3.1.5 problem 5

(a)

$p\left(n\right)$ is clearly non-negative. Also,

$$\sum _{n=0}^{\infty }p\left(n\right)=\frac{1}{2}\sum _{n=0}^{\infty }\frac{1}{2^n}=\frac{1}{2}\ast \frac{1}{1-\frac{1}{2}}=1.$$

Thus, $p\left(n\right)$ is a valid PMF.

(b)

$$F\left(x\right)=\sum _{n=0}^{\lfloor x\rfloor }p\left(n\right)=\frac{1}{2}\sum _{n=0}^{\lfloor x\rfloor }\frac{1}{2^n}=\frac{1}{2}\ast \frac{1-\frac{1}{2^{\lfloor x\rfloor +1}}}{1-\frac{1}{2}}=1-\frac{1}{2^{\lfloor x\rfloor +1}}$$

for $x\ge 0$ and $0$ for $x<0$.