# Probabilities, CDF

1. Sep 20, 2013

### fateswarm

1. The problem statement, all variables and given/known data

Given the function F(x) = 1 - e^(-ax) - axe^(-ax) for x>=0 and 0 elsewhere, for which values of 'a' does the function constitute a CDF?

2. Relevant equations

3. The attempt at a solution

I started with saying for that to occur, provided x1>x2 -> F(x1) > F(x2)

Though I've only managed to say for x1=0 and x2=1 -> a>0 but I'm not sure if it's an unequivocal answer.

How can I prove it decisively?

2. Sep 20, 2013

### Staff: Mentor

That should be ≥.

Did you check the derivative?

If all of them are fine, you are done.

3. Sep 20, 2013

### HallsofIvy

Staff Emeritus
If you want to show that F is an increasing function, then show that its derivative is never negative.

4. Sep 20, 2013

### fateswarm

Oh, good idea (+thanks). I'll look into it. I may return if I see an issue.

5. Sep 20, 2013

### fateswarm

I've got a bit of trouble. The derivative>0 appears to conclude to (a^2)xe^(-ax)>0 which appears to be true for any 'a' belonging in IR. It appears to be more direct to do 0<=F(x)<=1 which appears to conclude to 0<=(e^-ax)(1-ax)<=1 but how do I learn how to solve that inequality?

6. Sep 20, 2013

### Ray Vickson

Aren't you forgetting something very important? If $f(x) = a^2 x e^{-ax} 1_{\{x \geq 0 \}},$ then you need two things:
$$(1) \; f(x) \geq 0 \text{ for all }x; \text{ and }\\ (2)\; \int_{-\infty}^{\infty} f(x) \, dx = 1.$$
In the current case the integral goes from 0 to ∞ because f(x) = 0 for x < 0. So, if you have a = 0 do these conditions hold? What about if a < 0?

7. Sep 20, 2013

### fateswarm

Well I can I show that for a negative 'a' F will not be approaching 1 for x->+oo. The integral of f(x) between 0 and +oo appears to be 'a' times the expected value of any Exp(a) so it's 1 for any 'a', hrm.. provided a>0 (for any Exp(a)) so I guess that might be enough. It's a bit confusing since the exercise goes on to ask for the fx(x) in a later question so I'm unsure if it needs it exposed directly on this first question.