Question

3.10 (i) If X1, , Xn are i.i.d. according to the exponential density e-", r >0, show that (2.9.3) P [X(n)-log n < y]- e-e-v, -00 < y < oo. (ii) Show that the right side of (2.9.3) is a cumulative distribution function. (The distribution with this edf is called the ertreme value distribution.) (iii) Graph the cdf of X(n)-log n for n = 1, 2, 5 together with the mit e-e" (iv) Graph the densities corresponding to the cdf's ofi

Answer 1

$(i)~P(X_{(n)}-\log n<y)=P(X_{(n)}<y&plus;\log n)\\ =P(\max\{X_1,X_2,...,X_n\}<y&plus;\log n)\\ =P(X_1<y&plus;\log n,X_2<y&plus;\log n,...,X_n<y&plus;\log n)\\ =\prod_{i=1}^nP(X_i<y&plus;\log n)~since~X_i's~are~independent\\ =\prod_{i=1}^n\left(\int_0^{y&plus;\log n}e^{-x} \right )=(1-e^{-y-\log n})^n\\ =\left(1-\frac{1}{n}e^{-y} \right )^n\rightarrow e^{-e^{-y}}~as~n\rightarrow \infty\\ (Use~the~result:~(1&plus;a/n)^n\rightarrow e^a~as~n\rightarrow \infty).\\ (ii)~Let~F_n(y&plus;\log n)=(1-e^{-y-\log n})^n\\ Then~F_n(x)=(1-e^{-x})^n\\ We~easily~check~that~F_n(x)~is~continuous~function.\\ For~x_1<x_2,~-x_1>-x_2~or,~e^{-x_1}>e^{-x_2}~or~(1-e^{-x_1})^n<(1-e^{-x_2})^n\\ or,~F_n(x_1)<F_n(x_2).\\ Hence~F_n(.)~is~increasing~function.\\ F_n(\infty)=1~since~e^{-\infty}=0\\ F_n(-\infty)=0~Since~X_i's~follow~exponential~distribution~and\\ X_{(n)}=\max\{X_1,X_2,...,X_n\}~so~P(X_{(n)}<0)=0.\\ Therefore~F_n(x)~is~a~cdf.$

$(iii)~We~denote~cdf~of~X_{(n)}-\log n~by~F_n(x)=(1-e^{-x-\log n})^n~and\\ F=e^{-e^{-y}}$

R code:

x=seq(0,5,by=0.01)
n=c(1,2,5)
F=matrix(0, nrow=3,ncol=length(x))
for(i in 1:3)
{
for(j in 1:length(x))
{
F[i,j]=(1-exp(-x[j]-log(n[i])))^n[i]
}
}
p=exp(-exp(-x))
plot(x,F[1,],xlab="y",ylab="CDF",lwd=1,col=1,ylim=c(0,1),type="l")
lines(x,F[2,],col=2,lwd=2)
lines(x,F[3,],col=3,lwd=2)
lines(x,p,col=4,lwd=2)
legend('topleft', c(expression(F[1]),expression(F[2]),expression(F[5]),expression(F)),
lty=1, col=1:4, bty='n', cex=.75)

$(iv)~pdf~of~X_{(n)}-\log n~is\\ f_n=\frac{\mathrm{d} F_n}{\mathrm{d} x}=n(1-e^{-x-\log n})^{n-1}e^{-x-\log n},~x>0\\ =0~otherwise\\$

R code:

x=seq(0,5,by=0.01)
n=c(1,2,5)
f=matrix(0, nrow=3,ncol=length(x))
for(i in 1:3)
{
for(j in 1:length(x))
{
f[i,j]=(1-exp(-x[j]-log(n[i])))^(n[i]-1)*exp(-x[j]-log(n[i]))
}
}
p=exp(-exp(-x))
plot(x,f[1,],xlab="y",ylab="CDF",lwd=1,col=1,ylim=c(0,1),type="l")
lines(x,f[2,],col=2,lwd=2)
lines(x,f[3,],col=3,lwd=2)
legend('topright', c(expression(f[1]),expression(f[2]),expression(f[5])),
lty=1, col=1:3, bty='n', cex=.75)