# Find $\operatorname{Cov}(\frac{1}{C}, D)$ given that $\operatorname{Cov}(\frac{A}{BC}, D) = X$

Find $$\operatorname{Cov}(\frac{1}{C}, D)$$ given that $$\operatorname{Cov}(\frac{A}{BC}, D) = X$$, where $$\operatorname{Cov}$$ is the covariance and $$\operatorname{Cov}(A, B) = \operatorname{Cov}(A, C) = \operatorname{Cov}(A, D) = \operatorname{Cov}(B, C) = \operatorname{Cov}(B, D) = 0$$.

I'm an experimental physicist trying to combine some measurements to determine the covariance between $$\frac{1}{C}$$ and $$D$$, but I only know the covariance between $$\frac{A}{BC}$$ and $$D$$. $$A$$ and $$B$$ are fully uncorrelated and separate factors. I'm not really sure how to do this as I'm not that familiar with the algebra of covariances.

Is it the case that $$\operatorname{Cov}(\frac{A}{BC}, D) = \operatorname{Cov}(\frac{1}{C}, D)$$ in this case?

Thanks!

• Welcome to MSE! Please review the Meta Read and enhance your question to provide your motivation/attempts. Commented Feb 8, 2021 at 16:02
• You'll want to be more specific, there's no general formula for this. Commented Feb 8, 2021 at 16:19
• Okay, sorry. In this case, I will quickly edit to make it more specific. Commented Feb 8, 2021 at 16:29
• I do not think there is an answer to finding $\operatorname{Cov}(C, D)$ given that $\operatorname{Cov}(\frac{1}{C}, D) = x$, as scaling $C$ and $D$ together means you can achieve any magnitude for $\operatorname{Cov}(C, D)$ even before you consider distributional questions Commented Feb 8, 2021 at 17:08
• No - I am suggesting that your question may not have an answer, as a simpler version with $A$ and $B$ both constant and equal to $1$ also seems as if it may not have an answer Commented Feb 8, 2021 at 17:20