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Law of total expectation (VECTOR case)

  1. Nov 18, 2009 #1
    " 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!
  2. jcsd
  3. Nov 18, 2009 #2
    E(Y|X) and E(Y|X1,X2) are tricky to define for general random variables. Which definition are you working with?
  4. Nov 19, 2009 #3

    ∫ 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?

    Last edited: Nov 19, 2009
  5. Nov 19, 2009 #4
    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?
  6. Nov 20, 2009 #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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