# Monty Hall problem, with Thompson sampling

We all know the Monty Hall problem. Recently, Jason Rosenhouse published a book on that topic (entitled The Monty Hall Problem, The Remarkable Story of Math’s Most Contentious Brain Teaser). The game is more or less described by the following question

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

While I was preparing some slides for a lecture on Bayesian modeling and thinking, I wanted to find an illustration of what is sometimes called the Bayesian brain, that can be related to updates of beliefs, when we experience. And I was looking for examples of Thompson sampling. And actually, it is possible to learn that switching is the optimal strategy, in the Monty Hall problem, just by playing sequentially the game, and learning from previous strategies. The following code is used, to choose the door with the price (the car), and the one we first select

set.seed(1) n = 5000 listdoor = matrix(1:3,3,n) door = listdoor win = sample(1:3,size=n,replace=TRUE) pick1 = sample(1:3,size=n,replace=TRUE)

Then, the presenter picks one, that is neither the car, nor the one we chose initially. The following trick can be used, to get the list of available choices

door[win+(0:(n-1))*3] = NA door[,1:10] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] NA NA NA 1 NA NA 1 NA 1 NA [2,] 2 2 NA NA 2 2 2 NA NA 2 [3,] 3 NA 3 3 NA NA NA 3 NA NA door[pick1+(0:(n-1))*3] = NA door[,1:10] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] NA NA NA 1 NA NA 1 NA 1 NA [2,] 2 2 NA NA 2 2 2 NA NA 2 [3,] 3 NA 3 3 NA NA NA 3 NA NA

Then, the presenter picks one

presenter = apply(door,2, function(x) sample(x[!is.na(x)],size=1)) &gt; presenter[win != pick1] = apply(door,2,function(x) x[!is.na(x)])[win != pick1] presenter = unlist(presenter) presenter[1:10] [1] 3 2 3 1 2 2 2 3 1 2

Now, let us consider the  Monty Hall problem. We have two possible strategies. The first one is to keep the door we chose, initially

pick2a = pick1 gaina = (pick2a==win) mean(gaina) [1] 0.3392

As expected, on average, we win with (about) 1 chance out of 3. The second one is to (always) pick the other door (the one left). The code is close to the one we used before

door = listdoor door[pick1+(0:(n-1))*3] = NA door[presenter+(0:(n-1))*3] = NA pick2b = apply(door,2,function(x) x[!is.na(x)]) gainb = (pick2b==win) mean(gainb) [1] 0.6608

If you know Monty Hall problem the probability to win is now 2 chance out of 3 (which is what the maths tells us). That is what we have with simulations.

Now, what if we don’t know how to do the maths, and we don’t want to compute it? We can use Thompson sampling to explore, and exploit. In a general context, we have to choose among On a le choix entre $K$ alternatives (here $K=2$, since we can either keep our initial choice, or pick the other one), and the output is $\boldsymbol{X}=(X_1,\cdots, X_K)$, where $X_k\sim\mathcal{B}(\theta_k)$, but $\theta_k$ is unknow, and we will play the game, and learn. From previous computations, we know that $\theta_1=1/3$ while $\theta_2=2/3$.

We use some prior distribution, $\theta_k\sim\mathcal{B}eta(\alpha_k,\beta_k)$, since the Beta distribution is the conjugate of the Bernoulli. At time $t$, we draw $K$ (independent) Beta variables $B_k\sim\mathcal{B}eta(\alpha_k,\beta_k)$, and pick $k^\star = \displaystyle{\underset{k=1,\cdots,K}{\text{argmax}}\{B_k\}}$.  Here the code will be

set.seed(2) X = cbind(pick2a == win,pick2b == win)*1 AB1 = AB2 = tirage = matrix(NA,n,2) choix = rep(NA,n) k=1 AB1[k,] = AB2[k,] = c(1,1) for(k in 1:(n-1)){ tirage[k,] = c(rbeta(1,AB1[k,1],AB1[k,2]), rbeta(1,AB2[k,1],AB2[k,2])) choix[k] = which.max(tirage[k,]) if(choix[k] == 1){ AB1[k+1,] = AB1[k,] + c(X[k,1],1-X[k,1]) AB2[k+1,] = AB2[k,] } if(choix[k] == 2){ AB1[k+1,] = AB1[k,] AB2[k+1,] = AB2[k,] + c(X[k,2],1-X[k,2]) }}

Before showing some graphs, let us check that indeed, we select more the second strategy (which is here to select the other door)

AB1[n,] [1] 5 13 AB2[n,] [1] 3292 1693

Indeed, since the average of a Beta distribution, $\mathcal{B}eta(\alpha,\beta)$ is $\alpha/(\alpha+\beta)$

AB2[n,1]/(sum(AB2[n,])) [1] 0.6603811

i.e. the probability to win, with this second strategy is about $2/3$ (as obtained previously). We can visualize this on the animation below, with, in red the first strategy (keep your initial choice), in green the second one (select the other door), 0 and 1 respectively if we win, or not. Then we can visualize the evolution of $\alpha_2$ and $\beta_2$ on topc, and $\alpha_1$ and $\beta_1$ below (the index is time $t$). Finallly, we have the two variables $B_1$ and $B_2$ drawn,

Of course, another simulation would have given different $B_1$‘s and $B_2$‘s, but finally, we learn that the second strategy is better, and we learn it quite fast…

Here is another one (just to confirm)

So clearly, even if we don’t know which is the optimal strategy (keep our initial choice, or switch), a player who played that game about 30 times should be able to understand that switching should be a better strategy.

# Interpretability and explainability of predictive models

In 400 AD, in his Confessiones, Augustine wrote

quid est ergo tempus? si nemo ex me quaerat, scio; si quaerenti explicare velim, nescio

that can be translated as

What then is time? If no one asks me, I know what it is. If I wish to explain it to him who asks, I do not know.

To go a little further (because often, if we are asked to explain, we have some ideas), in A Study in Scarlet by Sir Arthur Conan Doyle, published in 1887, we have the following exchange, between Sherlock Holmes and Doctor Watson

– “I wonder what that fellow is looking for?” I asked, pointing to a stalwart, plainly-dressed individual who was walking slowly down the other side of the street, looking anxiously at the numbers. He had a large blue envelope in his hand, and was evidently the bearer of a message.
– “You mean the retired sergeant of Marines,” said Sherlock Holmes.

then, as it turns out that the person is indeed a sergeant in the navy (as is another character in the story, someone named Arthur Charpentier), Dr. Holmes asks him for an explanation, he wants to know how he arrived at this conclusion

– “How in the world did you deduce that?” I asked.
“Deduce what?” said he, petulantly.
“Why, that he was a retired sergeant of Marines.”
“I have no time for trifles,” he answered, brusquely; then with a smile, “Excuse my rudeness. You broke the thread of my thoughts; but perhaps it is as well. So you actually were not able to see that that man was a sergeant of Marines?”
“No, indeed.”
– “It was easier to know it than to explain why I knew it. If you were asked to prove that two and two made four, you might find some difficulty, and yet you are quite sure of the fact. Even across the street I could see a great blue anchor tattooed on the back of the fellow’s hand. That smacked of the sea. He had a military carriage, however, and regulation side whiskers. There we have the marine. He was a man with some amount of self-importance and a certain air of command. You must have observed the way in which he held his head and swung his cane. A steady, respectable, middle-aged man, too, on the face of him – all facts which led me to believe that he had been a sergeant.”

(to be honest, it is Liu Cixin who talks about it in The Three-Body Problem). For the record, this is the first story of the Holmes-Watson couple, which introduces Sherlock Holmes’ working method. For those who are familiar with the short stories, this narrative approach will be widely used thereafter: Sherlock Holmes states a fact, Dr. Watson is astonished and asks for an explanation, and Sherlock Holmes explains, point by point, how he arrived at this conclusion. This is a bit like the approach we try to implement when we build a predictive model: on the basis of the Titanic data, if we predict that such and such a person will die, and that such and such a person will survive, we want to understand why the model arrives at this conclusion.
Continue reading Interpretability and explainability of predictive models

# Could there be incentives to cycle through a red light?

This is of course a rhetorical question! Because cyclists must stop when the light is red! … But … there is always that moment, on a bicycle, when you stop, and  then you say to yourself

the worst part is that the lights are badly regulated, and I know that the next one will also be red once I will reach it … whereas if I had passed, I would have had the next one, and who knows, maybe a green wave afterwards?

Not having wanted to try the experiment with my bike, I wanted to try using simulations, on my computer.

