Conditional expectation on an indicator

[IniquiTrance]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I'm trying to show that,

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

by,

<br /> \begin{align*}<br /> \mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] &amp;= \frac{\mathbb{E}[(X+Y)1_{\{X+Y=0\}}]}{\mathbb{P}[X+Y=0]} \\<br /> &amp;= \frac{0}{(1-p)^2} \\<br /> &amp;= 0<br /> \end{align*}<br />

 

[haruspex]

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

I'm not familiar with the notation for the condition (1_{\{X+Y=0\}}). How does one read it? Can you post a link?

 

[IniquiTrance]

I'm not familiar with the notation for the condition (1_{\{X+Y=0\}}). How does one read it? Can you post a link?

It's the indicator of the event \{X+Y=0\}.

<br /> 1_{\{X+Y=0\}}=\begin{cases}1, \qquad \text{if }X+Y=0; \\ 0, \qquad \text{otherwise.}\end{cases}<br />

 

[Ray Vickson]

Homework Statement

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

Homework Equations

The Attempt at a Solution

I'm trying to show that,

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

by,

<br /> \begin{align*}<br /> \mathbb{E}[X+Y\vert 1_{\{X+Y=0\}}] &amp;= \frac{\mathbb{E}[(X+Y)1_{\{X+Y=0\}}]}{\mathbb{P}[X+Y=0]} \\<br /> &amp;= \frac{0}{(1-p)^2} \\<br /> &amp;= 0<br /> \end{align*}<br />

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?

 

[IniquiTrance]

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]

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.