It's time to roll the dice.
\(\mathtt{Definition}\)
令 \(X\) 为取值非负的随机变量，那么 \(X\) 的概率生成函数 \(\mathtt{Probability\ Generating\ Function}\) 为
\[\begin{aligned} G_x(z) = \sum_{k \ge 0} \mathrm{Pr}(X = k) z^k \end{aligned} \]根据上式可以得知该生成函数各项系数均非负，且其和为 \(1\) ，即是
\[\begin{aligned} G_x(1) = 1 \end{aligned} \]那么反过来，任何非负系数且满足 \(G_x(1) = 1\) 的幂级数 \(G_x\) 均为某随机变量的生成函数。
\(\mathtt{Average \ \& \ Variance}\)
\(\mathtt{Average}\)
\[\begin{aligned} \mathrm{E}(X) &= \sum_{k \ge 0} k \mathrm{Pr}(X = k) \\ &= \sum_{k \ge 0} \mathrm{Pr}(X = k) k \cdot 1^{k - 1} \\ &= G_{x}^{'} (1) \end{aligned} \]进而有
\[\begin{aligned} \mathrm{E}(X^{\underline{k} }) = G_x^{k}(1) (k \not = 0) \end{aligned} \]\(\mathtt{Variance}\)
\[\begin{aligned} \operatorname{D}(X)&=\operatorname{E}(X^2)-\operatorname{E}(X)^2\\ &=G_{X}^{''}(1)+G_{X}^{'}(1)-G_{X}^{'}(1)^2 \end{aligned} \]即是说在知道 \(G_{X}^{''}(1)\) 和 \(G_x^{'}(1)\) 的情况下就可以得到 \(\mathrm{D}(X)\) 。
\(\mathtt{Multiplication}\)
若两个随机变量 \(X, Y\) 互相独立，那么有
\[\begin{aligned} \operatorname{Pr}(X+Y=n)=&\sum_{k}\operatorname{Pr}(X=k \wedge Y=n-k)\\ =&\sum_{k}\operatorname{Pr}(X=k)\operatorname{Pr}(Y=n-k)\\ \end{aligned} \]写成卷积的形式
\[\begin{aligned} G_{X+Y}(z)=G_{X}(z)G_{Y}(z) \end{aligned} \] 标签：Function,Pr,Generating,20,begin,end,aligned,mathtt,operatorname From： https://www.cnblogs.com/Iridescent41/p/17528181.html