Let X be$Gamma(\alpha, \lambda)$Prove $(\lambda X - \alpha)/\sqrt{\alpha} \xrightarrow{d} N(0,1)$ as $\alpha \rightarrow \infty$ and $\lambda$is fixed

Question

First of all the continuity lemma is stated as follows:

Let $\mu_n, n=1,2, \dots$ be a sequence of distributions, and $\varphi$ the associated characteristic function.

1. If $\mu_n \xrightarrow{w} \mu$, then for all $t \in \mathbb{R}$ $\varphi_n(t) \rightarrow \varphi(t)$, where $\varphi$ is the characteristic function of $\mu$.

2. If for all $t \in \mathbb{R}$ $\varphi(t) := lim_{n \rightarrow \infty} \varphi_n(t)$ exists, and is continous at $t=0$, then $\varphi$ is the characteristic function of a distribution $\mu$, and $\mu_n \xrightarrow{w} \mu$

From here the question goes as follows:

Let X be $Gamma(\alpha, \lambda)$ distributed (notice that $\alpha$ might be a non-integer). Prove, via characteristic functions and the Continuity Lemma, that \begin{equation} \frac{\lambda X - \alpha}{\sqrt{\alpha}} \xrightarrow{d} Normal(0,1) \end{equation} as $\alpha \rightarrow \infty$ and $\lambda$ is fixed

Answer 1

Basically what you should do is find the MGF of $\frac{\lambda X -\alpha}{\sqrt{\alpha}}$ and then take the limit as $\alpha \to \infty$ and check that it is equal to $\exp{(t^2/2)}$

(I say MGF, but use CF)