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Conditional expectation on an indicator

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Let [itex]X[/itex] and [itex]Y[/itex] be independent Bernoulli RV's with parameter [itex]p[/itex]. Find,
    [tex]\mathbb{E}[X\vert 1_{\{X+Y=0\}}][/tex] and [tex]\mathbb{E}[Y\vert 1_{\{X+Y=0\}}][/tex]
    2. Relevant equations


    3. The attempt at a solution

    I'm trying to show that,

    [tex]\mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] = 0[/tex]

    by,

    [tex]
    \begin{align*}
    \mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] &= \frac{\mathbb{E}[(X+Y)1_{\{X+Y=0\}}]}{\mathbb{P}[X+Y=0]} \\
    &= \frac{0}{(1-p)^2} \\
    &= 0
    \end{align*}
    [/tex]
     
  2. jcsd
  3. Oct 25, 2014 #2

    haruspex

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    I'm not familiar with the notation for the condition ([itex] 1_{\{X+Y=0\}}[/itex]). How does one read it? Can you post a link?
     
  4. Oct 25, 2014 #3
    It's the indicator of the event [itex]\{X+Y=0\}[/itex].

    [tex]
    1_{\{X+Y=0\}}=\begin{cases}1, \qquad \text{if }X+Y=0; \\ 0, \qquad \text{otherwise.}\end{cases}
    [/tex]
     
  5. Oct 25, 2014 #4

    Ray Vickson

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    Basically, you are trying to show that the conditional distribution ##f(k) \equiv P(X= k|X+Y=0), \; k=0,1,2, \ldots## has mean zero; that is, that ##\sum_{k=0}^{\infty} k f(k) = 0##. How do you suppose that could happen?
     
  6. Oct 25, 2014 #5
    I think it is possible since both ##X## and ##Y## are Bernoulli, if their sum is 0, then ##\mathbb{P}[X=0|X+Y=0]=1##. Then, ##\sum_{k=0}^1kf(k)=0*1##.
     
  7. Oct 25, 2014 #6

    Ray Vickson

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    Rather than "thinking it is possible" you need to actually prove it. But, at least you are on the right track---that is, your thinking is OK, and just needs to be firmed up. It is not difficult once you see what needs to be established.
     
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