Limiting Distribution for Squared Sum of i.i.d RVs divided by Sum of Square iid RVs

central limit theoremlaw-of-large-numbersprobabilityprobability theory

I'm facing some difficulties in determining the limiting distribution of

$$\frac{\left( \sum_{i=1}^n X_i\right)^2}{\sum_{i=1}^n X_i^2}$$ as $n \to +\infty$ for a sequence of i.i.d. random variables $\left(X_n \right)_{n = 1}^{\infty}$ with zero mean and finite variance (denoted by $\sigma^2$).

I have $3$ approaches

  1. Express the term above as $\frac{n}{\sum_{i=1}^n X_i^2} \left( \frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \right)^2$ where the first term converges to $\frac{1}{\sigma^2}$ by SLLN and I was thinking of $\left( \frac{1}{\sigma\sqrt{n}} \sum_{i=1}^n X_i \right)^2$ would converge to $Z^2$, where Z is the standard normal by CLT… (wishful thinking) but I know multiplicity doesn't hold for weak convergence… (correction, multiplicity holds for continuous functions by continuous mapping theorem)

  2. Express the term above as $\frac{n}{\sum_{i=1}^n \left( \frac{X_i^2}{n} \right)} \left( \frac{\sum_{i=1}^n X_i}{n} \right)^2$ then the second term converges to $0$ by SLLN. As for the first term, I am arguing that we define a new sequence of i.i.d random variables $Y_i := \frac{X_i^2}{n}$ where $\mathbb{E}\left[Y_n\right] = \frac{\sigma^2}{n}$ so by SLLN, the first term converges a.s. to $\frac{n}{\sigma^2}$. Consequently, the limiting distribution is $n*0 = 0$ but it's strange…

  3. Express the term above as $n \frac{n}{\sum_{i=1}^n X_i^2} \left( \frac{\sum_{i=1}^n X_i}{n} \right)^2$ then similarly, $\frac{n}{\sum_{i=1}^n X_i^2}$ converges a.s. to $\frac{1}{\sigma^2}$, $\left( \frac{\sum_{i=1}^n X_i}{n} \right)^2$ converges a.s. to $0$ but we have that $n$ term which $\to +\infty$ pointwise $-$ resulting in a indeterminate limit.

Any hints would be greatly appreciated…

Best Answer

Hint:

Rewrite the random variable as

$$\frac{\left( \sum_{i=1}^n X_i\right)^2}{\sum_{i=1}^n X_i^2}=\frac{U_n^2}{V_n}\,,$$

where $$U_n=\sqrt n\cdot\frac1n\sum_{i=1}^n X_i$$

and $$V_n=\frac1n\sum_{i=1}^n X_i^2$$

$U_n^2$ converges weakly to a standard distribution, and $V_n$ converges in probability to a non-zero constant. So it is now a straightforward application of Slutsky's theorem to find the limiting distribution of $U_n^2/V_n$.

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