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Suppose $X_1,X_2,\ldots,X_n$ are I.I.D $N(0,1)$. Then, what is the approximate distribution of $\frac{1}{n}\sum_{i=1}^n X_i^2\,$?

I have the solution but none of it is making any sense to me.

Thank you for your help!

THISISIT453
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  • Sorry! My mistake, there should be no capital N's – THISISIT453 Dec 20 '15 at 19:31
  • If $X$ is N(0,1), then $X^2$ has mean $1$, and its variance is $E[(X^2-1)^2]=E[X^4-2X^2+1]$. You should calculate this expectation. Once you have it you can get an approximate distribution. – Ian Dec 20 '15 at 19:31
  • How would the mean be 1? How would I calculate the variance without any data? I know the variance is $\frac{2}{n}$ but I don't know how they got there – THISISIT453 Dec 20 '15 at 19:34
  • Well you know $E[-2X^2+1]=-2E[X^2]+1=-1$. You just need to compute $E[X^4]$, which is a calculus problem. – Ian Dec 20 '15 at 19:37
  • Wouldn't $E[X^4] = 1?$ – THISISIT453 Dec 20 '15 at 19:39
  • Nope. You need to actually compute the integral to see what you get. – Ian Dec 20 '15 at 19:41
  • I'm really confused, my knowledge of expectations is really limited, would you be able to offer another point? – THISISIT453 Dec 20 '15 at 19:42
  • You know the PDF is $\frac{1}{\sqrt{2 \pi}} e^{-x^2/2}$, so $E[X^4]=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2} dx$. Computing this requires a bit of integration by parts. – Ian Dec 20 '15 at 19:43
  • I see! Can you confirm that the answer is in fact $\frac{2}{n}$ please? – THISISIT453 Dec 20 '15 at 19:44
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    I think $E[X^4]$ will turn out to be $3$, in which case $\frac{2}{n}$ would be the correct variance. – Ian Dec 20 '15 at 19:44
  • Also how does this make any use of the Central Limit Theorem? – THISISIT453 Dec 20 '15 at 19:49
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    The calculations so far have nothing to do with the CLT, we have only computed the mean and variance. The CLT comes in when we say that if $n$ is reasonably large, our random variable has roughly normal distribution with that mean and variance. – André Nicolas Dec 20 '15 at 19:55
  • So we can say that by the CLT, $\frac{1}{n}\sum_{i=1}^n X_i^2\$ is approximately normal with mean 1 and variance $ \frac{2}{n} $ for a large n? – THISISIT453 Dec 20 '15 at 19:59
  • That is the way I had put it in the comment. It is a common but perhaps too casual way of putting it, since the CLT is a limit theorem and in principle says nothing about any specific $n$. And we did not say what roughly normal means. The result can be stated precisely, but for many applications the casual version is good enough. – André Nicolas Dec 20 '15 at 20:14

2 Answers2

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You can use some "heuristics" to answer this questions. If $X_{i}\sim \mathcal{N}(0,1)$, then by definition $X_i^2\sim \chi^2(1)$, where $EY=1$, and $Var(Y)=2$. Denote $Y_i = X_{i}^2$, hence $\frac{1}{n}\sum_{i}^nX_i^2 = \frac{1}{n}\sum_{i}^nY_i$. Thus, for large enough $n$, ($n\to \infty$), you can use the CLT to deduce \begin{align} \frac{1}{n}\sum_{i}^nY_i \xrightarrow{D}\mathcal{N}(1, 2/n). \end{align}

V. Vancak
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Standard references will tell you that $\sum\limits_{i=1}^n X_i^2$ has a chi-square distribution with $n$ degrees of freedom, and that distribution has expected value $n$ and variance $2n$. Consequently the distribution of $$ \frac{X_1^2+\cdots+X_n^2 - n}{\sqrt {2n}} $$ is approximately $N(0,1)$.

To prove the assertions about the expectation and variance, it is enough to deal with $X_1^2$ and then multiply the expectation and variance by $n$.

Let $\varphi$ be the standard normal density $$ \varphi(x) = \frac 1 {\sqrt{2\pi}} e^{-x^2/2}. $$ Then \begin{align} \operatorname{E}(X_1^2) & = \int_{-\infty}^\infty x^2 \varphi(x)\,dx \\[10pt] & = 2\int_0^\infty x^2 \varphi(x)\,dx & \text{by symmetry} \\[10pt] & = \sqrt{\frac 2 \pi} \int_0^\infty x \left( e^{-x^2/2} (x\,dx) \right) \\[10pt] & = \sqrt{\frac 2 \pi} \int_0^\infty \sqrt {2u}\, e^{-u}\,du \\[10pt] & = \frac 2 {\sqrt\pi} \int_0^\infty u^{\frac 3 2 -1} e^{-u}\,du \\[10pt] & = \frac 2 {\sqrt\pi} \Gamma\left( \frac 3 2 \right) = \frac 2 {\sqrt\pi}\,\frac 1 2 \Gamma\left( \frac 1 2 \right) = 1. \end{align}

Similarly $\operatorname{E}(X_1^4) = 3$. Therefore $$ \operatorname{var}(X_1^2) = \operatorname{E}(X_1^4) - (\operatorname{E}(X_1^2))^2 = 3 - 1 = 2. $$

Michael Hardy
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