[proofplan] We compute each [probability generating function](/page/Probability%20Generating%20Function) directly from the defining identity $G_X(z)=\mathbb E[z^X]=\sum_{k=0}^{\infty}\mathbb P(X=k)z^k$, with the convention $z^0=1$. For the Bernoulli law the sum has only two terms. For the binomial law the finite sum is exactly the binomial expansion of $(1-p+pz)^m$. For the Poisson and shifted geometric laws, the defining probability mass functions reduce the pgf to the exponential [power series](/page/Power%20Series) and a geometric series, respectively. [/proofplan] [step:Rewrite the probability generating function as a probability mass sum] Fix $z\in[0,1]$. Since every [random variable](/page/Random%20Variable) in the theorem takes values in $\{0,1,2,\dots\}$, the definition of the probability [generating function](/page/Generating%20Function) gives \begin{align*} G_X(z)=\sum_{k=0}^{\infty}\mathbb P(X=k)z^k. \end{align*} Here $z^0$ is interpreted as $1$, including when $z=0$. [guided] Fix $z\in[0,1]$. The point of using a probability generating function is that it packages the [probability mass function](/page/Probability%20Mass%20Function) into one power series. Since $X$ takes values only in $\{0,1,2,\dots\}$, the expectation of $z^X$ is the discrete expectation over these possible values: \begin{align*} G_X(z)=\mathbb E[z^X]=\sum_{k=0}^{\infty}z^k\mathbb P(X=k). \end{align*} We write this equivalently as \begin{align*} G_X(z)=\sum_{k=0}^{\infty}\mathbb P(X=k)z^k. \end{align*} The convention $z^0=1$ is part of the probability generating function convention, so the term corresponding to $X=0$ is always $\mathbb P(X=0)$, even at $z=0$. [/guided] [/step] [step:Evaluate the two point Bernoulli sum] Assume $X\sim\operatorname{Ber}(p)$ with $p\in[0,1]$. Then $\mathbb P(X=0)=1-p$ and $\mathbb P(X=1)=p$, while $\mathbb P(X=k)=0$ for every $k\ge2$. Therefore \begin{align*} G_X(z)=(1-p)z^0+pz^1. \end{align*} Using $z^0=1$, this becomes \begin{align*} G_X(z)=1-p+pz. \end{align*} [/step] [step:Apply the finite binomial expansion to the binomial law] Assume $X\sim\operatorname{Bin}(m,p)$ with $m\in\mathbb N$ and $p\in[0,1]$. The binomial probability mass function is \begin{align*} \mathbb P(X=k)=\binom{m}{k}p^k(1-p)^{m-k} \end{align*} for $k\in\{0,1,\dots,m\}$, and $\mathbb P(X=k)=0$ for $k>m$. Hence \begin{align*} G_X(z)=\sum_{k=0}^{m}\binom{m}{k}p^k(1-p)^{m-k}z^k. \end{align*} Regrouping the powers gives \begin{align*} G_X(z)=\sum_{k=0}^{m}\binom{m}{k}(pz)^k(1-p)^{m-k}. \end{align*} By the finite binomial expansion with terms $pz$ and $1-p$, \begin{align*} G_X(z)=(1-p+pz)^m. \end{align*} [/step] [step:Sum the Poisson probabilities using the exponential series] Assume $X\sim\operatorname{Poi}(\lambda)$ with $\lambda>0$. Then for every $k\in\{0,1,2,\dots\}$, \begin{align*} \mathbb P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}. \end{align*} Substituting this probability mass function into the pgf gives \begin{align*} G_X(z)=\sum_{k=0}^{\infty}e^{-\lambda}\frac{\lambda^k}{k!}z^k. \end{align*} Factoring out $e^{-\lambda}$ and combining powers, \begin{align*} G_X(z)=e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda z)^k}{k!}. \end{align*} The exponential power series gives \begin{align*} \sum_{k=0}^{\infty}\frac{(\lambda z)^k}{k!}=e^{\lambda z}. \end{align*} Therefore \begin{align*} G_X(z)=e^{-\lambda}e^{\lambda z}=\exp(\lambda(z-1)). \end{align*} [/step] [step:Sum the shifted geometric probabilities as a geometric series] Assume $\mathbb P(X=k)=p(1-p)^k$ for every $k\in\{0,1,2,\dots\}$, where $p\in(0,1]$. Substitution into the pgf gives \begin{align*} G_X(z)=\sum_{k=0}^{\infty}p(1-p)^kz^k. \end{align*} Combining powers, \begin{align*} G_X(z)=p\sum_{k=0}^{\infty}((1-p)z)^k. \end{align*} Since $z\in[0,1]$ and $p\in(0,1]$, we have $0\le (1-p)z<1$ if $p<1$, and $(1-p)z=0$ if $p=1$. Thus the geometric series formula applies, giving \begin{align*} \sum_{k=0}^{\infty}((1-p)z)^k=\frac{1}{1-(1-p)z}. \end{align*} Multiplying by $p$ yields \begin{align*} G_X(z)=\frac{p}{1-(1-p)z}. \end{align*} The four displayed formulas are exactly the asserted probability generating functions. [/step]