- #1

- 3

- 0

Let

Z = 1, if X < Y

Z = 0, otherwise

Find E(X|Z) and V(X|Z).

We should first find E(X|Z=z)

E(X|Z=z) = integral (from 0 to inf) of xf(x|z).

However, how do we find f(x|z) ?

- Thread starter covariance64
- Start date

- #1

- 3

- 0

Let

Z = 1, if X < Y

Z = 0, otherwise

Find E(X|Z) and V(X|Z).

We should first find E(X|Z=z)

E(X|Z=z) = integral (from 0 to inf) of xf(x|z).

However, how do we find f(x|z) ?

- #2

EnumaElish

Science Advisor

Homework Helper

- 2,304

- 124

E[X|Z=1] = E[X|X<Y].

E[X|Z=0] = E[X|X

Similarly for V[X|Z].

- #3

- 3

- 0

E[X|Z=1] = E[X|X<Y] = 1/9

E[X|Z=0] = E[X|X>Y] = 8/9

by integrating, and conditioning on the random variable Y.

So E(X) = E(E(X|Z)) = (1/9)(1/3) + (8/9)(2/3) = 17/27,

which contradicts the fact that E(X) = 1, for X exponential with mean 1.

I am wondering where is the error.

- #4

EnumaElish

Science Advisor

Homework Helper

- 2,304

- 124

How are you computing Pr{Z=0} and Pr{Z=1}?

- Last Post

- Replies
- 2

- Views
- 7K

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 3K

- Replies
- 2

- Views
- 640

- Last Post

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 3K

- Replies
- 4

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 7

- Views
- 3K