Expected remaining lifetime of a machine having rate $\Lambda$

Question

The lifetimes of 2 machines are independent exponential random variables having rates $\Lambda_1$ and $\Lambda_2$. One of the machines went out of order, what is the expected remaining lifetime of the other machine?

Now I do understand that the expected lifetime of second machine is not affected by the fact that first went out of order (because of memoryless property of exponential random variable). But how to get expected remaining lifetime?

Explanation and intuition would be appreciated.

Answer 1

Let $T_1, T_2$ be the lifetimes. So:

$$T_1\sim\mathcal{Exp}(\Lambda_1), T_2\sim\mathcal {Exp}(\Lambda_2), T_1\perp T_2$$

You wish to find: $\mathsf E(\max\{T_1,T_2\}-\min\{T_1,T_2\})$, which is the expected time between the earliest and latest failure.

Using the Law of Total Expectation, and partitioning on which machine fails first, the memoriless property tells us this is:

$$\begin{align}&\mathsf E(\max\{T_1,T_2\}-\min\{T_1,T_2\})\\[1ex]=~&\mathsf P(T_1>T_2)~\mathsf E(T_1-T_2\mid T_1>T_2)+\mathsf P(T_1\leq T_2)~\mathsf E(T_2-T_1\mid T_1\leq T_2)\\[1ex] =~& \mathsf P(T_1>T_2)~\mathsf E(T_1)+\mathsf P(T_1\leq T_2)~\mathsf E(T_2)\end{align}$$

Answer 2

If $T_1$, $T_2$ are the lifetimes, then $$T_1 + T_2 = (T_1\wedge T_2) + (T_1\vee T_2), $$ and hence $$(T_1\vee T_2) - (T_1\wedge T_2) = T_1 + T_2 - 2(T_1\wedge T_2).$$ It follows that \begin{align} \mathbb E\left[(T_1\vee T_2) - (T_1\wedge T_2)\right] &= \mathbb E\left[T_1 + T_2 - 2(T_1\wedge T_2)\right]\\ &= \mathbb E[T_1] + \mathbb E[T_2] - 2\mathbb E\left[T_1\wedge T_2\right]\\ &= \frac1{\Lambda_1} + \frac1{\Lambda_2} - \frac2{\Lambda_1+\Lambda_2}. \end{align}