Suppose $X$ is a continuous random variable, with $$P(X>a+b)=P(X>a)P(X>b)\qquad \forall a,b>0.$$ Prove that $X$ is exponentially distributed.

I know this random variable must have the PDF $f_X(x)=\lambda e^{-\lambda x}$; how do I prove it given the above?

  • $\begingroup$ Start with h(x)=P(X>x). Then h(a+b)=h(a)h(b). let g(x)=log(h(x)). So g(a+b)=g(a)+g(b). I suspect that g(x) must have a simple form like g(x)=kx (using continuity). You should be able to finish. $\endgroup$ – herb steinberg Apr 26 '18 at 23:53

This is too long for a comment

  1. From this you will see that any function that satisfies

$$ f(x + y) = f(x)f(y) $$

implies that

$$ f(x) = e^{\mu x} $$

  1. Call $f(y) = P(X > y)$ in the expression above, we must then have something like

$$ P(X > y) = e^{\mu y} $$

But since we require that this number vanishes as we approach infinity, this implies that $\mu$ must be negative, $\mu = -\lambda$, for $\lambda > 0$

$$ P(X > y) = e^{-\lambda y} = \int_y^{\infty}{\rm d}y' f_X(y') $$

  1. Applying the fundamental theorem of calculus you get

$$ f_X(x) = \lambda e^{-\lambda x} $$


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