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Martingales and Fn measurability

  1. Jul 10, 2006 #1
    By the definition of a martingale Mn, it is a sequence of random variables that are:
    a) integrable; i.e. E(|Mn|) < infinity
    b) Fn measurable where Fn are sigma fields
    and c) E(Mn+1 | Fn) = Mn

    now, I am having difficulty proving a) and b) for the following:
    1) Let Xi be a sequence of independent random variables with mean 0 and variance 1 for every i and Fn be the smallest sigma fields generated by X1, X2,...,Xn. Let Mn = Sn^2 - n where Sn = summaton of Xi from i =1 to n.

    How do I show that E(|Mn|) is finite?
    I got:
    =E(|Sn^2 - n|)
    =E(|Sn^2|) + E(|n|)
    =E(|Sn^2|) + n
    how do I show that the first term is finite?

    and how do I prove that Mn is Fn measurable? (I am having a hard time understanding Fn measurability, if anyone could explain it to me in a simple manner, I would be quite grateful. I do know that its definition is a r.v. X is F measurable if {w | X(w) >= a} is an element of F here a is a real number.

    Here, Mn = Sn^2 -n. I know n, being a constant is automatically Fn-measurable. How about the first term? Is the summation of Xi's automatically Fn measurable? what if its squared?

    On a related note, what if I define Mn = X1 * X2 * ... *Xn where Xi's are random variables and Fn is the smallest sigma field generated by X1 to Xn. How do I prove that Mn is Fn measurable?

    Two final question.
    Let X be a fixed random variable, with E(|X|) < infinity. Let Mn = E(X|Fn).
    1) Is E(X|Fn) automatically equal to X? How about E(|X| | Fn)? Is it equal to |X|? (since X is a fixed r.v)
    2) On a related note, the proof given to show Mn is a martingale is as follows:
    E(Mn+1 | Fn)
    = E(E(X|Fn+1)|Fn)

    in the third step is X = E(X|Fn+1) true because of the fact that X is fixed random variable, and hence a constant (same as question 1) or is there another reason?
  2. jcsd
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