1
$\begingroup$

Let $X_1, ..., X_n$ i.i.d., exponentially distributed with parameter $\alpha$. Consider $Y = \frac 1 n \sum_{i=1}^nX_i$. Find such constants $a(\alpha)$ and $\sigma^2(\alpha) > 0$ that $\sqrt{n}(Y \sin Y - a(\alpha)) \overset{\mathbb{d}}{\to} \mathcal{N}(0, \sigma^2(\alpha))$ as $n \rightarrow \infty$.

My attempt:

Let's denote $Y = \frac{S_n}{n}$. Using the law of large numbers, we get: $\frac{S_n}{n} \overset{P}{\to}a$ (where $a$ is the expectation of $X_i$) and therefore $\frac{S_n}{n} \sin (\frac{S_n}{n}) \overset{P}{\to} a \sin a.$ I suppose, we need to use the following lemma to finish the problem: $\xi_n \overset{d}{\to} \xi, \eta_n \overset{d}{\to} c = \text{const} \Rightarrow \xi_n + \eta_n \overset{d}{\to} \xi + c, \xi_n \eta_n \overset{d}{\to} \xi c$ and CLT. But I am stuck with how to do it.

$\endgroup$
1
  • $\begingroup$ By "exponentially distributed with parameter $\alpha$ do you mean the distribution is $$ e^{-x/\alpha} (dx/\alpha) \text{ for } x\ge0 $$ (so that the expected value is $\alpha$) or do you mean it is $$ e^{-\alpha x} (\alpha\,dx) \text{ for } x\ge0 $$ (so that the expected value is $1/\alpha\text{?}$ Both conventions are used. $\qquad$ $\endgroup$ Feb 15, 2017 at 1:16

1 Answer 1

1
$\begingroup$

Rewrite $\sqrt{n}(Y\sin Y-a\sin a)$ as $$ \sqrt{n}(Y\sin Y-a\sin a) = \sqrt{n}(Y-a)\cdot\dfrac{Y\sin Y-a\sin a}{Y-a}=A_n\cdot B_n, $$ where the product is assumed to be zero for $Y=a$.

For $A_n$ CLT is valid.

Then find the limit of $B_n$ in probability: $Y\xrightarrow{p}a$ implies $B_n\xrightarrow{p}(x\sin x)'$ at $x=a$. And finally use the lemma which you mentioned for product.

And rewrite $a$ in terms of $\alpha$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .