-cfbe-4cff-bfbb-57df9a2f42aa # ╠═6ca31280-a997-4e84-98bb-e7784d0c555c # ╠═9dcb122b-dd12-4e80-b391-f780e637d6fe -# ╟─ce7d57ed-4f31-4dcf-af3b-b37a2e2a9393 -# ╟─087994ce-a26d-40c4-87eb-ef9f0ce7f1fb -# ╟─9e5cc347-c74f-46a3-9534-c5ad812844bf -# ╠═47a43282-3892-4a9a-94b7-c359fa74e12b -# ╠═890eaeb5-c63f-470c-85eb-d77cec68b740 -# ╠═0d0297a3-60b6-47e5-b42b-bde99c09a16e -# ╠═c3cec28a-f472-46f5-9082-ddfdeb01ddf0 -# ╠═1911f031-d668-47df-995a-ca339ad4ae47 -# ╠═5c13b506-c05b-4258-95e7-93b648defa17 -# ╠═d8ae7b0a-afee-41c0-af0c-f8b1d30086f9 -# ╟─faa105b6-0700-4f4d-92fe-2bb72a4d6e44 -# ╠═b4f156df-0b9c-495a-b95b-000c7f166243 -# ╠═f7eb6faf-8884-4fea-9974-330d7fd555aa -# ╟─4decd959-aeb9-47d4-a381-14bbf4dbc5ab -# ╟─a465722f-49f9-4d37-9a2b-00d1120f5c06 -# ╠═b2999f6e-2dbd-44ab-a303-97234f665d33 -# ╠═5ea80b00-941a-4226-87cb-66da7ee7d76d # ╟─9740ea64-cd4f-46b1-a741-02e392280601 # ╟─187854bb-9e30-454d-9e03-cccf77aebb6b # ╟─747e3c0a-357a-448a-b479-d0fcbe44a6c0 diff --git a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl index d96a552..11e50b0 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl @@ -1,5 +1,5 @@ ### A Pluto.jl notebook ### -# v0.20.4 +# v0.20.21 using Markdown using InteractiveUtils @@ -93,53 +93,9 @@ varYcondX = missing # ╔═╡ ce9e2ce2-26e0-4f17-adf5-c922ba98239d missing # analytical answer of (missing) -# ╔═╡ ce7d57ed-4f31-4dcf-af3b-b37a2e2a9393 -md"## 2: Combinations" - -# ╔═╡ 087994ce-a26d-40c4-87eb-ef9f0ce7f1fb -md""" -Let `U ~ Uniform(0, 4)`, `V ∼ Normal(U, 1)` and `W ~ TriangularDist(0, 4, U)`. -1. Use sampling (n = 10_000) to make a histogram of `|V − W|`. -2. Estimate `P(V > W)` and `P(V * W >= 10)` -3. Are `V` and `W` independent? -""" - -# ╔═╡ 9e5cc347-c74f-46a3-9534-c5ad812844bf -md"### 1: Histogram" - -# ╔═╡ 47a43282-3892-4a9a-94b7-c359fa74e12b -@model function combinations() - U ~ missing - V ~ missing - W ~ missing -end - -# ╔═╡ b0502f20-17af-4ecc-be80-a26b3e42d57f -spVW = missing - -# ╔═╡ faa105b6-0700-4f4d-92fe-2bb72a4d6e44 -md"### 2: Probabilities" - -# ╔═╡ 97aec4a9-2954-4967-bc15-c0123bac2e75 -probVW1 = missing - -# ╔═╡ 30166840-d5f5-4a2a-acc2-cde76a87e95a -probVW2 = missing - -# ╔═╡ 4decd959-aeb9-47d4-a381-14bbf4dbc5ab -md"### 3: Independence" - -# ╔═╡ aa953baa-5105-49d7-82e6-94ca462624f7 -md""" -!!! hint - One way to prove dependence is showing that $E[V] \neq E[V \mid W \leq w]$ for at least one value $w$. - - If the expected value of $V$ can change based on some information about $W$, they can't be independent! -""" - # ╔═╡ 9740ea64-cd4f-46b1-a741-02e392280601 md""" -## 3: Dice +## 2: Dice """ # ╔═╡ 187854bb-9e30-454d-9e03-cccf77aebb6b @@ -214,7 +170,7 @@ md"### 3: Comparison" p_watercube_is_better = missing # ╔═╡ 34f3014f-f4d4-43d1-b46f-bdca73aee33f -md"## 4: Super eggs" +md"## 3: Super eggs" # ╔═╡ 372436c4-262f-49b8-b1cf-626b043542bf md""" @@ -265,7 +221,7 @@ missing # histogram 1 missing # histogram 2 # ╔═╡ ff06c070-50a2-43d0-9729-1c47e728ff52 -md"## 5: Birthdays" +md"## 4: Birthdays" # ╔═╡ 6ac2238a-16fd-4a8d-b779-8627d87367ed md""" @@ -317,16 +273,6 @@ end # ╠═aedd0fe8-da3e-4463-b0cf-7c4f9a22db52 # ╠═7ef53a87-e5df-4724-b896-3d1d46214c68 # ╠═ce9e2ce2-26e0-4f17-adf5-c922ba98239d -# ╟─ce7d57ed-4f31-4dcf-af3b-b37a2e2a9393 -# ╟─087994ce-a26d-40c4-87eb-ef9f0ce7f1fb -# ╟─9e5cc347-c74f-46a3-9534-c5ad812844bf -# ╠═47a43282-3892-4a9a-94b7-c359fa74e12b -# ╠═b0502f20-17af-4ecc-be80-a26b3e42d57f -# ╟─faa105b6-0700-4f4d-92fe-2bb72a4d6e44 -# ╠═97aec4a9-2954-4967-bc15-c0123bac2e75 -# ╠═30166840-d5f5-4a2a-acc2-cde76a87e95a -# ╟─4decd959-aeb9-47d4-a381-14bbf4dbc5ab -# ╟─aa953baa-5105-49d7-82e6-94ca462624f7 # ╟─9740ea64-cd4f-46b1-a741-02e392280601 # ╟─187854bb-9e30-454d-9e03-cccf77aebb6b # ╟─747e3c0a-357a-448a-b479-d0fcbe44a6c0 From b1377c14186a5bc5a308cc550da5f38299a5c0aa Mon Sep 17 00:00:00 2001 From: bspanoghe Date: Mon, 16 Mar 2026 18:40:17 +0100 Subject: [PATCH 8/8] P4 hotfixes intro - added section for using for-loops for many variables - made code for plotting distributions visible and added note - renamed `y_s` to `y_f` for consistency basics - added warning students need to add code cells now - changed important `hint`s to `tip`s to avoid blurring - changed naming convention for samples, now avoiding `sp` (SteekProef) advanced, review, P5 notebooks - changed naming convention for samples, now avoiding `sp` (SteekProef) --- .../P4_probmod/probmod_1-intro.jl | 87 ++++++++++++++----- .../P4_probmod/probmod_2-basics.jl | 57 ++++++------ .../P4_probmod/probmod_3-advanced.jl | 10 +-- .../P4_probmod/probmod_4-review.jl | 6 +- .../solved_notebooks/P5_mcmc/MCMC_1-intro.jl | 10 +-- .../solved_notebooks/P5_mcmc/MCMC_2-basics.jl | 6 +- .../P5_mcmc/MCMC_3-advanced.jl | 8 +- .../P4_probmod/probmod_1-intro.jl | 87 ++++++++++++++----- .../P4_probmod/probmod_2-basics.jl | 47 ++++++---- .../P4_probmod/probmod_3-advanced.jl | 4 +- .../student_notebooks/P5_mcmc/MCMC_1-intro.jl | 10 +-- .../P5_mcmc/MCMC_2-basics.jl | 4 +- 12 files changed, 220 insertions(+), 