Let $X$ be an intege valued rv and $Y=aX+b$ where $a,b > 0$ Let $X$ be an integer valued r.v. and $Y=aX+b$, where $a,b > 0$. If $P(s)$ be the probability generating function of $X$, then what is the probability generating function of $Y$?
If $Q(s)$ be the probability generating function of $Y$, I think that it will be of the following form:
$$Q(s) = \sum_{k=0}^{\infty}as^kp_k + \sum_{k=0}^{\infty}bs^k$$
$$Q(s) = aP(s) + \frac{b}{1-s}$$
Is this ok?
 A: So $X$ is discrete random variable so $Y=aX+b$ for $a,b>0$ is discrete random variable taking values over $\mathbb{Z}^{+}$.
By definition the Probability Generating Function for $Y$ is given by
\begin{align*}
Q_{Y}(s)&=\mathbb{E}[s^{Y}],\\&=\mathbb{E}[s^{aX+b}],\\ &=\color{blue}{s^{b}\mathbb{E}[s^{aX}]},\\&=\color{blue}{s^{b}P_{X}(s^{a})}.
\end{align*}
A: The other answer gave you the correct solution. I just want to explain why your solution does not work.
Simply put, if $p$ is the probability function of $X$ and $q$ of $Y$, then $Y = aX + B$ does not mean that
$$q_k = a p_k + b.
$$
It rather means that
$$q_{x} = \begin{cases}p_k &\text{ if }x=ak+b\\0 &\text{ otherwise}\end{cases}$$
(this simple form is because the $k\mapsto ak+b$ transform is injective, otherwise there would be a sum).
Then using that, you could write
$$Q(s) = \sum_{x : q_x\neq 0} q_x s^x = \sum_{k} q_{ak+b} s^{ak+b} = \sum_{k} p_k s^{ak+b} = s^bP(s^a)$$
Doing a change of variable to get from the first sum to the second sum.
You can see how easier it is to use the expectation notation as in the other answer.
