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If $X_1, ... , X_n$ be iid exponential with mean $1/\lambda$. Let $S_n = X_1 + ... + X_n$.

a) Show that $S_n$ is $\Gamma(n, 1/\lambda)$.

Each $X_i$ is $\Gamma(1, 1/\lambda)$ by the definition of an exponential distribution. We can look at the mgf of $S_n$.

$$M(t) = E(e^{t \sum X_i}) = E(e^{tX_1})...E(e^{tX_n}) = (\frac{1}{1-t/\lambda})...(\frac{1}{1-t/\lambda}) = (\frac{1}{1-t/\lambda})^n,$$

which is the mgf of $\Gamma(n, 1/\lambda)$. So $S_n \sim \Gamma(n, 1/\lambda)$.

b) Show that $\frac{(S_n - n/\lambda)}{\sqrt{n}}$ converges in distribution and identify the asymptotic distribution.

By the Central Limit Theorem, we know that $\frac{(S_n - n\mu)}{\sqrt{n}(1/\lambda^2)} = \frac{(S_n/n - \mu)}{(1/\lambda^2)(1/\sqrt{n})}$ converges in distribution to $N(0,1)$.

We see that $\frac{(S_n/n - 1/\lambda)}{(1/\lambda^2)(1/\sqrt{n})} = \frac{(S_n/\sqrt{n} - \sqrt{n}/\lambda)}{(1/\lambda^2)}(\frac{\sqrt{n}}{\sqrt{n}}) = \frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})}$

So, $$\frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})} \xrightarrow{D} N(0,1)$$ implies that $$ \frac{S_n - n/\lambda}{(\sqrt{n})} = (1/\lambda^2)\frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})} \xrightarrow{D} N(0, 1/\lambda^4).$$

c) Show that $\sqrt{n}(\sqrt{S_n/n} - \sqrt{1/\lambda})$ converges in distribution and identify the asymptotic distribution.

There is a theorem in our textbook (Introduction to Mathematical Statistics by Hogg) which states,

Let $\{X_n\}$ be a sequence of random variables such that $\sqrt{n}(X_n - \theta) \xrightarrow{D} N(0, \sigma^2)$. Suppose the function $g(x)$ is differentiable at $\theta$ and $g'(\theta) \not= 0$. Then $\sqrt{n}(g(X_n) - g(\theta)) \xrightarrow{D} N(0, \sigma^2(g'(\theta)^2)$.

From part b), we see that $$\frac{\sqrt{n}(S_n/n - 1/\lambda)}{(1/\lambda^2)} \rightarrow N(0,1).$$

So $$\sqrt{n}(S_n/n - 1/\lambda) \rightarrow N(0, 1/\lambda^4)$$.

Now we apply the theorem. We have $g(x)=\sqrt{x}$. So $g'(x)=1/(2\sqrt{x})$, and $g'(\frac{1}{\lambda}) = \frac{\sqrt{\lambda}}{2}$.

So by the theorem,

$$\sqrt{n}(\sqrt{S_n/n} - \sqrt{1/\lambda}) \rightarrow N(0, (\frac{\sqrt{\lambda}}{2\lambda^4})).$$

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    $\begingroup$ This is a little nit-picky, but in (b) you need to justify the applicability of the CLT. To see why this might be necessary, re-work your answers replacing the exponential distribution with a Cauchy distribution. $\endgroup$
    – whuber ♦
    Commented Jan 9, 2014 at 22:02
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    $\begingroup$ At the end of c), shouldn't you square the derivative of $g$? $\endgroup$
    – Alecos Papadopoulos
    Commented Jan 9, 2014 at 22:07
  • $\begingroup$ There are a few more algebraic mistakes in (c), too. Consider running a simulation for a tractable $\lambda$ (such as $\lambda=3$) and increasing values of $n$ (to establish near-convergence of the results) as a check of your formulas. It takes only a few minutes. $\endgroup$
    – whuber ♦
    Commented Jan 9, 2014 at 22:34
  • $\begingroup$ @whuber Thanks a lot. In our textbook, it says "Let $X_1, ... , X_n$ denote the observation that has mean $\mu$ and positive variance $\sigma^2$. Then the random variable $Y_n = (\sum^n_{i=1} X_i - n \mu)/\sqrt{n}\sigma = \sqrt{n}(\bar{X_n} - \mu)\sigma$ converges in distribution to a random variable which has a normal distribution with mean zero and variance 1." Since, in this case, the $X_i$ are iid and $\sigma^2 = n/\lambda^2 > 0$, we can use the CLT, right? $\endgroup$
    – Artus
    Commented Jan 10, 2014 at 12:16
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    $\begingroup$ Yes, that's correct: your comment supplies the justification for applying the CLT. $\endgroup$
    – whuber ♦
    Commented Jan 10, 2014 at 15:03

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To close this one: A slight algebraic correction in the calculation of the asymptotic variance (the OP forgot to square the derivative of the transformation function), gives that

$$\sqrt{n}\left(\sqrt{S_n/n} - \sqrt{1/\lambda}\right) \rightarrow N\left(0, \frac{1}{4\lambda^3}\right)$$

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