Random Variable Problem with unrestricted Parameters Worded Problem

Question

I have no idea how to go about solving (a) -> (c)

• For (a) Is $k=0.2$, because $\frac{k}{1-0.8}=1$
• Hence, $P(Z=z) = 0.2(0.8)^x$

But How do we determine the mean or variance with unrestricted z values.

Answer 1

(1)- yes.

(2)- in the general case

$\begin{align} \because \sum_{x=0}^\infty p (1-p)^x &= \frac{p}{1-(1-p)} & \forall |p|< 1:\text{ the series converges} \\ & = 1 \\[2ex] \therefore P(X=x) &= p (1-p)^x \;:\; x\in\{0,1,2,3,\ldots\} & \text{is a probability mass function} \\[2ex] \text{Then use the following:} \\[2ex] \mathsf{Var}[X] & := \mathsf{E}[X^2]-\mathsf{E}[X]\;^2 & \text{definition of variance} \\[2ex] \mathsf{E}[X] &= p \sum_{x=0}^\infty x(1-p)^x & \text{the expectation of }X \\[1ex] \mathsf{E}[X^2] & = p\sum_{x=0}^\infty x^2(1-p)^x & \text{the expectation of }X^2 \\[2ex] \therefore \ldots \end{align}$

Answer 2

Hint for the expectation value

$E(Z)=\sum_{z=0}^{\infty}k\cdot z\cdot 0.8^z$

index shift

$\sum_{z=1}^{\infty} k\cdot (z-1)\cdot \frac{0.8^z}{0.8}=\sum_{z=1}^{\infty}k\cdot z\cdot \frac{0.8^z}{0.8}- \sum_{z=1}^{\infty} k\cdot \frac{0.8^z}{0.8}$

$\sum_{z=1}^{\infty} k\cdot z\cdot \frac{0.8^z}{0.8}=\sum_{z=0}^{\infty} k\cdot z\cdot \frac{0.8^z}{0.8}=\frac{1}{0.8} E(Z)$

$\sum_{z=1}^{\infty} k\cdot \frac{0.8^z}{0.8}=k\cdot \sum_{z=0}^{\infty} 0.8^z$

$E(Z)=\frac{1}{0.8} E(Z)-k\cdot \sum_{z=0}^{\infty} 0.8^z$

This equation can be solved for $E(Z)$.