116 deletions(-) diff --git a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl index 062b891..411d1f4 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_1-intro.jl @@ -91,11 +91,11 @@ t_f = \frac{x_f}{\mathrm{cos}(\alpha \, v_0)} \, , ``` evaluating it's y-coordinate there ```math -y_d = \mathrm{sin}(α) \, v_0 \, t_f - \frac{g}{2} \, t_f^2 \, , +y_f = \mathrm{sin}(α) \, v_0 \, t_f - \frac{g}{2} \, t_f^2 \, , ``` and checking if it's in the range of our face ```math -y_d ∈ [1.5, 1.7] \, . +y_f ∈ [1.5, 1.7] \, . ``` """ @@ -145,6 +145,12 @@ plot( # ╔═╡ 314ba0ec-015f-488d-ac63-780b1dbf5bc2 md"The third is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." +# ╔═╡ 2878b32a-9e54-40b0-b151-031b95ad161b +md""" +!!! note + As you can see, to visualize a distribution you can simply call the `plot` function on the distribution (and set extra keyword arguments as desired). +""" + # ╔═╡ c070f3c9-dd80-49f9-b279-c77f46316116 md"## Defining the model" @@ -182,11 +188,11 @@ Our model can be defined as follows: # get the time it takes the droplet to travel the distance x_f t_f = x_f / (cos(α) * v0) # get the droplet's altitude at this time - y_s = sin(α)*v0*t_f - g/2 * t_f^2 + y_f = sin(α)*v0*t_f - g/2 * t_f^2 # to respect the ground, set altitude to 0 if it would be negative - y_s = max(0, y_s) + y_f = max(0, y_f) - return y_s + return y_f end # ╔═╡ 22144610-0c37-4501-b14c-83895d8b0eae @@ -235,22 +241,22 @@ md""" md"### Getting the model output" # ╔═╡ 46f17c07-9aec-4fee-8590-4eeb5a13b93e -md"To answer our question, we are interested in the `y_s` variable returned by the model. We can extract the output variable from our chain using the `generated_quantities` function:" +md"To answer our question, we are interested in the `y_f` variable returned by the model. We can extract the output variable from our chain using the `generated_quantities` function:" # ╔═╡ 071fc405-7a33-4dab-a69a-1e68a0fb35ad -y_s_samples = generated_quantities(splash_model, chain) +y_f_samples = generated_quantities(splash_model, chain) # ╔═╡ 20e99145-e80b-4259-9bca-f7e14c2d355e md"It's often useful to visualize the distribution of our variable of interest. We can use the `histogram` function for this:" # ╔═╡ 45d44cf5-2289-4947-8d4e-1dcddb5447de -histogram(y_s_samples, bins = 15) +histogram(y_f_samples, bins = 15) # ╔═╡ 8d179748-40aa-4e88-b277-9234fe938251 md"Checking whether the droplet altitude matches our face's can be done with a simple inequality operation:" # ╔═╡ 633a2b4c-379e-4e18-bbcd-b450930196a5 -face_hit_samples = [1.5 <= y_s <= 1.7 for y_s in y_s_samples] +face_hit_samples = [1.5 <= y_f <= 1.7 for y_f in y_f_samples] # ╔═╡ ad8a6ea2-30ba-4104-824f-12a7ea8e9718 md"Finally, we can estimate the probability of our face being hit by taking the mean amount of times we got hit in our sample." @@ -264,7 +270,7 @@ md"""And that's the answer to our question! If we're 1m behind the cyclist in fr # ╔═╡ bf5642f4-09d3-491d-8eb9-47fcdb67c977 md""" !!! note - Do you like your code sleek? You can also calculate this probability from your samples with just one line: `mean(1.5 .<= y_s_samples .<= 1.7)` + Do you like your code sleek? You can also calculate this probability from your samples with just one line: `mean(1.5 .<= y_f_samples .<= 1.7)` """ # ╔═╡ 3067af6e-df69-48a0-a024-a6e219dcfc7c @@ -343,10 +349,40 @@ md"It's also possible to skip the chain construction entirely and simply call yo # ╔═╡ 771b9cb1-d809-46de-9380-2db73ccd014d # get a sample of the output 500 times -y_s_samples_alternative = [splash_model() for _ in 1:500] +y_f_samples_alternative = [splash_model() for _ in 1:500] # ╔═╡ 0f1cccd8-937d-4fc2-8657-c9967db483b8 -histogram(y_s_samples_alternative, bins = 15) # yup, that's the same distribution +histogram(y_f_samples_alternative, bins = 15) # yup, that's the same distribution + +# ╔═╡ 7806eca0-a2ad-4910-8d6e-edc50073b37b +md"## For-loops for many variables" + +# ╔═╡ 64ebc3ce-83ca-4cee-ab93-cddf51138543 +md""" +Sometimes you need to define a large number of stochastic variables following the same distribution. Let's say, for example, you want to model the total mass of water being splashed on your face assuming that 10 droplets will hit you, each having a mass that is exponentially distributed with a mean of 30 mg. + +This can be done using a **for-loop** as follows: +""" + +# ╔═╡ bdf3c18c-b40b-4fb4-81e7-fe9d821cc166 +@model function splash_mass() + droplet_masses = zeros(10) # instantiate vector of 10 zeros + for i in 1:length(droplet_masses) + droplet_masses[i] ~ Exponential(30) # each element of the vector follows an Exponential distribution + end + total_mass = sum(droplet_masses) + + return total_mass +end + +# ╔═╡ 2cbe2750-38c1-4e7f-8d03-9bed7c16861a +mass_model = splash_mass(); + +# ╔═╡ ad8b48c5-f892-49b3-8f19-64d93424485c +mass_chain = sample(mass_model, Prior(), 1000); + +# ╔═╡ 8507f084-eff5-4351-800d-8eba7ca673bb +total_mass_samples = generated_quantities(mass_model, mass_chain) # ╔═╡ befb8f05-6ab7-4594-9cf1-71df47c1df03 md"## Essentials" @@ -376,25 +412,25 @@ let α ~ TriangularDist(0, pi/2) t_f = x_f / (cos(α) * v0) - y_s = sin(α)*v0*t_f - g/2 * t_f^2 - y_s = max(0, y_s) - return y_s + y_f = sin(α)*v0*t_f - g/2 * t_f^2 + y_f = max(0, y_f) + return y_f end # get samples splash_model = splash(x_f) chain = sample(splash_model, Prior(), 500); - y_s_samples = generated_quantities(splash_model, chain) + y_f_samples = generated_quantities(splash_model, chain) v0_samples = chain["v0"] α_samples = chain[:α] # calculate interesting things - prob_face_hit = mean([1.5 <= y_s <= 1.7 for y_s in y_s_samples]) + prob_face_hit = mean([1.5 <= y_f <= 1.7 for y_f in y_f_samples]) + y(x, α, v0) = sin(α)/cos(α)*x - g/2*(x / (cos(α) * v0))^2 trajectories = [ - x -> sin(α_samples[i]) / cos(α_samples[i]) * x - - g / 2 * ( x / (cos(α_samples[i])*v0_samples[i]) )^2 + x -> y(x, α_samples[i], v0_samples[i]) for i in 1:length(v0_samples) ] plot(trajectories, xlims = (0, 5), ylims = (0, 2), @@ -423,10 +459,11 @@ end # ╟─e8821ac2-baa3-4674-868c-3a5b8593cc95 # ╟─4811e1af-9ba4-49b3-a3a1-5d8ff8d5af92 # ╟─01156ef7-b194-4372-8bc8-de9fad5e12f1 -# ╟─f30290bd-9d31-4939-b5d6-f89f066775d2 +# ╠═f30290bd-9d31-4939-b5d6-f89f066775d2 # ╟─dab23eba-3eb1-434b-aef5-e08730e35114 -# ╟─299e402c-634a-4b18-b867-489bafaf4564 +# ╠═299e402c-634a-4b18-b867-489bafaf4564 # ╟─314ba0ec-015f-488d-ac63-780b1dbf5bc2 +# ╟─2878b32a-9e54-40b0-b151-031b95ad161b # ╟─c070f3c9-dd80-49f9-b279-c77f46316116 # ╟─84a08354-7bf3-4783-9285-a7e8850e798b # ╟─8f94567c-b87b-4eb4-8441-c6434da393a8 @@ -454,7 +491,7 @@ end # ╟─ad8a6ea2-30ba-4104-824f-12a7ea8e9718 # ╠═e1f9d1e9-8a0c-4b28-ab67-a89e3287b50f # ╟─1c9e2de1-21a0-4d4d-918d-9e2f0fc4d912 -# ╟─bf5642f4-09d3-491d-8eb9-47fcdb67c977 +# ╠═bf5642f4-09d3-491d-8eb9-47fcdb67c977 # ╟─3067af6e-df69-48a0-a024-a6e219dcfc7c # ╟─cbc04c5f-aa93-4340-a61e-1583d20d86b1 # ╟─d36784b8-9f3b-4434-88fe-03b75b94754e @@ -473,6 +510,12 @@ end # ╟─83a8b9b6-72da-4aff-8364-37846b527a2f # ╠═771b9cb1-d809-46de-9380-2db73ccd014d # ╠═0f1cccd8-937d-4fc2-8657-c9967db483b8 +# ╟─7806eca0-a2ad-4910-8d6e-edc50073b37b +# ╟─64ebc3ce-83ca-4cee-ab93-cddf51138543 +# ╠═bdf3c18c-b40b-4fb4-81e7-fe9d821cc166 +# ╠═2cbe2750-38c1-4e7f-8d03-9bed7c16861a +# ╠═ad8b48c5-f892-49b3-8f19-64d93424485c +# ╠═8507f084-eff5-4351-800d-8eba7ca673bb # ╟─befb8f05-6ab7-4594-9cf1-71df47c1df03 # ╟─52af5de0-b516-4947-84f7-dde63e4c513e # ╟─2ceede7e-8f0c-420c-bde2-f16b726cd204 diff --git a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl index 9eefb34..fc49a42 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_2-basics.jl @@ -49,13 +49,13 @@ dpmodel = doublepoisson() dpchain = sample(dpmodel, Prior(), 10_000) # ╔═╡ f6d90b53-0f16-4e43-b77c-8eb05b8c6943 -spY = dpchain[:Y]; +Y_samples = dpchain[:Y]; # ╔═╡ 90835b5d-9705-43d0-b449-09cde57d5394 plot(Poisson(10)) # ╔═╡ 431d9be9-edfc-48fb-9df7-6a2b961fe4b4 -histogram(spY) +histogram(Y_samples) # ╔═╡ 13989ba4-bcf8-4fdd-8aee-ab58c8905bc9 md"### 2: Probabilities" @@ -65,16 +65,16 @@ md""" !!! tip When comparing a vector of values to a single number, don't forget to use `.` to execute operations element-wise in Julia! - ✅ `spY .< 1` compares every element of `spY` to `1` + ✅ `Y_samples .< 1` compares every element of `Y_samples` to `1` - ❌ `spY < 1` compares an entire vector with a single number → errors :( + ❌ `Y_samples < 1` compares an entire vector with a single number → errors :( """ # ╔═╡ 09a87037-9f4f-4dd9-bb6c-fe717db39ea6 -probXY1 = mean(3 .< spY .<= 10) +probXY1 = mean(3 .< Y_samples .<= 10) # ╔═╡ 2e2f203a-34a6-4feb-8aa1-8a1c97a09165 -probXY2 = mean(spY.^2 .> 100) +probXY2 = mean(Y_samples.^2 .> 100) # ╔═╡ c243ca59-191d-4905-825e-6d7825a3c8a4 md"### 3: Variances" @@ -84,23 +84,29 @@ md""" !!! hint To create a sample of $X$ that is conditional on some value(s) of $Y$, you can start from a sample of $X$ and select only those elements for which the corresponding sample of $Y$ has the conditioned value(s). - In other words, you'll need to index `spX` based on `spY` (and vice versa for $\text{var}(Y ∣ X)$). + In other words, you'll need to index `X_samples` based on `Y_samples` (and vice versa for $\text{var}(Y ∣ X)$). """ # ╔═╡ 04cc5b42-7ab2-4055-a959-ba894c598f69 -spX = dpchain[:X]; +X_samples = dpchain[:X]; # ╔═╡ 597d97fb-cfbe-4cff-bfbb-57df9a2f42aa -varXcondY = var(spX[spY .== 15]) +varXcondY = var(X_samples[Y_samples .== 15]) # ╔═╡ 6ca31280-a997-4e84-98bb-e7784d0c555c -varYcondX = var(spY[spX .== 15]) +varYcondX = var(Y_samples[X_samples .== 15]) # ╔═╡ 9dcb122b-dd12-4e80-b391-f780e637d6fe var(Poisson(15)) # analytical answer of var(Y|X=15) # Given X, Y follows a Poisson distribution with mean X +# ╔═╡ c976dd77-6a72-4acc-9b07-342cc01fc7ee +md""" +!!! warning + Starting here, the entire structure of the answer will no longer be given. You are therefore expected to **add more code cells** yourself, using the `+` at the left in between two cells. +""" + # ╔═╡ 9740ea64-cd4f-46b1-a741-02e392280601 md""" ## 2: Dice @@ -127,7 +133,7 @@ md"### 1: Watercube" # ╔═╡ 13cdd6ee-04ec-4085-8083-3d52e88aa35c md""" -!!! hint +!!! tip Consider the humble `DiscreteUniform` distribution. Not sure how it works? Open the **🔍 Live Docs** at the bottom right of the screen for more information """ @@ -147,13 +153,13 @@ end watermodel = watercube() # ╔═╡ 97021710-4b26-4a1c-a794-265c9559ce4d -sp_w = [watermodel() for i in 1:2000] +watercube_samples = [watermodel() for i in 1:2000] # ╔═╡ c0cd75bd-2a46-4f62-a59a-9bb8d47d45f2 -p_watercube_kills = mean(sp_w .>= 50) +p_watercube_kills = mean(watercube_samples .>= 50) # ╔═╡ bc5f5d70-e269-45f9-867b-a00876ce8c40 -histogram(sp_w, bins = 30) +histogram(watercube_samples, bins = 30) # ╔═╡ 44a2cabc-aca9-4bb7-af9d-cd735c424d7b md"### 2: Dirtprism" @@ -172,19 +178,19 @@ end dirtmodel = dirtprism() # ╔═╡ 700cc645-1123-4c8a-821e-1ae4fe8b0674 -sp_d = [dirtmodel() for i in 1:2000] +dirtcube_samples = [dirtmodel() for i in 1:2000] # ╔═╡ 98d04790-e86f-438a-9389-cb9697934bbf -p_dirtprism_kills = mean(sp_d .>= 50) +p_dirtprism_kills = mean(dirtcube_samples .>= 50) # ╔═╡ d5413573-5202-4fa2-94f6-e92a613d63aa -histogram(sp_d, bins = 30) +histogram(dirtcube_samples, bins = 30) # ╔═╡ 49790a8f-9f53-4ba7-9543-d6a879b520e0 md"### 3: Comparison" # ╔═╡ 6e098cb6-eddd-4924-a184-c2578dc28473 -p_watercube_is_better = mean(sp_w .> sp_d) +p_watercube_is_better = mean(watercube_samples .> dirtcube_samples) # ╔═╡ 34f3014f-f4d4-43d1-b46f-bdca73aee33f md"## 3: Super eggs" @@ -214,7 +220,7 @@ md"### 1: Probability" # ╔═╡ 34aad3d6-9deb-4298-b613-bf3a616f2b31 md""" -!!! hint +!!! tip In this exercise, the output variable (the number of double-yolked eggs) is **also a random variable**! In other words, it also follows some distribution. When considering what distribution, consider that each of the $N$ eggs represents a "trial" with a $P$ chance of success for a double yolk. @@ -237,10 +243,10 @@ egg_model = eggs(); chain_egg = sample(egg_model, Prior(), 2000) # ╔═╡ 64d04ed9-6548-4251-8a4e-11b31e8f3143 -sp_egg = generated_quantities(egg_model, chain_egg); +egg_samples = generated_quantities(egg_model, chain_egg); # ╔═╡ 00fd5517-5232-4d88-8ab0-3a0eb925eb3e -p_multiple_double_eggs = mean(sp_egg .>= 2) +p_multiple_double_eggs = mean(egg_samples .>= 2) # a ~2% chance for two or more double eggs in one year # ╔═╡ 80e3544e-86ec-469c-be44-77f94fabfc28 @@ -250,10 +256,10 @@ md"### 2: Histograms" chicken_ages = chain_egg[:A]; # ╔═╡ fe87f9ee-a07c-4b4c-b89d-c3bb4ee7ee7f -histogram(sp_egg[chicken_ages .== 1], normalize = :probability) +histogram(egg_samples[chicken_ages .== 1], normalize = :probability) # ╔═╡ 9d9b4091-cf44-46e6-b81b-74663c9c8dfe -histogram(sp_egg[chicken_ages .== 5], normalize = :probability) +histogram(egg_samples[chicken_ages .== 5], normalize = :probability) # ╔═╡ ff06c070-50a2-43d0-9729-1c47e728ff52 md"## 4: Birthdays" @@ -301,10 +307,10 @@ end bday_model = birthdays(150) # ╔═╡ b0443045-74a1-4b48-8f6c-a1b5e15eace4 -sp_maxoccs = [bday_model() for i in 1:2000] +bday_samples = [bday_model() for i in 1:2000] # ╔═╡ cac00880-15fa-483a-a09f-9b6d1219b0cf -mean(sp_maxoccs .>= 3) +mean(bday_samples .>= 3) # ╔═╡ 13002efe-15f1-4096-be4f-671432a8991e md""" @@ -349,6 +355,7 @@ E_triplebday = p_triplebday * 365 # The expected value of the amount of triple b # ╠═597d97fb-cfbe-4cff-bfbb-57df9a2f42aa # ╠═6ca31280-a997-4e84-98bb-e7784d0c555c # ╠═9dcb122b-dd12-4e80-b391-f780e637d6fe +# ╟─c976dd77-6a72-4acc-9b07-342cc01fc7ee # ╟─9740ea64-cd4f-46b1-a741-02e392280601 # ╟─187854bb-9e30-454d-9e03-cccf77aebb6b # ╟─747e3c0a-357a-448a-b479-d0fcbe44a6c0 diff --git a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl index 80536cf..7ac3e93 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_3-advanced.jl @@ -105,10 +105,10 @@ petri_model = petrigrowth(8); chain_petri = sample(petri_model, Prior(), 2000); # ╔═╡ ce8b53c3-0af1-4e9a-ad80-6fd7a2bf020b -sp_splittable = generated_quantities(petri_model, chain_petri); +splittable_samples = generated_quantities(petri_model, chain_petri); # ╔═╡ e3092915-a083-425e-8fa1-b7bf370abc8a -prob_splittable = mean(sp_splittable) +prob_splittable = mean(splittable_samples) # ╔═╡ e4f42f16-5cce-4fc2-aa01-8971f37c710e md"### 3: Plot" @@ -215,13 +215,13 @@ md"### 2: Variance of Estimator" mean([cubemodel() for _ in 1:n]) # one sample of the estimated force # ╔═╡ a7e975a9-beb5-4169-9d22-3e970be2838b -estimated_force_sp = [mean([cubemodel() for _ in 1:n]) for _ in 1:2000]; # "n" samples of the estimated force +force_samples = [mean([cubemodel() for _ in 1:n]) for _ in 1:2000]; # "n" samples of the estimated force # ╔═╡ 7ff5cb25-47bd-432f-a404-359abaf44e3a -estimated_force_sp_var = var(estimated_force_sp) +force_samples_var = var(force_samples) # ╔═╡ 99c7f6d6-eada-491b-b966-fdb1195fc111 -histogram(estimated_force_sp) +histogram(force_samples) # ╔═╡ 5c8ba3b4-fee0-49bd-a5e4-4bef400807e0 md"#### Extra: The central limit