1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stats: Covariace of 2 Random Variables

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Let X ~ Exponential(3) and Y ~ Poisson(5). Assume X and Y are independent. Let Z = X + Y. Compute the Cov(X,Z).


    2. Relevant equations

    I know Cov(X, Z) = E(XZ) - E(X)E(Z). But how do I compute E(XZ) and E(Z) ?? Since for E(XZ), I would need the pdf/pmf (Exp is abs cts, while Poisson is discrete). Or can I do the following:


    3. The attempt at a solution

    Cov (X, Z) = Cov(X, X + Y)
    = Cov(X, X) + Cov(X,Y)
    = Var(X) + Cov(X,Y)
    Since X, Y indep., Cov(X,Y) = 0
    = Var(X)

    Is this the correct derivation? Any help would be greatly appreciated, as the exam is just around the corner. Thanks.
     
    Last edited: Dec 7, 2012
  2. jcsd
  3. Dec 7, 2012 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That is the answer I got. If you are not sure about the derivation, try writing it out in terms of expected values.
     
  4. Dec 7, 2012 #3
    Thank you for confirming. Although this problem is solved, I would like to know more. Before I was trying to Cov(X,Z) directly without subbing the 'X+Y', and problems rose there. Because

    Cov(X,Z) = E(XZ) - E(X)E(Z) then,

    E(XZ) = ∫Dxz fX,Z(x, z) dD, where fX,Z(x, z) is the prob. density function of X,Z. Is there any way we can find this function?

    E(XZ) = E(X)E(Z), iff X,Z are ind. But clearly Z is dependent of X. Hence this won't work. Is the above way, the only way?
     
  5. Dec 7, 2012 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Just as a matter of notation: does X~exponential(3) mean EX = 3 or EX = 1/3? (The second way is more common, but I have seen the first as well.)

    Anyway:
    [tex] Cov(X,Z) = E(XZ) - EX EZ = E X^2 + E(X Y) - (EX)^2 - EX EY. [/tex]
    What is E(X Y)? What do you get after simplification? (Hint: your final answer is OK.)

    Note: you used a result Cov(X,X+Y) = Cov(X,X) + Cov(X,Y) for uncorrelated X,Y, but you need to derive this first. By proceeding directly you can by-pass this.
     
  6. Dec 8, 2012 #5
    We use the mean as 1/3. E(XY) = E(X) E(Y) since X and Y are independent. So than it is equal to the Var(X). Thanks for both of your help. Much appreciated. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook