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Let $( X_1, X_2, \dots )$ be independent and identically distributed random variables. Denote $(\mu = \mathbb{E}[X_i])$ and $(\bar{X}_n = \frac{1}{n} \sum_{i=1}^{n} X_i)$.

Assume that $( \mu = 0)$ and $( \text{var}(X_i) = 1)$. Determine to what and in what sense the following random variables converge: $( n^{3/2} (\bar{X}_n)^2)$.

I tried this: From CLT we have that $( n^{1/2} \bar{X}_n)$ converge in distribution to a random variable with distribution N(0,1). Now I used delta-method for function g(x)=n$x^2$ which derivation 2nx is continuous on R and got that $( n^{3/2} (\bar{X}_n)^2)$ converges in distribution to a random variable with distribution N(0,0) so we get that $( n^{3/2} (\bar{X}_n)^2)$ converges in distribution to a random variable that is constant 0 almost surely.

Is this solution correct (if it is, is there another way how to do it?) or if it is not correct, how could this be solved? Can I use in delta-method a function that depends on n at all? And what would happen if, after substituting $\mu$, the function equaled something that contained n? Then we would send n to infinity, but the expression on the right-hand side would depend on n...

marek
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1 Answers1

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Take $X_{i}$ to be iid $N(0,1)$ variables. Then $(\overline{X}_{n})^{2}\sim |N(0,\frac{1}{n})|^{2}$. Thus $n^{3/2}\overline{X_{n}}^{2}\sim |N(0,n^{1/2})|^{2}$

$$E(\exp(it n^{3/2}(\overline{X_{n}})^{2}))=E(\exp(it\cdot |N(0,\sqrt{n})|^{2}))$$

But $N(0,\sqrt{n})\sim n^{1/4}N(0,1)$. So the above equals

$$E(\exp(it\sqrt{n}N(0,1)^{2}))=\frac{1}{\sqrt{1-2it\sqrt{n}}}\xrightarrow{n\to\infty} 0$$ for $t\neq 0$ and $1$ for $t=0$

(I have used the formula for the characteristic function of a Chi squared Random variable with degree $1$).

But this cannot be the characteristic function of any random variable (as it is not continuous ).

Thus the sequence $(n^{3/2}\bar{X_{n}}^{2})$ is not necessarily tight (i.e. there may not exist any subsequence of it which converges to some random variable) and hence, in general cannot converge to anything in distribution. (Hence other modes of convergence are ruled out as well).

NOTE:-If you don't want to use characteristic functions, then see that $P(n^{3/2}(\bar{X_{n}})^{2}>M)=P(|N(0,1)|^{2}>\frac{M}{\sqrt{n}})$

Now for each fixed $M>0$, we can find $N$, depending on $M$, such that $P(|N(0,1)|^{2}>\frac{M}{\sqrt{n}})\geq \frac{1}{2}$ for all $n\geq N$.

(This is simply because $P(|N(0,1)|^{2}>0)=1$ and $\frac{M}{\sqrt{n}}\to 0$ for any fixed $M$.)

Thus, for $\epsilon=\frac{1}{2}$, there does not exist any compact set $K$ such that $P(n^{3/2}(\bar{X_{n}})^{2}\notin K)<\epsilon$ for all $n$.

Thus, the sequence $n^{3/2}\bar{X_{n}}^{2}$ is not a tight sequence and hence cannot converge in distribution to any random variable.

Mr. Gandalf Sauron
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  • Thank you so much - I have a question though, is there another way how to do it without using characteristic function? And maybe, just to be sure, regarding the delta-method, I cannot use a function that depends on n at all (because then we would have a problem that I mentioned in the post above)? – marek Oct 03 '24 at 01:50
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    @marek The function you use for delta method must be a fixed function independent of n. If you don't want to use characteristic function, then you'll have to verify using basic analysis that $|N(0,\sqrt{n})|^2|$ is not a tight sequence. I'll add this into the answer. – Mr. Gandalf Sauron Oct 03 '24 at 07:45