# Conditional expectation of Poisson process given number of events

Let $$\{N(t), t\geq 0\}$$ be a Poisson process with rate $$\lambda$$, $$S_n$$ the instant of the $$n$$-th arrival and $$T_n$$ the $$n$$-th interarrival time, that is, $$T_n = S_n - S_{n-1}$$, $$n \geq 1$$.

Now consider the following result:

Theorem. Given that $$N(t) = n$$, the $$n$$ arrival times $$S_1, S_2, \dots, S_n$$ have the same distribution as the order statistics corresponding to $$n$$ independent random variables uniformly distributed on the interval $$(0,t)$$.

I would like to know how to calculate $$\mathbb{E}[S_4 | N(1) = 2]$$ using the theorem above.

I have already solved it using the memorylessness property of the exponential distribution, since $$T_i \sim Exponential(\lambda)$$, and it went like: you can call $$S_4 = 1 + T_3 + T_4$$, then \begin{align*} \mathbb{E}[S_4 | N(1) = 2] &= \mathbb{E}[1 + T_3 + T_4] \\ &= 1 + \mathbb{E}[T_3 + T_4] \\ &= 1 + \frac{2}{\lambda}, \end{align*} since $$(T_3 + T_4) \sim Gamma(2, \lambda)$$, so I know the result I should get.

My attempt: we can write $$S_4 = (T_1 + T_2) + T_3 + T_4 = S_2 + T_3 + T_4$$, so it follows \begin{align*} \mathbb{E}[S_4 | N(1) = 2] &= \mathbb{E}[S_2 + T_3 + T_4 | N(1) = 2] \\ &= \mathbb{E}[S_2 | N(1) = 2] + \mathbb{E}[T_3 + T_4] \\ &= \frac{1}{2} + \frac{2}{\lambda}, \end{align*} since increments are independent, $$(S_2|N(1) = 2) = \max \{U_1, U_2\}, U_i \sim Uniform(0,1)$$, and $$(T_3 + T_4) \sim Gamma(2, \lambda)$$.

What have I done incorrectly? I'd get the correct result if I wrote instead $$S_4' = S_1 + S_2 + T_3 + T_4$$, but that's absurd since \begin{align*} S_4' &= S_1 + S_2 + T_3 + T_4 \\ &= (T_1) + (T_1 + T_2) + T_3 + T_4 \\ &= T_1 + (T_1 + T_2 + T_3 + T_4) \\ &= T_1 + S_4. \end{align*}

In addition to that, in these MIT freely available online class notes from a "Discrete Stochastic Processes" course, on page $$92$$, we have equation $$(2.46)$$: \begin{align} \mathbb{E}[S_i|N(t) = n] = \frac{it}{n+1}, \end{align} which in my attempt would yield a completely different result: \begin{align} \mathbb{E}[S_2|N(1) = 2] = \frac{2}{3}. \end{align}

How to proceed?

Your mistake lies in the transition between this statement:

$$\mathbb{E}[S_4 | N(1) = 2] = \mathbb{E}[S_2 + T_3 + T_4 | N(1) = 2]$$

and this statement (the next line):

$$\mathbb{E}[S_4 | N(1) = 2] =\mathbb{E}[S_2 | N(1) = 2] + \mathbb{E}[T_3 + T_4]$$

in which you have disappeared the time gap between $$S_2$$ and $$1$$. You need to make $$\mathbb{E}[T_3 + T_4]$$ conditional upon $$S_2+T_3\geq 1$$, otherwise, the time calculation includes the possibility that the third (and fourth) arrivals will occur before $$t=1$$, but we know they can't, as $$N(1) = 2$$. Once you do that, the memoryless property will come into play, and you'll get your missing time back, as it's the expectation of the time between the second arrival and $$1$$.

Note also that $$\mathbb{E}[S_2 | N(1) = 2] \neq 1/2$$, so the line "$$=1/2 + 2/\lambda$$" is also incorrect. You can see this intuitively by noting that the maximum should have a larger expected value than that of (maximum + minimum)/2, which is evidently $$1/2$$ (the expectation of a $$U(0,1)$$ variate.) As you have in your final line, it equals $$2/3$$. We can calculate this easily enough by noting that the cumulative density function of a $$U(0,1)$$ variate is $$F(u) = u$$, so the c.d.f. of the maximum of two independent $$U(0,1)$$ variates, label it $$z$$, is $$F(z) = z^2$$, so $$f(z) = 2z$$, and work from there. (See the answers to Distribution of extremal values) for the derivation of the c.d.f. of the maximum.)

• No idea what was that comment about. Thank you for the answer! – user71487 Oct 4 '18 at 16:23