Poisson Probability Generating Function
For a non-negative integer-valued random variable $X$, the probability generating function (PGF) is defined as
$$p_X(z) = E[z^X], \quad |z| \leq 1.$$
Derive the PGF for $X \sim \text{Poisson}(\lambda)$. Then, with $\lambda = 2$, compute $p_X(1/2)$ to the nearest thousandth.
Hints
- Write out $E[z^X]$ explicitly as a sum using the Poisson PMF and LOTUS. Group all terms involving $z$ and $\lambda$ together.
- After factoring out $e^{-\lambda}$, the remaining sum is $\sum_{k=0}^{\infty} \frac{(\lambda z)^k}{k!}$. Does this series look familiar?
- The sum $\sum_{k=0}^{\infty} \frac{x^k}{k!} = e^x$ for all $x$. Substitute $x = \lambda z$ and simplify to get $p_X(z) = e^{\lambda(z-1)}$.
Worked Solution
How to Think About It: The PGF is just an expectation, so reach for LOTUS (Law of the Unconscious Statistician): write out $E[z^X]$ as a sum over all non-negative integers, weighted by the Poisson PMF. Once you group the terms, you will recognize the Taylor series for $e^x$ -- and the whole thing collapses to a clean closed form. This is one of those derivations worth memorizing because PGFs make compound Poisson sums trivial.
Quick Estimate: The Poisson(2) mean is 2, so $X$ is typically 0, 1, 2, or 3. At $z = 1/2$, $z^X$ is being down-weighted for large $X$. A rough bound: $E[z^X] \leq 1$ (since $z^X \leq 1$) and $E[z^X] \geq z^{E[X]} = (1/2)^2 = 0.25$ by Jensen (since $z^x$ is convex in $x$ for $z < 1$... actually concave -- so Jensen flips). The true answer should be between 0.25 and 1. Since $e^{-1} \approx 0.368$, we will land right in that range.
Approach: LOTUS + recognizing the $e^x$ Taylor series.
Formal Solution:
Using the Poisson PMF $P(X = k) = \dfrac{\lambda^k}{k!} e^{-\lambda}$ and LOTUS:
$$p_X(z) = E[z^X] = \sum_{k=0}^{\infty} z^k \cdot \frac{\lambda^k}{k!} e^{-\lambda} = e^{-\lambda} \sum_{k=0}^{\infty} \frac{(\lambda z)^k}{k!}.$$
The remaining sum is exactly the Maclaurin series for $e^{\lambda z}$, so:
$$\boxed{p_X(z) = e^{\lambda(z-1)}.}$$
With $\lambda = 2$ and $z = 1/2$:
$$p_X(1/2) = e^{2(1/2 - 1)} = e^{-1} \approx 0.368.$$
Answer: $p_X(z) = e^{\lambda(z-1)}$. For $\lambda = 2$, $z = 1/2$: $p_X(1/2) = e^{-1} \approx \mathbf{0.368}$.
Intuition
The Poisson PGF $e^{\lambda(z-1)}$ is one of the most useful formulas in applied probability. Its real power shows up when you compound distributions: if $N \sim \text{Poisson}(\lambda)$ and $X_1, X_2, \ldots$ are i.i.d. with PGF $p_X(z)$, then the random sum $S = X_1 + \cdots + X_N$ has PGF $p_S(z) = e^{\lambda(p_X(z)-1)}$. This is the backbone of compound Poisson processes used everywhere in insurance (aggregate claims), trading (order flow), and reliability (failure counts).
The derivation also illustrates a general pattern: whenever you need $E[g(X)]$ for a discrete distribution, write it as a weighted sum over the PMF, then hunt for a recognizable series. Recognizing the $e^x$ Taylor series here is exactly the same move as recognizing a geometric series or a binomial expansion -- these pattern-matches are the bread and butter of probability calculations in interviews.