theorem" diff --git a/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl b/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl index 44851cc..849a80a 100644 --- a/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl +++ b/exercises/solved_notebooks/P4_probmod/probmod_4-review.jl @@ -66,13 +66,13 @@ end needle_model = needles() # ╔═╡ 49a206fb-6390-45fa-900f-8031b77a2725 -sp_n = [needle_model() for i in 1:2000] +needle_samples = [needle_model() for i in 1:2000] # ╔═╡ b6009fc4-2f67-4b56-acc9-ccc1a013f5e9 -mean(sp_n) +mean(needle_samples) # ╔═╡ 63e0b367-6155-472d-8502-b405db10e979 -pi_est = 2 / mean(sp_n) +pi_est = 2 / mean(needle_samples) # ╔═╡ Cell order: # ╠═882b942c-a5ae-445d-a946-6eb8cddc423e diff --git a/exercises/solved_notebooks/P5_mcmc/MCMC_1-intro.jl b/exercises/solved_notebooks/P5_mcmc/MCMC_1-intro.jl index 7c7f57c..c73f3df 100644 --- a/exercises/solved_notebooks/P5_mcmc/MCMC_1-intro.jl +++ b/exercises/solved_notebooks/P5_mcmc/MCMC_1-intro.jl @@ -281,10 +281,10 @@ Taking the sampled values of the mutation rate from the chain and plotting a his """ # ╔═╡ b657217a-6ccc-4a41-b852-df4e39a7a10a -sp_alpha = mutation_chain[:α]; +alpha_samples = mutation_chain[:α]; # ╔═╡ 9cf08616-d599-42a3-82e8-e8c98853c1d8 -histogram(sp_alpha) # note the difference with the prior distribution! +histogram(alpha_samples) # note the difference with the prior distribution! # ╔═╡ e02c42dc-627c-4bc0-8ebb-fdb2b0f15b64 md"Plotting some sampled mutation rates from this distribution onto our data shows that they fit well:" @@ -295,17 +295,17 @@ begin ylabel = "Number of mutations", label = false, xlims = (0, 500), ylims = (0, 400), title = "A posteriori relationship between t and mean N" ); - plot!([x -> αᵢ*x for αᵢ in sp_alpha[1:10:end]], color = :purple, opacity = 0.1, label = false) + plot!([x -> αᵢ*x for αᵢ in alpha_samples[1:10:end]], color = :purple, opacity = 0.1, label = false) end # ╔═╡ 87059440-2919-4f6d-9d32-0df3ce75e2a2 md"And finally we can answer our first question:" # ╔═╡ 644cff58-68b9-4d4b-8896-617fcacc39c5 -mean(sp_alpha) +mean(alpha_samples) # ╔═╡ 5f2f5a78-ca90-4baf-8bf0-ff1fae88d785 -sqrt(var(sp_alpha)) +sqrt(var(alpha_samples)) # ╔═╡ bfbf811f-7b2f-473b-b132-0c7e965c8b0d md""" diff --git a/exercises/solved_notebooks/P5_mcmc/MCMC_2-basics.jl b/exercises/solved_notebooks/P5_mcmc/MCMC_2-basics.jl index a1bf6fa..8eee3ba 100644 --- a/exercises/solved_notebooks/P5_mcmc/MCMC_2-basics.jl +++ b/exercises/solved_notebooks/P5_mcmc/MCMC_2-basics.jl @@ -71,10 +71,10 @@ end molemodel = mole() # ╔═╡ b5bd5114-0a0c-4aa3-af00-4f3d05edc5e3 -spY = [molemodel() for i in 1:2000]; +Y_samples = [molemodel() for i in 1:2000]; # ╔═╡ 4e8b9000-cb61-4f01-9ba9-17276ad0335e -E_Y = mean(spY) +E_Y = mean(Y_samples) # ╔═╡ 8777b133-7d7c-4a85-b89c-2f00093e9984 md"### 3: Conditional expected value of X" @@ -89,7 +89,7 @@ molechain = sample(cond_mole, NUTS(), 2000); plot(molechain) # ╔═╡ b79dafee-4f6f-4301-9267-1b9798c125ea -spXcondY = molechain[:X]; +X_samplescondY = molechain[:X]; # ╔═╡ fd540765-95e5-4071-81f7-e689b06cad0c E_XcondY = mean(molechain[:X]) diff --git a/exercises/solved_notebooks/P5_mcmc/MCMC_3-advanced.jl b/exercises/solved_notebooks/P5_mcmc/MCMC_3-advanced.jl index 60facfa..82577cd 100644 --- a/exercises/solved_notebooks/P5_mcmc/MCMC_3-advanced.jl +++ b/exercises/solved_notebooks/P5_mcmc/MCMC_3-advanced.jl @@ -179,10 +179,10 @@ plot(petrichain) logfuns = generated_quantities(petrimodel, petrichain); # ╔═╡ c642574c-c01b-4398-aac9-43e514d7fa25 -sp_petri = [logfun(8.0) for logfun in logfuns] +petri_samples = [logfun(8.0) for logfun in logfuns] # ╔═╡ d15fa091-839c-4239-96e2-eaf7335ce620 -prob_splittable = mean((sp_petri .>= 1e4) .&& (sp_petri .<= 1e5)) +prob_splittable = mean((petri_samples .>= 1e4) .&& (petri_samples .<= 1e5)) # ╔═╡ 0ea95b67-d4da-4c5a-ad2e-05024ad074a3 md"### 2" @@ -227,8 +227,8 @@ let # so we dont need to rename all variables petrimodel = petrigrowth🌟(5) | (P_obs = 21_000,) petrichain = sample(petrimodel, NUTS(), 2_000) logfuns = generated_quantities(petrimodel, petrichain); - sp_petri = [logfun(8.0) for logfun in logfuns] - prob_splittable = mean((sp_petri .>= 1e4) .&& (sp_petri .<= 1e5)) + petri_samples = [logfun(8.0) for logfun in logfuns] + prob_splittable = mean((petri_samples .>= 1e4) .&& (petri_samples .