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Let $X$, $Y$ be two independent normal random variable, say $N(0,1)$. And I want to estimate $cov(X,Y)$. What is the distribution of the $\widehat{cov}(X,Y)$?

  • Of course, I can calculate the true value directly with $cov(X,Y)=0$.
  • Suppose I have data $X_1,\cdots,X_n \sim N(0,1)$ and $Y_1,\cdots,Y_n \sim (another\ independent) N(0,1)$
  • Consider the estimator $\widehat{cov}(X,Y)=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})$

Then my question is what is the distribution of $\widehat{cov}(X,Y)$?

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  • $\begingroup$ I did a simulation and the result is very similar to a normal distribution. Is is possible to prove that it is normal, with, like, CLT? $\endgroup$ – breezeintopl Apr 4 '17 at 13:45
  • $\begingroup$ The normal distribution is about $N(0,(\frac{1}{\sqrt{n}})^2)$. $\endgroup$ – breezeintopl Apr 4 '17 at 14:00
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The answer seems to be given here (see $f(x_{12})$). Although I can't see the reference.

It's not quite as ugly as it seems -- it resembles the chi-square distribution, but is also modulated by a Bessel function. Note the simplification when $\rho = 0$, as in your question.

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