[Math] Actuarial Exam P problem (expected value of a max function.)

probabilitystatistics

Question
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 $X = \max(T, 2)$.

I understand that the density function of $T$ is $f(t)=\frac{1}{3} e^{-\frac{t}{3}}$ and I'm pretty sure that $\max(T,2)= T$ if $T>2$ and $2$ elsewhere.

Here's where I'm stuck:
$$\text{E}(X)=\text{E}(\max(T,2))=\text{?}$$

I'm not sure how to figure this part out.

Best Answer

Comment. Here is a simulation in R statistical software with very nearly the correct numerical answer. A direct method based on $X$ is used. You can compare the result (to two or three places) with your analytic answer.

t = rexp(10^6, rate = 1/3)
x = pmax(x,2)
mean(x)       
## 3.538123          # aprx E(X) from simulation

mean(t > 2)
## 0.51334           # aprx P(T > 2) from simulation
1 - pexp(2, 1/3)
## 0.5134171         # exact P(T > 2)

A little over half of the devices survive beyond 2.

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