# Proving a random variable is an exponential distribution

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?

• 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. – 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

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