Central limit theorem for sequence of Gamma-distributed random variables.

central limit theoremconvergence-divergencegamma distributionprobability theoryweak-convergence

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.

If $X \sim \Gamma(\alpha,\lambda)$ and $Y \sim \Gamma(\beta,\lambda)$ are independent, then $X+Y \sim \Gamma(\alpha+\beta,\lambda)$.

Hints: Let $(Y_i)_{i \in \mathbb{N}}$ be a sequence of independent identically distributed random variables such that $Y_i \sim \Gamma(\alpha,\lambda)$.

  1. Show that $\tilde{X}_n := \sum_{i=1}^n Y_i$ satisfies $\tilde{X}_n \sim \Gamma(n \alpha,\lambda)$.
  2. Apply the central limit theorem to prove that $$\frac{1}{\sqrt{n}} \left( \tilde{X}_n - \frac{n \alpha}{\lambda} \right) \stackrel{d}{\to} Z$$ for $Z \sim N(0,\sigma^2)$ with $\sigma^2 = \text{var}(Y_1)$; here $\stackrel{d}{\to}$ denotes convergence in distribution.
  3. Use the fact that $\tilde{X}_n$ equals in distribution $X_n$ for each $n \in \mathbb{N}$ to conclude from Step 2 that $$\frac{1}{\sqrt{n}} \left( X_n - \frac{n \alpha}{\lambda} \right) \stackrel{d}{\to} Z$$
  4. Compute $\sigma^2 = \text{var}(Y_1)$ (...or look it up, e.g. on wikipedia).
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