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Homework Help: How to find this particular probability?

  1. Feb 25, 2014 #1
    1. The problem statement, all variables and given/known data

    Suppose there are three statistically i.i.d continuous random variables X1, X2, X3 each are uniformly distributed in the range [0,1]. How to find the probability P(X1+X2<X3)?

    2. Relevant equations
    The below given equations are the steps to the solution. But I can't figure out how the limits of integral comes this way.

    [itex] \int_0^1 \int_0^{x_3}\int_0^{x_3-x_2} \,dx_1\,dx_2\,dx_3 =\int_0^1 \int_0^{x_3} (x_3-x_2) dx_2\,dx_3 = \int_0^1 x_3^2 - \frac{x_3^2}{2}\,dx_3 = \frac16 = 0.1\overline 6[/itex]

    3. The attempt at a solution

    I tried this using a software called MATLAB by generating three pseudo random variables (1000 samples) and finding X1+X2−X3 and plotting its CDF through a MATLAB tool called dfittool. I got the answer around 0.169. But how do I do this theoretically? Especially how to figure out the limits in those integrals?
  2. jcsd
  3. Feb 25, 2014 #2

    Ray Vickson

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    Your questions make no sense: you have already obtained the answer theoretically, and you have already written the limits of integration.
  4. Feb 25, 2014 #3
    @Ray Vickson : Yes I have got those limits from someone else, but never told how they come?
  5. Feb 25, 2014 #4

    D H

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    They come from two different concerns, that x1+x2<x3 and that each xi must be between 0 and 1. Those integration limits represent the intersection of those concerns.
  6. Feb 25, 2014 #5


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    You are told that the three variables all lie in [0, 1]. The limits on the outer integral, with respect to [itex]x_3[/itex] must be constants so must be 0 and 1. The next inner integral can have limits depending on [itex]x_3[/itex]. Since we have [itex]x_1+ x_2< x_3[/itex] and [itex]x_1[/itex] can be 0, [itex]x_2[/itex] can go from 0 to [itex]x_3[/itex]. Finally, [itex]x_1+ x_2< x_3[/itex] means that [itex]x_1< x_3- x_2[/itex] so the inmost integral has limits of 0 to [itex]x_3- x_2[/itex].
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