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conditional prob question...

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  • shar_deo
    If x and y are gaussian, x ~ N(0, R) y ~ N(0, V) x and y are uncorrelated. and z = x + y, then (z | x) ~ N(x, V) Now if z = f(x) + y where f(x) is some
    Message 1 of 2 , Aug 29, 2003
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      If x and y are gaussian,
      x ~ N(0, R)
      y ~ N(0, V)

      x and y are uncorrelated.

      and z = x + y, then
      (z | x) ~ N(x, V)

      Now if z = f(x) + y
      where f(x) is some complicated nonlinear function.
      Then can I still say,
      (z | x) ~ N(f(x), V) ??

      If true, some sort of proof/theorem/argument would also be helpful.
      Thanks.

      shar
    • Noel Vaillant
      Hi Shar, ... Strictly speaking, this is not true (choose x to be N(0,1), and y=ex where e is a bernouilli variable with values in {-1,1} independent of x). For
      Message 2 of 2 , Sep 3, 2003
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        Hi Shar,

        > If x and y are gaussian,
        > x ~ N(0, R)
        > y ~ N(0, V)
        >
        > x and y are uncorrelated.
        >
        > and z = x + y, then
        > (z | x) ~ N(x, V)

        Strictly speaking, this is not true (choose x to be N(0,1), and y=ex
        where e is a bernouilli variable with values in {-1,1} independent of
        x). For the statement to be true, you *either* need to replace the
        word "uncorrelated" by "independent", *or*, "x and y are gaussian"
        by "x and y are jointly gaussian". At the end of the day, if you want
        your statement to be true, you need the independence between x and y,
        and being "uncorrelated" will imply independence, only in the case
        when x and y are *jointly* gaussian...

        So we assume that x and y are jointly normal and uncorrelated...


        > Now if z = f(x) + y
        > where f(x) is some complicated nonlinear function.
        > Then can I still say,
        > (z | x) ~ N(f(x), V) ??


        Yes you can (as long as f is a borel measurable map), and the proof
        is as follows: we want to know the conditional distribution of z
        knowing x, and for this purpose we compute the conditional
        characteristic function of z knowing x, i.e. for u in R:

        E[exp(iuz)|x]=E[exp(iu(f(x)+y))|x]
        =exp(iuf(x))E[exp(iuy)|x] (*)
        =exp(iuf(x))E[exp(iuy)] (**)
        =exp(iuf(x))exp(-u^2V/2) (***)

        Equality (*) stems from the measurability of exp[(iuf(x))] with
        respect to s(x) [the sigma-algebra generated by x]. Equality (**)
        stems from the independence of exp(iuy) and x. Equality (***) stems
        from the assumption y~N(0,V).

        We conclude that (see e.g. Lemma 6.13 p.85 of Karatzas and Shreve's
        book, brownian motion and stochastic calculus) that conditionally on
        x, the law of z is normal with mean f(x) and variance V.

        Regards. Noel.
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