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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.

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    My apologies for the formatting, this is my first question and I haven't learned formatting yet. – 2017-01-02
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    I was going to comment on the formatting, but it's okay to not know it at first. Consider Googling around for MathJax and LaTeX tutorials. – 2017-01-02
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    _To start,_ can you evaluate $\int_2^\infty xf(x)\,dx \approx 2.567 ?$ – 2017-01-03

2 Answers 2

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Hint: $\Pr(X=2) = \Pr(T\leq2)$

Addendum: There are two common ways to ensure $X\geq a$ (here $a=2$): left-truncating, and left-censoring. I will mention both, but I believe the latter is what you want in this case.

For truncating, take $X=T|T\geq a$. Samples of $X$ are obtained by taking samples of $T$ and discarding them if they are less than $a$. The CDF would be $$F_X(x)=\begin{cases}0,& x

For censoring, take $X=T\vee a$ (the greater of $T$ and $a$). Samples of $X$ are obtained by taking samples of $T$ and replacing them by $a$ if they are less than $a$. The CDF would be $$F_X(x) =\begin{cases}0,& x

The answer: Since you want the left-censored variety, with $a=2$, you then get $$\operatorname{E}[X] = 2\cdot\Pr(X=2) + \int_2^{\infty}x\cdot f_X(x)\;dx$$ $$=2\int_0^2\tfrac13e^{-x/3}\;dx + \int_2^{\infty}x\cdot\tfrac13e^{-x/3}\;dx$$ $$= 2 + 3e^{-2/3}\approx \boxed{3.5403}$$

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    @BruceET : I bet so -- I didn't read carefully and thought this was a discrete distribution :O Whoops! Will modify hint accordingly – 2017-01-03
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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.