[proofplan] The probability generating function $G_X(z) = \sum_{k=0}^{\infty} \mathbb{P}(X = k) z^k$ is a power series in $z$. A power series is uniquely determined by its coefficients, and the coefficients of $G_X$ are exactly the probabilities $\mathbb{P}(X = k)$. We recover each probability by differentiating $G_X$ and evaluating at $z = 0$. [/proofplan] [step:Express $G_X$ as a power series with coefficients $\mathbb{P}(X = k)$] Let $X$ be a random variable taking values in $\{0, 1, 2, \ldots\}$. By definition, the probability generating function of $X$ is the [power series](/page/Power%20Series) \begin{align*} G_X(z) = \sum_{k=0}^{\infty} \mathbb{P}(X = k) \, z^k. \end{align*} This series converges absolutely for all $|z| \le 1$, since $\sum_{k=0}^{\infty} \mathbb{P}(X = k) |z|^k \le \sum_{k=0}^{\infty} \mathbb{P}(X = k) = 1$. [/step] [step:Recover each probability by differentiating and evaluating at $z = 0$] Since $G_X$ is a power series with radius of convergence at least $1$, it is infinitely differentiable on $(-1, 1)$ and its derivatives are obtained by term-by-term differentiation. The $k$-th derivative is \begin{align*} G_X^{(k)}(z) = \sum_{j=k}^{\infty} j(j-1)\cdots(j-k+1)\, \mathbb{P}(X = j)\, z^{j-k}. \end{align*} Evaluating at $z = 0$, every term with $j > k$ vanishes because it contains a positive power of $z$, leaving only the $j = k$ term: \begin{align*} G_X^{(k)}(0) = k!\, \mathbb{P}(X = k). \end{align*} Therefore \begin{align*} \mathbb{P}(X = k) = \frac{G_X^{(k)}(0)}{k!} \quad \text{for each } k \ge 0. \end{align*} [/step] [step:Conclude that equal PGFs imply equal distributions] If two random variables $X$ and $Y$ taking values in $\{0, 1, 2, \ldots\}$ satisfy $G_X(z) = G_Y(z)$ for all $z \in [0, 1]$, then equality of the power series forces equality of all coefficients: \begin{align*} \mathbb{P}(X = k) = \frac{G_X^{(k)}(0)}{k!} = \frac{G_Y^{(k)}(0)}{k!} = \mathbb{P}(Y = k) \quad \text{for all } k \ge 0. \end{align*} Hence $X$ and $Y$ have the same distribution. [/step]