A continuous random variable is memoryless if and only if it is an exponential random variable

Question

I'm reviewing some probability, and I remembered this quite interesting claim.

A continuous random variable is memoryless if and only if it is an exponential random variable.

Obviously, "$\Longleftarrow$" is easy to prove.

But how does one prove "$\Longrightarrow$"?

I suppose, suppose we have an absolutely continuous CDF $F_{X}$ for a random variable $X$ which is memoryless, and define $S_{X} = 1 - F_{X}$. Then we have for $s > t$, $$\mathbb{P}\left(X > s \mid X > t\right) = \dfrac{S_{X}(s)}{S_{X}(t)} = S_{X}(s-t)\text{.}$$ What to do from here, though, I'm not sure.

I think you should prove that the relation above implies that $S(t)= e^{-a t}$, for some positive a. — Kolmo, Jan 20 '16 at 17:02
$S(t+s)=S(t)*S(s)$, then plug t=s=1. This should give you the way to imply a form for S for all rationals. Continuity and the fact that S is a tail cdf should do the rest. — Kolmo, Jan 20 '16 at 17:27
The memoryless property is just the probability interpretation of the calculus/analysis fact that $S(s+t)=S(s)S(t)$ characterizes the exponential function (up to scale). — snarfblaat, Jan 20 '16 at 22:13

EditPiAf · Accepted Answer · 2018-04-26T16:08:04.027

In the case where $S_X > 0$, the equality in the OP rewrites as \begin{aligned} \ln \frac{S_X (s)}{S_X (t)} = \ln S_X (s-t) \, . \end{aligned} The function $T = \ln \circ S_X$ satisfies $$ T(s) - T(t) = T(s-t) $$ for all $s$ and $t$. Therefore, $T$ is a linear function, and $S_X = \exp \circ T$ is the exponential of a linear function. The properties of the CDF $F_X = 1-S_X$ reduce the possibilities, so that $F_X$ has no other choice than being the CDF of the exponential distribution. Since the CDF characterizes the distribution of a random variable, one concludes that $X\sim \mathcal{E}(\lambda )$ for some $\lambda>0$.

A continuous random variable is memoryless if and only if it is an exponential random variable

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