# Probabilities, CDF

fateswarm

## Homework Statement

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?

## 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?

## Answers and Replies

Mentor
I started with saying for that to occur, provided x1>x2 -> F(x1) > F(x2)
That should be ≥.

What about F(x)≥0? F(x)≤1?
Did you check the derivative?

If all of them are fine, you are done.

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

1 person
fateswarm
If you want to show that F is an increasing function, then show that its derivative is never negative.

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

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?

$$(1) \; f(x) \geq 0 \text{ for all }x; \text{ and }\\ (2)\; \int_{-\infty}^{\infty} f(x) \, dx = 1.$$