I have found in this Pdf, theorem 2, that asymptotically: $(\frac{\sum X_i}{N}, \frac{\sum X_i^{2}}{N})$ converges in distribution to a bivariate random variable with mean $(\mu_1,\mu_2)$ and covariance matrix $A$. I don't know how to prove it using the multivariate CLT theorem? Hints are most apreciated.
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1$\begingroup$ Please tell us what the $X_i$ are and what assumptions you are making about them. $\endgroup$– whuber ♦Commented Apr 4, 2018 at 15:19
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$\begingroup$ @whuber, I think that the only assumption here is that $X_i$'s are iid. $\endgroup$– NooobCommented Apr 4, 2018 at 15:31
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1$\begingroup$ You will need more than that for the conclusion to hold. At a minimum, that common distribution must have a finite fourth moment. $\endgroup$– whuber ♦Commented Apr 4, 2018 at 17:36
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$\begingroup$ @whuber, my bad, it says on the document that the $X_i$'s has finite $4$-th moments. $\endgroup$– NooobCommented Apr 4, 2018 at 18:44
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Hints: The proof is a straightforward calculation. Rewrite the quantity of interest $$ \sqrt n(\hat\mu_{1,n} - \mu_1,\hat\mu_{2,n}-\mu_2) $$ in the form $$ \frac1{\sqrt n}\sum_{i=1}^N{\bf X_i}\tag1 $$ where ${\bf X}_i$ is shorthand for the 2-vector $(X_i-\mu_1,X_i^2-\mu_2)$. The summands ${\bf X}_i$ in (1) are iid with mean zero and finite covariance matrix, so the multivariate CLT applies immediately. It remains to verify the covariance matrix of the limiting distribution, which equals the covariance matrix of the generic summand $(X-\mu_1,X^2-\mu_2)$. A sample calculation: $$\operatorname{Var}(X^2-\mu_2)\stackrel{(a)}=E[(X^2-\mu_2)^2]=E[X^4-2\mu_2X^2+\mu_2^2]=:\mu_4-\mu_2^2$$ where step (a) uses the fact that $\operatorname{Var}(Y)=E(Y^2)$ when $E(Y)=0$.
answered Apr 4, 2018 at 16:47
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$\begingroup$ Aren't there any conditions to verify, as has @whuber commented? $\endgroup$– NooobCommented Apr 5, 2018 at 10:06
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$\begingroup$ Why the $\frac{1}{\sqrt{n}}$ in equation $(1)$? $\endgroup$– NooobCommented Apr 5, 2018 at 11:36
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$\begingroup$ @Nooob The $\frac1{\sqrt n}$ appears because there is a $\frac1n$ inside $\hat\mu_{1,n}$ and $\hat\mu_{2,n}$ . The conditions for the CLT are satisfied because the $X_i$ have a finite fourth moment. $\endgroup$ Commented Apr 5, 2018 at 17:30
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$\begingroup$ I see now. Is this all what is needed to be done in order to show that joint asymptotic distribution is normal ?! I feel stupid. $\endgroup$– NooobCommented Apr 6, 2018 at 17:36
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$\begingroup$ @Nooob The calculation is straightforward once you see that the conditions for the CLT are satisfied. The key step is recognizing equation (1). But you have to get past the notation -- it isn't obvious how to get to (1) until you play around with what you've been given. $\endgroup$ Commented Apr 6, 2018 at 17:53