# Probability of even/odd using probability generating functions

## Probability Generating Functions

A probability generating function for a discrete random variable $$X$$ taking values $$\{0,1,\ldots\}$$ is defined as $G(z) = E[z^X] = \sum_{j = 0}^\infty z^j P(X = j)$

Which is defined for all complex $$z$$ in which the sum converges. It always converges for $$|z| < 1$$ but the radius might be larger depending on our distribution. Clearly we have that $$G(0) = 0$$. Also we can see that $G(1) = \sum_{j = 0}^\infty P(X = j) = 1$ since we are summing over the entire sample space.

### Probability of Even/Odd

This is problem 1.11 in Ross’s Stochastic Processes Second Edition.

We have that \begin{aligned} G(-1) &= \sum_{j = 0}^\infty (-1)^j P(X = j)\\ &= P(X = 0) - P(X = 1) + P(X = 2) - \cdots\\ G(1) &= P(X = 0) + P(X = 1) + P(X = 2) + \cdots\\ G(-1) + G(1) &= 2P(X = 0) + 2P(X = 2) + 2 P(X = 4) + \cdots \end{aligned}

Assuming that 0 is considered even it follows that \begin{aligned} P(X \text{ is even}) = \frac{G(1) + G(-1)}{2} = \frac{1 + G(-1)}{2} \end{aligned} We also have that $P(X \text{ is odd}) = 1 - \frac{1 + G(-1)}{2} = \frac{1 - G(-1)}{2}$

#### Binomial

Assume that $$X$$ is a binomial with parameters $$n$$ and $$p$$ $P(X = x) = {n \choose x} p^x (1 - p)^{n - x}$ Then we can compute the probability generating function

\begin{aligned} G(z) = E[z^X] &= \sum_{j = 0}^\infty z^j P(X = j)\\ &= \sum_{j = 0}^\infty z^j {n \choose j} p^j (1 - p)^{n - j}\\ &= \sum_{j = 0}^n {n \choose j} (zp)^j (1 - p)^{n - j}\\ &= (zp + 1 - p)^n && \text{Binomial theorem} \end{aligned}

pgf_binom <- function(z, n, p) {
(z * p + 1 - p)^n
}

z <- seq(0, 1, length.out = 1000)
n <- 10
p <- 1/3
plot(z, pgf_binom(z, n, 3/4), type = "l")

Note that the sum converges for all real $$z$$. This is a good point to check that our result matches that $$G(1) = 1$$. So $$G(-1) = (1 - 2p)^n$$. Thus, $P(X \text{ is even}) = \frac{1 + (1 - 2p)^n}{2}$

#### Poisson

Let $$X$$ be a Poisson random variable with mean $$\lambda$$ $P(X = k) = e^{-\lambda} \frac{\lambda^k}{k!}$

Then \begin{aligned} G(z) = E[z^X] &= \sum_{j = 0}^\infty z^j e^{-\lambda} \frac{\lambda^j}{j!}\\ &= e^{-\lambda} \sum_{j = 0}^\infty \frac{(\lambda z)^j}{j!}\\ &= e^{-\lambda} e^{\lambda z}\\ &= e^{-\lambda + \lambda z} \end{aligned} Note that the sum converges for all real $$z$$. Again as an exercise check that $$G(1) = 1$$. It follows that $G(-1) = e^{-2\lambda}$ So the probability of a Poisson random variable being even is $P(X \text{ is even}) = \frac{1 + e^{-2\lambda}}{2}$

#### Geometric

Assume now that $$X$$ is geometric with parameter $$p$$ with $$X \in \{1,2,\ldots\}$$. $P(X = k) = p (1 - p)^{k - 1}$ Then the probability generating function for $$X$$ is

\begin{aligned} G(z) = E[z^X] &= \sum_{j = 1}^\infty z^j P(X = j)\\ &= \sum_{j = 1}^\infty z^j p (1 - p)^{j - 1}\\ &= pz \sum_{j = 1}^\infty (z(1 - p))^{j - 1}\\ &= pz \sum_{j = 0}^\infty (z(1 - p))^{j}\\ &= \frac{pz}{1 - z(1 - p)} && \text{ for } |z| < \frac{1}{1 - p} \end{aligned} Since $$1 - p < 1$$, then the sum converges for $$z = -1$$. It follows that $G(-1) = \frac{- p}{2 - p}$ So the probability that the geometric random variable $$X$$ is even is given by

\begin{aligned} P(X \text{ is even}) &= \frac{1 + \frac{- p}{2 - p}}{2}\\ &= \frac{1 - p}{2 - p} \end{aligned}