Conditional Expectation

  • Thread starter JamesF
  • Start date
  • #1
14
0
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.

Homework Statement


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.

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

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

The Attempt at a Solution


I've searched the book and the web, but all I find is the formula for conditional expectation for [tex] E(X | Y = y) [/tex] for joint distributions and the like. Is my formula correct?
 

Answers and Replies

  • #2
284
3
You know that [tex]\mathbb{E}[X|Y]=\frac{\mathbb{E}[X \mathbf{1}_Y]}{\mathbb{P}(Y)}[/tex] so your formula looks correct.
 
Last edited:
  • #3
statdad
Homework Helper
1,495
36
Think this way: if you know [tex] Y \le z[/tex], then the truncated distribution has density

[tex]
g(y \mid Y \le z) = \frac{f(y)}{F(z)}
[/tex]

so the expectation is

[tex]
\int_0^z y g(y \mid Y \le z) \, dy = \frac{\int_0^z y f(y) \, dy}{F(z)}
[/tex]

exactly as you have it.
 

Related Threads on Conditional Expectation

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
0
Views
910
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
999
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
12
Views
2K
Replies
19
Views
2K
Top