Results 1 to 2 of 2

Math Help - joint probability density function

  1. #1
    Newbie
    Joined
    Dec 2008
    Posts
    18

    joint probability density function

    If some could help me out with this problem I' really appreciate it, what I'm having trouble with on the second one is the bounds, If x plus y have to be less than or equal to 3 then shouldn't the bounds just be 0<x<2 and 1<y<3 because x can't be 3 and y cant be 0?

    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor undefined's Avatar
    Joined
    Mar 2010
    From
    Chicago
    Posts
    2,340
    Awards
    1
    Quote Originally Posted by jason0 View Post
    If some could help me out with this problem I' really appreciate it, what I'm having trouble with on the second one is the bounds, If x plus y have to be less than or equal to 3 then shouldn't the bounds just be 0<x<2 and 1<y<3 because x can't be 3 and y cant be 0?

    The bounds you mentioned won't work because for example they include the point (x, y) = (2, 3) where x+y = 5 > 3. (I'm not being very careful to distinguish between > and \geq, etc., because it's not important for these kinds of problems, and the meaning should be clear.)

    Geometrically the inequality x+y < 3 represents the region under the line x+y = 3 in the xy plane. So in D, this is the region under the diagonal going from the upper left to the lower right corner of the square.

    Here's what I get:

    \int_1^3\int_0^{3-y}f(x,y)\,dx\,dy
    =\frac{1}{12}\int_1^3\int_0^{3-y}(x+y)\,dx\,dy
    =\frac{1}{12}\int_1^3\big[\frac{x^2}{2}+xy\big]_{x=0}^{x=3-y}\,dy
    =\frac{1}{12}\int_1^3\bigg(\frac{(3-y)^2}{2}+(3-y)y\bigg)\,dy
    =\frac{1}{24}\int_1^3\big((3-y)^2+2(3-y)y\big)\,dy
    =\frac{1}{24}\int_1^3\big(y^2-6y+9+6y-2y^2\big)\,dy
    =\frac{1}{24}\int_1^3\big(-y^2+9\big)\,dy
    =\frac{1}{24}\big[-\frac{y^3}{3}+9y\big]_1^3
    =\frac{1}{24}\bigg(-\frac{27}{3}+9(3)-(-\frac{1}{3}+9)\bigg)
    =\frac{1}{24}\big(-9+27+\frac{1}{3}-9\big)
    =\frac{1}{24}\bigg(\frac{28}{3}\bigg)
    =\frac{7}{18}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Joint Probability Density Function
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: August 4th 2010, 03:06 AM
  2. Replies: 5
    Last Post: December 5th 2009, 11:30 PM
  3. Help with Joint Probability Density Function
    Posted in the Advanced Statistics Forum
    Replies: 8
    Last Post: November 29th 2009, 08:40 AM
  4. Joint probability density function
    Posted in the Statistics Forum
    Replies: 1
    Last Post: May 5th 2009, 05:09 PM
  5. Joint Probability Density Function
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 5th 2008, 08:36 PM

Search Tags


/mathhelpforum @mathhelpforum