Question

Answer 1

4.2) The CDF of exponential distribution is $F_X\left ( x \right )=1-e^{-\lambda x};x>0$ . Solving for ,

x =-1 In [1-Ex (X)]

By inverse transform method, $X=-\frac{1}{\lambda }\ln \left [ 1-\xi \right ]$ are random variables from the exponential distribution if
Unif (0,1)
. But we know the distribution of $1-\xi \sim Unif\left ( 0,1 \right )$ . Hence,

$X_i=-\frac{1}{\lambda }\ln \left ( \xi _i \right )$ are exponential random variables if $\xi _i$ are standard uniform random variables.

3.4)The expectation is

E(X)xf (x)dx 0 0 0o 0 E(X) = 0-0+- E(X) =-

The second moment is (use the result above)

0 0 2 2 2

The proofs are complete.

The variance is

Var (X) E (X2)-E (X)' Var (X) = 2-2 Var (X) =