# Covariance

Discussion in 'P1.T2. Quantitative Methods (20%)' started by cidare, Apr 11, 2011.

1. ### cidareNew Member

Hi David,
There is a question about covariance, and unfortunately i couldn't realize why the answer cannot be A.

Q: If Cov(c, a + b) = 0 , which of the following must be true?
A. Cov(c, a) = 0, Cov(c, b) = 0
B. Cov(c, a) * Cov(c, b) = 0
C. Cov(c, a) <> 0, Cov(c, b) <> 0
D. None of the above
Explanation:
We know that Cov(x, y +z) = Cov(x, y) + Cov(x, z).
Then, cov(c, a +b) = cov(c, a) + cov(c, b).
Meaning that cov(c, a) + cov(c, b) = 0. So if cov(c, a) = 1, cov(c, b) = –1, A and B are incorrect;
if cov(c, a) = cov(c, b) = 0, C is incorrect.

C.

2. ### David Harper CFA FRMDavid Harper CFA FRM (test)Staff Member

Hi cidare,

I think it's harder due to the a/b/c notation which does distinguish between variables and constants, but i guess they are all variables. I think the explain is getting at: imagine a and b are perfectly negatively correlated, for example:
if a = 5, then b = -5
if a = 6, then b = -6; i.e., a 45 degree line with perfect -1.0 slope

now make (c) perfectly correlated to (a), just for extreme example. (c) is nevertheless independent of (a+b) because (a+b) is always 0; i.e., (a+b) is a constant and therefore uncorrelated to c.

The math applies covariance property for more than two variables; see Properies under http://en.wikipedia.org/wiki/Covariance

Hope that helps, David