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For two random variables $X>0$ and $Y>0$, is there a way to express $\textrm{Cov}(\frac{1}{Y},\frac{X}{Y})$ as a function of $\textrm{Cov}(X,Y)$ or $\textrm{Corr}(X,Y)$?

rasmodius
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Liu Mendoza Perez
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    One simple property of covariance should make it obvious why such a formula cannot exist: when you scale $Y$ by a positive factor $\lambda,$ $\operatorname{Cor}(X,Y)$ is unchanged, $\operatorname{Cov}(X,Y)$ is multiplied by $\lambda,$ but $\operatorname{Cov}(1/Y,X/Y)$ is multiplied by $\lambda^{-2}.$ At the same time, scaling $X$ will multiply both the covariances by $\lambda.$ Finally, it's possible the covariance of $1/Y$ with anything else will be infinite: see https://stats.stackexchange.com/questions/299722. It's impossible to construct a function that has all these properties. – whuber Mar 09 '21 at 22:21

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No, there is not. Assume all moments exist and $X$ and $Y$ are independent. Then both the covariance and the correlation are $0$. But, the expression is a nonzero value if $E[X]\neq 0$ and $Y$ is not a constant random variable. $$\begin{align}\operatorname{cov}(1/Y,X/Y)&=E[X/Y^2]-E[1/Y]E[X/Y]=E[X](E[1/Y^2]-E[1/Y]^2)\\&=E[X]\operatorname{var}(1/Y)\end{align}$$

gunes
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