Question

Suppose X = Exp(1) and Y= -ln(x)

(a)Find the cumulative distribution function of Y .

(b) Find the probability density function of Y .

(c) Let X1, X2, ... , Xk be i.i.d. Exp(1), and let Mk = max{X1,..... , Xk)(Maximum of X1, ..., Xk). Find the probability density function of Mk.(Hint: P(min(X1, X2, X3) > k) = P(X1 >= k, X2 >= k, X3 >= kq, how about max ?)

(d) Show that as k → 00, the CDF of Z Mk – In(k) is the CDF of Y. Hint: Note that as k → 0, (1+)* = eº.

Answer 1

Solution,

a)

cdf of Y :

Fy(y) = p <y) = pl-InX <y)

  $[X>0=>\infty lnX<\infty ]=>\infty <-lnX<\infty ]$

     

EpInX -Y) = p X > e-

  
- ſ fx(a)dx = - T e-de ey
  $=e^{-e}^{-y}$ ,  $u_{-\infty }<y<\infty$

b)

pdf of Y : $f_{Y}(y)=\frac{d}{dy}F_{Y}(y)$ $=\left\{\begin{matrix} e^{-e} ^{-y}& , -\infty <y<\infty \\ 0&,else \end{matrix}\right.$

c)

cdf of M_x : $F_{M_{k}}(y)=p(M_{k}\leq y)=p(max\left \{ X_{1},..,X_{k}} \right \}\leq y)$

$=[p(X_{1}\leq y)]^{k}$$[X_{1},..., X_{k} are iid, Max {X_{1}, ,X_{k}}\leq a=>llX_{i}\leq a]$

$=\left [ \int_{0}^{y} f_{X_{1}}(x)\right ]^{k}=\left [ \int_{0}^{y}e^{-x}dx\right ]^{k}=(1-e^{-y})^{k}, y>0$

pdf of M_k : $f_{M_{k}}(y)=\frac{d}{dy}F_{M_{k}}(y)=\left\{\begin{matrix} k(1-e^{-y})^{k-1}e^{-y} &,y>0 \\ 0 & ,else \end{matrix}\right.$ d)

cdf of Z : $F_{(z)}(z)=p(Z\leq z)=p(M_{k}-lnk\leq z)=p(M_{k}\leq lnk+z)$ $=F_{M_{k}}(lnk+z)=(1-e^{-lnk+z})^{k}=\left ( 1-\frac{e^{-z}}{k} \right )^{k}$$\underset{k\rightarrow \infty }{\rightarrow}e^{-e^{-z}}$

= F_Y(z) [ a = -e^-z

the hint]