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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!

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    $\begingroup$ Welcome to MSE! Please review the Meta Read and enhance your question to provide your motivation/attempts. $\endgroup$
    – Jessie
    Commented Feb 8, 2021 at 16:02
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    $\begingroup$ You'll want to be more specific, there's no general formula for this. $\endgroup$ Commented Feb 8, 2021 at 16:19
  • $\begingroup$ Okay, sorry. In this case, I will quickly edit to make it more specific. $\endgroup$
    – Pronitron
    Commented Feb 8, 2021 at 16:29
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    $\begingroup$ 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 $\endgroup$
    – Henry
    Commented Feb 8, 2021 at 17:08
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    $\begingroup$ 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 $\endgroup$
    – Henry
    Commented Feb 8, 2021 at 17:20

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