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Let $X,Y$ are two random variables which are not necessarily independent. It is easy to get $\mathbb{E}(X)$ ann $\mathbb{E}(Y)$. I want to know: is there some approximation to $\mathbb{E}(\frac{X}{Y})$?

[Update] The background is that I want to calculate the expectation of Pearson product-moment correlation coefficient - $\mathbb{E}(\rho_{xy})$.

The Pearson product-moment correlation coefficient is $\rho_{xy} = \frac{Cov_{xy}}{\sigma_x\sigma_y}$

It is easily to get $\mathbb{E}(Cov_{xy})$ and $\mathbb{E}(\sigma_x\sigma_y)$, so I want to know a approximation to get $\mathbb{E}(\rho_{xy})$ from the above two value.

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The expected value of a constant is the constant itself. For random variables $X$ and $Y$, $Cov_{xy}$ by which I assume you mean $E[(X-E[X])(Y-E[Y])]$ is a constant. So, when you write $E[Cov_{xy}]$, its value is just $Cov_{xy}$, unless there is some other random variable with respect to which you are computing the correlation. That is, are there three random variables $X,Y,Z$ and $Cov_{xy}$ is the conditional covariance of $X$ and $Y$ conditioned on $\{Z = \alpha\}$, say, and you are now trying to find the unconditional covariance? – Dilip Sarwate Jan 15 '12 at 13:29
@DilipSarwate Please see the second answer of… – Fan Zhang Jan 15 '12 at 13:43
I have looked at the question and answer on stats.SE. There the issue was estimating the variance, covariance, correlation coefficient, etc. given $n$ i.i.d. samples from $(X,Y)$ with unknown joint distribution, and you got answers for that. Here you are not talking about $n$ i.i.d. samples, or estimating $Cov_{xy}$ from such samples, etc. but just about two random variables $X$ and $Y$. I ask again, what is the meaning of $E[Cov_{xy}]$? – Dilip Sarwate Jan 15 '12 at 15:34
up vote 2 down vote accepted

If $X$ and $Y$ have a joint density $f(x,y)$, $E[X/Y] = \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{x}{y} \ f(x,y)\ dx\ dy$ (assuming that converges absolutely). Similarly, if they have a joint probability mass function $p(x,y)$, $E[X/Y] = \sum_x \sum_y \frac{x}{y} p(x,y)$.

Let $\mu = E[Y]$. If you can treat $Y - \mu$ as small compared to $\mu$ (in particular if there is $c>0$ such that $c < Y < 2 \mu - c$ almost surely) then $$E[X/Y] = \sum_{j=0}^\infty (-1)^j \frac{E[X (Y - \mu)^j]}{ \mu^{j+1}}$$ so you could use a partial sum of that as an approximation.

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