A device that continuously measures and records seismic activity is placed in a remote region. The time, $T,$ to failure of this device is exponentially distributed with mean $3$ years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is $X = \max(T, 2)$. Determine $E(X)$.
My attempt:
$$E[X] = E[X\mid T\ge 2] \cdot P(T \ge 2) + E[X\mid T< 2] \cdot P(T< 2)$$
$$= E[T\mid T\ge 2] \cdot P(T \ge 2) + E[2\mid T< 2] \cdot P(T< 2) $$
Using the memoryless property of the Exponential distribution, we have:
$$ = E(T) \cdot P(T\ge 2) + E(2) \cdot P(T<2)$$
$$ = 3[1-P(T<2)] + 2\cdot P(T<2)$$
(Edited based on commments)
Now, $P(T<2)=\int_0^2 \frac{1}{3} e^{-t/3} \; dt = 1- e^{-2/3}$. These calculations do not lead to the correct answer. Can someone please point out what I did incorrectly?
As pointed out in the answer by Michael Hardy, for the benefit of anyone who visits this post in the future, this is how the solution should go:
$$E[T\mid T\ge 2] \cdot P(T \ge 2) + E[2\mid T< 2] \cdot P(T< 2) = \int_2^\infty \frac{t}{3} e^{-t/3}\; \text{d}t + 2 \cdot [1- e^{-2/3}]$$
$$= 5e^{-2/3} + 2 – 2e^{-2/3} = \boxed{3e^{-2/3} +2}$$
Best Answer
You're making a commonplace mistake: \begin{align} \require{cancel} \text{wrong:} & \xcancel{\operatorname E(T\mid T>2) = \operatorname E(T)} \\[10pt] \text{right: } & \operatorname E( T-2\mid T>2) = \operatorname E(T) \end{align}