Distribution of arrival times in a Poisson process

I have a Poisson process so its inter-event times are exponentially distributed.

Suppose I fix a time T, (=0, say), and ask what the distribution of the time of arrival of the first event after T is. How do I find this?

Could I reason as follows? Suppose the time of first event after T is t, and the time of the event previous to that is s. Then, s < T < t, and t-s is exponentially distributed.

Now p(t = a) = p(t-T = a-T) = p(t-s > a-T).

But I am worried that the last equality may not be true.

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The time of the event before $T$ is irrelevant. The waiting time $t$ for the first event after time $T$ has exponential distribution with parameter $\lambda$.

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Thanks! That seemed the right answer in view of the memory-less property, but I wasn't sure of the logic. – Thad Jun 27 '12 at 12:39
Yes, that's what memorylessness means: every instant the future begins anew, the atom can die but does not age. – André Nicolas Jun 27 '12 at 13:01
It is also the case that the random time between $T$ and the last previous arrival has that same distribution, so $s-t$ is distributed as the sum of two independent exponentially distributed random variables. A random scheme for choosing some particular interarrival time $X$, that is biased in favor of longer ones, may make $\Pr(X\in\text{any particular set})$ quite different from what it would be if you didn't have that biased way of picking which interarrival time it is. It may fail to be the probability you'd get from an exponential distribution. That is what happens in this instance. – Michael Hardy Jun 27 '12 at 18:57

Your notation is a bit confusing since you're using lower-case letters for random variables and a capital letter for a constant, whereas it's is conventional to do it the other way around.

Since $t$ is a continuous random variable and $a$ is a constant, $\Pr(t=a)=0$, and $\Pr(t-s>a-T)>0$.

Now here we have something paradoxical: $t-s$ is not exponentially distributed, but rather has the distribution of the sum of two independent exponential random variables, and its expected value is twice as big as the expected interarrival time. If you have some random scheme for choosing some particular interarrival time $X$, that is biased in favor of longer ones (or has any of various other possible sorts of biases), then $\Pr(X\in\text{any particular set})$ may be quite different from what it would be if you didn't have that biased way of picking which interarrival time it is. It may fail to be the probability you'd get from an exponential distribution. And that is what happens in this instance. This is the "inspection paradox". When you pick a time $T$, you're more likely to pick a long interval between arrivals than a short one, so on average you get a longer time than the average interarrival time.

But $s-t$ is not exponentially distributed. If you have some random scheme for choosing some particular interarrival time $X$, that is biased in favor of longer ones (or has any of various other possible sorts of biases), then $\Pr(X\in\text{any particular set})$ may be quite different from what it would be if you didn't have that biased way of picking which interarrival time it is. It may fail to be the probability you'd get from an exponential distribution. And that is indeed what happens in this instance. – Michael Hardy Jun 27 '12 at 18:54