I have a question from my practice actuary exam... I understand one method of arriving at the answer, however the alternative method is giving me a bit of confusion!
I have the lifetime of seismic activity recorder as an exponential random variable with mean lifetime of 3 years.
The recorder will not be serviced for the first two years, so the time to discovery of it's failure is $X=\max[T,2]$.
Determine $E[x]$
So... I would expect the answer to be: (sorry for my lack of mathematica)
[Edit: MathJaxed it for you.]
$$2 + \int_2^\infty \frac 1 3 \cdot t \cdot e^{-t/3} \operatorname{d}t$$
or... $E[T \mid _2 ^\infty] + 2$
However, the guide instead does the following:
$$2 + \int_2^\infty \frac 1 3 \underbrace{\cdot 3\cdot}_{\color{blue}{\text{E(t)}}} e^{-t/3}\operatorname{d}t$$
So, they are multiplying by the mean, 3, instead of t, as you would with a normal expected value.
Why is this- and when do I need to do this instead of multiplying by $t$?
Any clarification would be really helpful!
This should probably only be a comment, but it has gotten too long. You can account in two ways: separately count the contributions to the expectation from the cases where the device lasts under $2$ years and over $2$ years, or count $2$ years always and only more if the device lasts over $2$ years.
Your expectation of the answer is incorrect; you should have $\displaystyle\int_0^2 \frac13 \cdot 2 \cdot \exp(-t/3) \; dt$ instead of $2$ for the first term. You get $2$ only if the device fails in the first $2$ years, which doesn't happen with probability $1$. The second term stays the same, and this is the first method above.
I believe the guide you refer to is using the memoryless property of exponential random variables to use the second method above. It might be easier to write as $\mathbf{E}(t) \displaystyle\int_2^{\infty} \frac13 \exp(-t/3) \; dt$ for the second term. This method of accounting gives the first two years always (the $2$) . If the device has lasted more than $2$ years, the mean lifetime is $3$ years, with an exponential distribution.
It was probably too late for me to be attempting this; I hope this is helpful.