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Probabilities, CDF

  • Thread starter fateswarm
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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?
 

Answers and Replies

  • #2
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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
HallsofIvy
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If you want to show that F is an increasing function, then show that its derivative is never negative.
 
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  • #4
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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
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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
Ray Vickson
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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?
 
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  • #7
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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.
 

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