The Problem: Fix $0<\lambda<\infty$ and for integers $n>\lambda$ let $X_n\thicksim\text{Binom}(n,\lambda/n)$ and $X\thicksim\text{Poisson}(\lambda)$. Prove the limit in distribution $X_n\overset{d}\longrightarrow X$ using moment generating functions.
My Attempt: First we compute the moment generating function of $X\thicksim\text{Poisson}(\lambda)$. For all $t\in\mathbb R$ we have that
\begin{align}
M_X(t)&=E\left[e^{tX}\right]=\sum_{k=0}^\infty\frac{e^{tk}\lambda^k e^{-\lambda}}{k!}\\
&=e^{-\lambda}\sum_{k=0}^\infty\frac{(e^t\lambda)^k}{k!}\\
&=e^{\lambda(e^t-1)}.
\end{align}
On the other hand, using the binomial theorem yields for all $t\in\mathbb R$ that
\begin{align}
M_{X_n}(t)&=\sum_{k=0}^ne^{tk}\binom{n}{k}\left[\frac{\lambda}{n}\right]^k\left[1-\frac{\lambda}{n}\right]^k\\
&=\left[\frac{e^t\lambda}{n}+1-\frac{\lambda}{n}\right]^n\\
&=\left[1+\frac{\lambda(e^1-1)}{n}\right]^n.
\end{align}
The above moment generating functions are finite for, say, all $t\in(-1,1)$ and in addition since
$$\lim_{n\to\infty}M_{X_n}(t)=\lim_{n\to\infty}\left[1+\frac{\lambda(e^1-1)}{n}\right]^n=e^{\lambda(e^t-1)}=M_X(t),$$
also holds for all $t\in(-1,1)$, the continuity theorem for moment generating functions implies that $X_n\overset{d}\longrightarrow X.$
Do you agree with my proof above?
Any feedback is most welcomed and appreciated. Thank your for your time.
Best Answer
I followed each step and they all look good to me, except there is a typo in the statement of the definition of the moment generating function for $X_n$ ($k$ should be $n-k$ for the exponent of $1-\lambda/n$. Also, the radius of convergence of both moment generating functions is all of $\mathbb{R}$, so why have you specified $t \in (-1,1)$?