# Exercice 1 : IQ
Knowing that IQ is a normal measure of mean 100 and standard deviation 15, what is the probability of having an IQ
more than 120? less than 100?
```{r}
pnorm(120, mean = 100, sd = 15, lower.tail = F, log.p = FALSE)
```
ou après quelques calculs :
```{r}
1 - pnorm(4/3)
```
Visualisation :
```{r}
library(ggplot2)
```
```{r}
QI.sup.120<-function(x){ ifelse(x>120,dnorm(x,mean=100,sd=15),NA) }
ggplot(data.frame(x=c(20, 180)),aes(x)) + stat_function(fun = dnorm,args = list(mean=100,sd=15)) + stat_function(fun =QI.sup.120 , geom = "area", fill = "coral", alpha = 0.3) + geom_text(x = 127, y = 0.003, size = 4, fontface = "bold", label = paste0(round(pnorm(120,mean=100,sd=15,lower.tail = F),2))) + scale_x_continuous(breaks = c(80,100,120,130)) + geom_vline(xintercept=120,colour="coral")
```
# Exercice 2 : Bias of the maximum likelihood estimator of the variance
Show that the maximum likelihood estimator of the variance is biased and propose an unbiased estimator.
Demo :
$$
\mathbb{E}[ \, \hat{\sigma}^2 \, ] = \mathbb{E}[\, \dfrac{1}{n} \sum_{i=1}^n (X_i - \bar X)^2] \\
= \mathbb{E}[\, \dfrac{1}{n} \sum_{i=1}^n X_i^2 - \bar{X}^2]\\
= \sigma^2 + \mu^2 - \dfrac{\sigma^2}{n} - \mu^2
$$
# Exercice 3 : Extreme values
Consider the Fisher irises. Find flowers whose measured widths and lengths are exceptionally large or small.
```{r}
library(tidyverse)
```
```{r}
data(iris)
head(iris)
```
```{r}
parameters <-
iris %>%
select(-"Species") %>%
gather(factor_key = TRUE) %>%
group_by(key) %>%
summarise(mean= mean(value), sd= sd(value)) %>%
mutate(min=mean - 2*sd,max=mean + 2*sd)
parameters
```
```{r}
#flower.outliers <-(apply( X=t((t(iris[,1:4]) < parameters$min) + (t(iris[,1:4]) > parameters$max)),MARGIN = 1,FUN = function(x) if(x) return(1) else(0)))
flower.outliers <- t((t(iris[,1:4]) < parameters$min) + (t(iris[,1:4]) > parameters$max))
flower.outliers <- rowSums(flower.outliers)
ggplot(iris,aes(x=Sepal.Length,y=Sepal.Width))+
geom_point(colour=as.numeric(iris$Species),size= flower.outliers*2 + 1 )
```
# Exercice 4 : Equiprobability Ellipses
Generate 1000 observations of a two-dimensional normal distribution $\mathcal{N}(\mu, \Sigma)$
$$
\mu = \left(\begin{array}{c}
0 \\
0
\end{array}\right) \\
\Sigma = \left(\begin{array}{cc}
2 & 1\\
1 & 0.75
\end{array}\right)
$$
After, draw the ellipses of equiprobability of the multiples of 5%.
```{r}
#par(mfrow=c(1,3)) # partage l'affichage en 2
sigma<-matrix(c(2,1,1,0.75),2,2)
mu <- c(0,0)
cholesky_sigma =chol(sigma)
t(chol(sigma)) %*% chol(sigma)
#Y<- t(t(chol(sigma)) %*% t(matrix(rnorm(2000),1000,2)) + mu)
Y<- matrix(rnorm(2000),1000,2) %*% chol(sigma) + mu
plot(Y,xlab="x",ylab="y",pch='.')
```
```{r}
Q<-qchisq(p=seq(0.05,0.95,by=0.1),df=2)
x<-seq(-4,4,length=100)
y<-seq(-4,4,length=100)
sigmainv<-solve(sigma)
a<-sigmainv[1,1]
b<-sigmainv[2,2]
c<-sigmainv[1,2]
z<-outer(x,y,function(x,y) (a*x**2+b*y**2+2*c*x*y)) ## Fonction is t(y) %*% y
image(x,y,z)
contour(x,y,z,col="blue4",levels=Q,labels=seq(from=0.05,to=0.95,by=0.1),add=T)
```
```{r}
persp(x,y,1/(2*pi)*det(sigma)**(-1/2)*exp(-0.5*z),col="cornflowerblue",zlab="f(x)")
```