Find the moment generating function of $r_n Y_n$

Question

Let $(Y_n)_{n\geq 1}$ be a sequence of geometrical random variables with parameter $r_n$ such that $r_n\to 0$ as $n\to \infty$.

Find the moment generating function of $r_n Y_n$. You can assume that $r_n$ is a function on $\mathbb{R}$ in $C^1 ([0,g])$ for $g\geq 1$

This is what I did \begin{equation} \begin{split} M_{r_nY_n}(t)&=\sum_{k=0}^{\infty} e^{tk}P(r_nY_n=k) \\ &=\sum_{k=0}^{\infty} e^{tk}P(Y_n=\frac{k}{r_n})=\sum_{k=0}^{\infty} e^{tk}r_n(1-r_n)^{\frac{k}{r_n}-1}\\ &=\frac{r_n}{1-r_n} \sum_{k=0}^{\infty} e^{tk}(1-r_n)^{\frac{k}{r_n}}= \frac{r_n(1-r_n)^{\frac{1}{r_n}-1}}{1-r_n} \sum_{k=0}^{\infty}e^{tk}(1-r_n)^k \\ &=(1-r_n)^{\frac{1}{r_n}-2}r_n \sum_{k=0}^{\infty}[e^{t}(1-r_n)]^k\\ &= \frac{(1-r_n)^{\frac{1}{r_n}-2}r_ne^t}{1-(1-r_n)e^t}\\ \notag \end{split} \end{equation}

I was wondering if this was the right way to do this?

Answer 1

No. $Y_n$ takes integer values, whereas $r_nY_n$ doesn’t generally take integer values, so it makes no sense to sum over probabilities that it’s an integer. You want

$$ M_{r_nY_n}(t)=\sum_{k=0}^\infty\mathrm e^{tr_nk}\mathsf P(Y_n=k)\;. $$

(Note also that there’s an excess factor $\mathrm e^t$ in your last line.)