Expected value of failure time given the component has survived until some time

Question

I have a component with exponential failure function, i.e. the CDF is ($T$ being the time of failure): $$F(t):= P(T \leq t) = 1 - e^{-\lambda t}$$

I want to find the expected time of failure of the component, given it has survived until time $\tau$, i.e., $E(T | T \geq \tau)$.

I understand that finding the expected value of a random variable generally can be achieved by integrating $1-CDF$ thus:

$$E(X) = \int_{x=0}^{x=\infty} (1-CDF(x)) dx$$

So, the expected time of failure generally would be:

$$E(T) = \int_{t=0}^{t=\infty} (1-F(t)) dt$$

(In my case, this would simply be $1$, as it is an exponential distribution).

I have tried to approach the problem by integrating the CDF after the specified time ($\tau$) and then offseting by $\tau$, however I am unsure if this is the right way to approach it:

$$E(T|T\geq \tau) = \tau + \int_{t=\tau}^{t=\infty} (1-CDF(t)) dt$$

What is the correct way to compute $E(T|T\geq \tau)$?

Answer 1

Your definition of expected value is at least non-standard, and may be wrong. For a continuous random variable $X$ with probability density function $f$,

$$E(X)=\int_{-\infty}^\infty xf(x)\ dx$$

Given the CDF, the PDF can be easily found by

$$f(x) = \frac{d}{dx}F(x)$$

So in your specific example, we have

$$f(t) = \lambda e^{-\lambda t}, t>0$$

$$E(T) = \int_0^\infty t\lambda e^{-\lambda t}\ dt=\frac{1}{\lambda}$$

Now, we can use Bayes' rule to find a CDF for the random variable with conditional.

$$F_{T\geq\tau}(t)=P(T\leq t|T\geq \tau) = \frac{P(\tau \leq T\leq t)}{P(T\geq \tau)} = \frac{F(t)-F(\tau)}{1-F(\tau)},t\geq\tau$$

$$=\frac{e^{-\lambda \tau}-e^{-\lambda t}}{e^{-\lambda \tau}}$$

And we can differentiate to get a new PDF

$$f_{T\geq\tau}(t) = \frac{\lambda e^{-\lambda t}}{e^{-\lambda \tau}}=\lambda e^{-\lambda (t-\tau)},t>\tau$$

And finally the expected value is

$$E(T) = \int_\tau^\infty t\lambda e^{-\lambda (t-\tau)}\ dt=\int_0^\infty (u+\tau)\lambda e^{-\lambda (u)}\ du=\tau+\frac{1}{\lambda}$$

This gives a very nice result for the exponential distribution shown, owing to its memoryless property. However this process should work (albeit with more difficulty) for other distributions.

Answer 2

By definition,

$$E[T \mid T >\tau] = \frac{\int_{\tau}^\infty t f_T(t) \, dt}{\int_{\tau}^\infty f_T(t) \, dt}$$

where

$$f_T(t)=F'(t)=\lambda e^{-\lambda t}$$

Now compute two integrals:

$$\int_{\tau}^\infty t f_T(t) \, dt=\int_{\tau}^\infty t \lambda e^{-\lambda t} \, dt=\frac{e^{-\lambda \tau}(1+\lambda \tau)}{\lambda}$$

$$\int_{\tau}^\infty f_T(t) \, dt=\int_{\tau}^\infty \lambda e^{-\lambda t} \, dt=e^{-\lambda \tau}$$

Final answer:

$$E[T \mid T >\tau] =\frac{1+\lambda \tau}{\lambda}$$