# Expected Value

• Sep 16th 2010, 08:04 AM
cuteylion
Expected Value
Hi,

Suppose E(U|X) = X^2. Suppose that E(X) = 0, Var(X) = 1 and E(X^3) = 0

What is the E(UX)?

Thanks alot for helping!
• Sep 16th 2010, 08:48 AM
Moo
Hello,

$\displaystyle E[UX]=E[E[UX|X]]=E[XE[U|X]]=E[X^3]$

:)
• Sep 16th 2010, 10:17 PM
aman_cc
@ Moo - Thanks for your response.

But I am not confortable with all the theorms/or rather properties of Expectation (in case of joint distributions). I basically get confused and at times rely on intuition.

Will it be possible for you to reccomend me
1. A source/book I can read to understand this better
2. For e.g. Why is
E(Y) = E(E(Y|X))
And then how you got
E[E[UX|X]]=E[XE[U|X]] - I understand it intuitively but not rigorously.

Thanks
• Sep 16th 2010, 11:50 PM
matheagle
E(Y) = E(E(Y|X))
is easy to prove, just write the conditional expectation and integrate/sum a second time
• Sep 17th 2010, 12:08 AM
aman_cc
Thanks.

But where I'm weak is a better understanding of Joint Distributions. I tried referring some online resources but couldn't get a very clear idea. And, I'm not attending any university so little hard for me to find out where can you read up on it. Thanks
• Sep 17th 2010, 01:06 AM
cuteylion
• Sep 17th 2010, 07:26 AM
matheagle
In the continuous setting....

$\displaystyle E(Y|X)=\int yf_{Y|X}(y)dy$

$\displaystyle =\int y{f_{X,Y}(x,y)\over f_X(x)}dy$

now take the expectation wrt x...

$\displaystyle E(E(Y|X))=\int E(Y|X)f_X(x)dx$

and wave the Fubini wand...

$\displaystyle E(E(Y|X))=\int \int y{f_{X,Y}(x,y)\over f_X(x)}dyf_X(x)dx$

$\displaystyle =\int \int yf_{X,Y}(x,y)dydx=E(Y)$
• Sep 17th 2010, 11:31 AM
Moo
aman_cc,

I have a very formal proof of the fact that $\displaystyle E[E[X|Y]]=E[X]$ but it's kind of long, and very theoretical. You also need measure theory knowledge. Well I don't think it's a relevant one...

The general idea is to say that in a probability space $\displaystyle \Omega,\mathcal A,\mathbb P)$ and given a sigma-algebra $\displaystyle \mathcal B$, and X a random variable $\displaystyle \in L^1(\Omega,\mathcal A,\mathbb P)$, then there exists a unique random variable $\displaystyle Z=E[X|\mathcal B$ such that for any $\displaystyle B\in \mathcal B,\int_B Z d\mathbb P=\int_B Xd\mathbb P$.
In particular, for $\displaystyle B=\Omega$, we get the desired equality.

Note that when we write $\displaystyle E[X|Y]$, it's in fact $\displaystyle E[X|\sigma(Y)$, where $\displaystyle \sigma(Y)$ is the sigma-algebra generated by $\displaystyle Y$. So it's in place of $\displaystyle \mathcal B$ in the above paragraph.

Also note that with the Aussie singer's method, matheagle, we have to assume that these random variables have a pdf.

For the second point, the proof doesn't look that difficult, it's just long and painful... I don't know how to explain it with words...
• Sep 19th 2010, 09:36 PM
aman_cc