# Conditional expectation on an indicator

1. Oct 25, 2014

### IniquiTrance

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

I'm trying to show that,

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

by,

\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*}

2. Oct 25, 2014

### haruspex

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

3. Oct 25, 2014

### IniquiTrance

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

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

4. Oct 25, 2014

### Ray Vickson

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?

5. Oct 25, 2014

### IniquiTrance

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, $\sum_{k=0}^1kf(k)=0*1$.

6. Oct 25, 2014

### Ray Vickson

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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