# Conditional Expectation

1. Mar 18, 2009

### JamesF

This result isn't in our book, but it is in my notes and I want to make sure it's correct. Please verify if you can.

1. The problem statement, all variables and given/known data
I have two I.I.D random variables. I want the conditional expectation of Y given Y is less than some other independent random variable Z.

$$E(Y \, \vert \, Y < z) = \dfrac{\int_0^{z} y \cdot f(y) \, dy}{F(z)}$$

Where f(y) is the pdf of Y and F(z) is the cdf for Z

3. The attempt at a solution
I've searched the book and the web, but all I find is the formula for conditional expectation for $$E(X | Y = y)$$ for joint distributions and the like. Is my formula correct?

2. Mar 19, 2009

### Focus

You know that $$\mathbb{E}[X|Y]=\frac{\mathbb{E}[X \mathbf{1}_Y]}{\mathbb{P}(Y)}$$ so your formula looks correct.

Last edited: Mar 19, 2009
3. Mar 19, 2009

Think this way: if you know $$Y \le z$$, then the truncated distribution has density
$$g(y \mid Y \le z) = \frac{f(y)}{F(z)}$$
$$\int_0^z y g(y \mid Y \le z) \, dy = \frac{\int_0^z y f(y) \, dy}{F(z)}$$