Hints Lack of Memory implies Exponential Distributiom

Question

Let $X$ be a continuous Random Variable with a continuous Distribution function and with values in $(0,\infty)$, and with a denisty w.r.t the lebesgue measure.

I have already been able to prove that $X$ ~ $\lambda e^{-\lambda x}$ implies $P(X \geq x+t |X \geq x)=P(X \geq t)$

But I am stuck on the converse so that I can prove "$\iff$".

Any hints?

Answer 1

Let $G(x) = P(x>x)$ denote the complimentary CDF of a memoryless random variable. Using Baye's rule:

$P(X \geq x+t|X \geq x)$

$=\frac{P(X\geq x+t, X \geq x)}{P(X \geq x)}$

$= \frac{P(X \geq x+t)}{P(X \geq x)} = P(X \geq t)$

implying: $G(x+t)=G(x)G(t)$.

Assuming $X \geq 0$, we have $G(0)=1$. Now,

$G'(x) = \lim_{t \rightarrow 0} \frac{G(x+t)-G(x)}{t}$

$= \lim_{t \rightarrow 0} \frac{G(x)G(t)-G(x)}{t}$

$=G(x)\lim_{t \rightarrow 0} \frac{G(t)-1}{t}$

$=G(x)\lim_{t \rightarrow 0}\frac{G(t)-G(0)}{t} $

$ = G(x)G'(0)$

Thus, you have $\frac{d}{dx} G(x) = cG(x)$, with $c = G'(0)$. You can easily solve this differential equation (with appropriate boundary conditions) to show that $G(x)$ must take an exponential form.

Answer 2

Let $X$ be a memoryless random variable with continuous distribution function satisfying $P(X \geq x+t|X \geq x) = P(X \geq t)$. This gives $P(X \geq x+t) = P(X \geq x) P(X \geq t)$ for all $x,t$.

Define the function $G(y) = P(X \geq y)$. $G$ then satisfies $G(x)G(t) = G(x+t)$ on $(0,\infty)$. What can you say about $G$? Remember the exponential functional equation. Now, how is $G$ related to the distribution function of $X$? Does this give the answer?