Probability generating functions for a linear change of variable?

Question

I am struggling to understand, why, in textbook exercise on probability generating functions it says that if $X$ is a random variable (rv) with probability generating function (pgf) $G_{X}(s)$, then for rv $Z = X + k$ for positive integer $k$ the pgf is given by $$G_{Z}(s) = s^{k}G_{X}(s).$$

Surely if we assume that the rv $X$ takes integer values $0,1,2,,,$ with probability $p_{0}, p_{1}, p_{2},...$then the rv $Z$ takes integer values $k, k+1, k+2,...$ with corresponding probability $p_{0}, p_{1}, p_{2},...$, and thus the corresponding generating functions should be identical? The only thing that changes is the integer starting position (it is shifted by $k$) for the distribution of rv?

Answer 1

Since $\mathbb P(X\geqslant0)=1$ we have \begin{align} G_Z(s) &= \sum_{n=0}^\infty \mathbb P(Z=n)s^n\\ &=\sum_{n=0}^\infty \mathbb P(X+k=n)s^n\\ &=\sum_{n=0}^\infty \mathbb P(X=n-k)s^n\\ \end{align} and with the change of variables $m=n-k$, this is equal to $$ \sum_{m=0}^\infty \mathbb P(X=m)s^{m+k} = s^k\sum_{m=0}^\infty \mathbb P(X=m)s^m = s^kG_X(s). $$ Your intuition was correct. The factor of $s^k$ is precisely what accounts for the shift of support from $\{0,1,\ldots\}$ to $\{k,k+1,\ldots\}$. Note that for example, the coefficient of $s^k$ is $\mathbb P(X=0)$, when in the probability generating function for $X$ the coefficient of $s^k$ was $\mathbb P(X=k)$.

Answer 2

Let $G_X(s)$ be the probability generating function (PGF) of the random variable $X$. Then if $Z = X + k$ for some fixed $k$, we simply have $$G_Z(s) = \operatorname{E}[s^Z] = \operatorname{E}[s^{X + k}] = \operatorname{E}[s^k s^X] = s^k \operatorname{E}[s^X] = s^k G_X(s).$$