# Show that the distribution of $\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3)$ is normal

Let $$X_1,\ldots,X_n$$ be i.i.d. variables with $$\mathbb{E}[X_i]=0$$ and $$\mathbb{V}[X_i]=3$$ and assume that $$\mathbb{E}[X^4_i]<\infty$$, show that

$$\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3)$$

follows a normal distribution.

Textbook says this is TRUE, but I don't see how. I can show that the quantity is bounded above by a normal (via delta method+fourth moment) or I can show that part of this quantity is normal (again by dm) but then the quantity $$-\sqrt{N}3$$ should always diverge.

• This conclusion is not true. I suspect you are omitting some crucial aspects of the textbook question, such as that this follows a standard Normal distribution in the limit as $n$ grows arbitrarily large. Could you quote the text exactly?
– whuber
Sep 30 '19 at 16:48
• This seems like a question that wants you to figure out aspects of the central limit theorem. The distribution of $\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3)$ is not equal to a normal distribution but will be as close as you wish to a normal distribution (for large $n$). .......... The point of the textbook question is that you are gonna work with the CLT. Where do the mean and the variance fit in (and also as a playfull twist you need to obtain the mean and variance of $X_i^2$ based on what is said about the moments of $X_i$)? Sep 30 '19 at 19:54

This is not true in general, but as $$n$$ grows, the distribution of this random variable can be made as close as desired to normal, by the central limit theorem. Probably the author had in mind that for large values of $$n,$$ this can for practical purposes be taken to be normal.

Since $$\operatorname E(X)=0$$ you have $$\operatorname{var}(X) = \operatorname E(X^2),$$ so $$\operatorname E(X^2)=3.$$

And $$\frac{ \sum_{i=1}^n (X_i^2-\operatorname E(X_i^2))}{\operatorname{sd}\left( \sum_{i=1}^n (X_i^2-\operatorname E(X_i^2))\right) } = \frac 1 {\operatorname{sd}(X_1) \sqrt n} \sum_{i=1}^n (X_i^2 - 3),$$ so the central limit theorem can then be applied.

The fact that $$\operatorname E(X^4)<+\infty$$ is used by seeing that it entails that $$\operatorname{var}(X_1^2) < +\infty.$$

To see that it's not exactly normal, consider the case were $$X_i= \pm \sqrt 3,$$ each with probability $$1/2.$$ You get a discrete distribution; hence not exactly a normal distribution.

• That example which you give is a fun illustration but also it seems to be a trivial case; It has $E[X^4] =0$. You might say that the sum follows $$1/\sqrt {n} \sum_i ( X_i^2 - 3) \rightarrow \mathcal {N}(0,0)$$ And you may actually replace $\rightarrow$ by $\sim$. This makes 'not exactly normal' a complicated topic involving ideas about degenerate distributions and the meaning of $\mathcal {N}(0,0)$. Oct 1 '19 at 7:40
• @MartijnWeterings : Where in the world did you get $\operatorname E(X^4) = 0 \text{ ?} \qquad$ Oct 2 '19 at 20:44
• $X^2 =3$ it has only a single value. But I see now that indeed the central moments are zero and the raw moment is not. But anyway it still holds that (if you allow a normal with zero variance) you could say $$1/\sqrt {n} \sum_i ( X_i^2 - 3) \rightarrow \mathcal {N}(0,0)$$ because it depends on the moments of $X_i^2 -3$, which will be all zero. Oct 3 '19 at 5:21
• @MartijnWeterings : What is given is that $\operatorname E(X^2)=3,$ not that $X^2$ is constantly equal to $3.$ And if it were, that would make $\operatorname E(X^4) = 3^4,$ not $0. \qquad$ Oct 4 '19 at 19:13
• @MartijnWeterings : Correction: It would mean $\operatorname E(X^4) = 3^2.$ At any rate, it is not consistent with the given information to say that $X$ is constant. Somehow you became confused about that. $\qquad$ Oct 4 '19 at 19:23

according to Lindeberg-Levy CLT https://en.wikipedia.org/wiki/Central_limit_theorem

let $$S_n = \frac{\sum_{k=1}^n X_k}{n}$$

Suppose {X1, X2, …} is a sequence of i.i.d. random variables with E[Xi] = µ and Var[Xi] = σ2 < ∞. Then as n approaches infinity, the random variables √n(Sn − µ) converge in distribution to a normal N(0,σ2)

$${\displaystyle {\sqrt {n}}\left(S_{n}-\mu \right)\ {\xrightarrow {d}}\ N\left(0,\sigma ^{2}\right).}$$

so let's mark $$X_i^2 = Y_i$$ $$\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3) = \frac{1}{\sqrt{n}}\sum_{i=1}^n(Y_i-3)= \sqrt{n}(\sum_{i=1}^nY_i-3)$$ now let's show that $$EY_i = 3$$ and $$VarY_i = 1$$

$$VarX_i = EX_i^2 - E^2X_i => EY_i = EX_i^2 = VarX_i + E^2X_i = 3$$

$$VarY_i = VarX_i^2 = EX_i^4 - E^2X_i^2 = EX_i^4 - 9 < \infty$$

thus

$$\sqrt{n}(\sum_{i=1}^nY_i-3)\xrightarrow {d} N(0, VarY_i)$$