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Problem. Let $X_1,X_2,...$ be independent and identically distributed random variables such that $EX_1^4<\infty$ and $0<V(X_1)=\sigma^2$. Put $$T_n = \frac{1}{n}\sum_{j=1}^n \left(X_j-\bar{X_n}\right)^2$$ where $\bar{X_n}=n^{-1}\sum_{j=1}^n X_j$. Then we have that $$\sqrt n \left(T_n-\sigma^2\right) \longrightarrow_d N(0,V\left(X_1-EX_1\right)^2) \quad\text{(convergence in distribution}).$$

My attempt is as follows: It suffices to show that $$\frac{\sqrt n \left(T_n-\sigma^2\right)}{\sqrt{V(X_1-EX_1)^2}} \longrightarrow_d N(0,1)$$ which I aim to show with the Lindeberg-Feller CLT.

Putting $$S_n=\sqrt n\left(T_n-\sigma^2\right)=\sum_{j=1}^n \underbrace{\frac{(X_j-\bar{X_n})^2-\sigma^2}{\sqrt n}}_{= Y_{nj}},$$ we note that $$V(S_n)=V\left(X_1-\bar{X_n}\right)^2.$$ This is asymptotically equivalent to $V\left(X_1-EX_1\right)^2$ since $\lim_{n\to \infty} \bar{X_n} = EX_1$ almost everywhere by the law of large numbers. Similarly, $E(S_n)$ is asymptotically equivalent to $0$.

Then the claim rephrases to $$\frac{S_n}{\sqrt{V(S_n)}}\longrightarrow_d N(0,1)$$ for which it suffices to check the Lindeberg condition, or, in fact, the Ljapunov condition which implies the former. So, if for some $\delta>0$, we have that $$\lim_{n\to \infty} \frac{1}{V(S_n)^{1+\delta/2}} \sum_{j=1}^n E|Y_{nj}|^{2+\delta}=0,$$ then we are done. This is clearly satisfied if we chose, say, $\delta=2$.

Is my solution okay? Where was I too imprecise? There ought to be a simpler solution since this is an old exam problem. Do you see an easier approach to the problem?

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You argument is sensible, however, there is are hidden pitfalls.

The main problem is that $Y_{nj} = \big((X_j - \overline{X}_n)^2 - \sigma^2\big)/\sqrt{n}$ are not independent, so you cannot use the CLT directly. Nevertheless, you can note that $T_n = \overline{X^2} - (\overline{X})^2$ and use the classical vector CLT for vectors $Y_j = (X_j^2, X^{\vphantom{2}}_j)$. You will need the following equality: $$ T_n - \sigma^2 = (\overline{X^2} - E[X^2]) - \big((\overline{X})^2 - (E[X])^2\big) = (\overline{X^2} - E[X^2]) - \big(\overline{X} - E[X]\big)\big(\overline{X} + E[X]\big) $$ and Slutsky's theorem to identify the required limit distribution.

Another problem is that Lyapunov condition will not hold (as the $(4+\delta)$th moment is not guaranteed to be finite). However, if you use the classical (vector) CLT, this is not needed: the finiteness of $4$th moment is enough.

Update: it is possible to simplify things by noting that $T_n = \overline{(X-\mu)^2} - (\overline{X-\mu})^2$. Then $\sqrt{n}(T_n - \sigma^2) = \sqrt{n}\big(\overline{(X-\mu)^2}-\sigma^2) - \sqrt{n} (\overline{X-\mu})^2$. The second term converges to $0$ in probability (e.g. by Chebyshev), while the first weakly converges by the classical (scalar) CLT; therefore, the sum converges weakly by Slutsky's theorem.

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  • $\begingroup$ Great, thank you - that cleared it up. $\endgroup$ – iwriteonbananas Sep 7 '15 at 6:54

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