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Homework Help: Conditional Expectation

  1. Mar 18, 2009 #1
    This result isn't in our book, but it is in my notes and I want to make sure it's correct. Please verify if you can.

    1. The problem statement, all variables and given/known data
    I have two I.I.D random variables. I want the conditional expectation of Y given Y is less than some other independent random variable Z.

    [tex] E(Y \, \vert \, Y < z) = \dfrac{\int_0^{z} y \cdot f(y) \, dy}{F(z)} [/tex]

    Where f(y) is the pdf of Y and F(z) is the cdf for Z

    3. The attempt at a solution
    I've searched the book and the web, but all I find is the formula for conditional expectation for [tex] E(X | Y = y) [/tex] for joint distributions and the like. Is my formula correct?
     
  2. jcsd
  3. Mar 19, 2009 #2
    You know that [tex]\mathbb{E}[X|Y]=\frac{\mathbb{E}[X \mathbf{1}_Y]}{\mathbb{P}(Y)}[/tex] so your formula looks correct.
     
    Last edited: Mar 19, 2009
  4. Mar 19, 2009 #3

    statdad

    User Avatar
    Homework Helper

    Think this way: if you know [tex] Y \le z[/tex], then the truncated distribution has density

    [tex]
    g(y \mid Y \le z) = \frac{f(y)}{F(z)}
    [/tex]

    so the expectation is

    [tex]
    \int_0^z y g(y \mid Y \le z) \, dy = \frac{\int_0^z y f(y) \, dy}{F(z)}
    [/tex]

    exactly as you have it.
     
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