Properties of pgf's I am currently studying pdf's, and there is this property that I am keep using which allows me to solve the problems correctly, however I do not have a justified reason why I can use this property. Given the definition of PGF as the following:
$$G_{X}(t)=\sum_{k=0}^{\infty}t^kP(X=k)$$
Does the property below hold true?:
$$G_{aX+b}(t)=\sum_{k=0}^{\infty}t^{ak+b}P(X=k)=t^bG_{X}(t^a)$$
I know this should hold true as I have used it many times, but I do not get how I can just power $t$ to $ak+b$ mathematically. If this holds true, can anyone help me explain how this holds?
 A: Assume $X : \Omega \to \mathbb{N}$ and consider the function
$$f: \Omega \to \mathbb{R}; \quad f(\omega) = t^{aX(\omega) + b}. $$
First note that the PGF can generally be written as $G_X(t) = \mathbb{E}(t^X).$
So we have 
\begin{align}
G_{aX + b}(t) &= \mathbb{E}(f) = \mathbb{E} (t^{aX + b})
\\
&= t^b \mathbb{E} \big( (t^a)^X\big)  
\\
&= t^b \sum_{k=0}^\infty (t^a)^k P(X = k)
\\
&= \sum_{k=0}^\infty t^{ak + b} P(X = k).
\end{align}
A: For $t\in (-1,1]$ the PGF always exists. By definition
$$ G_{aX+b} (t) = \sum_m t^m P(aX+b=m) = \sum_k t^{ak+b} P(X=k)= t^b G_X(t^a)$$ 
for this to hold $a>0$ is needed. In case $a<0$ take $|t|>1$. 
A: $G_{aX+b}(t)=\sum_{k=0}^{\infty}t^kP(aX+b=k)$
$\qquad =\sum_{k=0}^{\infty}t^kP(aX=k-b)$
$\qquad =\sum_{k=-b}^{\infty}t^{k+b}P(aX=k)$
$\qquad =\sum_{k=0}^{\infty}t^{k+b}P(aX=k)$
$\qquad =\sum_{k=0}^{\infty}t^{k+b}P(X=\frac{k}{a})$
$\qquad =\sum_{k=0}^{\infty}t^{ak+b}P(X=k)$
A: Let $G_{X}(t)=Et^{X}$ be the probability generating function of $X$. Then note that
$$
G_{aX+b}(t)=Et^{aX+b}=E[t^{b}(t^a)^X]=t^bE[(t^a)^X]=t^{b}G_X(t^a)
$$
