# Memorylessness of the Exponential Distribution

$T$ is the time required to repair a machine. We have that $T$ is exponentially distributed with a mean of $\frac{1}{2}$ hours.

For the first part of the question, I am asked to find the probability that repair time exceeds $\frac{1}{2}$ hours. I find
$$P(T> \frac{1}{2})= \frac{1}{e}$$

I am a bit stuck on the second part to the question:
“What is the probability that a repair takes at least 12.5 hours given that its duration exceeds 12 hours”?

I was thinking the answer is

$$P(T\geq 12.5 \, | \, T>12)=P(T\geq .5) = \frac{1}{e} \, \text{ ,}$$

since exponential distribution is memoryless. Is this correct?

#### Solutions Collecting From Web of "Memorylessness of the Exponential Distribution"

Yes, this is correct, this is indeed what is meant by memorylessness.

In general, for all $s\geq 0$, $t \geq 0$, we have

$$P(T>s+t\mid T>s) = P(T>t) = e^{-\lambda t}.$$

We can prove the memorylessness by transforming the expression on the LHS using the definition of conditional probability, as follows

\begin{align} P(X>s+t\mid X>s) &= \frac{P\big(X>s+t\cap X>s\big)}{P(X>s)} = \frac{P(X>s+t)}{P(X>s)} \\ &= \frac{\int_{s+t}^{\infty}\lambda e^{-\lambda x}dx}{\int_{s}^{\infty}\lambda e^{-\lambda x}dx}= \frac{0- -e^{-\lambda (s+t)}}{0- -e^{-\lambda s}} = e^{-\lambda t} = P(X>t) \end{align}

Accreditation: This is a slight modification of this answer by Tianyu Zheng