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Probability that sum of two random variables is greater than 1

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Let us choose at random a point from the interval (0,1) and let the random variable [itex]X_1[/itex] be equal to the number which corresponds to that point. Then choose a point at random from the interval (0,[itex]x_1[/itex]), where [itex]x_1[/itex] is the experimental value of [itex]X_1[/itex]; and let the random variable [itex]X_2[/itex] be equal to the number which corresponds to this point.

    Compute [itex]P(X_1 + X_2 >= 1)[/itex]

    2. Relevant equations
    The joint pdf is [itex]1/x_1 , 0<x_2<x_1<1[/itex]

    3. The attempt at a solution
    Many. For one, set [itex]Y=X_1+X_2[/itex]. Then find [itex]P(Y>=1)[/itex]. Then evaluate [itex]\int_0^1\int_{1-x_2}^1 1/x_1 dx_1dx_2[/itex]. Evaluating this integral gives me zero.

    The other solutions I come up with end up giving me [itex]ln(0)[/itex], which is undefined. Any suggestions on how to approach this?
     
  2. jcsd
  3. Oct 15, 2013 #2

    mfb

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    How can you get zero from the integral? 1/x_1 is always positive, the result cannot be zero.
    I don't see how you consider the sum in this integral.

    for a given x_1, what is the probability that the sum is larger than 1?
    With this probability, you can use a one-dimensional integral to get the answer.
     
  4. Oct 15, 2013 #3

    Ray Vickson

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    You could use a conditioning argument, where you condition on ##x_1##. In other words, find ##P(X_1+X_2 \geq 1|X_1 = x_1)##, etc.

    Alternatively, you can use the joint density, but you need to apply the correct integration. To do that, first draw a picture to help you get the region of integration correctly
     
  5. Oct 15, 2013 #4

    D H

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    As has been noted, that integral does not evaluate to zero. (It evaluates to one.)

    Much more importantly, that is the wrong integral. Look what happens when x2 is, for example, 3/4. The integration limits for x1 are 1/4 to 1. When x1 is 1/4, the maximum possible value for x2 is 1/4. An x2 value of 3/4 is not possible.

    Bottom line: Your integral covers more than the sample space.

    If you want to use this approach, I suggest drawing a picture. You want the portion of the sample space for which x1+x2≥1.
     
  6. Oct 15, 2013 #5
    Thanks a lot for the help everyone. Once I thought carefully about summing all values of [itex]x_1[/itex], the proper limits of the integrals was a lot more clear.
     
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