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How to analytically express cov(X,Y), when:

X=C*A/(A+B)

and 

Y=C*B/(A+B)

Here C, A and B are independent variables with normal distributions.

More specifically I would like to express cov(X,Y) using the expected means of each of the input variables C, A and B, as well as their standard deviations or variances. 

A is a normal distribution with an expected mean E(A) and a variance var(A)
B is a normal distribution with an expected mean E(B) and a variance var(B)
C is a normal distribution with an expected mean E(C) and a variance var(C)

Thank you so much in advance!

Astrid Marie
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  • if this is HW please add the self-study tag – Antoine Jun 09 '16 at 09:42
  • Thank you for asking but it is for my research at work. I have really searched a lot the last few days for the answer, but I can only find examples where X and Y are linear combinations of random variables like X=aA+bB+cC (with small a, b and c as constants) but I cannot find solutions where X is a function that is the product of a combination of A, B and C. – Astrid Marie Jun 09 '16 at 10:18
  • I think that would help if you could show some of those examples you refer to and then explain how it is different in your case. – Antoine Jun 09 '16 at 10:27
  • Thanks for the tip. The question ressembles a bit this http://stats.stackexchange.com/questions/137571/covariance-of-product , but the example in the link is in matrix notation and it is not clear for me if Xi and Xj in the example are correlated, so I cannot used this directly. – Astrid Marie Jun 09 '16 at 10:36
  • `Here C, A and B are independent variables with a normal distribution` ... the same Normal distribution (e.g. standard Normal), or with differing means and variances? – wolfies Jun 09 '16 at 11:11
  • Thank you for the comment, I was not being clear. A, B and C have different means and variances. – Astrid Marie Jun 09 '16 at 11:13
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    Given that this involves the ratio of Normals, convergence may be an issue – wolfies Jun 09 '16 at 11:24
  • OK. Just for curiosity, would it be easier/possible to solve a simplified problem, and express cov(X,Y) if X=A*C and Y=B*C? – Astrid Marie Jun 09 '16 at 11:29
  • Well, that's a different question, but by quick calculation, the answer would be: $\text{Cov}( A C, BC) = \mu _a \mu _b \sigma _c^2$ – wolfies Jun 09 '16 at 18:23

1 Answers1

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One can find a general solution, without assuming Normality. In particular, if $A$, $B$ and $C$ are independent, and noting that the Covariance operator is the {1,1} central moment, then $\text{Cov}( \frac{C A}{A+B},\frac{C B}{A+B})$ is:

enter image description here

where I am using a developmental version of the CentralMomentToCentral function in mathStatica (alas, not in any public release yet).

The reason the solution does not converge (assuming Normal parents) is because $A+B$ is Normal, and thus $E[\frac{1}{A+B}]$ is the expectation of the inverse of a Normal variable, which is known not to converge (see, for instance, Mean and variance of inverse of a normal RV)

wolfies
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  • Thank you so much Wolfies! It is extremely helpful to see your answer, and I really appreciate that you took your time to work with this problem. Now I try to implement the result in my calculations and compare with the simulations :-) The mathStatica seems like a really powerful tool. – Astrid Marie Jun 10 '16 at 12:58
  • Just one little question, I see that the C in your code has a different colour than A and B. Is there any reason for this? – Astrid Marie Jun 10 '16 at 13:21