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.)

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

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