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.
Best Answer
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 $\lambda>0$.