<= 1e5)) println("The new `prob_splittable` is ", prob_splittable) plot(petrichain) end diff --git a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl index 062b891..411d1f4 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_1-intro.jl @@ -91,11 +91,11 @@ t_f = \frac{x_f}{\mathrm{cos}(\alpha \, v_0)} \, , ``` evaluating it's y-coordinate there ```math -y_d = \mathrm{sin}(α) \, v_0 \, t_f - \frac{g}{2} \, t_f^2 \, , +y_f = \mathrm{sin}(α) \, v_0 \, t_f - \frac{g}{2} \, t_f^2 \, , ``` and checking if it's in the range of our face ```math -y_d ∈ [1.5, 1.7] \, . +y_f ∈ [1.5, 1.7] \, . ``` """ @@ -145,6 +145,12 @@ plot( # ╔═╡ 314ba0ec-015f-488d-ac63-780b1dbf5bc2 md"The third is an **input variable**: when cycling, we can choose how far away we stay from the cyclist in front of us. The x-position of our face $x_f$ is therefore an input to the model." +# ╔═╡ 2878b32a-9e54-40b0-b151-031b95ad161b +md""" +!!! note + As you can see, to visualize a distribution you can simply call the `plot` function on the distribution (and set extra keyword arguments as desired). +""" + # ╔═╡ c070f3c9-dd80-49f9-b279-c77f46316116 md"## Defining the model" @@ -182,11 +188,11 @@ Our model can be defined as follows: # get the time it takes the droplet to travel the distance x_f t_f = x_f / (cos(α) * v0) # get the droplet's altitude at this time - y_s = sin(α)*v0*t_f - g/2 * t_f^2 + y_f = sin(α)*v0*t_f - g/2 * t_f^2 # to respect the ground, set altitude to 0 if it would be negative - y_s = max(0, y_s) + y_f = max(0, y_f) - return y_s + return y_f end # ╔═╡ 22144610-0c37-4501-b14c-83895d8b0eae @@ -235,22 +241,22 @@ md""" md"### Getting the model output" # ╔═╡ 46f17c07-9aec-4fee-8590-4eeb5a13b93e -md"To answer our question, we are interested in the `y_s` variable returned by the model. We can extract the output variable from our chain using the `generated_quantities` function:" +md"To answer our question, we are interested in the `y_f` variable returned by the model. We can extract the output variable from our chain using the `generated_quantities` function:" # ╔═╡ 071fc405-7a33-4dab-a69a-1e68a0fb35ad -y_s_samples = generated_quantities(splash_model, chain) +y_f_samples = generated_quantities(splash_model, chain) # ╔═╡ 20e99145-e80b-4259-9bca-f7e14c2d355e md"It's often useful to visualize the distribution of our variable of interest. We can use the `histogram` function for this:" # ╔═╡ 45d44cf5-2289-4947-8d4e-1dcddb5447de -histogram(y_s_samples, bins = 15) +histogram(y_f_samples, bins = 15) # ╔═╡ 8d179748-40aa-4e88-b277-9234fe938251 md"Checking whether the droplet altitude matches our face's can be done with a simple inequality operation:" # ╔═╡ 633a2b4c-379e-4e18-bbcd-b450930196a5 -face_hit_samples = [1.5 <= y_s <= 1.7 for y_s in y_s_samples] +face_hit_samples = [1.5 <= y_f <= 1.7 for y_f in y_f_samples] # ╔═╡ ad8a6ea2-30ba-4104-824f-12a7ea8e9718 md"Finally, we can estimate the probability of our face being hit by taking the mean amount of times we got hit in our sample." @@ -264,7 +270,7 @@ md"""And that's the answer to our question! If we're 1m behind the cyclist in fr # ╔═╡ bf5642f4-09d3-491d-8eb9-47fcdb67c977 md""" !!! note - Do you like your code sleek? You can also calculate this probability from your samples with just one line: `mean(1.5 .<= y_s_samples .<= 1.7)` + Do you like your code sleek? You can also calculate this probability from your samples with just one line: `mean(1.5 .<= y_f_samples .<= 1.7)` """ # ╔═╡ 3067af6e-df69-48a0-a024-a6e219dcfc7c @@ -343,10 +349,40 @@ md"It's also possible to skip the chain construction entirely and simply call yo # ╔═╡ 771b9cb1-d809-46de-9380-2db73ccd014d # get a sample of the output 500 times -y_s_samples_alternative = [splash_model() for _ in 1:500] +y_f_samples_alternative = [splash_model() for _ in 1:500] # ╔═╡ 0f1cccd8-937d-4fc2-8657-c9967db483b8 -histogram(y_s_samples_alternative, bins = 15) # yup, that's the same distribution +histogram(y_f_samples_alternative, bins = 15) # yup, that's the same distribution + +# ╔═╡ 7806eca0-a2ad-4910-8d6e-edc50073b37b +md"## For-loops for many variables" + +# ╔═╡ 64ebc3ce-83ca-4cee-ab93-cddf51138543 +md""" +Sometimes you need to define a large number of stochastic variables following the same distribution. Let's say, for example, you want to model the total mass of water being splashed on your face assuming that 10 droplets will hit you, each having a mass that is exponentially distributed with a mean of 30 mg. + +This can be done using a **for-loop** as follows: +""" + +# ╔═╡ bdf3c18c-b40b-4fb4-81e7-fe9d821cc166 +@model function splash_mass() + droplet_masses = zeros(10) # instantiate vector of 10 zeros + for i in 1:length(droplet_masses) + droplet_masses[i] ~ Exponential(30) # each element of the vector follows an Exponential distribution + end + total_mass = sum(droplet_masses) + + return total_mass +end + +# ╔═╡ 2cbe2750-38c1-4e7f-8d03-9bed7c16861a +mass_model = splash_mass(); + +# ╔═╡ ad8b48c5-f892-49b3-8f19-64d93424485c +mass_chain = sample(mass_model, Prior(), 1000); + +# ╔═╡ 8507f084-eff5-4351-800d-8eba7ca673bb +total_mass_samples = generated_quantities(mass_model, mass_chain) # ╔═╡ befb8f05-6ab7-4594-9cf1-71df47c1df03 md"## Essentials" @@ -376,25 +412,25 @@ let α ~ TriangularDist(0, pi/2) t_f = x_f / (cos(α) * v0) - y_s = sin(α)*v0*t_f - g/2 * t_f^2 - y_s = max(0, y_s) - return y_s + y_f = sin(α)*v0*t_f - g/2 * t_f^2 + y_f = max(0, y_f) + return y_f end # get samples splash_model = splash(x_f) chain = sample(splash_model, Prior(), 500); - y_s_samples = generated_quantities(splash_model, chain) + y_f_samples = generated_quantities(splash_model, chain) v0_samples = chain["v0"] α_samples = chain[:α] # calculate interesting things - prob_face_hit = mean([1.5 <= y_s <= 1.7 for y_s in y_s_samples]) + prob_face_hit = mean([1.5 <= y_f <= 1.7 for y_f in y_f_samples]) + y(x, α, v0) = sin(α)/cos(α)*x - g/2*(x / (cos(α) * v0))^2 trajectories = [ - x -> sin(α_samples[i]) / cos(α_samples[i]) * x - - g / 2 * ( x / (cos(α_samples[i])*v0_samples[i]) )^2 + x -> y(x, α_samples[i], v0_samples[i]) for i in 1:length(v0_samples) ] plot(trajectories, xlims = (0, 5), ylims = (0, 2), @@ -423,10 +459,11 @@ end # ╟─e8821ac2-baa3-4674-868c-3a5b8593cc95 # ╟─4811e1af-9ba4-49b3-a3a1-5d8ff8d5af92 # ╟─01156ef7-b194-4372-8bc8-de9fad5e12f1 -# ╟─f30290bd-9d31-4939-b5d6-f89f066775d2 +# ╠═f30290bd-9d31-4939-b5d6-f89f066775d2 # ╟─dab23eba-3eb1-434b-aef5-e08730e35114 -# ╟─299e402c-634a-4b18-b867-489bafaf4564 +# ╠═299e402c-634a-4b18-b867-489bafaf4564 # ╟─314ba0ec-015f-488d-ac63-780b1dbf5bc2 +# ╟─2878b32a-9e54-40b0-b151-031b95ad161b # ╟─c070f3c9-dd80-49f9-b279-c77f46316116 # ╟─84a08354-7bf3-4783-9285-a7e8850e798b # ╟─8f94567c-b87b-4eb4-8441-c6434da393a8 @@ -454,7 +491,7 @@ end # ╟─ad8a6ea2-30ba-4104-824f-12a7ea8e9718 # ╠═e1f9d1e9-8a0c-4b28-ab67-a89e3287b50f # ╟─1c9e2de1-21a0-4d4d-918d-9e2f0fc4d912 -# ╟─bf5642f4-09d3-491d-8eb9-47fcdb67c977 +# ╠═bf5642f4-09d3-491d-8eb9-47fcdb67c977 # ╟─3067af6e-df69-48a0-a024-a6e219dcfc7c # ╟─cbc04c5f-aa93-4340-a61e-1583d20d86b1 # ╟─d36784b8-9f3b-4434-88fe-03b75b94754e @@ -473,6 +510,12 @@ end # ╟─83a8b9b6-72da-4aff-8364-37846b527a2f # ╠═771b9cb1-d809-46de-9380-2db73ccd014d # ╠═0f1cccd8-937d-4fc2-8657-c9967db483b8 +# ╟─7806eca0-a2ad-4910-8d6e-edc50073b37b +# ╟─64ebc3ce-83ca-4cee-ab93-cddf51138543 +# ╠═bdf3c18c-b40b-4fb4-81e7-fe9d821cc166 +# ╠═2cbe2750-38c1-4e7f-8d03-9bed7c16861a +# ╠═ad8b48c5-f892-49b3-8f19-64d93424485c +# ╠═8507f084-eff5-4351-800d-8eba7ca673bb # ╟─befb8f05-6ab7-4594-9cf1-71df47c1df03 # ╟─52af5de0-b516-4947-84f7-dde63e4c513e # ╟─2ceede7e-8f0c-420c-bde2-f16b726cd204 diff --git a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl index 11e50b0..8a31b13 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_2-basics.jl @@ -45,11 +45,17 @@ end # ╔═╡ 42c18a70-efb3-436b-83bb-b586280d4a4e dpmodel = doublepoisson(); +# ╔═╡ a21b2f44-c8d7-4838-926f-8fa39c36884e +dpchain = missing + # ╔═╡ b22a18d5-f70b-42a5-a09e-3de515148a6d -spY = missing +Y_samples = missing + +# ╔═╡ a11629df-31d2-4ac7-bf8e-80b910cab2fb +missing # plot of X # ╔═╡ 34bd60df-272f-49d7-9346-fb4d125fe89b -histogram(spY) +missing # histogram of Y # ╔═╡ 13989ba4-bcf8-4fdd-8aee-ab58c8905bc9 md"### 2: Probabilities" @@ -59,9 +65,9 @@ md""" !!! tip When comparing a vector of values to a single number, don't forget to use `.` to execute operations element-wise in Julia! - ✅ `spY .< 1` compares every element of `spY` to `1` + ✅ `Y_samples .< 1` compares every element of `Y_samples` to `1` - ❌ `spY < 1` compares an entire vector with a single number → errors :( + ❌ `Y_samples < 1` compares an entire vector with a single number → errors :( """ # ╔═╡ 39e4f3eb-b7a9-4ece-846f-eb02dbd77860 @@ -78,11 +84,11 @@ md""" !!! hint To create a sample of $X$ that is conditional on some value(s) of $Y$, you can start from a sample of $X$ and select only those elements for which the corresponding sample of $Y$ has the conditioned value(s). - In other words, you'll need to index `spX` based on `spY` (and vice versa for $\text{var}(Y ∣ X)$). + In other words, you'll need to index `X_samples` based on `Y_samples` (and vice versa for $\text{var}(Y ∣ X)$). """ # ╔═╡ 263048e5-206c-499e-836c-bfe489ed9b74 -spX = missing; +X_samples = missing; # ╔═╡ aedd0fe8-da3e-4463-b0cf-7c4f9a22db52 varXcondY = missing @@ -93,6 +99,12 @@ varYcondX = missing # ╔═╡ ce9e2ce2-26e0-4f17-adf5-c922ba98239d missing # analytical answer of (missing) +# ╔═╡ 19e16f77-ea34-45f6-83eb-8d256e5fd10d +md""" +!!! warning + Starting here, only part of the answer's structure will be given. You are therefore expected to **add more code cells** yourself, using the `+` at the left in between two cells. +""" + # ╔═╡ 9740ea64-cd4f-46b1-a741-02e392280601 md""" ## 2: Dice @@ -119,7 +131,7 @@ md"### 1: Watercube" # ╔═╡ 4fefa78b-746d-4e25-baee-d791eb22d930 md""" -!!! hint +!!! tip Consider the humble `DiscreteUniform` distribution. Not sure how it works? Open the **🔍 Live Docs** at the bottom right of the screen for more information """ @@ -134,11 +146,11 @@ md""" return dicesum end -# ╔═╡ 02bc37e2-5cf6-404a-b3fe-b2120671adb2 -watermodel = watercube() +# ╔═╡ 5fe1080e-8c2c-4c6b-84b4-3857dc1f399e +watercube_samples = missing # samples of dicesum for watercube # ╔═╡ b5886255-5c1d-4d84-b7ee-6c690fa526dc -p_watercube_kills = missing +p_watercube_kills = missing # probability that watercube kills the monster # ╔═╡ 84e06162-e3f1-4fd7-baf6-5095172413d2 missing # histogram @@ -150,15 +162,12 @@ md"### 2: Dirtprism" @model function dirtprism() # check the "For-loops for many variables" section from