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Math Help - Simple conditional continuous distribution Q

  1. #1
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    Simple conditional continuous distribution Q

    Hi all,

    Been going through a question I have the answer to but no working from an exercise book... can't see the logic. Wonder if anyone can lend an eye?

    Q:Let (X,Y) be a random point chosen uniformly on the region R = {(x,y): |x| + |y| \le 1}

    a) Sketch R
    b) Find the marginal densities of X & Y
    c) Find the conditional density of Y given X

    My answers:

    a) simply a kite shape with vertices at (-1,-1),(-1,1),(1,1) & (1,-1)
    b) Evaluate joint density function:

    By observation, the area is a tilted square with sides 1 x1 => area = 1 =>  f_{XY}(xy) dy = 1

    f_{X}(x) = \int_{-\infty}^{\infty} f_{XY}(xy) dy

    = [y]_{0}^{1-|y|}

    = 1-|y| over -1\le x \le 1, now this makes sense to me, as we integrate of the 0 and (1-mod(y)) as limits given by initial conditions. By symetary, marginal of Y is the same except Y instead of X in the expression.

    c)

    conditional density f_{Y/X}(y/x) = {f_{YX}(y/x) \over f_{X}(x)}

    surely this gives the reciprocol of the marginals already found ie:

    f_{Y/X}(y/x) = {1 \over 1-|y|}

    The answer in the back of the book is:

    f_{Y/X}(y/x) = {1 \over 2-2|y|}

    over 1-|x| \le y \le 1-|x|

    Now I can understand the range of y, but cant seem to see where the conditional comes from..... would make my day if someone says its a typo lol

    Thanks for reading.
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  2. #2
    MHF Contributor matheagle's Avatar
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    What do you mean by a kite shape density?
    I don't follow how (-1,-1) satisfies that inequality.
    Do you mean the points (1,0), (-1,0), (0,1), (0,-1)?
    AND how can the marginal density of X, a function of only x, have y's in it?
    A limit of dy cannot have y's in it.
    Last edited by matheagle; February 21st 2011 at 04:50 PM.
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  3. #3
    MHF Contributor matheagle's Avatar
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    I get the square on my 4 points, which has 4 triangles each with area equal to .5.
    So the total area is 2, hence the joint density is .5.

    Now on 0<x<1, we have f_X(x)=\int^{1-x}_{-1+x} .5dy=1-x

    You also need to compute this density on -1<x<0.
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  4. #4
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    Apoligies for the typos- even though there were some fundamental maths errors!

    Shape of polynomial is square with vertices (1,0)(0,1)(-1,0)(0,-1) and area root 2 X root 2 = 2 => density function is half as explained already.

    To finish off what you started matheagle:

    Between -1 \le x \le 0 the boundries of y are the line y =  x + 1 for upper bound and y = -x - 1 for lower bound; therefore

    f_{X}(x) = \int_{-x-1}^{x+1} 0.5 dy = x+1

    Now representing this over the entire region, is the answer in the book:

    f_{X}(x) = 1 - |x| over -1 \le x \le 1.... now that the density function is correctly stated as 1/2, I undertsnad the answers.

    It wasnt a typo in the text book afterall!

    Thanks a lot Matheagle.
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