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Thread: Probability Density Function and Cumulative Distribution function

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    Probability Density Function and Cumulative Distribution function




    Suppose you have a random variable X and Y with Joint PDF define by:
    f(x,y)= Ky -1<= x <=1, 0<y<|x|
    f(x,y) = 0 otherwise
    I found K=3.
    How would you find
    P[\frac{2X}{Y} < 1]
    and
    f_{Y|X}(y|x)
    matter of fact it would be nice to know what it means.
    Thanks.

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    Re: Probability Density Function and Cumulative Distribution function

    a) The probability is found by integrating over the region

    \mathcal{S} = \left\{(x, y): \frac {2x} {y} < 1 \right\}

    in the following double integration, with the integrand as the joint pdf:

    \iint_\mathcal{S} f_{X,Y}(x,y) d(x,y)

    In practice of course it would be easier for you to visualize this region in the 2-D plane so that you can break it piece-wisely and translate it in terms of upper and lower limits of x and y.


    b) This is a conditional pdf. You check the definition first and see if you got any difficulty.

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    Re: Probability Density Function and Cumulative Distribution function


    visually, 2x<y just left the negative half of the dominion, that is 1/2

    analitically, in the positive y<x, so 2x<y<x no chance
    if x<0 then 2x<y<-x, but as 0<y<-x then you have just the second integral that gave u K=3

    f()y/x = f(x,y) / f(x) -> f(x) = 3/2 x^2, you can make the integral from 0 to |x| without problem because later you have x^2, then you test f(x) from -1 to 1

    then f()y/x=2 y / x^2 0<y<|x| then you test that this a prob func

    i did this problem mostly because of the first part, not without some difficulties, i think i achieved a valid solution
    Last edited by amilsan; 11-16-2015 at 09:03 PM. Reason: forgot to mention f()y/x limits

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