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Math Help - marginal densities

  1. #1
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    marginal densities

    Suppose f(x,y) is a prob density function: f(x,y) = 1/k -2<=x<y<=2

    Find the value of k such that f(x,y) is a valid probability distribution function?

    Find the marginal density of X and the marginal density of Y?

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  2. #2
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    Re: marginal densities

    Hey JDAWES.

    Hint: To be a valid PDF what does the integral over the whole space have to be?
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  3. #3
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    Re: marginal densities

    To be a valid PDF the integral has to equal 1 right? I have an answer but I don't know if it is right!

    k= 1+0.5x-0.25x^2. I integrated with respect to first x and then y. The first integral i got x/k and then subbed in y and -2. Then i integrated again and got y^2/2k - 2y/k and then input limits of 2 and x and ended up with the above!
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  4. #4
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    Re: marginal densities

    Quote Originally Posted by JDAWES View Post
    To be a valid PDF the integral has to equal 1 right? I have an answer but I don't know if it is right!

    k= 1+0.5x-0.25x^2. I integrated with respect to first x and then y. The first integral i got x/k and then subbed in y and -2. Then i integrated again and got y^2/2k - 2y/k and then input limits of 2 and x and ended up with the above!
    That makes no sense. The integral over an area cannot be a function of either variable. The "outer" integral must have constants as limits. Your region is described as "-2<=x<y<=2" which is the same as saying -2\le x\le 2, x\le y\le 2, which would give an integral \int_{x= -2}^2 \int_{y= x}^2 \frac{1}{k} dydx. Or it can be given as -2\le y\le 2, -2\le y\le x, which would give an integral \int_{y= -2}^2\int_{x= -2}^y \frac{1}{k}dxdy.

    A simpler way to do this is to note that since k is a constant, the integral \int\int \frac{1}{k}dydx= \frac{1}{k}\int\int dA= \frac{1}{k}A. That is, the integral of just dxdy (or dydx) over a region in the plane is the area of that region. Here the region is a triangle of with vertices at (-2, -2), (-2, 2), and (2, 2). That triangle is a right triangle with legs of length 4 and so has area (1/2)(4)(4)= 8. Your condition is 8/k= 1.

    The "marginal density", of X, is the probatility of x over all possible values of y. It is given by the first integral \int_{y= x}^2 \frac{1}{k}dy, a function of x. The marginal density of Y is the other way: \int_{x= -2}^y \frac{1}{k} dx, a function of y.
    Last edited by HallsofIvy; January 14th 2013 at 08:30 AM.
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