# Finding the mean of an exponential distribution that starts at time T

Given an exponential distribution $$\lambda e^{-\lambda t}$$ for $$t \geq 0$$, the mean is $$1/\lambda$$. I'm wondering how one would compute the mean if the process starts at a certain time and we start observing it at a time $$T$$ later. Intuitively given the exponential distribution is memoryless I think the mean would be $$1/\lambda$$ from the time after we start observing (so the mean would occur $$1/\lambda$$ time units after T). If this is correct, I'm not sure how to prove this; what would the integral of the expectation would look like? Is it the same as the one we perform if we start observing at t = 0?

## 1 Answer

If you have some probability function $$P(t)$$ for $$t\geq0$$, then the mean value of $$t$$ is given by

$$\left=\int_0^\infty tP(t)\text{d}t$$

Because you are shifting the start time, you need to adjust the normalization constant; you find this by solving

$$A\int_T^\infty \text{e}^{-\lambda t}dt = 1$$

where $$A$$ is your new normalization constant. We find that $$A$$ is given by

$$A = \lambda \text{e}^{\lambda T}$$

so your distribution becomes

$$P(t)=\lambda\text{e}^{\lambda (T+t)}$$

thus the integral you need to compute is

$$\left=\lambda\text{e}^{\lambda T}\int_T^\infty t~\text{e}^{-\lambda t}~\text{d}t=T+\frac{ 1}{\lambda}$$

so your intuition is correct, this is just shifted forward in time by T.

• Given the exponential distribution is memoryless (P( t > r + s | t > r) = P(t > s)), why does it make sense that the mean does not remain the same relative to the point in time where the observation is recorded? May 6 '19 at 3:30