# Thread: Transformations and Distribution Function

1. ## Transformations and Distribution Function

Hey guys,

Consider the following short example of transformations.

Let the joint density of be given by the unit square, i.e.

Then the Cumulative Distribution Function of is given by:

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.

2. ## Re: Transformations and Distribution Function

Note that the CDF is

As what you said, the support of is a unit square in the plane.

And the region is the bottom-left region divided by the line which has a slope of .

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

E.g. when or , will not intersect with the unit square and thus you have the trivial solution.

And when , 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.

3. ## The Following User Says Thank You to BGM For This Useful Post:

JohnK (10-12-2013)

4. ## Re: Transformations and Distribution Function

Thanks but I have to say that the second case, that is , 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 ?

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