Suppose that $X_ n \sim \text {Gamma}\ (n\alpha , \lambda)$ for all $n \ge 1$, for fixed $\alpha,\lambda >0.$ Show that
$$\frac {1} {\sqrt n} \left (X_n – \frac {n \alpha} {\lambda} \right ) \implies Z\ ,$$
where $Z \sim \mathcal N (0,{\sigma}^2)$ for some $\sigma.$ Calculate $\sigma.$
I have constructed a sequence random variables $\{S_n \}$ in the following way $:$
$S_1=X_1,S_2=X_2-X_1,S_3=X_3-X_2, \cdots , S_n = X_n – X_{n-1} , \cdots.$ Then I observed that $\sum\limits_{k=1}^{n} S_k = X_n$ for all $n \ge 1$. Also I observed that $\Bbb E(S_n) = \frac {\alpha} {\lambda}$ and $\Bbb {Var} (S_n) = \frac {\alpha} {{\lambda}^2}$ for all $n \ge 1$. If moreover $S_n$'s can be shown to be independent then they are i.i.d. random variables with mean $\frac {\alpha} {\lambda}$ and variance $\frac {\alpha} {{\lambda}^2}$. Then by central limit theorem we can say that
$$\frac {1} {\sqrt n} \left (X_n – \frac {n \alpha} {\lambda} \right ) \implies Z$$ as $n \rightarrow \infty$ where $Z \sim \mathcal N (0,\frac {\alpha} {{\lambda}^2})$. Hence ${\sigma}^2 = \frac {\alpha} {{\lambda}^2}$ i.e. $\sigma =\frac {\sqrt \alpha} {\lambda}.$
But how do I prove that $S_i$'s are i.i.d. random variables keeping in mind the fact that $X_i$'s are independent random variables? Please help me in this regard.
Thank you very much.
Best Answer
This answer uses the the following fact.
Hints: Let $(Y_i)_{i \in \mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i \sim \Gamma(\alpha,\lambda)$.