1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Conditional Expectations (Stochastic Calculus)

  1. Jul 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Let (X_n; for all counting number n) be a sequence of independent random
    variables. We focus on the random walk S_n := X_1 + . . . + X_n and set
    F_n = 'sigma-algebra' of (S_1, . . . , S_n).

    1. Compute E[S_(n+1) \ F_n]
    2. For any z belonging to the complex plane C, show that:

    E[ z^S_(n+1) \ F_n] = z^S_n E(z^X_(n+1))


    2. Relevant equations

    -N.A.- (Hoping for your consideration)

    3. The attempt at a solution

    Here's what I've done:

    1. E[S_(n+1) \ F_n] = E[S_n + X_(n+1) \ F_n]

    = E[S_n \ F_n] + E[X_(n+1) \ F_n]

    = S_n + E[X_(n+1) \ F_n] <- S_n is F_n - measurable

    The last line is my final answer. Should I leave it as is? Is there
    something I can do with the second term?

    2. Note that X_(n+1) is independent of X_1, X_2, ..., X_n. Also,
    S_(n+1) = Sn + X_(n+1). Thus,

    z^S_(n+1) = z^{Sn + X_(n+1)}

    = (z^Sn)(z^X_(n+1))

    Hence, E[ z^S_(n+1) \ F_n] = E[(z^Sn)(z^X_(n+1)) \ F_n]
    Because S_n is F_n - measurable,

    E[ z^S_(n+1) \ F_n] = (z^S_n)E[z^X_(n+1)\ F_n]

    = z^S_n E(z^X_(n+1))

    ...because X_(n+1) is independent of X_1, X_2, ..., X_n.

    Please inform me if there are some mistakes in my solution. Hoping for your
    consideration.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?



Similar Discussions: Conditional Expectations (Stochastic Calculus)
Loading...