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Conditional expectation on an indicator

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1. Homework Statement

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. Homework 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]
 

haruspex

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

Ray Vickson

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1. Homework Statement

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. Homework 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]
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?
 
190
0
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?
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##.
 

Ray Vickson

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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##.
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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