I'm having a lot of trouble with this problem:
Let $(X_n)_{n\ge1}$ a sequence of i.i.d. random variables with expected value equal to $1$ and variance equal to $1$. Find the distribution when $n$ goes to infinity of:
$$n^{3/2}\left(\frac{X_1 + \cdots + X_n – n}{3(X_1 + \cdots + X_n)^2}\right)$$
Not sure how to proceed. I have the feeling it's normal distribution with mean $0$ but not sure about the variance or if this is even correct.
Best Answer
Write the expression as $$ Y_n=\frac{X_{1}+\dotsb+X_n-n}{\sqrt{n}}\biggr/\frac{3(X_1+\dotsb+X_n)^2}{n^2} $$ and use Slutsky's theorem. Specifically, $$ \frac{X_{1}+\dotsb+X_n-n}{\sqrt{n}}\stackrel{d}{\to} Z\sim N(0,1) $$ by the central limit theorem and $$ \frac{3(X_1+\dotsb+X_n)^2}{n^2}\stackrel{\text{a.s.}}{\to} 3 $$ by the SLLN and in particular in probability. Hence Slutsky's theorem implies that $Y_n\stackrel{d}{\to} Z/3. $