
Probability question
For two random variables, the "distance" between two distributions is defined to be the maximum, $\displaystyle \text{ max all x}F_{1}(x)F_{2}(x)$ over the range which $\displaystyle F_1$ and $\displaystyle F_2$ are defined, where $\displaystyle f(x)$ is the cumulative distribution function. Find the distance between the following two distributions:
$\displaystyle \text{(i) uniform on the interval }[0,1]$
$\displaystyle \text{(ii) pdf is }f(x)=\frac{1}{(x+1)^2}\text{ for }0<x<\infty$
So:
$\displaystyle F_1=x$
$\displaystyle F_2=1\frac{1}{1+x}$
$\displaystyle F_1(x)F_2(x)=\Big{}x1+\frac{1}{1+x}\Big{}$
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 $\displaystyle [0,\infty)$ shouldn't absolute value function be
$\displaystyle x+\frac{1}{x+1}1$?
After that, I thought that I take the derivative, which is
$\displaystyle 1\frac{1}{(x+1)^2}$
Set it equal to 0, and find the values x=c such that $\displaystyle f'(c)=0 \text{ or }f'(c)\text{ doesn't exist and }f(c) \text{ exists}$

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 $\displaystyle x=0,1$, or a critical point; critical points occur where $\displaystyle 1\frac{1}{(x+1)^2}=0$ or $\displaystyle x=0$.
So, I guess:
$\displaystyle f(0)=1\frac{1}{(0+1)^2}=0$
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.