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

  1. cidare

    cidare New 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
    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.

    Thanks for your help.
  2. David Harper CFA FRM

    David Harper CFA FRM David 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

    Hope that helps, David

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