I have an exercise of this type that I just can not solve
"Are $x$ and $y$ be independent random variables, $X$-Poisson($a$), $Y$-Poisson($b$). Find the probability generating function of the random variable $z = 2x+3y+4$"
help me :(
The only thing I could do is to put into practice a property of the probability generating function: $$G_z(u)= E(u^z) = E(u^{2x+3y+4}) = E(u^{2x} \cdot u^{3y} \cdot u^4) )= u^4 \cdot E(u^{2x}) \cdot E(u^{3y})$$
First you can find the probability generating function for a Poisson distribution.
If $X \sim \mathcal{P}(a)$ then : $$\begin{array}{rcl} E(t^X) & = & \sum_{n=0}^{\infty}e^{-a} \frac{a^n}{n!} t^n\\ & = & e^{-a} \sum_{n=0}^{\infty}\frac{(at)^n}{n!} \\ \Longleftrightarrow E(t^X) & = & e^{-a + at} \end{array} $$
You should now be able to find the probability generating function of Z with what you wrote : $$E(t^Z) = t^4 E(t^{2X}) E(t^{3Y})$$ by substituting $t$ by $t^2$ and $t^3$ in the equation above.
You should find : $$E(t^Z) = t^4e^{-a-b + at^2 + bt^3}$$