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Bivariate probability

  1. May 9, 2009 #1
    Let the joint probability density function of random variables X and Y be given by

    f(x,y) = 2 if 0 <= y <= x <= 1
    and 0 otherwise

    a) calculate the marginal probability density functions.

    f(x) = 2x
    f(y) = 2-2y

    b) find E(X) and E(Y).

    E(X) = 2/3
    E(Y) = 1/3

    c) calculate P(X<1/2) , P(X<2Y) and P(X=Y).

    P(X<1/2) = integral from 0 to 1/2 of 2x dx = 1/4

    but as for the other 2 I have no idea on how to do them. do they also involve the marginal distribution functions? Any help as always is greatly appreciated.

    Thanks.
     
  2. jcsd
  3. May 9, 2009 #2

    jbunniii

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    These are correct, except remember that the equations are valid only for [tex]0 \leq x \leq 1[/tex] and [tex]0 \leq y \leq 1[/tex]. Both f(x) and f(y) are zero for x and y outside those intervals.

    Yep.

    Correct.

    Try calculating P(X<2Y) directly from the joint probability density function. All you need to do is work out what the region "X < 2Y" looks like and how to perform a double integral over that region.

    For P(X=Y) my suggestion is similar: what does the region "X=Y" look like in the X-Y plane, and can you immediately identify the probability without even integrating?
     
  4. May 10, 2009 #3
    Thanks for the help..

    I can see by looking at the graph that the answer should be 1/2 but i dont know how to set up the integrals, i know it involves the f(x,y) and a double integral over dx and dy.

    as for P(X=Y) is that 0 (does this have anything to do with it being continuous)?
     
  5. May 10, 2009 #4

    jbunniii

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    Notice that the original joint density function is zero unless Y <= X. What does this say about the probability that X < 2Y?

    Yes and yes. The line X = Y is a region with zero area (it's just a line), so in order for the probability to be nonzero, there would have to be a concentration of probability on that line, which requires at least one non-continuous random variable.
     
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