Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expected Value, Expected Variance,covariance

  1. Oct 7, 2007 #1
    Can someone help me with this problem????

    The joint probability mass function of X and Y, p(i,j)=P{X=i,Y=j}, is given as follows
    p(-1,-2)=1/9, p(-1,-1)=1/18, p(-1,0)=1/12, p(-1,1)=0,
    p(0,-2)=1/12, p(0,-1)=1/9, p(0,0)=0, p(0,1)=1/8,
    p(1,-2)=0, p(1,-1)=1/8, p(1,0)=1/4, p(1,1)=1/18,

    a) Compute the E[X], Var(X), and Cov(X,Y)
    b) Calculate P{X,Y=k} for k=-2,-1,0,1,2
    c) Evaluate E[Y|X=k] for k=-1,0,1

    here is what I attempted to do:
    E[X]=E[X1] + E[X2]+.............E[Xn]=np
    Var(X)= E[X^2]-(E[X])^2
    Cov=(X,Y)=E[(X-E[X])(Y-E[Y])]
    =E[XY-YE[X]-XE[Y]+E[X]E[Y]]
    =E[XY]-E[Y]E[X]-E[X]E[Y]+E[X]E[Y]
    =E[XY]-E[X]E[Y]

    E[X]=E[X|Y=-2]= 1/9(-1)+1/12(0)+0(1)=-1/9
    E[X|Y=-1]=1/18(-1)+1/9(0)+1/8(1)=5/72
    E[X|Y=0]=1/12(-1)+0(0)+1/4(1)=1/6
    E[X|Y=1]=0(-1) + 1/8(0) +1/18(1)=1/18
    E[X]=-1/9+5/72+1/6+1/18=13/72

    now the variance Var(X)
    (-1/9)^2+(5/72)^2+(1/6)^2+(1/18)^2=1/81+25/5184+1/36+1/324
    =83/1728
    Var=83/1728-169/5184=5/324

    E[Y]= [Y|X=-1]=1/9(-2)+1/18(-1)+1/12(0)+0(1)=-5/18
    [Y|X=0]=1/12(-2)+1/9(-1)+0(0)+1/8(1)=-11/72
    [Y|X=1]=0(-2)+1/8(-1)+1/4(0)+1/18(1)=-5/72
    E[Y]=-5/18 -11/72-5/72=-41/72

    Cov=(83/1728x5/324)-(13/72x-41/72)=37/5000

    Could you check part A ...........I really need help with part b and c
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted