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I have a non-negative continuous random variable $X$ and I am given $P(x>a+b)=P(x>a)P(x>b)$. I need to prove that the probability distribution is exponential. I already know that $F(x)=0$ if $x \leq 0$ where $F(x)=P(X \leq x)$ so I should be able to show that otherwise $F(x)=1- \exp(-ax)$ for some constant a.

Daryl
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Niel
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  • Can i say that the exponential distribution fits and such it must be unique fit? Seems such a rough way of phrasing it. – Niel May 25 '14 at 23:22
  • Have you considered the "memorylessness" of the exponential distribution? (Try rearranging the given equation to one that describes conditional probability.) – White Shirt May 25 '14 at 23:23
  • In essence, what you want to know is: if $f:(0,\infty) \to \mathbb R$ satisfies $f(a+b) = f(a) + f(b)$ (which you have assumed is the case if I set $f(a) = \log(1-F(a))$, and if $f$ satisfies some very mild extra hypothesis (e.g. measurable, or monotone), then $f$ is linear. This has been dealt with in many places. But I just cannot think of the correct phrase which will let google find it. Anyway, its is not really a problem in probability. Probably, in the form I present it, it has been answered elsewhere on stackexchange. – Stephen Montgomery-Smith May 25 '14 at 23:23
  • i think this might help me then, reading it now http://math.stackexchange.com/questions/152944/what-can-we-say-about-functions-satisfying-fa-b-fafb-for-all-a-b-i – Niel May 25 '14 at 23:28
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    Cauchy functional equation. – Robert Israel May 25 '14 at 23:29
  • Also: Remarks on the Functional Equation f(x+y) = f(x)+f(y), Edwin Hewitt and Herbert S. Zuckerman, Pages 121-123 – Stephen Montgomery-Smith May 25 '14 at 23:29
  • i see now my problem can be rephrased as f(x+y)=f(x)*f(y) , f(0)=1 – Niel May 25 '14 at 23:35
  • For rational numbers that would yield f(x)=exp(k*x) where k is constant and the function is continuous so it would have to be so on all positive numbers. – Niel May 25 '14 at 23:39
  • and the constant is negative since it is for probability distributions. I still have to write it out decently, but i think i have got it. Thank you. – Niel May 25 '14 at 23:42

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