19 Lecture 18, Feburary 16, 2024
19.1 Mean of Poisson distribution
Let \(Z\sim Poi(\mu)\). Then the expectation of \(Z\) is \(E[Z] = \mu\).
Proof. Suppose \(X \sim Poi(\mu)\). Then we have \[\begin{align*} E[X] &= \sum\limits_\text{x \in X(S)} x\cdot f(x) && \text{definition}\\ &= \sum_{x=0}^\infty x \cdot e^{-\mu} \frac{\mu^x}{x!} && \text{P.F.}\\ &= \sum_{x=1}^\infty x \cdot e^{-\mu} \frac{\mu^x}{x!}\\ &= \mu \sum_{x=1}^\infty e^{-\mu} \frac{\mu^{x-1}}{(x-1)!} && \text{let $Y=x-1$}\\ &= \mu \underbrace{\sum_{y=0}^\infty e^{-\mu} \frac{\mu^{y}}{y!}}_{=1\text{ as sum of p.f.}}\\ \end{align*}\]
19.2 Mean of Hypergeometric and Negative binomial
19.3 Note that the expectation does not always exist
Example 19.1 Consider a random variable \(X\) with probability function \(f(x)=\frac{1}{x}\) for \(x=2, 4, 8, 16,\dots\) and 0 otherwise.
The rv \(X\) can only take values \(x\) of the form \(x=2^n\). Note that we can write \(P(X=2^n) = 2^{-n}\) for \(n=1,2,\dots\). Thus \[\sum\limits_{\text{ all x}} f(x) = \sum_{n=1}^\infty P(X=2^n)= \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n=\frac{1}{1-1/2} -1 = 1.\] Then \[ E[X] = \sum_{n=1}^\infty 2^n \left(\frac{1}{2^n}\right)=\sum_{n=1}^\infty 1 = \infty,\]