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?
Best Answer
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.)