From SOA sample #270
The lifetime of a machine part is exponentially distributed with a mean of five years.
Calculate the mean lifetime of the part, given that it survives less than ten years.
To find the conditional distribution, I must define $$P(X|X<10)={P(X=x\cap X<10)\over P(X<10)}$$
The denominator is $1-e^{-2}$. Regarding the numerator, the SOA solution says it is simply the exponential distribution of mean $5$, ie. $\frac1{5}e^{-\frac x{5}}$. This confuses me, because how do we take into account the second part of the numerator, the condition of $X<10$? Can we simply define this expression $\frac1{5}e^{-\frac x{5}}$, as only applying when $0<x<10$, without modifying the expression in any way?
The answer here seems to do it a different way.
Best Answer
You need to consider the probability distribution $\color{blue}{restricted}$ to $[0,10)$ but besides this you need to $\color{blue}{normalize}$ it by $P([0,10)) = 1-\frac{1}{e^2}$.
So, the probability density of the conditioned random variable is: $$f_{X|X<10}(x)= \begin{cases} 0 & x \geq 10 \\ \frac{1}{5\left(\color{blue}{1-\frac{1}{e^2}}\right)} e^{-\frac{x}{5}} & 0 \color{blue}{\leq x <10} \\ 0 & x < 10 \end{cases}$$
Now, $$E(X|X<10) = \int_0^{\infty}xf_{X|X<10}(x)\, dx = \frac{1}{5\left(\color{blue}{1-\frac{1}{e^2}}\right)} \int_0^{\color{blue}{10}} x e^{-\frac{x}{5}} \, dx$$