Let us assume that in my city, there are red lights every 250 m, and they go from red to green in 1 minute, and similarly from green to red. Then, I can use a simple loop, where I compute the time from home, each time I pass a light: either the light is green, I go through, and I do not wait; or the light is red, then I wait, until the end of the minute. Here is the code. First, I define the constant parameter, with lights every 250 m, light turning either red or green every 1 minute, and I assume a constant speed here, of 15/60 km per min (or 15km per hour).

v = 15/60 d = .250 t = 1

Suppose that the office is at 10 km from home. The code to compute the time is simply (I do add a random noise on the time it takes to reach the next light, I will get back to that later on)

Dist = seq(0,10,by=.25) Time1=rep(NA,length(D)) Time1[1]==0 for(k in 2:length(Dist)){ noise=rnorm(1,sd=.05) Time1[k]=Time1[k-1]+d/v+noise if((floor(Time1[k])%%2==0)&amp;(k&lt;length(Dist))) Time1[k]=ceiling(Time1[k]) }

I can now visualize myself on my bike

T=seq(0,60,by=1/60) colr = c("red","green") plot(T,T*v,col="white") for(k in 1:40){ points(T,rep(d*k,length(T)),cex=.3,pch=15,col=colr[rep(1:2,each=t*60)]) } points(Time1,Dist,pch=19,cex=.5)

Here, it took me about 51 min to reach the office. And each time I have a red light, I stop and wait. Here is the distribution of the time it will take, as a function of my speed

simul = function(v=.250,sd=.05){ Dist = seq(0,10,by=.25) Time1=Time2=rep(NA,length(D)) Time1[1]=Time2[1]=0 for(k in 2:length(Dist)){ noise=rnorm(1,sd=sd) Time1[k]=Time1[k-1]+d/v+noise if((floor(Time1[k])%%2==0)&amp;(k&lt;length(Dist))) Time1[k]=ceiling(Time1[k]) } max(Time1) } S=function(v=.250,sd=.05,n=1000){ Vectorize(function(x) simul(v=x,sd=sd))(rep(v,n)) } vit = seq(.2,.3,length=101) MS = Vectorize(function(x) S(v=x,sd=.05))(vit) qMSsup = apply(MS,2,function(x) quantile(x,.95)) qMSinf = apply(MS,2,function(x) quantile(x,.05)) MSmed = apply(MS,2,function(x) quantile(x,.5)) MSmean = apply(MS,2,mean) par(mfrow = c(1,1)) plot(vit,MSmed,type="l") polygon(c(vit,rev(vit)),c(qMSinf,rev(qMSsup)),col="light blue",border=NA) lines(vit,MSmed,lwd=2)

Obviously, on average, the faster I cycle, the shorter the ride will be. Of course, 40 min is not a (real) lower bound : if I go much faster, it can be a rather quick ride

Now, consider a simple rhetorical alternative. What if I decide to go through a red light (after checking that there is no danger), say, with 1 chance of out 20 ?

simul_random = function(v=.250,sd=.05){ Dist = seq(0,10,by=.25) Time1=rep(NA,length(D)) Time1[1]=0 for(k in 2:length(Dist)){ noise=rnorm(1,sd=sd) Time1[k]=Time1[k-1]+d/v+noise red = sample(c(0,1),size=1,prob = c(.95,.05)) if((floor(Time1[k])%%2==0)&amp;(k&lt;length(Dist))&amp;(red == 0)) Time1[k]=ceiling(Time1[k]) } max(Time1) }

Here, 1 chance out of 20 could mean that on a 10 km ride, I wil always stop : first, I need a red light, and then 95% of the time, I stop. Here is the average distribution, with confidence bands

S_random=function(v=.250,sd=.05,n=1000){ Vectorize(function(x) simul_random(v=x,sd=sd))(rep(v,n)) } MS3 = Vectorize(function(x) S_random(v=x,sd=.05))(vit) qMSsup3 = apply(MS3,2,function(x) quantile(x,.95)) qMSinf3 = apply(MS3,2,function(x) quantile(x,.05)) MSmed3 = apply(MS3,2,function(x) quantile(x,.5))

It is possible to compare the two actually (on average times)

plot(MSmed,MSmed3,type="l") abline(a=0,b=1,col="red",lty=2) plot(MSmed,(MSmed-MSmed3)/MSmed*100,type="l")

which means that, somehow I can save some time, maybe from 3% to 5% if I do not cycle to fast, otherwise probably less than 2%. I would not claim, here, that it is worth it.

Consider another rhetorical alternative. What if I decide to go through the red light only if it is during the very first second ?

simul_1sec = function(v=.250,sd=.05){ Dist = seq(0,10,by=.25) Time1=rep(NA,length(D)) Time1[1]=0 for(k in 2:length(Dist)){ noise=rnorm(1,sd=sd) Time1[k]=Time1[k-1]+d/v+noise if((floor(Time1[k])%%2==0)&amp;(k&lt;length(Dist))&amp;(Time1[k]-floor(Time1[k])&gt;1/60)) Time1[k]=ceiling(Time1[k]) } max(Time1) }

Here is the time it takes to go to the office

and here is the potential gain

We can now clearly see the impact of those nonlinearities : with my average 15 km per hour speed, I can save 8% of the time if I go through red during the very first second. Of course, I will never do such a think, but mathematically speaking, it is stricking. Here is the gain for the first 2 seconds (out of a full minute)

so the gain can be about 12%. And now, what if we remove the noise – or more precisely, what if the standard deviation become 0.001 instead of 0.05. Here is the first distribution of the time (when I stop at each red light, and wait)

If I do not go fast enough, I will stop at almost each light, and wait until the end of the minute: going at 20 km per hour or 24 km per hour is exactly the same. But if I can go slightly faster than 25km per hour, that is awesome, and I have my green wave. Now, her is the graph I get when I cycle through the red light only in the very first second

This is now the gain

I can save up to 40% of the time, and more realistically, my 55 min ride could now be a 40 min ride. Which is substantial.

But on that one, observe that another strategy is also possible : what if I do not cycle through red light, but I simply cycle +1% faster ?

simul_faster = function(v=.250,sd=.05){ Dist = seq(0,10,by=.25) Time1=rep(NA,length(D)) Time1[1]=0 for(k in 2:length(Dist)){ noise=rnorm(1,sd=sd) Time1[k]=Time1[k-1]+d/(v*1.01)+noise if((floor(Time1[k])%%2==0)&amp;(k&lt;length(Dist))&amp;(Time1[k]-floor(Time1[k])&gt;1/60)) Time1[k]=ceiling(Time1[k]) } max(Time1) }

We have a similar result here, since riding 1% faster can actually help me save up to 50% of my time ! and (more realistically) if I ride 1% faster with a large random noise in the time between two light, I get almost the same as previously (when passing through the red light)

Again, my point here is not that we should cycle through a red light, of course not. But there may be accumulation of (small) nonlinear effects that might have a major impact at the end. And I believe that the best way to avoid this is to offer a green wave to cyclists, assuming that they ride at a reasonnable speed…

# From multinomial regression to binary classification on some Siamese data

There are two kinds of people in the world: people who think there are two kinds of people in the world and people who don’t

(borrowed from Menand (2018)). Because things are always simpler when we face only binary choice, aren’t they? But consider here the case were multiple options are possible, and let us see if we cannot get back to simpler binary choices. Consider a collection of observations $(y_i,\boldsymbol{x}_i)$ where $y_i$ is some categorical variable, $y_i\in\mathcal{A}$ where $\mathcal{A}=\lbrace A_1,\cdots,A_\kappa \rbrace$, with $\kappa$ possible categories. Let $\mathcal{I}_k=\lbrace i:y_i\in A_k \rbrace$.

In a classical multinomial logistic regression, suppose that $A_1$ is the reference, then $$\mathbb{P}[Y=A_j|\boldsymbol{X}=\boldsymbol{x}]=\frac{\exp[\boldsymbol{x}^\top\boldsymbol{\beta}_j]}{1+\exp[\boldsymbol{x}^\top\boldsymbol{\beta}_2]+\cdots+\exp[\boldsymbol{x}^\top\boldsymbol{\beta}_k]}$$With a lot a categories, and a small number of observations, inference can be complicated, and non-robust.

• the Siamese dataset

The name Siamese I use, here, comes from Siamese Networks. Or sort of… As we say in French, it is an « histoire de l’homme qui a vu l’homme qui a vu l’ours » (story of the man who saw the man who saw the bear). A few years ago, a student tried to explain to me the idea of Siamese Networks and this is what I understood. I might be completely wrong, but the idea I got from it did make sense, in my mind at least. That is the story of that blog post…

The idea of the siamese algorithm will be to consider all pairs of observations, $(y_i,\boldsymbol{x}_i)$ and $(y_j,\boldsymbol{x}_j)$ :

1. $\tilde y_{i,j}=\boldsymbol{1}(y_i=y_j)$ indicating if individuals $i$ and $j$ are in the same category
2. $\tilde{\boldsymbol{x}}_{i,j}$ is a collection of $p-1$ variables,
• $\tilde {x}_{k:i,j}={x}_{k:i}-{x}_{k:j}$ if $x_k$ is continuous, or $\tilde {x}_{k:i,j}=|{x}_{k:i}-{x}_{k:j}|$ (we can use another metric, e.g. $\tilde {x}_{k:i,j}=|{x}_{k:i}-{x}_{k:j}|^2$, and this is why I decided to use some GAM model in the logistic regression on the Siamese dataset)
• $\tilde {x}_{k:i,j}=({x}_{k:i},{x}_{k:j})\in\mathcal{X}_k\times\mathcal{X}_k$ if $x_k$ is a categorical variables (taking values in the set $\mathcal{X}_k$), or $\tilde {x}_{k:i,j}=\boldsymbol{1}({x}_{k:i}\neq{x}_{k:j})\in\{0,1\}$

