
Probability question
For two random variables, the "distance" between two distributions is defined to be the maximum, over the range which and are defined, where is the cumulative distribution function. Find the distance between the following two distributions:
So:
From this point I'm not sure of what to do. Any help?

Re: Probability question
Hey downthesun01.
You will have to look at two situations: the first is when x is in [0,1] and the second is when x > 1. You will have two situations and you need to find the maximum value of both situations separately.
In the second situation F1 will be a constant (i.e. 1).

Re: Probability question
Thanks. I see what you're saying about the two situations. How do I go about finding the maximum value of each situation? Am I taking the derivative of each situation?

Re: Probability question
Expand out the absolute value function so that it's analytic (i.e. x = x if x > 0 and x if x < 0) and then use standard calculus to find the maximum point.

Re: Probability question
Hmm.. For the situation involving [0,1], I've tried, but I'm getting x= 0, 1 while the book gives x=0, 1
The solution in the manual lacks any real intermediate steps, I'm left kind of stumped right now

Re: Probability question
You can't get 1 since the range is in [0,infinity). Can you show us how you got 1 as a solution?

Re: Probability question
I can't find where I wrote everything down at the moment, but since the range is shouldn't absolute value function be
?
After that, I thought that I take the derivative, which is
Set it equal to 0, and find the values x=c such that

Re: Probability question
You can only use that function in the interval [0,1]. In the interval (1,infinity) you will use 1  1/(1+x)  1 = 1/(x+1) = dist(F1,F2) for x > 1.

Re: Probability question
Sorry, I meant [0,1]
The manual's solution says that it's maximized at either , or a critical point; critical points occur where or .
So, I guess:
I don't see how they got 1 though

Re: Probability question
That takes care of x = 0.
For x = 1 case we look at maximizing 1  1/(x+1)^2  1 or 1/(x+1)^2
If x > 1 then 1/(x+1)^2 = 1/(x+1)^2.
We can show that the function is decreasing for all x > 1 since d/dx 1/(x+1)^2 = 2/(x+1)^3 < 0 for x >= 1.
This implies that the maximum value for x >= 1 is at x = 1 and thus we have the second solution of x = 1.