# Confidence vs. Credibility Intervals

Tomorrow, for the final lecture of the Mathematical Statistics course, I will try to illustrate – using Monte Carlo simulations – the difference between classical statistics, and the Bayesien approach.

The (simple) way I see it is the following,

• for frequentists, a probability is a measure of the the frequency of repeated events, so the interpretation is that parameters are fixed (but unknown), and data are random
• for Bayesians, a probability is a measure of the degree of certainty about values, so the interpretation is that parameters are random and data are fixed

Or to quote Frequentism and Bayesianism: A Python-driven Primer,  a Bayesian statistician would say “given our observed data, there is a 95% probability that the true value of $\theta$ falls within the credible region” while a Frequentist statistician would say “there is a 95% probability that when I compute a confidence interval from data of this sort, the true value of $\theta$ will fall within it”.

To get more intuition about those quotes, consider a simple problem, with Bernoulli trials, with insurance claims. We want to derive some confidence interval for the probability to claim a loss. There were $n$ = 1047 policies. And 159 claims.

Consider the standard (frequentist) confidence interval. What does that mean that $$\overline{x}\pm\sqrt{\frac{\overline{x}(1-\overline{x})}{n}}$$is the (asymptotic) 95% confidence interval? The way I see it is very simple. Let us generate some samples, of size $n$, with the same probability as the empirical one, i.e. $\widehat{\theta}$ (which is the meaning of “from data of this sort”). For each sample, compute the confidence interval with the relationship above. It is a 95% confidence interval because in 95% of the scenarios, the empirical value lies in the confidence interval. From a computation point of view, it is the following idea,

> xbar <- 159
> n <- 1047
> ns <- 100
> M=matrix(rbinom(n*ns,size=1,prob=xbar/n),nrow=n)

I generate 100 samples of size $n$. For each sample, I compute the mean, and the confidence interval, from the previous relationship

> fIC=function(x) mean(x)+c(-1,1)*1.96*sqrt(mean(x)*(1-mean(x)))/sqrt(n)
> IC=t(apply(M,2,fIC))
> MN=apply(M,2,mean)

Then we plot all those confidence intervals. In red when they do not contain the empirical mean

> k=(xbar/n<IC[,1])|(xbar/n>IC[,2])
> plot(MN,1:ns,xlim=range(IC),axes=FALSE,
+ xlab="",ylab="",pch=19,cex=.7,
+ col=c("blue","red")[1+k])
> axis(1)
> segments(IC[,1],1:ns,IC[,2],1:
+ ns,col=c("blue","red")[1+k])
> abline(v=xbar/n)

Now, what about the Bayesian credible interval ? Assume that the prior distribution for the probability to claim a loss has a $\mathcal{B}(\alpha,\beta)$ distribution. We’ve seen in the course that, since the Beta distribution is the conjugate of the Bernoulli one, the posterior distribution will also be Beta. More precisely

$\mathcal{B}\left(\alpha+\sum x_i,\beta+n-\sum x_i\right)$

Based on that property, the confidence interval is based on quantiles of that (posterior) distribution

> u=seq(.1,.2,length=501)
> v=dbeta(u,1+xbar,1+n-xbar)
> plot(u,v,axes=FALSE,type="l")
> I=u<qbeta(.025,1+xbar,1+n-xbar)
> polygon(c(u[I],rev(u[I])),c(v[I],
+ rep(0,sum(I))),col="red",density=30,border=NA)
> I=u>qbeta(.975,1+xbar,1+n-xbar)
> polygon(c(u[I],rev(u[I])),c(v[I],
+ rep(0,sum(I))),col="red",density=30,border=NA)
> axis(1)

What does that mean, here, that we have a 95% credible interval. Well, this time, we do not draw using the empirical mean, but some possible probability, based on that posterior distribution (given the observations)

> pk <- rbeta(ns,1+xbar,1+n-xbar)

In green, below, we can visualize the histogram of those values

> hist(pk,prob=TRUE,col="light green",
+ border="white",axes=FALSE,
+ main="",xlab="",ylab="",lwd=3,xlim=c(.12,.18))

And here again, let us generate samples, and compute the empirical probabilities,

> M=matrix(rbinom(n*ns,size=1,prob=rep(pk,
+ each=n)),nrow=n)
> MN=apply(M,2,mean)