The original dataset was a $n\times p$ matrix, and (if there are no categorical variable), it becomes a $n(n-1)/2\times p$ matrix. The key point is that if the original variable $y_i$ was multinomial, $y_{i,j}$ is now binomial. For instance, if our initial dataset was the following, with two covariates, one continuous and one categorical

its siamese counterpart is the following

• Classification step

On the dataset $(\tilde y_{i,j},\tilde {x}_{k:i,j})_{i,j}$, fit a logistic regression, $$\mathbb{P}[\tilde Y|\tilde{\boldsymbol{X}}=\tilde{\boldsymbol{x}}]=\frac{\exp[\tilde{\boldsymbol{x}}^\top\boldsymbol{\beta}]}{1+\exp[\tilde{\boldsymbol{x}}^\top\boldsymbol{\beta}]}$$(or any classification model – CART, random forest, etc). But that is the easy part (unless $n$ is large, because the siamese dataset has (roughly) $n^2/2$ rows). The difficult task is the prediction

• Prediction step

Consider a new input variable $\boldsymbol{x}_{\cdot}$, and define its siamese version, $\tilde{\boldsymbol{x}}_{\cdot}=(\tilde{\boldsymbol{x}}_{\cdot,j})_j$, i.e. a database with $n$ rows. Then compute
$$p_{\cdot,j}=\mathbb{P}[\tilde Y|\tilde{\boldsymbol{X}}=\tilde{\boldsymbol{x}}_{\cdot,j}]=\frac{\exp[\tilde{\boldsymbol{x}}_{\cdot,j}^\top\boldsymbol{\beta}]}{1+\exp[\tilde{\boldsymbol{x}}_{\cdot,j}^\top\boldsymbol{\beta}]}$$ where $p_{\cdot,j}$ is the probability that $(y_j,\boldsymbol{x}_{j})$ and $(y_{\cdot},\boldsymbol{x}_{\cdot})$ are in the same category, as well as $$p_{i,j}=\mathbb{P}[\tilde Y|\tilde{\boldsymbol{X}}=\tilde{\boldsymbol{x}}_{i,j}]=\frac{\exp[\tilde{\boldsymbol{x}}_{i,j}^\top\boldsymbol{\beta}]}{1+\exp[\tilde{\boldsymbol{x}}_{i,j}^\top\boldsymbol{\beta}]}$$
Let $\boldsymbol{p}_{\cdot}=(p_{\cdot,j})$, and similarly $\boldsymbol{p}_{i}=(p_{i,j})$. Then several techniques can be used to predict $y_{\cdot}$.

1. $\widehat{y}_{\cdot}=y_{j^\star}$ where ${j^\star}=\underset{j=1,\cdots,n}{\text{argmax}}\{p_{\cdot,j}\}$: the predicted class is the one of the observation the most likely to be other same class
2. $\widehat{y}_{\cdot}=y_{j^\star}$ where ${j^\star}=\underset{\ell=1,\cdots,k}{\text{argmax}}\{\overline{p}_{\ell}\}$, where$$\overline{p}_{\ell} = \frac{1}{n_{\ell}}\sum_{j\in\mathcal{I}_s} \boldsymbol{1}(y_j =y_{\ell}),\text{ where }\mathcal{I}_s=\lbrace i:p_{\cdot,i}>s\rbrace$$consider only probabilities sufficiently high to be considered, and predict the most important class (majority rule)
3. $\widehat{y}_{\cdot}=y_{j^\star}$ where $${j^\star}=\underset{i=1,\cdots,n}{\text{argmax}}\{\theta_i\}$$ where $$\theta_i=\cos(\boldsymbol{p}_{\cdot},\boldsymbol{p}_{i})=\displaystyle{\frac{\boldsymbol{p}_{\cdot}\cdot\boldsymbol{p}_{i}}{\|\boldsymbol{p}_{\cdot}\|\|\boldsymbol{p}_{i}\|}}$$
4. $\widehat{y}_{\cdot}=y_{j^\star}$ where $${j^\star}=\underset{i=1,\cdots,n}{\text{argmax}}\{KL_{\cdot|i}\}$$ and $KL_{\cdot|i}=\displaystyle{\sum_{j=1}^n p_{\cdot,j}\log\frac{p_{\cdot,j}}{p_{i,j}}}$ (but one can select another metric)
5. $\widehat{y}_{\cdot}=y_{j^\star}$ where ${j^\star}=\underset{j\in\mathcal{J}}{\text{argmax}}\{p_{\cdot,j}\}$ and $\mathcal{J}$ is a sample of $k$ observations, chosen randomly, one in each group (one-shot procedure): the predicted class is the one of the observation the most likely to be other same class

Heuristically, it can be related to some $k$ nearest neighbors strategy: we give the attribute that most neighbors have. The total distance is a weighted sum of the componentwise distances (for the logistic regression).

• Simulation study

In order to test that technique, let us generate some multinomial model where $y$ has 10 possible labels, with 6 (independent) covariates $x_1,\cdots,x_6$, and $\mathbb{P}[Y=A_k|\boldsymbol{X}=\boldsymbol{x}]\propto \exp[\boldsymbol{x}^\top\boldsymbol{\beta}_k]$ (where coefficients $\boldsymbol{\beta}_k$ where generated randomly) for $k\in\{1,2,\cdots,10\}$ (there were 10 categories) and with $n=700$ observations.

n=700 X1=rnorm(n) X2=rnorm(n) X3=rnorm(n) X4=rnorm(n) X5=rnorm(n) X6=rnorm(n) X=cbind(1,X1,X2,X3,sqrt(abs(X4)),X5*X1,X6) k = 10 PARAM = matrix(rnorm(k*6),k,6) PARAM[,1]=PARAM[,1]-1 PARAM=cbind(PARAM,0) P=matrix(NA,n,k-1) for(j in 1:(k-1)) P[,j] = X %*% (PARAM[j,])+rnorm(n) P=cbind(P,0) S=apply(exp(P),1,sum) Pb = exp(P)/S tirage = function(i){ sample(1:10,size=1,prob = Pb[i,]) } Y = LETTERS[Vectorize(tirage)(1:n)] dbase = data.frame(Y=as.factor(Y),X1,X2,X3,X4,X5)

In the paragraph previously, I suggested to take the most likely one. Being wrong means that it was not the first choice. But perhaps being the second or the third choice is not that bad, actually. So in my simulations, I look at the proportion of predictions were our prediction is the good one (top 1), if the true one is either the most likely or the second most likely (top 2), or in the top 3. That will be on my $x$-axis. I draw some line, but we simply have three points (top 1, top 2 and top 3). I compute the proportion of good prediction, using cross-validation techniques (10-fold). The black lines are one of the methods described above. The red one is the standard multinomial model (with a logistic link function). For the Siamese model, I tried several models. I tried a logistic regression, and some smooth version (GAM) on top

and a classification tree, on the left, as well as some random forest on the right, below.

It looks like the multinomial approach performs always better than any Siamese one… and to be honest, I am disappointed.

Here is the code I did use when I considered a logistic regression on the Siamese dataset,

