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Thread: Transformations and Distribution Function

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    Transformations and Distribution Function




    Hey guys,

    Consider the following short example of transformations.

    Let the joint density of x\ \text{and}\ y be given by the unit square, i.e.

    \displaystyle f_{X,Y}\left( x,y \right)=\begin{cases} 1\ 0<x<1\ \text{and}\ 0<y<1 \\ 0\ \text{elsewhere}\end{cases}

    Then the Cumulative Distribution Function of z=x+y is given by:

    \displaystyle F_{Z}\left( z \right) = \begin{cases}0\ \text{for}\ z<0 \\ \int_0^z \int_0^{z-x} dydx\ \text{for}\ 0\leq z < 1 \\ 1-\int_{z-1}^1 \int_{z-x}^1 dydx\ \text{for}\ 1\leq z<2 \\ 1\ \text{for}\ 2\leq z \end{cases}\

    I understand of course why the CDF is 0 and 1 but for the 2 cases in the middle I have been struggling to understand why we partition the CDF like that. In general what is the intuition behind the above result? I am very confused so any help is greatly appreciated, thanks.

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    Re: Transformations and Distribution Function

    Note that the CDF is F_Z(z) = \Pr\{X + Y \leq z\}

    As what you said, the support of (X, Y) is a unit square in the x-y plane.

    And the region \{(x, y): x + y \leq z\} is the bottom-left region divided by the line L: x + y = z which has a slope of -1.

    So you just need to know the position of L for different values of z. It will be easier if you can draw it out (at least in your mind)

    E.g. when z < 0 or z > 2, L will not intersect with the unit square and thus you have the trivial solution.

    And when z = 1, the line is just the diagonal line. Then you should understand why you need to split case at this value.

    For the upper and lower limit it is just a result from "combining" the inequality, and you should be figure it out when you have the diagram.

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    JohnK (10-12-2013)

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    Re: Transformations and Distribution Function


    Thanks but I have to say that the second case, that is 1-\int_{z-1}^1 \int_{z-x}^1 dydx, is still far from evident to me. Could you please be a little more specific on that one?

    Also, isn't that partition arbitrary? The function is continuous so what if we chose z=0.5?
    Last edited by JohnK; 10-10-2013 at 06:54 PM.

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