# Probability question

• Jun 25th 2013, 02:12 AM
downthesun01
Probability question
For two random variables, the "distance" between two distributions is defined to be the maximum, $\text{ max all x}|F_{1}(x)-F_{2}(x)|$ over the range which $F_1$ and $F_2$ are defined, where $f(x)$ is the cumulative distribution function. Find the distance between the following two distributions:

$\text{(i) uniform on the interval }[0,1]$
$\text{(ii) pdf is }f(x)=\frac{1}{(x+1)^2}\text{ for }0

So:
$F_1=x$
$F_2=1-\frac{1}{1+x}$
$|F_1(x)-F_2(x)|=\Big{|}x-1+\frac{1}{1+x}\Big{|}$

From this point I'm not sure of what to do. Any help?
• Jun 25th 2013, 02:21 AM
chiro
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).
• Jun 25th 2013, 03:07 AM
downthesun01
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?
• Jun 25th 2013, 04:15 AM
chiro
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.
• Jun 25th 2013, 04:26 AM
downthesun01
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
• Jun 25th 2013, 04:39 PM
chiro
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?
• Jun 25th 2013, 10:13 PM
downthesun01
Re: Probability question
I can't find where I wrote everything down at the moment, but since the range is $[0,\infty)$ shouldn't absolute value function be

$x+\frac{1}{x+1}-1$?

After that, I thought that I take the derivative, which is

$1-\frac{1}{(x+1)^2}$

Set it equal to 0, and find the values x=c such that $f'(c)=0 \text{ or }f'(c)\text{ doesn't exist and }f(c) \text{ exists}$
• Jun 26th 2013, 12:19 AM
chiro
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.
• Jun 26th 2013, 01:58 AM
downthesun01
Re: Probability question
Sorry, I meant [0,1]

The manual's solution says that it's maximized at either $x=0,1$, or a critical point; critical points occur where $1-\frac{1}{(x+1)^2}=0$ or $x=0$.

So, I guess:

$f(0)=1-\frac{1}{(0+1)^2}=0$

I don't see how they got 1 though
• Jun 26th 2013, 02:15 AM
chiro
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.