set.seed(1) kfold = sample(rep(1:10,n/10))   KFOLDglm = function(i){ i_test=which(kfold==i) i_calibration=which(kfold!=i) y=credit[i_calibration,"Y"] tirage = function(){ v=c(sample(i_calibration[y==levels(y)[1]],size=1), sample(i_calibration[y==levels(y)[2]],size=1), sample(i_calibration[y==levels(y)[3]],size=1), sample(i_calibration[y==levels(y)[4]],size=1), sample(i_calibration[y==levels(y)[5]],size=1), sample(i_calibration[y==levels(y)[6]],size=1), sample(i_calibration[y==levels(y)[7]],size=1), sample(i_calibration[y==levels(y)[8]],size=1), sample(i_calibration[y==levels(y)[9]],size=1), sample(i_calibration[y==levels(y)[10]],size=1)) names(v)=levels(y) return(v) }   LogisticModel &lt;- multinom(Y ~ ., data = credit[i_calibration,], trace=FALSE)   comparaisonx = function(base,x=base[1,]){ mix_base = base for(j in 1:ncol(base)){ xj = as.numeric(x[j])-base[,j] mix_base[,j] = (xj) } mix_base } comparaisony = function(base,y=base[1]){ as.factor(base == y) } creditx = credit[,-which(names(credit) == "Y")] nc=length(i_calibration) B=comparaisonx(base = creditx[i_calibration[2:nc],],x=creditx[i_calibration[1],]) B$Y=comparaisony(base = credit[i_calibration[2:nc],"Y"],y=credit[i_calibration[1],"Y"]) for(i in 2:(nc-1)){ B0=comparaisonx(base = creditx[i_calibration[(i+1):nc],],x=creditx[i_calibration[i],]) B0$Y=comparaisony(base = credit[i_calibration[(i+1):nc],"Y"],y=credit[i_calibration[i],"Y"]) B=rbind(B,B0) } credit_mix = B   OneShotLogisticModel &lt;- glm(Y ~ ., data = credit_mix, family=binomial) A_ref = table(credit[i_calibration,"Y"])/length(i_calibration)   vect_oneshot = function(i){ B2=comparaisonx(base = creditx[i_calibration,],x=creditx[i,]) predict(OneShotLogisticModel,type="response",newdata=B2) }   prediction_oneshot = function(i,type=1){ B2=comparaisonx(base = creditx[i_calibration,],x=creditx[i,]) p=predict(OneShotLogisticModel,type="response",newdata=B2) y=credit[i_calibration,"Y"] base = data.frame(p,y) base = base[rev(order(base$p)),] if(type==1){T = table(base$y[1:11]) return(names(which.max(T)))} if(type==2){return(base$y[1])} if(type==3){A=table(base$y[1:10])/10 T=A/A_ref return(names(which.max(T)))} if(type==4){ costheta = rep(NA,length(i_calibration)) for(j in 1:length(i_calibration)){ vecteur_proba = vect_oneshot(i_calibration[j]) costheta[j] = sum(vecteur_proba*p)/(sqrt(sum(vecteur_proba^2))*sqrt(sum(p^2))) } return(y[which.max(costheta)])} if(type==5){ kl = rep(NA,length(i_calibration)) for(j in 1:length(i_calibration)){ vecteur_proba = vect_oneshot(i_calibration[j]) kl[j] = sum(p*log(vecteur_proba/p)) } return(y[which.max(as.vector(kl))])} if(type==6){ ## one shot : tirer au hasard un de chaque, et dire lequel est plus credible !   y=credit[i_calibration,"Y"] tirage = function(){ v=c(sample(i_calibration[y==levels(y)[1]],size=1), sample(i_calibration[y==levels(y)[2]],size=1), sample(i_calibration[y==levels(y)[3]],size=1), sample(i_calibration[y==levels(y)[4]],size=1), sample(i_calibration[y==levels(y)[5]],size=1), sample(i_calibration[y==levels(y)[6]],size=1), sample(i_calibration[y==levels(y)[7]],size=1), sample(i_calibration[y==levels(y)[8]],size=1), sample(i_calibration[y==levels(y)[9]],size=1), sample(i_calibration[y==levels(y)[10]],size=1)) names(v)=levels(y) return(v) } pd=rep(NA,101) for(ix in 1:101){ ids = tirage() B2=comparaisonx(base = creditx[ids,],x=creditx[i,]) p=predict(OneShotLogisticModel,type="response",newdata=B2) pd[ix]=levels(y)[which.max(p)] } levels(y)[which.max(table(pd))] } } PRED0=as.character(credit[i_test,"Y"]) PRED1=as.character(predict(LogisticModel,type = "class", newdata=credit[i_test,])) PRED21=as.character(Vectorize(function(i) prediction_oneshot(i,type=1)) (i_test)) PRED22=as.character(Vectorize(function(i) prediction_oneshot(i,type=2))(i_test)) PRED23=as.character(Vectorize(function(i) prediction_oneshot(i,type=3))(i_test)) PRED24=as.character(Vectorize(function(i) prediction_oneshot(i,type=4))(i_test)) PRED25=as.character(Vectorize(function(i) prediction_oneshot(i,type=5))(i_test)) PRED26=as.character(Vectorize(function(i) prediction_oneshot(i,type=6))(i_test)) B=data.frame(PRED0,PRED1,PRED21,PRED22,PRED23,PRED24,PRED25,PRED26) B }   s=1/100;setTxtProgressBar(pb, s*2) for(i in 2:10){ PREDICTION = rbind(PREDICTION,KFOLDglm(i)) s=s+1/100;setTxtProgressBar(pb, s*2)} for(j in 1:5) PREDICTION[,j]=as.character(PREDICTION[,j]) L=list() v=mean(PREDICTION[,1]!=PREDICTION[,2]) names(v)="logistic" L[["logistic"]]=v v=c(mean(PREDICTION[,1]!=PREDICTION[,3]), mean(PREDICTION[,1]!=PREDICTION[,4]), mean(PREDICTION[,1]!=PREDICTION[,5]), mean(PREDICTION[,1]!=PREDICTION[,6]), mean(PREDICTION[,1]!=PREDICTION[,7]), mean(PREDICTION[,1]!=PREDICTION[,8])) names(v)=c("top10","max","10norm","cos","KL","OS") L[["glm"]]=v

# Some general thoughts on Partial Dependence Plots with correlated covariates

The partial dependence plot is a nice tool to analyse the impact of some explanatory variables when using nonlinear models, such as a random forest, or some gradient boosting.The idea (in dimension 2), given a model $m(x_1,x_2)$ for $\mathbb{E}[Y|X_1=x_1,X_2=x_2]$. The partial dependence plot for variable $x_1$ is model $m$ is function $p_1$ defined as $x_1\mapsto\mathbb{E}_{\mathbb{P}_{X_2}}[m(x_1,X_2)]$. This can be approximated, using some dataset using $$\widehat{p}_1(x_1)=\frac{1}{n}\sum_{i=1}^n m(x_1,x_{2,i})$$My concern here what the interpretation of that plot when there are some (strongly) correlated covariates. Let us generate some dataset to start with

n=1000 library(mnormt) r=.7 set.seed(1234) X = rmnorm(n,mean = c(0,0),varcov = matrix(c(1,r,r,1),2,2)) Y = 1+X[,1]-2*X[,2]+rnorm(n)/2 df = data.frame(Y=Y,X1=X[,1],X2=X[,2])

As we can see, the true model is here is $y_i=\beta_0+\beta_1 x_{1,i}+\beta_2x_{2,i}+\varepsilon_i$ where $\beta_1 =1$ but the two variables are positively correlated, and the second one has a strong negative impact. Note that here

reg = lm(Y~.,data=df) summary(reg)   Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) 1.01414 0.01601 63.35 &lt;2e-16 *** X1 1.02268 0.02305 44.37 &lt;2e-16 *** X2 -2.03248 0.02342 -86.80 &lt;2e-16 ***

If we estimate a wrongly specified model $y_i=b_0+b_1 x_{1,i}+\eta_i$, we would get

reg1 = lm(Y~X1,data=df) summary(reg1)   Coefficients: Estimate Std. Error t value Pr(&gt;|t|) (Intercept) 1.03522 0.04680 22.121 &lt;2e-16 *** X1 -0.44148 0.04591 -9.616 &lt;2e-16 ***

Thus, on the proper model, $\widehat{\beta}_1\sim+1.02$ while $\widehat{b}_1\sim-0.44$  on the mispecified model.

Now, let us look at the parial dependence plot of the good model, using standard R dedicated packages,

library(pdp) pdp::partial(reg, pred.var = "X1", plot = TRUE, plot.engine = "ggplot2")

which is the linear line $y=1+x$, that corresponds to $y=\beta_0+\beta_1x$.

library(DALEX) plot(DALEX::single_variable(DALEX::explain(reg, data=df),variable = "X1",type = "pdp"))

which corresponds to the previous graph. Here, it is also possible to creaste our own function to compute that partial dependence plot,

pdp1 = function(x1){ nd = data.frame(X1=x1,X2=df$X2) mean(predict(reg,newdata=nd)) } that will be the straight line below (the dotted line is the theoretical one $y=1+x$, vx=seq(-3.5,3.5,length=101) vpdp1 = Vectorize(pdp1)(vx) plot(vx,vpdp1,type="l") abline(a=1,b=1,lty=2) which is very different from the univariate regression on $x_1$ abline(reg1,col="red") Actually, the later is very consistent with a local regression, only on $x_1$ library(locfit) lines(locfit(Y~X1,data=df),col="blue") Now, to get back to the definition of the partial dependence plot, $x_1\mapsto\mathbb{E}_{\mathbb{P}_{X_2}}[m(x_1,X_2)]$, in the context of correlated variable, I was wondering if it would not make more sense to consider some local version actually, something like $x_1\mapsto\mathbb{E}_{\mathbb{P}_{X_2|X_1}}[m(x_1,X_2)]$. My intuition was that, somehow, it did not make any sense to consider any $X_2$ while $X_1$ was fixed (and equal to $x_1$). But it would make more sense actually to look at more valid $X_2$‘s given the value of $X_1$. And a natural estimate could be some $k$ neareast-neighbors, i.e. $$\tilde{p}_1(x_1)=\frac{1}{k}\sum_{i\in\mathcal{V}_k(x)}^n m(x_1,x_{2,i})$$ where $\mathcal{V}_k(x)$ is the set of indices of the $k$ $x_i$‘s that are the closest to $x$, i.e. lpdp1 = function(x1){ nd = data.frame(X1=x1,X2=df$X2) idx = rank(abs(df$X1-x1)) mean(predict(reg,newdata=nd[idx&lt;50,])) } vlpdp1 = Vectorize(lpdp1)(vx) lines(vx,vlpdp1,col="darkgreen",lwd=2) Surprisingly (?), this local partial dependence plot gives a curve that corresponds to the simple regression… # 3rd Insurance Data Science Conference Registrations and call for abstracts, for the 3rd Insurance Data Science Conference, organised on-line 16 – 18 June 2021 (PM in Europe, AM in America), are now open. See https://insurancedatascience.org/ for more details… # Lilliefors, Kolmogorov-Smirnov and cross-validation In statistics, Kolmogorov–Smirnov test is a popular procedure to test, from a sample $\{x_1,\cdots,x_n\}$ is drawn from a distribution $F$, or usually $F_{\theta_0}$, where $F_{\theta}$ is some parametric distribution. For instance, we can test $H_0:X_i\sim\mathcal{N(0,1)}$ (where $\theta_0=(\mu_0,\sigma_0^2)=(0,1)$) using that test. More specifically, I wanted to discuss today $p$-values. Given $n$ let us draw $\mathcal{N}(0,1)$ samples of size $n$, and compute the $p$-values of Kolmogorov–Smirnov tests n=300 p = rep(NA,1e5) for(s in 1:1e5){ X = rnorm(n,0,1) p[s] = ks.test(X,"pnorm",0,1)$p.value }

We can visualise the distribution of the $p$-values below (I added some Beta distribution fit here)

library(fitdistrplus) fit.dist = fitdist(p,"beta") hist(p,probability = TRUE,main="",xlab="",ylab="") vu = seq(0,1,by=.01) vv = dbeta(vu,shape1 = fit.dist$estimate[1], shape2 = fit.dist$estimate[2]) lines(vu,vv,col="dark red", lwd=2)

It looks like it is quite uniform (theoretically, the $p$-value is uniform). More specifically, the $p$-value was lower than 5% in 5% of the samples

[note: here I compute ‘mean(p<=.05)’ but I have some trouble with the ‘<‘ and ‘>’ symbols, as always]

mean(p&lt;=.05) [1] 0.0479

i.e. we wrongly reject $H_0:X_i\sim\mathcal{N(0,1)}$ is 5% of the samples.

