How does one prove that the unique continuous distribution with the memoryless property is the exponential distribution? i.e. Suppose we know that a continuous random variable $X$ satisfies

$$P\{X > t+s \mid X > s \} = P\{X > t \}$$

Then how do we show that $X \sim \text{Exponential}(\lambda)$ for some $\lambda > 0$?


  • 1
    $\begingroup$ This is equivalent to $P(X>t+s)=P(X>t)P(X>s)$ which tells you that $P(X>t)$ is a decaying exponential. $\endgroup$ – A.S. Feb 20 '16 at 16:30
  • $\begingroup$ I always find it irritating when someone writes $\text{Exponential}(\lambda)$ as if that were not at best ambiguous. Sometimes $X\sim\text{Exponential}(\lambda)$ means $\Pr(X>x) = e^{-\lambda x}$ for $x\ge0$ (so the intensity is $\lambda$ and the expected value is $1/\lambda$) and sometimes it means $\Pr(X>x) = e^{-x/\lambda}$ for $x\ge0$ (so the intensity is $1/\lambda$ and the expected value is $\lambda$). $\qquad$ $\endgroup$ – Michael Hardy Feb 20 '16 at 17:12
  • $\begingroup$ @Michael I shared your sentiment initially, but it looks like today - esp. in modern works - $\lambda$ is predominantly/exclusively referring to rate/intensity while $\beta$ is to denote the mean - but it might be field/application dependent. $\endgroup$ – A.S. Feb 21 '16 at 3:21
  • $\begingroup$ math.stackexchange.com/questions/1801830/… $\endgroup$ – StubbornAtom Dec 6 '19 at 9:44

Let $g(t) = \Pr(X>t)$. Then memoryless tells us that $g(t+s)=g(t)g(s)$. Consequently $g(t_1+t_2+t_3)=g((t_1+t_2)+t_3) = g(t_1+t_2) g(t_3) = g(t_1) g(t_2) g(t_3)$, and it should become clear why $g(t_1+\cdots+t_n)=g(t_1)\cdots g(t_n)$ for every finite number $n$. Thus $$ g(n) = g(1+\cdots+1) = g(1)\cdots g(1) = (g(1))^n, $$ so $g$ is an exponential function of $n$ as long as $n$ is a positive integer. So what about non-integers? $$ \left(g\left(\frac 1 8 \right)\right)^8 = g\left(\frac 1 8\right)\cdots g\left(\frac 1 8\right) = g\left(\frac 1 8 + \cdots + \frac 1 8 \right) = g(1) $$ so $$ g(1)^{1/8} = g\left(\frac 1 8 \right) $$ and it becomes clear why $g(t)=g(1)^t$ if $t=\left\{1\text{ over a positive integer}\right\}$. Next let's try $11/8$: \begin{align} g\left(\frac{11}8\right) & = g\left( \frac 1 8 + \cdots + \frac 1 8 \right) \\[10pt] & = \left(g\left(\frac 1 8 \right)\right)^{11} & & \text{because of what was shown about integers above} \\[10pt] & = \left( g(1)^{1/8} \right)^{11} & & \text{because of what was shown about $1/8$ above} \\[10pt] & = (g(1))^{11/8}, \end{align} so $g(t)$ behaves like an exponential function with base $g(1)$ when $t$ is a positive rational number.

Finally, what about irrational numbers $t$? Notice that $g(t) = \Pr(X>t)$ is squeezed between $\Pr(X>a)$ for all positive rational numbers $a$ that are $>t$ and $\Pr(X>b)$ for all positive rational numbers $b$ that are $<t$. And also $g(1)^t$ is squeezed between all numbers $g(1)^a$ for rational numbers $a$ that are $>t$ and all numbers $g(1)^b$ for rational numbers $b$ that are $<t$. One can deduce then that $g(t)$ is still equal to $g(1)^t$.

Hence we have $$ g(t) = g(1)^t \text{ for all real }t\ge0. $$


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