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

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.

• Regarding your last paragraph: the $X_i$ are not i.i.d., as their distributions depend on $i$. – Quaternion Dec 1 '18 at 8:29
• Sorry I mean to say that they are independent. Thanks for pointing this out @Quaternion. I will edit my question soon. – Dbchatto67 Dec 1 '18 at 8:36
• @Dbchatto67 I don't think your variance of $S_n$ is calculated correctly. Note that $\textrm{V}(S_n) = \textrm{V}(X_n-X_{n-1}) = \textrm{V}(X_n)+\textrm{V}(X_{n-1}) = \frac{\alpha n}{\lambda^2} + \frac{\alpha (n-1)}{\lambda^2}$ – Lundborg Dec 1 '18 at 9:06
• In the statement of your problem you do not mention independence of the $X_i$... – saz Dec 1 '18 at 9:20
• I'm not sure if introducing the $S_n$'s and using the CLT is a good idea. Have you tried directly computing the limit while using the Stirling approximation for the $\Gamma(\alpha n)$ term? – Quaternion Dec 1 '18 at 9:21

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).
• So $Z \sim \mathcal N \left (0, \frac {\alpha} {{\lambda}^2} \right )$. Then $\sigma = \frac {\sqrt \alpha} {\lambda}$. Am I right @saz? – Dbchatto67 Dec 1 '18 at 11:28
• @Dbchatto67 Yes, that's right. – saz Dec 1 '18 at 11:50
• Could you take a look at my last question? would really appreciate it – badatmath Dec 1 '18 at 14:10
• @itry To which end? There is an answer and you accepted it... so where is the problem? – saz Dec 1 '18 at 17:23
• I mean the one with the bounty about cramers theorem – badatmath Dec 1 '18 at 17:24