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I'm reading the proof for the strong law of large numbers. It says:

Let $X_1,X_2,\ldots$ be a sequence of independent and i.i.d. random variables with finite mean $\mu$ and finite variance $\sigma^2$. Then:
$$\overline{X}_n=\frac{X_1+x_2+\cdots+x_n}{n}\underset{n\to\infty}{\stackrel{\text{a.s.}}{\longrightarrow}} \mu. $$

The proof reads:

Without loss of generality we can assume that $\mu =0$. Now $$\mathbb E\left(\left[\overline{X_n}\right]^2\right)=\frac{1}{n^2}\sum\limits_{1\leq j,k\leq n}\mathbb E\left(X_jX_k\right) = \frac{1}{n^2}\sum\limits_{j=1}^n\mathbb E\left(X_j^2\right) = \frac{\sigma ^2}{n}.$$ Therefore: $$\lim\limits_{n\to\infty}\mathbb E\left(\left[\overline{X_n}\right]^2\right)=0.$$

Here comes the notation I don't understand and have never seen before:

Moreover $\mathbb E\left(\left[\overline{X}_n\right]^2\right)=\dfrac{\sigma^2}{n}$ implies that: $$\sum\limits_{n=1}^\infty\mathbb E\left(\left[\overline{X}_{n^2}\right]^2\right) = \sum\limits_{n=1}^\infty\frac{\sigma^2}{n^2}<\infty.$$

I know that

$$\sum\limits_{n=1}^\infty\frac{\sigma^2}{n^2}=\sum\limits_{n=1}^\infty \frac{1}{n}\mathbb E\left(\left[\overline{X}_n\right]^2\right),$$

So this $_{n^2}$ would be adding that factor $\dfrac{1}{n}$ into the summation.
Any idea of what $\overline{X}_{n^2}$ could mean? If you have seen this before I appreciate you share your knowledge about this notation.

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$$\overline{X}_{n^2}=\frac{1}{n^2}\sum_{i=1}^{n^2}X_i$$


Indeed since you have already proved that $$\mathbb E\left(\left[\overline{X_n}\right]^2\right)= \frac{1}{n^2}\sum\limits_{j=1}^n\mathbb E\left(X_j^2\right) = \frac{\sigma ^2}{n}$$ then you have that $$\mathbb E\left(\left[\overline{X}_{n^2}\right]^2\right)= \frac{1}{n^4}\sum\limits_{j=1}^{n^2}\mathbb E\left(X_j^2\right) = \frac{1}{n^4}\cdot n^2σ^2=\frac{\sigma ^2}{n^2}$$ as the author of proof claims.

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    $\begingroup$ Yes, I was writing the answer. Just noticed that after I posted the question. It seemed strange because there would be a $\frac{1}{n^4}$, however we can cancel $n^2$ after... $\endgroup$ – Vladimir Vargas Nov 22 '14 at 0:40
  • $\begingroup$ @VladimirVargas Yes indeed it cancels out! I edited also my answer to add this calculation so that you can verify it! $\endgroup$ – Jimmy R. Nov 22 '14 at 0:44

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