Slides 1 – loi binomiale

Allez, on reprend les cours, après un peu plus de deux semaines d’interruption… Mais je reprends tranquillement, en revenant sur des rappels que j’avais fait au premier cours, et dont on aura finalement besoin très bientôt ! On revient sur la loi de Bernoulli, la loi binomiale et la loi multinomiale. Et on parlera inférence, et tests. Les slides sont en ligne (slides 1) et la vidéo aussi (slides 1)

Il y a une coquille dans les slides (7 et 8) mais je n’ai pas interrompu la vidéo et j’essayais de continuer comme si de rien n’était. Pour la proportion \widehat{p}, on a juste la loi asymptotique Gaussienne (ou une transformation de la loi Binomiale à distance finie, mais pas de loi de Student, désolé pour le copier-coller depuis des vieux slides qui parlaient d’estimation de moyenne \overline{x} pour des observations supposées Gaussiennes)

J’essaye de limiter le nombre d’erreurs, mais ces cours à distance me prennent beaucoup de temps, et je n’ai pas le courage pour refaire les vidéos, pour l’instant. Toutes mes excuses.

 

 

Reprise des cours (à venir)

Lundi, les cours de la session d’hiver reprennent… à distance. Je vais mettre en ligne une série de capsules vidéos pour finir le cours, sur les GLM. J’ai mis en ligne une première vidéo (slides 0) pour annoncer le plan. J’ai fait des slides rapidement (ça changera des cours que je faisais au tableau) et je fais des enregistrement unique, sans montage, histoire de mettre en ligne rapidement le cours en ligne. Le pdf des slides est aussi en ligne (slides 0). J’essayerais de mettre les vidéos en ligne au fur et à mesure. Je ne suis pas particulièrement fier de moi, mais quitte à perdre du temps à faire le guignol devant la caméra, autant que ça serve au plus grand nombre.

Ah oui, il y a probablement des coquilles, voire des erreurs dans les slides… les commentaires sont ouverts pour me faire part de toute suggestion !

Il était prévisible que Léonard DiCaprio ne survive pas au naufrage du Titanic

Vendredi dernier, j’ai mis en ligne un petit quiz pour prédire qui a survécu au naufrage du Titanic, en 1912. J’avais mis en ligne une sous-base avec quelques passagers

loc_fichier = "http://freakonometrics.free.fr/titanic.RData"
download.file(loc_fichier, "titanic.RData")
load("titanic.RData")
str(base)

et plus précisément

  • Survived: Passenger survival indicator (1 if survived)
  • Pclass: Passenger class
  • Sex: Sex of the passenger
  • Age: Age of the passenger
  • SibSp: Number of siblings/spouses aboard
  • Parch: Number of parents/children aboard
  • Embarked: Port of Embarkation
  • Name: Name of passenger

La dernière variable ne sert pas à grand chose, donc on la vire,

base = base[,1:7]

On va maintenant pouvoir répondre aux questions.

Continue reading Il était prévisible que Léonard DiCaprio ne survive pas au naufrage du Titanic

Reinforcement Learning in Economics and Finance

With Romuald Elie and Carl Remlinger we recently uploaded on ArXiv a paper on Reinforcement Learning in Economics and Finance

Reinforcement learning algorithms describe how an agent can learn an optimal action policy in a sequential decision process, through repeated experience. In a given environment, the agent policy provides him some running and terminal rewards. As in online learning, the agent learns sequentially. As in multi-armed bandit problems, when an agent picks an action, he can not infer ex-post the rewards induced by other action choices. In reinforcement learning, his actions have consequences: they influence not only rewards, but also future states of the world. The goal of reinforcement learning is to find an optimal policy — a mapping from the states of the world to the set of actions, in order to maximize cumulative reward, which is a long term strategy. Exploring might be sub-optimal on a short-term horizon but could lead to optimal long-term ones. Many problems of optimal control, popular in economics for more than forty years, can be expressed in the reinforcement learning framework, and recent advances in computational science, provided in particular by deep learning algorithms, can be used by economists in order to solve complex behavioral problems. In this article, we propose a state-of-the-art of reinforcement learning techniques, and present applications in economics, game theory, operation research and finance.

Qui a survécu au naufrage du Titanic?

also known as quiz numéro 4 du cours STT5100 (de la session d’hiver). Vendredi dernier, alors que nous terminions le cours vers midi, François Legault a décrété l’état d’urgence sanitaire pour une quinzaine de jours. Le cours est donc sur la glace depuis une semaine, et (pour l’instant) encore une bonne semaine.

Les deux derniers cours ont été l’occasion de voir comment faire, et comment analyser, une régression logistique. Histoire de briser un peu le confinement que l’on connaît depuis une semaine, j’ai envoyé ce matin un quiz aux étudiants, et j’en profite pour le mettre en ligne (la correction arrivera dans le milieu de la semaine prochaine) pour tous ceux qui voudraient se pratiquer un peu sur les modèles de classification…

J’ai mis en ligne une base de données avec des informations sur les passagers du Titanic, qui est un paquebot qui a fait naufrage il y a plus d’un siècle (certains ont peut être vu le documentaire)

loc_fichier = "http://freakonometrics.free.fr/titanic.RData"
download.file(loc_fichier, "titanic.RData")
load("titanic.RData")
str(base)

On a dans la base

  • Survived: Passenger survival indicator (1 if survived)
  • Pclass: Passenger class
  • Sex: Sex of the passenger
  • Age: Age of the passenger
  • SibSp: Number of siblings/spouses aboard
  • Parch: Number of parents/children aboard
  • Embarked: Port of Embarkation
  • Name: Name of passenger

