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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?

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

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

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

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

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

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  • $\begingroup$ 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? $\endgroup$
    – Yandle
    May 6 '19 at 3:30

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