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A proving on random variables

  1. Sep 11, 2004 #1
    Hello, I need some help on the independence of random variables...
    "How do I prove that if X and Y are two independent random variables, then U=g(X) and V = h(Y) are also independent?"

    - Isn`t that if random variables X and Y are independent, it implies
    that f(x,y) = g(x)h(y) and vice versa? Also, note that g(x) and h(y) are
    two marginals. But what I don`t understand is that what does it mean to
    have U = g(X) to be a capital "X"?

    - {then U=g(X) and V = h(Y) are also independent} what am I supposed to
    show in this proof? And lastly, what is my first step/strategy in proving
    this? Hope you can give me hints.. =)
  2. jcsd
  3. Sep 11, 2004 #2

    matt grime

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    U is g of the random variable X. You do this all the time, such as when working out the variance: it is
    E(X^2)-E(X)^2, So there's a function of a random variable there (X^2).

    Are these continuous of discrete R.V.'s? Not that it matters two much. If it 's continuous look at the pdf of the joind dist. since g is a function of X alone, and h a function of Y alone the double integrals INT dxdy split as int dx int dy. If discrete replace integrals with sums.
  4. Sep 11, 2004 #3
    Sir matt grime/anyone... =]
    I hope someone can guide me.
    I want to prove first the continuous. So, the joint pdf can be described as

    f[g(X),h(Y)] = INTaINTb g(x)h(y) dx dy -> am I right here?
    where a and b are arbitrary intervals.
    = INTa h(y) [INTb g(x)dx] dy -> h(y) is treated as a constant.
    = [INTb g(x)dx] [INTa h(y)dy] -> [INTb g(x)dx] is now a constant
    = g(X) h(Y)

    I believe I got screwed up in my notations... is this the proof? I hope it is.. but can someone help me edit this... will I use u's and v's here?... I think not.

    For the discrete case...

    f(g(X), h(Y)) = P(U = g(X), V = h(Y)) = P(U = g(X)) P(V = h(Y)) = g(X)h(Y)?

    Is this the right proof? I hope someone can help me.. =]
  5. Sep 12, 2004 #4
    I believe you made a mistake here, it's not

    [tex] f[g(X),h(Y)] = g(x) \cdot h(y)[/tex]

    but it's


    f(X,Y) = g(x) \cdot h(y)


    You were on the right track, but it should be

    [tex] f(U,V) = g(U)h(V) = g(g(x)) \cdot h(h(y)) [/tex]

    I believe that U and V (g(x) and h(y), respectively) should be independent since Y cannot influence g(x) and X cannot influence h(y) since X and Y are independent. I just don't know how to prove it in mathematical notation, but it's worth a try =)
    Last edited: Sep 12, 2004
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