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Thread: Problem on Random Variables

  1. #1
    Junior Member
    Oct 2008

    Problem on Random Variables

    I'm working on the following problem for a probability class.

    Let $\displaystyle X_1, X_2,...$ be independent random variables with a common continuous distribution function. Let $\displaystyle B$ be the $\displaystyle \omega$-set where $\displaystyle X_m(\omega)=X_n(\omega)$ for some pair $\displaystyle m,n$ of distinct integers, and show that $\displaystyle P(B)=0$.

    Part of my problem is that this section we're studying on random variables introduces so many new terms that I haven't been able to see how they work together very well. I understand that $\displaystyle P(B)=0$ means that $\displaystyle X_m(\omega)\ne X_n(\omega)$ for any $\displaystyle m\ne n$ except on a set of probability (measure) 0, but am struggling mostly with the first sentence. I'm not asking for a full fledged solution to the problem, but I would be grateful for any assistance on understanding the interplay between the other terms involved and advice on how to start.
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  2. #2
    Moo is offline
    A Cute Angle Moo's Avatar
    Mar 2008
    P(I'm here)=1/3, P(I'm there)=t+1/3

    You can try to find the pdf of $\displaystyle X_m-X_n$ (by first finding the pdf of $\displaystyle (Y,Z)=(X_m-X_n,X_n)$ thanks to a Jacobian transformation and then integrating)

    Then, you'll get something like $\displaystyle g(y)=\int f(y+z)f(z) ~dz$ for the pdf of Y.

    $\displaystyle P(Y=0)=\int_{\{0\}} \int f(y+z)f(z) ~dz ~dy$

    Reverse the integration order (what's in the integral is positive so we can apply Fubini-Tonelli's theorem) and you'll get 0, because you integrate over a set of Lebesgue measure 0.

    Otherwise, maybe you can have a look into the properties of convolution... But right now I have to go to school... Good luck !
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