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Homework Help: Finding conditional variance?

  1. Sep 28, 2005 #1
    the discrete prob distribution

    X/Y - G - D
    0 - 0,1 - 0,15
    1 - 0,1 - 0,3
    2 - 0,05 - 0,3


    this is what i have so far:
    E[X|Y=D]=0,2
    E[X|Y=g]=0,9
    E[X]=0,725
    E[X^2|Y=D]=0,3
    E[X^2|Y=G]=1,5
    Var(X|Y=G)=0,69
    Var(X|Y=D)=0,26


    i.e. [X]=0,2*0,25 + 0,9*0,75=0,725

    is the previous correct and how do i find Var(X)?
    the conditional variance forumal is:
    Var(X)=E[Var(X|Y)] + Var(E[X|Y])
     
  2. jcsd
  3. Sep 28, 2005 #2
    :confused: I am not sure but it seems easier to calculate Var[X]= E[X^2]- (E[X])^2
    you already have E[X]. you can try to calculate E[X^2] by conditioning on Y.
     
  4. Sep 28, 2005 #3

    how do i do that?

    secondly, i think i've might have made a mistake. i think i forgot to calculate the conditional probability mass function.

    that would mean distribution, given Y=G) is i.e.
    0 - 0,2
    1 - 0,4
    2 - 04

    so then E[X|Y=G]=0,4+0,8=1,2

    what is it? 1,2 or 0,9? i think the former 'result' is correct. so forget the other results, these are the corrected(?) ones

    E[X|Y=G]=1,2
    E[X|Y=D]=0,8
    E[X^2|Y=G]=2
    E[X^2|Y=D]=1,2
    Var(X|Y=G)=0,56
    Var(X|Y=D)=0,56
    E[X]=1,1

    i want to find Var(X)...
     
  5. Oct 1, 2005 #4

    EnumaElish

    User Avatar
    Science Advisor
    Homework Helper

    Your notation is confusing -- at least for me. What does the table
    stand for?
     
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