I have two random variables $X$ and $Y$, both with zero mean. $\newcommand{\E}{\mathrm{E}}$ $\newcommand{\Var}{\mathrm{Var}}$ $\newcommand{\Cov}{\mathrm{Cov}}$

Let's suppose I only know their covariance, which is, in this case, simply $\mathrm{E}[XY]$.

Can I easily calculate $\mathrm{E}\left[\frac{X}{Y}\right]$ from $\mathrm{E}[XY]$?

If not, what other information would I need to calculate $\mathrm{E}\left[\frac{X}{Y}\right]$?

EDIT: I add some assumptions: $X$ and $Y$ are Gaussian and their covariance is $\neq 0$. Thus, referring to @j-delaney 's answer, I should be in the case of Correlated central normal ratio.

The Correlated central normal ratio is a Cauchy distribution for which the mean is not defined (thus $\mathrm{E}\left[\frac{X}{Y}\right]$ is not defined). The $x_0$ parameter of the Cauchy distribution, in my specific case, should be $E[XY]/E[Y^2]$


2 Answers 2


You will have to know the full joint distribution of $X$ and $Y$ in order to calculate $$E[X/Y] = \int (x/y) p(x,y) ~dx dy. $$

Note that $E[X/Y]$ might not even be defined - this is the case for example when $X$ and $Y$ are normally distributed, and the ratio has a Cauchy distribution which has no mean.

See also Ratio distribution.

  • $\begingroup$ Thank you for your link to Ratio distribution. I should be specifically in the case of Correlated central normal ratio $\endgroup$ Commented Feb 11, 2022 at 12:43

Intuitively, take the singular density in $\mathbb{R}^2$ that is only nonzero along some line $X = a Y$, $var(Y) = b$, $Y$ always nonzero, and that has $E[X] = E[Y] = 0$, as you required. Then:

$$ E\left[\frac{X}{Y}\right] = E[a] = a $$


$$ E[XY] = a\,E[Y^2] = a \; var(Y) = ab. $$

So you cannot compute $E\left[\frac{X}{Y}\right]$ from the covariance alone.

I don't think that there are easily obtainable information which, would you avail yourself of them in addition to $E[XY]$, were to give you $E[X/Y]$. But you can of course always construct some (silly) additional information satisfying your request, e.g. the knowledge of $E[X/Y - XY]$ (just kidding).


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