Recall that we say $X$ ~ $Geometric(p)$ if $P(X=k)=p(1-p)^{k-1}$ for each $k=1,2,3,...$ The mean, variance, and moment generating function are $E(X)=\frac{1}{p}$ $var(X)=\frac{1-p}{p^2}$ $M_X(s)=\frac{p}{e^{-s}-(1-p)}$
Let $N$ ~ $Geometric(p)$. Let $Y$ denote the sum of $N$ independent identically distributed random variables $X_i$ ~ $Geometric(p)$.
I have the following solution and would appreciate some input as to whether or not my work along with my answers is correct.
(a) What are $E(Y|N)$ $E(Y)$ $var(Y|N)$ $var(E(Y|N)$ and $var(Y)$
$$E(Y|N)=E(X_1+...+X_n|N)=\frac{1}{p}N$$ $$E(Y)=E(E(Y|N))=E(\frac{1}{p}N)=\frac{1}{p}E(N)=\frac{1}{p^2}$$ $$var(Y|N)=E(Y^2|N)-(E(Y|N))^2=\frac{1}{p^2}N-(\frac{1}{p}N)^2=\frac{N-N^2}{p^2}$$ $$var(E(Y|N))=var(\frac{1}{p}N)=\frac{1}{p}var(N)=\frac{1}{p}\frac{1-p}{p^2}=\frac{1-p}{p^3}$$ $$var(Y)=var(E(Y|N))+E(var(Y|N))=\frac{1-p}{p^3}+E(\frac{N-N^2}{p^2})=\frac{1-p}{p^3}+\frac{1}{p^2}E(N)-\frac{1}{p^2}E(N^2)=\frac{1-p}{p^3}+\frac{1}{p^2}\frac{1}{p}-\frac{1}{p^2}E(N^2)$$ $$E(N^2)=Var(N)+E(N)^2=\frac{1-p}{p^2}+\frac{1}{p^2}=\frac{2-p}{p^2}$$ $$var(Y)=\frac{1-p}{p^3}+\frac{1}{p^2}\frac{1}{p}-\frac{1}{p^2}\frac{2-p}{p^2}=\frac{2-p}{p^3}-\frac{2+p}{p^4}$$
(b) What is the moment generating function of $Y$?
$$M_Y(s)=E(e^{sk})=\sum_{k=1} ^\infty p(1-p)^{k-1}e^{sk}=$$
Not sure what to do from here.
Hint 1: $\mathsf {Var}(N)=\mathsf E(N^2)-\mathsf E(N)^2$
Hint 2: $\mathsf M_Y(s) ~{=~ \mathsf E(e^{sY}) \\= \mathsf E\big(\mathsf E(e^{sY}\mid N)\big)\\=~\mathsf E\big(\mathsf E(e^{s\sum_{k=1}^N X_k}\mid N)\big)\\ =~\mathsf E\big(\prod_{k=1}^N\mathsf E(e^{sX_k})\big) \qquad\text{(why fore?)} \\ =~ \mathsf E\Big(\big(\mathsf E(e^{sX_1})\big)^N\Big)\\=~\mathsf E\big(e^{N\ln\mathsf E(e^{s X_1})}\big)\\~\vdots}$