Here, there is 95% chance that those empirical means lie in the credible interval, defined using quantiles of the posterior distribution. We can actually visualize all those means : in black the mean used to generate the sample, and then, in blue or red, the averages obtained on those simulated samples,

> abline(v=qbeta(c(.025,.975),1+xbar,1+
+ n-xbar),col="red",lty=2)
> points(pk,seq(1,40,length=ns),pch=19,cex=.7)
> k=(MN<qbeta(.025,1+xbar,1+n-xbar))|
+ (MN>qbeta(.975,1+xbar,1+n-xbar))
> points(MN,seq(1,40,length=ns),
+ pch=19,cex=.7,col=c("blue","red")[1+k])
> segments(MN,seq(1,40,length=ns),
+ pk,seq(1,40,length=ns),col="grey")

More details and exemple on Bayesian statistics, seen with the eyes of a (probably) not Bayesian statistician in my slides, from my talk in London, last Summer,

# Confidence interval for predictions with GLMs

Consider a (simple) Poisson regression . Given a sample where , the goal is to derive a 95% confidence interval for given , where is the prediction. Hence, we want to derive a confidence interval for the prediction, not the potential observation, i.e. the dot on the graph below

> r=glm(dist~speed,data=cars,family=poisson)
> P=predict(r,type="response",
+ newdata=data.frame(speed=seq(-1,35,by=.2)))
> plot(cars,xlim=c(0,31),ylim=c(0,170))
> abline(v=30,lty=2)
> lines(seq(-1,35,by=.2),P,lwd=2,col="red")
> P0=predict(r,type="response",se.fit=TRUE,
+ newdata=data.frame(speed=30))
> points(30,P1$fit,pch=4,lwd=3) i.e. Let denote the maximum likelihood estimator of . Then where is Fisher information of (from standard maximum likelihood theory). Recall that where computation of those values is based on the following calculations In the case of the log-Poisson regression Let us get back to our initial problem. • confidence interval for the linear combination A first idea to get a confidence interval for is to get a confidence interval for (by taking exponential values of bounds, since the exponential is a monotone function). Asymptotically, we know that thus, an approximation for the variance matrix of will be based on , obtained by plugging estimators of the parameters. Then, since as an asymptotic multivariate distribution, any linear combination of the parameters will also be normal, i.e. has a normal distribution, centered on , with variance where is the variance of . All those quantities can be easily computed. First, we can get the variance of the estimators > i1=sum(predict(reg,type="response")) > i2=sum(cars$speed*predict(reg,type="response"))
> i3=sum(cars$speed^2*predict(reg,type="response")) > I=matrix(c(i1,i2,i2,i3),2,2) > V=solve(I) Hence, if we compare with the output of the regression, > summary(reg)$cov.unscaled
(Intercept)         speed
(Intercept)  0.0066870446 -3.474479e-04
speed       -0.0003474479  1.940302e-05
> V
[,1]          [,2]
[1,]  0.0066871228 -3.474515e-04
[2,] -0.0003474515  1.940318e-05

Based on those values, it is easy to derive the standard deviation for the linear combination,

> x=30
+ newdata=data.frame(speed=x))
> P2
$fit 1 5.046034$se.fit
[1] 0.05747075

$residual.scale [1] 1 > sqrt(V[1,1]+2*x*V[2,1]+x^2*V[2,2]) [1] 0.05747084 > sqrt(t(c(1,x))%*%V%*%c(1,x)) [,1] [1,] 0.05747084 And once we have the standard deviation, and normality (at least asymptotically), confidence intervals are derived, and then, taking the exponential of the bounds, we get confidence interval > segments(30,exp(P2$fit-1.96*P2$se.fit), + 30,exp(P2$fit+1.96*P2$se.fit),col="blue",lwd=3) Based on that technique, confidence intervals are no longer centered on the prediction. But who cares ? • delta method Actually, those who like to use “more or less” expressions for confidence intervals will not like non centered intervals. So, an alternative is to use the delta method. Instead of writing (again) something on the theory, we can use a package which computes that method, > estmean=t(c(1,x))%*%coef(reg) > var=t(c(1,x))%*%summary(reg)$cov.unscaled%*%c(1,x)
> library(msm)
> deltamethod (~ exp(x1), estmean, var)
[1] 8.931232
> P1=predict(r,type="response",se.fit=TRUE,
+ newdata=data.frame(speed=30))
> P1
$fit 1 155.4048$se.fit
1
8.931232