As discussed previously on the blog, in many cases, we do care about the distribution, and not really the parameters, so we wish to test something like $H_0:X_i\sim\mathcal{N(\mu,\sigma^2)}$, for some $\mu$ and $\sigma^2$. Therefore, a natural idea can be to test $H_0:X_i\sim\mathcal{N(\hat\mu,\hat\sigma^2)}$, for some estimates of $\mu$ and $\sigma^2$. That’s the idea of Lilliefors test. More specifically, Lilliefors test suggests to use , Kolmogorov–Smirnov statistics, but corrects the $p$-value. Indeed, if we draw many samples, and use Kolmogorov–Smirnov statistics and its classical $p$-value to test for $H_0:X_i\sim\mathcal{N(\hat\mu,\hat\sigma^2)}$,

n=300 p = rep(NA,1e5) for(s in 1:1e5){ X = rnorm(n,0,1) p[s] = ks.test(X,"pnorm",mean(X),sd(X))$p.value } we see clearly that the distribution of $p$-values is no longer uniform fit.dist = fitdist(p,"beta") hist(p,probability = TRUE,main="",xlab="",ylab="") vu = seq(0,1,by=.01) vv = dbeta(vu,shape1 = fit.dist$estimate[1], shape2 = fit.dist$estimate[2]) lines(vu,vv,col="dark red", lwd=2) More specifically, if $x_i$‘s are actually drawn from some Gaussian distribution, there are no chance to reject $H_0$, the $p$-value being almost never below 5% mean(p&lt;=.05) [1] 0.00012 Usually, to interpret that result, the heuristics is that $\hat\mu$ and $\hat\sigma^2$ are both based on the sample, while previously $0$ and $1$ where based on some prior knowledge. Somehow, it reminded me on the classical problem when mention when we introduce cross-validation, which is Goodhart’s law When a measure becomes a target, it ceases to be a good measure i.e. we cannot assess goodness of fit using the same data as the ones used to estimate parameters. So here, why not use some hold-out (or cross-validation) procedure : split the dataset in two parts, $\{x_1,\cdots,x_k\}$ (with $k) to estimate parameters $\mu$ and $\sigma^2$ and then use $\{x_{k+1},\cdots,x_n\}$ and Kolmogorov–Smirnov statistics on it to test if $x_i$‘s are drawn from some Gaussian distribution. More precisely, will the $p$-value computed using the standard Kolmogorov–Smirnov procedure be ok here. Here, I tried two scenarios, $k/n$ being either $1/3$ or $2/3$, p = matrix(NA,1e5,4) for(s in 1:1e5){ X = rnorm(n,0,1) p[s,1] = ks.test(X,"pnorm",0,1)$p.value p[s,2] = ks.test(X,"pnorm",mean(X),sd(X))$p.value p[s,3] = ks.test(X[1:200],"pnorm",mean(X[201:300]),sd(X[201:300]))$p.value p[s,4] = ks.test(X[201:300],"pnorm",mean(X[1:200]),sd(X[1:200]))$p.value } Again, we can visualize the distributions of $p$-values, in the case where $1/3$ of the data is used to estimate $\mu$ and $\sigma^2$, and $2/3$ of the data is used to test fit.dist = fitdist(p[,3],"beta") hist(p[,3],probability = TRUE,main="",xlab="",ylab="") vu=seq(0,1,by=.01) vv=dbeta(vu,shape1 = fit.dist$estimate[1], shape2 = fit.dist$estimate[2]) lines(vu,vv,col="dark red", lwd=2) and in the case where $2/3$ of the data is used to estimate $\mu$ and $\sigma^2$, and $1/3$ of the data is used to test fit.dist = fitdist(p[,4],"beta") hist(p[,4],probability = TRUE,main="",xlab="",ylab="") vu=seq(0,1,by=.01) vv=dbeta(vu,shape1 = fit.dist$estimate[1], shape2 = fit.dist$estimate[2]) lines(vu,vv,col="dark red", lwd=2) Observe here that we (wrongly) reject too frequently $H_0$, since the $p$-values are below 5% in 25% of the scenarios, in the first case (less data used to estimate), and 9% of the scenarios, in the second case (less data used to test) mean(p[,3]&lt;=.05) [1] 0.24168 mean(p[,4]&lt;=.05) [1] 0.09334 We can actually compute that probability as a function of $k/n$ n=300 p = matrix(NA,1e4,99) for(s in 1:1e4){ X = rnorm(n,0,1) KS = function(p) ks.test(X[1:(p*n)],"pnorm",mean(X[(p*n+1):n]),sd(X[(p*n+1):n]))$p.value p[s,] = Vectorize(KS)((1:99)/100) }

The evolution of the probability is the following

prob5pc = apply(p,2,function(x) mean(x&lt;=.05)) plot((1:99)/100,prob5pc)

so, it looks like we can use some sort of hold-out procedure to test for $H_0:X_i\sim\mathcal{N(\mu,\sigma^2)}$, for some $\mu$ and $\sigma^2$, using Kolmogorov–Smirnov test with $\mu=\hat\mu$ and $\sigma^2=\hat\sigma^2$ but the proportion of data used to estimate those quantities should be (much) larger that the one used to compute the statistics. Otherwise, we clearly reject too frequently $H_0$.

# Insurance Pricing Game

Would you like to put your data science skills to the test?

Imperial College London, Universite du Quebec à Montreal (UQAM), and actuarial institutes in Singapore, the UK, including the IFoA, and Australia, ASTIN, the Casualty Actuarial Society are co-organising a global data science competition.

Would you like to accurately predict the cost of insurance by putting your data science skills to the test? We are hosting two competitions with separate datasets, a loss prediction competition on Kaggle with synthetic workers’ compensation data, and a pricing competition in a simulated market hosted on AI Crowd with real-world motor insurance contracts. Codes can be either in R or python. The competition is being sponsored by a number of different organisations, with a total of US$12,000 in cash prizes to be won. For more information about how to take part please visit www.pricing-game.com # Trees and forests For my ACT6100 weekly quiz, I usually generate some datasets, and then ask students to compare various predictive algorithms. Last week, it was about classification trees and random forests. And students were surprised to have such differences (they had to estimate the probability to have a specific label, for the barycenter of the covariates). Usually, I use the following to generate some (here 12) covariates that could be correlated library(FactoMineR) n=279 library(clusterGeneration) library(mnormt) k=12 S=genPositiveDefMat("unifcorrmat",dim=k) X=round(rmnorm(n,varcov=S$Sigma)+8,2) rownames(X)=1:n colnames(X)=LETTERS[1:k]

