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,

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

 

[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 $\{X+Y=0\}$.

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

 

[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,

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

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.