$residual.scale [1] 1 The delta method gives us (asymptotic) normality, so once we have a standard deviation, we get the confidence interval. > segments(30,P1$fit-1.96*P1$se.fit,30, + P1$fit+1.96*P1$se.fit,col="blue",lwd=3) Note that those quantities – obtained with two different approaches – are rather close here > exp(P2$fit-1.96*P2$se.fit) 1 138.8495 > P1$fit-1.96*P1$se.fit 1 137.8996 > exp(P2$fit+1.96*P2$se.fit) 1 173.9341 > P1$fit+1.96*P1\$se.fit
1
172.9101
• bootstrap techniques

And a third method (but far from what I expect to teach on that course) is to use bootstrap techniques to about those results based on asymptotic normality (we have only 50 observations). The idea is to sample from out dataset, and to run a log-Poisson regression on those new samples, and to repeat a lot of time,

# Intervalle de confiance pour une proportion

Je voulais poster un petit billet puisque c’est la deuxième fois que l’on me pose une question similaire. Pour déterminer un intervalle de confiance pour une proportion, on connaît tous la formule

La question qui m’avait été posée est simple: que se passe-t-il si n est petit, et si – empiriquement – on a trouvé une proportion de 100% ? Dit autrement: “on fait 5 lancers de “pile” ou “face” et que on obtient 5 fois “pile”: que peut-on dire sur p, la probabilité de tomber sur “pile” ?”

On utilisant la formule précédente, on est un peu bloqué…

• La première réponse pourrait être fréquentiste, et consiste à proposer une lecture duale de l’intervalle de confiance, et des abaques.

Le graphique de droite montre, en trait violet, l’intervalle de confiance approché, c’est à dire avec l’intervalle de confiance indiqué ci-dessus. La courbe bleue, encadrant la zone verte correspond à l’intervalle de confiance calculé à partir des quantiles d’une loi binomiale de paramètre la fréquence empirique observée. Ici, on aurait envie de dire que p a 95% de chance d’être supérieure à 47.8%.

En fait, cette “lecture duale” des abaques revient à construire – me semble-t-il – l’intervalle de confiance de Clopper-Pearson. L’intervalle de confiance a ici la forme suivante

Notons que c’est cette technique qui est utilisée sous R, via la fonction binom.test du package de base. Ici, on obtient

> binom.test(x, n, p = 0.5,conf.level = 0.95)
Exact binomial test
data: x and n
number of successes = 5, number of trials = 5, p-value = 0.0625
alternative hypothesis: true probability of success is not equal to 0.5
95 percent confidence interval:
0.4781762 1.0000000
sample estimates:
probability of success
1

…. ce qui correspond exactement à l’intervalle de confiance calculé précédemment. On retrouve d’ailleurs cette fonction dans binconf i.e.

> binconf(x = 5, n = 5, alpha = .05, method = "exact")
PointEst Lower Upper
1 0.4781762 1

qui donne (c’est rassurant) toujours la même chose. On peut aussi programme l’intervalle de confiance proposé par Heldge Blacker, qui donne ici un intervalle de confiance de la forme [50%; 100%]. Pour aller plus loin, je pourrais renvoyer au papier de Lawrence Brown, Tony Cai et Anirban DasGupta, dans Statistical Science sur ce sujet.

• Une seconde réponse pourrait être une approche bayésienne, qui d’ailleurs correspond exactement à l’expérience de Bayes (1763) ou Laplace (1786).

On suppose que p est une variable aléatoire. On se donne une information a priori, et on peut se demander ce que devient la loi a posteriori de p, sachant que 5 “piles” ont été observés. On suppose que les tirages sont indépendants, et donc la probabilité d’avoir k fois “piles” sur n tirages suit une loi binomiale, conditionnellement à p, i.e.

Le plus simple est de considérer comme loi a priori la loi conjuguée de la loi binomiale, à savoir une loi beta. Dans ce cas, si p suit a priori une loi B(a,b), alors la loi a posteriori de p, sachant que x “pile” ont été observés est une loi B(a+x,b+n-x).

La courbe en rouge est la densité de la loi a priori (uniforme ou beta de paramètres 2 et 2), et la courbe bleue, la loi a posteriori, avec un intervalle de confiance à 95%. Sur l’exemple de droite, par exemple, on a 95% de chance que p soit compris entre 47,3% et 96,8%.