Does the truncated exponential distribution preserves the memoryless property?

Question

The right truncated exponential distribution would be defined as

$$ f(x=t \mid x<T) = \frac{\lambda e^{-\lambda t}}{1-e^{-\lambda T}} $$

right?

I would like to know if this distribution preserves the memoryless property. I think that the answer is no, because the memoryless property characterizes completely the exponential distribution, so no other distribution (including truncated exponential) have it. But this is counterintuitive to me, and I cannot derive a mathematical prove for that, any help is appreciated

Since $X$ is a continuous r.v., it won't admit singletons, so $P(X = t \mid x < T) = 0$. — GNUSupporter 8964民主女神地下教會, Apr 26 '18 at 15:16

score 1 · Accepted Answer · answered Apr 26 '18 at 15:25

1

In order for a distribution to be memoryless, you need $$P(X>a+b|X>b)=P(X>a)$$ to hold for all $a,b>0$. Now, take $a=b=T/2$. Then $P(X>T/2)$ is nonzero, but $P(X>T|X>T/2)=0$ because $X$ is never greater than $T$.

answered Apr 26 '18 at 15:25

Mike Earnest

75,930

Does the truncated exponential distribution preserves the memoryless property?

1 Answers1