Then I need to generate some data, based on some covariates (5 out of 12), with various strengths

idx = sample(1:k,size=5) u = sample(c(-(4:1),1:4),5) beta = rep(0,k) beta[idx] = u U = X%*%beta U = U-min(U) U = U/max(U)*6-3 p = exp(( U))/(1+exp((U ))) Y = rbinom(n,size=1,prob=p) df = data.frame(Y=as.factor(Y),X) levels(df$Y)=levels=c("blue","red") We can run a classification tree library(rpart) arbre = rpart(Y~., data=df) and a random forest, library(randomForest) set.seed(1) arbres = randomForest(Y~., data=df) Here are the partial plots for 4 of the explanatory variables that actually have an impact partialPlot(arbres,pred.data = df, x.var = "A") Predictions for the “average” point of the dataset is here (parbre = predict(arbre,newdata=data.frame(t(apply(df[,-1],2,mean))),type = "prob")) blue red 1 0.8064516 0.1935484 (parbres = predict(arbres,newdata=data.frame(t(apply(df[,-1],2,mean))),type = "prob")) blue red 1 0.422 0.578 attr(,"class") [1] "matrix" "votes" and there is a substantial difference, with a probability of 19% with a single tree, 58% with 500 trees (the default value of the function). To understand why we can have such a difference, we should not only focus on the bagging stratgy, but look at the variability of the predictions, obtained with trees, B=1e4 parbres = rep(NA,B) m=data.frame(t(apply(df[,-1],2,mean))) for(b in 1:B){ idx = sample(1:nrow(df),size=nrow(df),replace=TRUE) arbre = rpart(Y~., data=df[idx,]) parbres[b] = predict(arbre,newdata=m,type = "prob")[2] } hist(parbres) Surprisingly, we have here a bimodal function for $\hat{y}$ which is either very small for some trees, of very large for others. On average, we have a value close to 55%… I think I will use more that generative algorithm for future quiz… # Sharing pictures from holidays in the Canadian Rockies (with R) My kids have a very popular blog (at least among their grandmothers) where they frequently post pictures from everyday’s life (since they live 5000km from them), as well as pictures taken from holidays. This afternoon, I tried to used the popupImage function from the leaflet package to post pictures, on a map (to explain where we spent our holiday this summer). This post is just to keep tracks of that code. First, we need to load the appropriate R packages library(leaflet) library(mapview) Then, we take a picture, and we locate it, for instance Mirror Lake (on the trail to Lake Agnes). Since leaflet uses openstreetmap, I recommend to use it also for location (and not google maps… coordinates can be slightly different) df=data.frame(lat =51.41603, long=-116.23946, nom = "Miror Lake",photo="http://freakonometrics.free.fr/jaspeR/_DSC5967.jpg") I guess you can also use the metadata if you take pictures with a cell phone, and you add the location… but I am (very) old fashioned, and still use a camera to take pictures. Then you can add a dozen pictures df=rbind(df, data.frame(lat =51.4164, long=-116.2442, nom = "Lake Agnes",photo="http://freakonometrics.free.fr/jaspeR/_DSC6003.jpg")) df=rbind(df, data.frame(lat =51.3215642,long=-116.193718, nom="Moraine Lake",photo="http://freakonometrics.free.fr/jaspeR/_DSC5957.jpg")) From that dataframe, we need two kinds of information: the location, and the url of the picture, data_df=df[,c("lat","long")] images = as.character(df$photo)

Then we can create the leaflet map (sorry for typos, but wordpress converts the > symbol into some “&gt;” characters… which makes R pipe operator hard to read)

m = leaflet(data_df) %&gt;% addTiles() %&gt;% addCircleMarkers( fillOpacity = 0.8, radius = 5, lng = ~long, lat =~lat, popup = popupImage(images) )

and export it (in a nice html file)

library(htmlwidgets) saveWidget(m, file="jaspR.html")

# Regression discontinuity model for TV series

In September, we are usually happy to see our favorite TV series back on air… Or not? Because, admit it, if we are happy to see those characters back, most of the time, we are disappointed, too. So why not look at the data, to confirm this feeling? Nazareno Andrade shared some nice codes to get IMDB ratings in a nice csv file (you can either use the large csv file, or run your own codes)

download.file("https://github.com/nazareno/imdb-series/raw/master/data/series_from_imdb.csv", destfile="series_from_imdb.csv") base = read.csv("series_from_imdb.csv")

It is a large dataset, with more than 64,000 episodes of almost 890 TV series,

str(base) 'data.frame': 64018 obs. of 18 variables: $series_name: Factor w/ 889 levels "'Allo 'Allo!",..: 137 137 137 137 137 137 137 137 137 137 ...$ episode : Factor w/ 54090 levels "-30-","¡Viva los muertos!",..: 32314 7446 16 7176 17748 9562 1379 36218 17845 5553 ... $series_ep : int 1 2 3 4 5 6 7 8 9 10 ...$ season : int 1 1 1 1 1 1 1 2 2 2 ... $season_ep : int 1 2 3 4 5 6 7 1 2 3 ...$ user_rating: num 8.9 8.7 8.7 8.2 8.3 9.2 8.8 8.7 9.2 8.3 ...

Just pick a TV series, for instance Dan Harmon’s Community,

sbase = base[base$series_name=="Community",] We can plot the evolution of the rating over the 110 episodes. sbase=sbase[!duplicated(sbase[,c(1,2,4,5)]),] sbase$series_ep=1:nrow(sbase)

()since there could be some problem with the data (such as duplicates, let us clean it quickly)

plot(sbase$series_ep,sbase$UserRating,xlab=sbase$series_name[1]) idx=c(0,which(diff(sbase$season)!=0),nrow(sbase)) abline(v=idx+.5,lty=2,col=colr[2]) a = unique(sbase$season) for(u in a){ ssbase = sbase[sbase$season==u,] reg = lm(UserRating~series_ep,data=ssbase) lines(ssbase$series_ep,predict(reg),col=colr[3],lwd=2) } The vertical lines are here to visualize the seasons. On issue is that the lenght can vary with time. Consider Linwood Boomer’s Malcom in The Middle, sbase = base[base$series_name=="Malcolm in the Middle",]

or Craig Thomas and Carter Bays’s How I Met Your Mother,

sbase = base[base$series_name=="How I Met Your Mother",] On those two, the evolution is rather stable. Look at AMC’s The Walking Dead, sbase = base[base$series_name=="The Walking Dead",]

Now, look at Howard Gordon and Alex Gansa’s Homeland,

sbase = base[base$series_name=="Homeland",] There is an issue here with the last episode of season4, “Long Time Coming“, that has a very poor rating. If we remove that point, we get the thin line. Note that the regression line is always increasing. For Michael Hirst’s Vickings, we have sbase = base[base$series_name=="Vicking",]

If we look more carefully on the previous graph, for five seasons (out of six), we have a positive slope. Well, to be honest, it is not significantly positive most of the time, but still. Out of 80 shows, and a total of 583 seasons, the slope is postive 75% of the time (433) and negative 25% of the time (150).

BASE = NULL L80 = unique(base$series_name) for(j in 1:length(L)){ sbase=base[base$series_name==L[j],] sbase=sbase[!duplicated(sbase[,c(1,2,4,5)]),] sbase=sbase[sbase$season&gt;0,] sbase$series_ep=1:nrow(sbase) a=unique(sbase$season) a=a[!is.na(a)] for(u in a){ ssbase=sbase[sbase$season==u,] reg=lm(UserRating~series_ep,data=ssbase) pente = NA if((!is.na(coefficients(reg)[2]))&amp;(!is.na((summary(reg)$coefficients[2,4])))){ if((summary(reg)$coefficients[2,4]&lt;.05)&amp;(coefficients(reg)[2]&gt;0)) pente="positive" if((summary(reg)$coefficients[2,4]&lt;.05)&amp;(coefficients(reg)[2]&lt;0)) pente="negative" sdf=data.frame(nom=sbase$series_name[1],season=u,slope=coefficients(reg)[2],inf=confint(reg)[2,1],sup=confint(reg)[2,2],signe=pente) BASE=rbind(BASE,sdf)} }} str(BASE) 'data.frame': 583 obs. of 6 variables: $nom : Factor w/ 80 levels "Friends","Game of Thrones",..: 1 1 1 1 1 1 1 1 1 1 ... mean(BASE$slope&gt;0) [1] 0.7427101 table(BASE$signe) negative positive 15 144 Most of the time, the slope is not significant. To be more specific, 72% of the time, the slope is not significant. But when it is, 90% of the time, it is positive (144 seasons). Let us look at other TV series, for instance Joel Surnow and Robert Cochran’s 24, sbase = base[base$series_name=="24",]

Álex Pina’s La Casa de Papel,

sbase = base[base$series_name=="La Casa de Papel",] Steven Knight’s Peaky Blinders, sbase = base[base$series_name=="Peaky Blinders",]

or David Simon’s The Wire,

sbase = base[base$series_name=="The Wire",] The slope is increasing over almost all seasons. But a major drawback is that when we get back to our show, for a new season, we usually get disapointed. More specifically, we can quantify the difference in red below that can be estimated using sbase12 = sbase[sbase$season%in%c(a[ij],a[ij+1]),] seuil = sbase12$series_ep[which(diff(sbase12$season)!=0)]+.5 s = function(x) (x-seuil)*(x&gt;seuil) reg = lm(UserRating~series_ep+s(series_ep)+I(series_ep&gt;seuil),data=sbase12)

Here we have

summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 8.45000 0.16338 51.719 2e-16 *** series_ep 0.10000 0.03235 3.091 0.008598 ** s(series_ep) 0.02000 0.04218 0.474 0.643291 I(series_ep)TRUE.  -1.01778 0.20486 -4.968 0.000257 ***

so the drop of 1 point (out of 10) cannot be claimed as being significant. That is the idea of regression discontinuity.

If we loop again over all our series, we have 485 pairs of consecutive seasons. As expected, in 75% of the casse, from season $t-1$ to season $t$, we observe a negative rupture. As previously, in 70% of the cases, it is not significat (with linear models before and after), and when it is significant, it is negative in 96% of the cases ! But an alternative can be to use nonparametric models, on both sides.

