Show that $S_n= \sqrt{n} \sum_{i=1}^{n} (X_i - \theta) / \sum_{i=1}^{n} X_i^2$ converges in distribution

Question

Studying for quals, looking for some help on this.

Let $X_i$, where $i = 1,2, \ldots, n$, be iid with common pdf $f(x) = \frac{1}{\theta} e^{-x/\theta}$ for $x>0$ and $0$ otherwise.

Let $S_n = \sqrt{n} \sum_{i=1}^{n} (X_i - \theta) / \sum_{i=1}^{n} X_i^2$. Show that $S_n$ converges in distribution and derive its limiting distribution.

What I know:

• $E(X_i) = \theta$.
• $Var(X_i) = \theta^2$
• Convergence in Probability is stronger than Convergence in Distribution
• Convergence in Probability definition is $\lim_{n \rightarrow \infty} P([X_n - X | \ge \epsilon ] = 0$ (also know equiv. definitions).
• If $X_n \rightarrow X$ and $Y_n \rightarrow Y$ in probability then $X_nY_n \rightarrow XY$ in probability.

I left out the convergence in distribution definitions because I think that what I need to do is show that $\sqrt{n} \sum_{i=1}^{n} (X_i - \theta)$ converges in probability and $\sum_{i=1}^{n} X_i^2$ converges in probability, then $S_n = \sqrt{n} \sum_{i=1}^{n} (X_i - \theta) / \sum_{i=1}^{n} X_i^2$ converges in probability and this converges in distribution.

I have no idea how to go about finding the limiting distribution.

Any help would be most appreciated...

-IdleMathGuy

Answer 1

Hint:

Since $$S_n = \frac{\sqrt n\left(\frac1n\sum_{i=1}^n X_i-\theta\right)}{\frac1n\sum\limits_{i=1}^n X_i^2}\,,$$

this is a simple application of Slutsky's theorem where the numerator converges in law to a normal distribution using central limit theorem and the denominator converges in probability to a constant using the (weak) law of large numbers.