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law of total expectation (VECTOR case)

 
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Nov18-09, 01:57 AM   #1
 

law of total expectation (VECTOR case)


" The law of total expectation is: E(Y) = E[E(Y|X)].
It can be generalized to the vector case: E(Y) = E[E(Y|X1,X2)].

Further extension:
(i) E(Y|X1) = E[E(Y|X1,X2)|X1]
(ii) E(Y|X1,X2) = E[E(Y|X1,X2,X3)|X1,X2] "

====================

I understand the law of total expectation itself, but I don't understand the generalizations to the vector case and the extensions.

1) Is E(Y|X1,X2) a random variable? Is E(Y|X1,X2) a function of both X1 and X2? i.e. E(Y|X1,X2)=g(X1,X2) for some function g?

2) Are (i) and (ii) direct consequences of the law of total expectation? (are they related at all?) I don't see how (i) and (ii) can be derived as special cases from it...can somone please show me how?

Any help is much appreciated!
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Nov18-09, 03:26 PM   #2
 
Quote by kingwinner View Post
I understand the law of total expectation itself, but I don't understand the generalizations to the vector case and the extensions.

1) Is E(Y|X1,X2) a random variable? Is E(Y|X1,X2) a function of both X1 and X2? i.e. E(Y|X1,X2)=g(X1,X2) for some function g?

2) Are (i) and (ii) direct consequences of the law of total expectation? (are they related at all?) I don't see how (i) and (ii) can be derived as special cases from it...can somone please show me how?
E(Y|X) and E(Y|X1,X2) are tricky to define for general random variables. Which definition are you working with?
Nov19-09, 12:45 PM   #3
 
Quote by bpet View Post
E(Y|X) and E(Y|X1,X2) are tricky to define for general random variables. Which definition are you working with?
E(Y|X=x)=

∫ y f(y|x) dy
-∞
for continuous random variables X and Y.
(similarly for discrete).

General definition:
E(Y|A)=E(Y I_A)/E(I_A)=E(Y I_A)/P(A)
where I_A is the indicator function of A.

If it's too hard to show it in general, can you please show me how can we derive (i) and (ii) from the law of total expectation for the case of CONTINUOUS random variables?

Thanks!
Nov19-09, 03:37 PM   #4
 

law of total expectation (VECTOR case)


Quote by kingwinner View Post
... for the case of CONTINUOUS random variables?
Even then it's tricky - try some examples first with X1 iid X2 and then with X1=X2 and you'll see how those definitions break. The only robust proofs I've seen work with implicit definition for E(Y|X) etc. Maybe someone else here can suggest a simpler way?
Nov20-09, 08:26 PM   #5
 
First of all, is E(Y|X1,X2) a function of X1 and X2??

Is E(Y|X1) = E[E(Y|X1,X2)|X1] a special case of the law of total expectation E(Y) = E[E(Y|X)]?
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