the intro notebook! - dicesum = missing # consider the `sum` function + dicesum = missing return dicesum end -# ╔═╡ 83afb3c1-9e6a-4d18-b0c7-05ed0173df40 -dirtmodel = dirtprism() - # ╔═╡ d9152416-8a7b-480c-ba9f-7ab15404b7a6 -p_dirtprism_kills = missing +p_dirtprism_kills = missing # probability that dirtprism kills the monster # ╔═╡ 90e058d7-b3fe-4c42-a652-3c42bf9d851a missing # histogram @@ -197,7 +206,7 @@ md"### 1: Probability" # ╔═╡ a0341046-c14a-494d-a9bd-a60c207c9e76 md""" -!!! hint +!!! tip In this exercise, the output variable (the number of double-yolked eggs) is **also a random variable**! In other words, it also follows some distribution. When considering what distribution, consider that each of the $N$ eggs represents a "trial" with a $P$ chance of success for a double yolk. @@ -261,7 +270,9 @@ end # ╟─def27d26-2205-4a66-94f3-eddbc17483bf # ╠═ff38df99-f843-414d-8e45-b46e06a65c22 # ╠═42c18a70-efb3-436b-83bb-b586280d4a4e +# ╠═a21b2f44-c8d7-4838-926f-8fa39c36884e # ╠═b22a18d5-f70b-42a5-a09e-3de515148a6d +# ╠═a11629df-31d2-4ac7-bf8e-80b910cab2fb # ╠═34bd60df-272f-49d7-9346-fb4d125fe89b # ╟─13989ba4-bcf8-4fdd-8aee-ab58c8905bc9 # ╟─aba42086-224f-44a0-b616-f8f651afdd18 @@ -273,18 +284,18 @@ end # ╠═aedd0fe8-da3e-4463-b0cf-7c4f9a22db52 # ╠═7ef53a87-e5df-4724-b896-3d1d46214c68 # ╠═ce9e2ce2-26e0-4f17-adf5-c922ba98239d +# ╟─19e16f77-ea34-45f6-83eb-8d256e5fd10d # ╟─9740ea64-cd4f-46b1-a741-02e392280601 # ╟─187854bb-9e30-454d-9e03-cccf77aebb6b # ╟─747e3c0a-357a-448a-b479-d0fcbe44a6c0 # ╟─eda95f45-083a-4e65-b57e-bd9890da1f9c # ╟─4fefa78b-746d-4e25-baee-d791eb22d930 # ╠═36477423-5628-4ed3-b54d-9a050557f6b7 -# ╠═02bc37e2-5cf6-404a-b3fe-b2120671adb2 +# ╠═5fe1080e-8c2c-4c6b-84b4-3857dc1f399e # ╠═b5886255-5c1d-4d84-b7ee-6c690fa526dc # ╠═84e06162-e3f1-4fd7-baf6-5095172413d2 # ╟─a1b933ac-5d1b-4800-a6e8-e942846b19d8 # ╠═6b009ba3-83a8-4176-86d4-dd9f70ed29ec -# ╠═83afb3c1-9e6a-4d18-b0c7-05ed0173df40 # ╠═d9152416-8a7b-480c-ba9f-7ab15404b7a6 # ╠═90e058d7-b3fe-4c42-a652-3c42bf9d851a # ╟─49790a8f-9f53-4ba7-9543-d6a879b520e0 diff --git a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl index e685e50..f61e298 100644 --- a/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl +++ b/exercises/student_notebooks/P4_probmod/probmod_3-advanced.jl @@ -169,14 +169,14 @@ md"### 2: Variance of Estimator" @bind n Slider(10:10:200, show_value = true) # ╔═╡ 55cf127f-5534-4926-955a-487ea9553b70 -estimated_force_sp = missing +force_samples = missing # multiple samples of your estimated force, using `n` samples # ╔═╡ 45f0813b-c4c9-4f13-8d66-1e58293c4422 missing # histogram of estimated force samples # ╔═╡ f6a67c1e-3677-4e51-b6dd-5d11a5146ea7 -estimated_force_sp_var = missing +force_samples_var = missing # variance of the estimated force # ╔═╡ Cell order: diff --git a/exercises/student_notebooks/P5_mcmc/MCMC_1-intro.jl b/exercises/student_notebooks/P5_mcmc/MCMC_1-intro.jl index 46f856a..c40109a 100644 --- a/exercises/student_notebooks/P5_mcmc/MCMC_1-intro.jl +++ b/exercises/student_notebooks/P5_mcmc/MCMC_1-intro.jl @@ -281,10 +281,10 @@ Taking the sampled values of the mutation rate from the chain and plotting a his """ # ╔═╡ b657217a-6ccc-4a41-b852-df4e39a7a10a -sp_alpha = mutation_chain[:α]; +alpha_samples = mutation_chain[:α]; # ╔═╡ 9cf08616-d599-42a3-82e8-e8c98853c1d8 -histogram(sp_alpha) # note the difference with the prior distribution! +histogram(alpha_samples) # note the difference with the prior distribution! # ╔═╡ e02c42dc-627c-4bc0-8ebb-fdb2b0f15b64 md"Plotting some sampled mutation rates from this distribution onto our data shows that they fit well:" @@ -295,17 +295,17 @@ begin ylabel = "Number of mutations", label = false, xlims = (0, 500), ylims = (0, 400), title = "A posteriori relationship between t and mean N" ); - plot!([x -> αᵢ*x for αᵢ in sp_alpha[1:10:end]], color = :purple, opacity = 0.1, label = false) + plot!([x -> αᵢ*x for αᵢ in alpha_samples[1:10:end]], color = :purple, opacity = 0.1, label = false) end # ╔═╡ 87059440-2919-4f6d-9d32-0df3ce75e2a2 md"And finally we can answer our first question:" # ╔═╡ 644cff58-68b9-4d4b-8896-617fcacc39c5 -mean(sp_alpha) +mean(alpha_samples) # ╔═╡ 5f2f5a78-ca90-4baf-8bf0-ff1fae88d785 -sqrt(var(sp_alpha)) +sqrt(var(alpha_samples)) # ╔═╡ bfbf811f-7b2f-473b-b132-0c7e965c8b0d md""" diff --git a/exercises/student_notebooks/P5_mcmc/MCMC_2-basics.jl b/exercises/student_notebooks/P5_mcmc/MCMC_2-basics.jl index 8c9f21e..e49044f 100644 --- a/exercises/student_notebooks/P5_mcmc/MCMC_2-basics.jl +++ b/exercises/student_notebooks/P5_mcmc/MCMC_2-basics.jl @@ -71,7 +71,7 @@ end molemodel = mole() # ╔═╡ 61b8ef63-a613-4dcb-8f7a-936e0d59b862 -spY = missing +Y_samples = missing # ╔═╡ 4e8b9000-cb61-4f01-9ba9-17276ad0335e E_Y = missing @@ -89,7 +89,7 @@ molechain = missing missing # plot molechain # ╔═╡ af4811ce-c6e7-458a-8f89-0562355b3eb0 -spXcondY = missing +X_samplescondY = missing # ╔═╡ fd540765-95e5-4071-81f7-e689b06cad0c E_XcondY = missing