To illustrate, consider David Benioff and D. B. Weiss’s Game of Thrones,

sbase = base[base$series_name=="Game of Thrones",] but let us remove the last season (no spoiler here, but clearly not worst watching) Consider for instance the drop between season 1 and season 2, library(rdd) sbase12=sbase[sbase$season%in%c(1,2),] lmr=RDestimate(UserRating~series_ep,data=sbase12,cutpoint=mean(range(sbase12$series_ep))) plot(lmr) This is very consistent with what we observed with our linear regressions actually, seuil=10.5 s = function(x) (x-seuil)*(x&gt;seuil) reg = lm(UserRating~series_ep+s(series_ep)+I(series_ep&gt;seuil),data=sbase12) summary(reg) Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 8.70000 0.15458 56.281 2e-16 *** series_ep 0.07273 0.02491 2.919 0.01003 * s(series_ep) 0.01455 0.03523 0.413 0.68520 I(series_ep)TRUE -0.94000 0.20316 -4.627 0.00028 *** Here, the drop of one point is significant… So, your favorite show had an outstanding finale ? and you can’t wait to watch the new season… Well, statistically, it’s very likely that you will be disapointed by the first episode of the forthcoming season… # Testing for Covid-19 in the U.S. For almost a month, on a daily basis, we are working with colleagues (Romuald, Chi and Mathieu) on modeling the dynamics of the recent pandemic. I learn of lot of things discussing with them, but we keep struggling with the tests. Paul, in Montréal, helped me a little bit, but I think we will still have to more to get a better understand. To but honest, we stuggle with two very simple questions • how many people are tested on a daily basis ? Recently, I discovered Modelling COVID-19 exit strategies for policy makers in the United Kingdom, which is very close to what we try to do… and in the document two interesting scenarios are discussed, with, for the first one, “1 million ‘reliable’ daily tests are deployed” (in the U.K.) and “5 million ‘useless’ daily tests are deployed”. There are about 65 millions unhabitants in the U.K. so we talk here about 1.5% people tested, on a daily basis, or 7.69% people ! It could make sense, but our question was, at some point, is that realistic ? where are we today with testing ? In the U.S. https://covidtracking.com/ collects interesting data, on a daily basis, per state. url = "https://raw.githubusercontent.com/COVID19Tracking/covid-tracking-data/master/data/states_daily_4pm_et.csv" download.file(url,destfile="covid.csv") base = read.csv("covid.csv") Unfortunately, there is no information about the population. That we can find on wikipedia. But in that table, the state is given by its full name (and the symbol in the previous dataset). So we new also to match the two datasets properly, url="https://en.wikipedia.org/wiki/List_of_states_and_territories_of_the_United_States_by_population" download.file(url,destfile = "popUS.html") #pas contaminé 2/3 R=3 library(XML) tables=readHTMLTable("popUS.html") T=tables[[1]][3:54,c("V3","V4")] names(T)=c("state","pop") url="https://en.wikipedia.org/wiki/List_of_U.S._state_abbreviations" download.file(url,destfile = "nameUS.html") tables=readHTMLTable("nameUS.html") T2=tables[[1]][13:63,c(1,4)] names(T2)=c("state","symbol") T=merge(T,T2) T$population = as.numeric(gsub(",", "", T$pop, fixed = TRUE)) names(base)[2]="symbol" base = merge(base,T[,c("symbol","population")]) Now our dataset is fine… and we can get a function to plot the number of people tested in the U.S. (cumulated). Here, we distinguish between the positive and the negative, drawing = function(st ="NY"){ sbase=base[base$symbol==st,c("date","positive","negative","population")] sbase$DATE = as.Date(as.character(sbase$date),"%Y%m%d") sbase=sbase[order(sbase$DATE),] par(mfrow=c(1,2)) plot(sbase$DATE,(sbase$positive+sbase$negative)/sbase$population,ylab="Proportion Test (/population of state)",type="l",xlab="",col="blue",lwd=3) lines(sbase$DATE,sbase$positive/sbase$population,col="red",lwd=2) legend("topleft",c("negative","positive"),lwd=2,col=c("blue","red"),bty="n") title(st) plot(sbase$DATE,sbase$positive/(sbase$positive+sbase$negative),ylab="Ratio of positive tests",ylim=c(0,1),type="l",xlab="",col="black",lwd=3) title(st)}

drawing("NY")

As at now, 4% of the entiere population got tested… over 6 weeks…. The graph on the right is the proportion of people who tested positive… I won’t get back on that one here today, I keep it for our work. In New Jersey, we got about 2.5% of the entiere population tested, overall,

drawing("NJ")

Let us try a last one, Florida

drawing("FL")

As at today, it is 1.5% of the population, over 6 weeks. Overall, in the U.S. less than 0.1% people are tested, on a daily basis. Which is far from the 1.5% in the U.K. scenarios. Now, here come the second question,

• what are we actually testing for ?

On that one, my experience in biology is… very limited, and Paul helped me. He mentioned this morning a nice report, from a lab in UC Berkeley

One of my question was for instance, if you get tested positive, and you do it again, can you test negative ? Or, in the context of our data, do we test different people ? are some people tested on a regular basis (perhaps every week) ? For instance, with antigen tests (Reverse Transcription Quantitative Polymerase Chain Reaction (RT-qPCR) – also called molecular or PCR – Polymerase Chain Reaction – test) we test if someone is infectious, while with antibody test (using serological immunoassays that detect viral-specific antibodies — Immunoglobin M (IgM) and G (IgG) — also called serology test), we test for immunity. Which is rather different…

I have no idea what we have in our database, to be honest… and for the past six weeks, I have seen a lot of databases, and most of the time, I don’t know how to interpret, I don’t know what is measured… and it is scary. So, so far, we try to do some maths, to test dynamics by tuning parameters “the best we can” (and not estimate them). But if anyone has good references on testing, in the context of Covid-19 (for instance on specificity, sensitivity of all those tests) I would love to hear about it !

# On the “correlation” between a continuous and a categorical variable

Let us get back on the Titanic dataset,

loc_fichier = "http://freakonometrics.free.fr/titanic.RData" download.file(loc_fichier, "titanic.RData") load("titanic.RData") base = base[!is.na(base$Age),] On consider two variables, the age $x$ (the continuous one) and the survivor indicator $y$ (the qualitative one) X = base$Age Y = base$Survived It looks like the age might be a valid explanatory variable in the logistic regression, summary(glm(Survived~Age,data=base,family=binomial)) Coefficients: Estimate Std. Error z value Pr(&gt;|z|) (Intercept) -0.05672 0.17358 -0.327 0.7438 Age -0.01096 0.00533 -2.057 0.0397 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 964.52 on 713 degrees of freedom Residual deviance: 960.23 on 712 degrees of freedom AIC: 964.23 The significance test here has a $p$-value just below $4%$. Actually, one can relate it with the value of the deviance (the null deviance and the residual deviance). Recall that$$D=2\big(\log\mathcal{L}(\boldsymbol{y})-\log\mathcal{L}(\widehat{\boldsymbol{\mu}})\big)$$while$$D_0=2\big(\log\mathcal{L}(\boldsymbol{y})-\log\mathcal{L}(\overline{y})\big)$$Under the assumption that $x$ is worthless, $D_0-D$ tends to a $\chi^2$ distribution with 1 degree of freedom. And we can compute the $p$-value dof that likelihood ratio test, 1-pchisq(964.52-960.23,1) [1] 0.03833717 (which is consistent with a Gaussian test). But if we consider a nonlinear transformation summary(glm(Survived~bs(Age),data=base,family=binomial)) Coefficients: Estimate Std. Error z value Pr(&gt;|z|) (Intercept) 0.8648 0.3460 2.500 0.012433 * bs(Age)1 -3.6772 1.0458 -3.516 0.000438 *** bs(Age)2 1.7430 1.1068 1.575 0.115299 bs(Age)3 -3.9251 1.4544 -2.699 0.006961 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 964.52 on 713 degrees of freedom Residual deviance: 948.69 on 710 degrees of freedom which seems to be “more significant” 1-pchisq(964.52-948.69,3) [1] 0.001228712 So it looks like the variable $x$ is interesting here. To visualize the non-null correlation, one can consider the condition distribution of $x$ given $y=1$, and compare it with the condition distribution of $x$ given $y=0$, ks.test(X[Y==0],X[Y==1]) Two-sample Kolmogorov-Smirnov test data: X[Y == 0] and X[Y == 1] D = 0.088777, p-value = 0.1324 alternative hypothesis: two-sided i.e. with a $p$-value above $10%$, the two distributions are not significatly different. F0 = function(x) mean(X[Y==0]&lt;=x) F1 = function(x) mean(X[Y==1]&lt;=x) vx = seq(0,80,by=.1) vy0 = Vectorize(F0)(vx) vy1 = Vectorize(F1)(vx) plot(vx,vy0,col="red",type="s") lines(vx,vy1,col="blue",type="s") (we can also look at the density, but it looks like that there is not much to see) An alternative is discretize variable $x$ and to use Pearson’s independence test, k=5 LV = quantile(X,(0:k)/k) LV[1] = 0 Xc = cut(X,LV) table(Xc,Y) Y Xc 0 1 (0,19] 85 79 (19,25] 92 45 (25,31.8] 77 50 (31.8,41] 81 63 (41,80] 89 53 chisq.test(table(Xc,Y)) Pearson's Chi-squared test data: table(Xc, Y) X-squared = 8.6155, df = 4, p-value = 0.07146 The $p$-value is here $7%$, with five categories for the age. And actually, we can compare the $p$-value pvalue = function(k=5){ LV = quantile(X,(0:k)/k) LV[1] = 0 Xc = cut(X,LV) chisq.test(table(Xc,Y))$p.value} vk = 2:20 vp = Vectorize(pvalue)(vk) plot(vk,vp,type="l") abline(h=.05,col="red",lty=2)

which gives a $p$-value close to $5$%, as soon as we have enough categories. In the slides of the course (STT5100), I claim that actually, the age is an important variable when trying to predict if a passenger survived. Test mentioned here are not as conclusive, nevertheless…

