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Math Help - cdf of a transformation of two random variables

  1. #1
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    cdf of a transformation of two random variables

    X and Y have the following joint pdf.

    f(x,y)=\frac{1}{4} for -1<x<1,-1<y<1, zero elsewhere.

    Find the cdf of Z=X+Y

    Thanks in advance
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  2. #2
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    Quote Originally Posted by akolman View Post
    X and Y have the following joint pdf.

    f(x,y)=\frac{1}{4} for -1<x<1,-1<y<1, zero elsewhere.

    Find the cdf of Z=X+Y

    Thanks in advance
    F(z) = \Pr(Z < z) = \Pr(X + Y < z) = \Pr(Y < -X + z).

    A diagram is essential to see why there are two cases to consider and where the integral terminals come from in each case.


    Case 1: z \leq 0 (that is, -2 < z \leq 0)

    F(z) = \int_{x = -1}^{x = z + 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....


    Case 2: z > 0 (that is, 0 < z \leq 2)

    F(z) = \int_{x = -1}^{x = z - 1} \int_{y = -1}^{y = 1} \frac{1}{4} \, dy \, dx + \int_{x = z-1}^{x = 1} \int_{y = -1}^{y = -x + z} \frac{1}{4} \, dy \, dx = \, ....


    As I've said, a diagram is essential to see where these results have come from. The calculations are left for you to do.
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