Solutions - 5150-08

(a)

(1) Express $F_X(x)$ via the CDF of $U$:

For $x<0$: $F_X(x)=0$ since $X=-\ln (1-U)\ge 0$.

For $x\ge 0$:

$$\begin{array}{c}{F}_X(x)=P[-\ln (1-U)\le x]\gg \gg P[1-U\ge e^{-x}]\\ \\ \gg \gg P[U\le 1-e^{-x}]\end{array}$$

Since $0\le 1-e^{-x}\le 1$ for $x\ge 0$, this equals $1-e^{-x}$:

$$F_X(x)=\{\begin{array}{cc}0& x<0\\ 1-e^{-x}& x\ge 0\end{array}$$

---

(b)

Differentiate $F_X(x)$:

$$f_X(x)=\{\begin{array}{cc}{e}^{-x}& x\ge 0\\ 0& \text{otherwise}\end{array}$$

This is the exponential PDF with parameter $a=1$.

---

(c)

Since $X\sim \text{Exp}(1)$,

$$E[X]=1$$