# Modeling Pandemics (3)

In Statistical Inference in a Stochastic Epidemic SEIR Model with Control Intervention, a more complex model than the one we’ve seen yesterday was considered (and is called the SEIR model). Consider a population of size $N$, and assume that $S$ is the number of susceptible, $E$ the number of exposed, $I$ the number of infectious, and $R$ for the number recovered (or immune) individuals, \displaystyle{\begin{aligned}{\frac {dS}{dt}}&=-\beta {\frac {I}{N}}S\\[8pt]{\frac {dE}{dt}}&=\beta {\frac {I}{N}}S-aE\\[8pt]{\frac {dI}{dt}}&=aE-b I\\[8pt]{\frac {dR}{dt}}&=b I\end{aligned}}Between $S$ and $I$, the transition rate is $\beta I$, where $\beta$ is the average number of contacts per person per time, multiplied by the probability of disease transmission in a contact between a susceptible and an infectious subject. Between $I$ and $R$, the transition rate is $b$ (simply the rate of recovered or dead, that is, number of recovered or dead during a period of time divided by the total number of infected on that same period of time). And finally, the incubation period is a random variable with exponential distribution with parameter $a$, so that the average incubation period is $a^{-1}$.

Probably more interesting, Understanding the dynamics of ebola epidemics suggested a more complex model, with susceptible people $S$, exposed $E$, Infectious, but either in community $I$, or in hospitals $H$, some people who died $F$ and finally those who either recover or are buried and therefore are no longer susceptible $R$.

Thus, the following dynamic model is considered\displaystyle{\begin{aligned}{\frac {dS}{dt}}&=-(\beta_II+\beta_HH+\beta_FF)\frac{S}{N}\\[8pt]\frac {dE}{dt}&=(\beta_II+\beta_HH+\beta_FF)\frac{S}{N}-\alpha E\\[8pt]\frac {dI}{dt}&=\alpha E+\theta\gamma_H I-(1-\theta)(1-\delta)\gamma_RI-(1-\theta)\delta\gamma_FI\\[8pt]\frac {dH}{dt}&=\theta\gamma_HI-\delta\lambda_FH-(1-\delta)\lambda_RH\\[8pt]\frac {dF}{dt}&=(1-\theta)(1-\delta)\gamma_RI+\delta\lambda_FH-\nu F\\[8pt]\frac {dR}{dt}&=(1-\theta)(1-\delta)\gamma_RI+(1-\delta)\lambda_FH+\nu F\end{aligned}}In that model, parameters are $\alpha^{-1}$ is the (average) incubation period (7 days), $\gamma_H^{-1}$ the onset to hospitalization (5 days), $\gamma_F^{-1}$ the onset to death (9 days), $\gamma_R^{-1}$ the onset to “recovery” (10 days), $\lambda_F^{-1}$ the hospitalisation to death (4 days) while $\lambda_R^{-1}$ is the hospitalisation to recovery (5 days), $\eta^{-1}$ is the death to burial (2 days). Here, numbers are from Understanding the dynamics of ebola epidemics (in the context of ebola). The other parameters are $\beta_I$ the transmission rate in community (0.588), $\beta_H$ the transmission rate in hospital (0.794) and $\beta_F$ the transmission rate at funeral (7.653). Thus

epsilon = 0.001 Z = c(S = 1-epsilon, E = epsilon, I=0,H=0,F=0,R=0) p=c(alpha=1/7*7, theta=0.81, delta=0.81, betai=0.588, betah=0.794, blambdaf=7.653,N=1, gammah=1/5*7, gammaf=1/9.6*7, gammar=1/10*7, lambdaf=1/4.6*7, lambdar=1/5*7, nu=1/2*7)

If $\boldsymbol{Z}=(S,E,I,H,F,R)$, if we write $$\frac{\partial \boldsymbol{Z}}{\partial t} = SEIHFR(\boldsymbol{Z})$$where $SEIHFR$ is

SEIHFR = function(t,Z,p){ S=Z[1]; E=Z[2]; I=Z[3]; H=Z[4]; F=Z[5]; R=Z[6] alpha=p["alpha"]; theta=p["theta"]; delta=p["delta"] betai=p["betai"]; betah=p["betah"]; gammah=p["gammah"] gammaf=p["gammaf"]; gammar=p["gammar"]; lambdaf=p["lambdaf"] lambdar=p["lambdar"]; nu=p["nu"]; blambdaf=p["blambdaf"] N=S+E+I+H+F+R dS=-(betai*I+betah*H+blambdaf*F)*S/N dE=(betai*I+betah*H+blambdaf*F)*S/N-alpha*E dI=alpha*E-theta*gammah*I-(1-theta)*(1-delta)*gammar*I-(1-theta)*delta*gammaf*I dH=theta*gammah*I-delta*lambdaf*H-(1-delta)*lambdaf*H dF=(1-theta)*(1-delta)*gammar*I+delta*lambdaf*H-nu*F dR=(1-theta)*(1-delta)*gammar*I+(1-delta)*lambdar*H+nu*F dZ=c(dS,dE,dI,dH,dF,dR) list(dZ)}

We can solve it, or at least study the dynamics from some starting values

library(deSolve) times = seq(0, 50, by = .1) resol = ode(y=Z, times=times, func=SEIHFR, parms=p)

For instance, the proportion of people infected is the following

plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="",col="red") lines(resol[,"time"],resol[,"H"],col="blue")

# Modeling pandemics (2)

When introducing the SIR model, in our initial post, we got an ordinary differential equation, but we did not really discuss stability, and periodicity. It has to do with the Jacobian matrix of the system. But first of all, we had three equations for three function, but actually$$\displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}$$so it means that our problem is here simply in dimension 2. Hence\displaystyle {\begin{aligned}&X={\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&Y={\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\mu+\gamma)I\end{aligned}}and therefore, the Jacobian of the system is$$\begin{pmatrix}\displaystyle{\frac{\partial X}{\partial S}}&\displaystyle{\frac{\partial X}{\partial I}}\\[9pt]\displaystyle{\frac{\partial Y}{\partial S}}&\displaystyle{\frac{\partial Y}{\partial I}}\end{pmatrix}=\begin{pmatrix}\displaystyle{-\mu-\beta\frac{I}{N}}&\displaystyle{-\beta\frac{S}{N}}\\[9pt]\displaystyle{\beta\frac{I}{N}}&\displaystyle{\beta\frac{S}{N}-(\mu+\gamma)}\end{pmatrix}$$We should evaluate the Jacobian at the equilibrium, i.e. $$S^\star=\frac{\gamma+\mu}{\beta}=\frac{1}{R_0}$$and$$I^\star=\frac{\mu(R_0-1)}{\beta}$$We should then look at eigenvalues of the matrix.

Our very last example was

times = seq(0, 100, by=.1) p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[,"time"],resol[,"I"],type="l",xlab="time",ylab="")

We can compute values at the equilibrium

mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"] N=1 S = (gamma + mu)/beta I = mu * (beta/(gamma + mu) - 1)/beta

and the Jacobian matrix

J=matrix(c(-(mu + beta * I/N),-(beta * S/N), beta * I/N,beta * S/N - (mu + gamma)),2,2,byrow = TRUE)

Now, if we look at the eigenvalues,

eigen(J)$values [1] -0.024975+0.6318831i -0.024975-0.6318831i or more precisely $2\pi/b$ where $a\pm ib$ are the conjuguate eigenvalues 2 * pi/(Im(eigen(J)$values[1])) [1] 9.943588

we have a damping period of 10 time lengths (10 days, or 10 weeks), which is more or less what we’ve seen above,

The graph above was obtained using

p = c(mu = 1/100, N = 1, beta = 50, gamma = 10) start_SIR = c(S=0.19, I=0.01, R = 0.8) resol = ode(y=start_SIR, t=times, func=SIR, p=p) plot(resol[1:1e5,"time"],resol[1:1e5,"I"],type="l",xlab="time",ylab="",lwd=3,col="red") yi=resol[,"I"] dyi=diff(yi) i=which((dyi[2:length(dyi)]*dyi[1:(length(dyi)-1)])&lt;0) t=resol[i,"time"] arrows(t[2],.008,t[4],.008,length=.1,code=3)

If we look carefully. at the begining, the duration is (much) longer than 10 (about 13)… but it does converge towards 9.94

plot(diff(t[seq(2,40,by=2)]),type="b") abline(h=2 * pi/(Im(eigen(J)\$values[1]))

So here, theoretically, every 10 weeks (assuming that our time length is a week), we should observe an outbreak, smaller than the previous one. In practice, initially it is every 13 or 12 weeks, but the time to wait between outbreaks decreases (until it reaches 10 weeks).