1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41
|
Exercice 3:
```{r minimum}
u<- rexp(1000, 1/20)
v<- rexp(1000, 1/30)
z<- 0:1
m<- 0:1
som<- 0
vingt<-0
for (i in 1:1000) {
z[i]<- min(u[i], v[i])
m[i]<- max(u[i], v[i])
if (u[i]>v[i]) {
som <- som+1
}
if (u[i]<20 || v[i]<20){
vingt<-vingt+1
}
}
hist(z)
newz <- rexp(1000, 1/12)
hist(newz)
som/1000
vingt/1000
plot(ecdf(m))
```
```{}
9.b. P(Y<X) = Int(0; inf)Int(0; x)p(y,x)dydx = 20/(20+30) = 2/5
10.b Fx(t)=20exp(-20t) Fy(t)=30exp(-30t) d'où Fm(t)=max(20exp(-20t), 30exp(-30t))
c. Pm(t)=2/5 * 20exp(-20t) + 3/5 * 30exp(-30t)
d.
```{r fm}
newm <- 2/5*rexp(1000, 1/20)+3/5*rexp(1000, 1/30)
plot(hist(newm))
```
``` |
Partager