CDF: Prove a>0 for F(x) = 1 - e^(-ax) - axe^(-ax)

In summary: If all of them are fine, you are done.If you want to show that F is an increasing function, then show that its derivative is never negative.
  • #1
18
0

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?

Homework Equations




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?
 
Physics news on Phys.org
  • #2
fateswarm said:
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.
 
  • #3
If you want to show that F is an increasing function, then show that its derivative is never negative.
 
  • Like
Likes 1 person
  • #4
HallsofIvy said:
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.
 
  • #5
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
fateswarm said:
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?

Aren't you forgetting something very important? If ##f(x) = a^2 x e^{-ax} 1_{\{x \geq 0 \}},## then you need two things:
[tex](1) \; f(x) \geq 0 \text{ for all }x; \text{ and }\\
(2)\; \int_{-\infty}^{\infty} f(x) \, dx = 1.[/tex]
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?
 
  • Like
Likes 1 person
  • #7
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.
 

1. What is CDF?

CDF stands for cumulative distribution function and it is a function used in statistics to describe the probability distribution of a continuous random variable.

2. What does "a>0" mean in the equation?

In this equation, "a>0" means that the value of a is greater than zero. This is important because it ensures that the function F(x) is always positive and therefore follows the properties of a cumulative distribution function.

3. How is the equation used to prove that a>0?

The equation F(x) = 1 - e^(-ax) - axe^(-ax) is used to prove that a>0 by showing that the function is always positive for any value of x. This is done by taking the derivative of the function and setting it equal to zero, which results in the value of a being greater than zero.

4. What is the significance of proving a>0 for F(x) = 1 - e^(-ax) - axe^(-ax)?

Proving that a>0 for this function is important because it ensures that the function follows the properties of a cumulative distribution function. This allows for the accurate calculation of probabilities and other statistical measures.

5. Can the equation be used for any value of a?

Yes, the equation F(x) = 1 - e^(-ax) - axe^(-ax) can be used for any value of a as long as a>0. However, the specific value of a may affect the shape and behavior of the cumulative distribution function.

Suggested for: CDF: Prove a>0 for F(x) = 1 - e^(-ax) - axe^(-ax)

Back
Top