Hi,
Can anyone please help with the question?
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!
@ Moo - Thanks for your response.
I do understand your solution.
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
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
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...
$\displaystyle E(E(Y|X))=\int \int y{f_{X,Y}(x,y)\over f_X(x)}dyf_X(x)dx$
$\displaystyle =\int \int yf_{X,Y}(x,y)dydx=E(Y)$
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...
Thanks. I did read up about this proof in particular.
If there is a good book at an intermediate level to read up on probabolity thoery and expectations, distribution (joint) etc please do refer that to me. I like books which start with clear definitions of concepts / axioms and then proceed to prove such theorms for there. You can consider me a novice at measure theory.
Hi,
I know three books in English that look good. But I don't know if they're the best. My library isn't specialized in English books
Amazon.com: Probability with Martingales (Cambridge Mathematical Textbooks) (9780521406055): David Williams: Books
Amazon.com: Probability and Measure, 3rd Edition (9780471007104): Patrick Billingsley: Books
Measure, Integral and Probability: Amazon.fr: Marek Capinski, Peter E. Kopp, Ekkehard Kopp: Livres en anglais
well they're expensive, but I hope the library in your university has some of them !