La dernière variable ne sert pas à grand chose, donc on la vire,

base = base[,1:7]

J’ai ensuite posé une série de questions,

  1. quelle proportion de passagers ont survécu ?
  2. quelle proportion de passagers de première classe ont survécu ?
  3. quelle est la statistique de test pour un test d’indépendance du chi-deux [vu en cours vendredi dernier] entre le fait de survivre (ou pas) et la classe ?
  4. ensuite il faut faire une régression logistique, et prévoir la probabilité de survie pour deux passagers (fictifs) du paquebot
newbase = data.frame(
  Pclass = as.factor(c(1,3)),
  Sex = as.factor(c("female","male")),
  Age = c(17,20),
  SibSp = c(1,0),
  Parch = c(2,0),
  Embarked = as.factor(c("S","S")),
  Name = as.factor(c("Winslet, Miss. Kate","DiCaprio, Mr. Leonardo")))

Je ne peux mettre le formulaire de réponse en ligne ici, mais les commentaires sont ouverts pour ceux qui voudraient d’entraîner un peu pendant cette période un peu longue de confinement…

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}andI^\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)])<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).

Modeling pandemics (1)

The most popular model to model epidemics is the so-called SIR model – or Kermack-McKendrick. Consider a population of size N, and assume that S is the number of susceptible, I the number of infectious, and R for the number recovered (or immune) individuals, \displaystyle {\begin{aligned}&{\frac {dS}{dt}}=-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-\gamma I,\\[6pt]&{\frac {dR}{dt}}=\gamma I,\end{aligned}}so that \displaystyle{{\frac{dS}{dt}}+{\frac {dI}{dt}}+{\frac {dR}{dt}}=0}which implies that S+I+R=N. In order to be more realistic, consider some (constant) birth rate \mu, so that the model becomes\displaystyle {\begin{aligned}&{\frac {dS}{dt}}=\mu(N-S)-{\frac {\beta IS}{N}},\\[6pt]&{\frac {dI}{dt}}={\frac {\beta IS}{N}}-(\gamma+\mu) I,\\[6pt]&{\frac {dR}{dt}}=\gamma I-\mu R,\end{aligned}}Note, in this model, that people get sick (infected) but they do not die, they recover. So here, we can model chickenpox, for instance, not SARS.

The dynamics of the infectious class depends on the following ratio:\displaystyle{R_{0}={\frac {\beta }{\gamma +\mu}}} which is the so-called basic reproduction number (or reproductive ratio). The effective reproductive ratio is R_0S/N, and the turnover of the epidemic happens exactly when R_0S/N=1, or when the fraction of remaining susceptibles is R_0^{-1}. As shown in Directly transmitted infectious diseases:Control by vaccination, if S/N<R_0^{-1} the disease (the number of people infected) will start to decrease.

Want to see it  ? Start with

mu = 0
beta = 2
gamma = 1/2

for the parameters. Here,  R_0=4. We also need starting values

epsilon = .001
N = 1
S = 1-epsilon
I = epsilon
R = 0

Then use the ordinary differential equation solver, in R. The idea is to say that \boldsymbol{Z}=(S,I,R) and we have the gradient \frac{\partial \boldsymbol{Z}}{\partial t} = SIR(\boldsymbol{Z})where SIR is function of the various parameters. Hence, set

p = c(mu = 0, N = 1, beta = 2, gamma = 1/2)
start_SIR = c(S = 1-epsilon, I = epsilon, R = 0)

The we must define the time, and the function that returns the gradient,

times = seq(0, 10, by = .1)
SIR = function(t,Z,p){
S=Z[1]; I=Z[2]; R=Z[3]; N=S+I+R
mu=p["mu"]; beta=p["beta"]; gamma=p["gamma"]
dS=mu*(N-S)-beta*S*I/N
dI=beta*S*I/N-(mu+gamma)*I
dR=gamma*I-mu*R
dZ=c(dS,dI,dR)
return(list(dZ))}

To solve this problem use

library(deSolve)
resol = ode(y=start_SIR, times=times, func=SIR, parms=p)

We can visualize the dynamics below

par(mfrow=c(1,2))
t=resol[,"time"]
plot(t,resol[,"S"],type="l",xlab="time",ylab="")
lines(t,resol[,"I"],col="red")
lines(t,resol[,"R"],col="blue")
plot(t,t*0+1,type="l",xlab="time",ylab="",ylim=0:1)
polygon(c(t,rev(t)),c(resol[,"R"],rep(0,nrow(resol))),col="blue")
polygon(c(t,rev(t)),c(resol[,"R"]+resol[,"I"],rev(resol[,"R"])),col="red")

We can actually also visualize the effective reproductive number is R_0S/N, where

R0=p["beta"]/(p["gamma"]+p["mu"])

The effective reproductive number is on the left, and as we mentioned above, when we reach 1, we actually reach the maximum of the infected,

plot(t,resol[,"S"]*R0,type="l",xlab="time",ylab="")
abline(h=1,lty=2,col="red")
abline(v=max(t[resol[,"S"]*R0&gt;=1]),col="darkgreen")
points(max(t[resol[,"S"]*R0&gt;=1]),1,pch=19)
plot(t,resol[,"S"],type="l",xlab="time",ylab="",col="grey")
lines(t,resol[,"I"],col="red",lwd=3)
lines(t,resol[,"R"],col="light blue")
abline(v=max(t[resol[,"S"]*R0&gt;=1]),col="darkgreen")
points(max(t[resol[,"S"]*R0&gt;=1]),max(resol[,"I"]),pch=19)

And when adding a \mu parameter, we can obtain some interesting dynamics on the number of infected,

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="")