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This question is a conceptual question about understanding the answer to another problem. (original problem here: Average waiting time in a Poisson process)

The original problem asked for an average waiting time $E(S)$ where $S = \inf\{t \geq 0 ; N(t+a) = N(t)\}$. Here, $N(t)$ is a Poisson process and $a$ is a constant such that $a \geq 0$.

Using the first success time $T_1$, we see that $S = 0$ whenever $T_1 > a$. When $T_1 < a$, due to the memoryless property of the Poisson process we see that S is related to itself since we have more or less started over the waiting.

We arrive at:

$$E(S) = E(E(S|T_1);T_1\leq a) = E(T_1 + E(S);T_1 \leq a) = E(T_1;T_1 \leq a) + E(S)P(T_1 \leq a)$$

From there, one can fairly easily derive an expression for $E(S)$.

My question is about one of the steps in the above equation. How do I know, algebraically and intuitively, that $E(S) = E(E(S|T_1);T_1\leq a)$?

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CLL
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  • $$E(S)=E(S;T\le a)+E(S;T> a)\qquad E(S;T>a)=0\qquad E(S;T\le a)=E(E(S|T);T\le a)$$ – Did Mar 10 '15 at 20:35
  • Okay, I nearly understand. This implies that $S = E(S|T)$ when $T \leq a$ - perhaps this is a gap in my understanding, but I can't figure out why $S = E(S|T)$ when $T \leq a$ is true. Could you expand on that? – CLL Mar 10 '15 at 20:42
  • Nobody said that S=E(S|T) on A=[T≤a] but that E(S;A)=E(E(S|T);A), which follows from the very definition of E(S|T). Remember this definition? – Did Mar 11 '15 at 00:07
  • Yes, that makes sense - thank you! – CLL Mar 11 '15 at 01:41

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