1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.