# 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 [closed]

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 $$$$\frac{\lambda X - \alpha}{\sqrt{\alpha}} \xrightarrow{d} Normal(0,1)$$$$ as $$\alpha \rightarrow \infty$$ and $$\lambda$$ is fixed

## closed as off-topic by Lee David Chung Lin, YuiTo Cheng, Cesareo, marwalix, Adrian KeisterMay 8 at 13:29

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## 1 Answer

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)