Calculating inter-arrival times and arrival times of a Poisson process For a practice exam in stochastic processes I have to answer the following questions.
Let $\{N(t): t\geq 0\}$ be a poisson process with rate $\lambda$. Let $T_n$ denote the n-th inter-arrival time and $S_n$ the arrival time of the n-th event. Let $t>0$. Find:
a) $P(N(t)=1,N(2t)=2,N(3t)=3)$
b) $E[S_{10} |S_4=3]$
c) $E[T_2|T_1<T_2<T_3]$
d) $E[N(t)N(2t)]$
e) $E[N(t) |N(2t)=5]$
I believe I was able to answer the first two but from then I don't know how to answer the question. Also for a I have a bit of a different answer then my teacher but I think he made a mistake.
So this is what I've got at the moment
a) $P(N(t)=1,N(2t)=2,N(3t)=3)$
$=P(N(t)=1,N(2t)-N(t)=1,N(3t)-N(2t)=1)$
$=P(N(t)=1)P(N(2t)-N(t)=1)P(N(3t)-N(2t)=1)$
$=P(N(t)=1)P(N(t)=1)P(N(t)=1)$
$=(e^{-\lambda} )^3$
b)$E[S_{10} |S_4=3]=3+ \frac{6}{\lambda}$
c)$E[T_2|T_1<T_2<T_3]$
$=E[T_1+ \Delta_2|T_1<T_2<T_3]$ , with $\Delta_2=T_2-T_1$
$=E[T_1 |T1=min\{T_1,T_2,T_3\}] + E[\Delta_2|T_1<T_2<T_3]$
But from this point I have no clue on how to proceed. I think I have to condition on the fact that $T_2$ is the minimum of $T_2,T_3$ but I still have to take into account $T_1$. So how should I do this.
For question d  I think I should again condition but I don't know how to do this. Obviously they are independent thus I cannot calculate them apart. Perhaps I can write $N(2t)$ in terms of $N(t)$ and $N(2t-t)$ but I don't know how this would help me.
For question e I wonder if I should split it in every possible combination, thus $N(t)=0,1,2,3,4,5$ and calculate that but again I don't know how to do it properly.
 A: (a) Your answer is right until the last step. Because $N(t)$ has a Poisson distribution with parameter $\lambda t$, it should be:
$$P(N(t)=1)\cdot P(N(t)=1)\cdot P(N(t)=1) = (\lambda t \;e^{-\lambda t})^3.$$
(b) Your answer is right.
(c)
\begin{eqnarray*}
E(T_2\mid T_1\lt T_2\lt T_3) &=& \int_{t_2}{t_2 P(T_2=t_2\mid T_1\lt T_2\lt T_3)\;dt_2} \\
&=& \int_{t_2}{t_2\dfrac{P(T_2=t_2\cap T_1\lt T_2\lt T_3)}{P(T_1\lt T_2\lt T_3)}\;dt_2} \\
&=& 6\int_{t_2=0}^{\infty}{t_2 e^{-\lambda t_2} \int_{t_1=0}^{t_2}{e^{-\lambda t_1} \int_{t_3=t_2}^{\infty}{e^{-\lambda t_3}\;dt_3}\;dt_1}\;dt_2} \\
&& \qquad\qquad\text{since, by symmetry, $P(T_1\lt T_2\lt T_3) =1/6$} \\
&=& \dfrac{6}{\lambda} \int_{t_2=0}^{\infty}{t_2 e^{-\lambda t_2} \int_{t_1=0}^{t_2}{e^{-\lambda t_1-\lambda t_2} \;dt_1}\;dt_2} \\
&=& \dfrac{6}{\lambda^2} \int_{t_2=0}^{\infty}{t_2 \left( e^{-2\lambda t_2} - e^{-3\lambda t_2} \right) \;dt_2} \\
&=& \dfrac{5}{6\lambda^4}.
\end{eqnarray*}
(d)
\begin{eqnarray*}
E(N(t)\cdot N(2t)) &=& E(N(t)\cdot (N(t) + (N(2t)-N(t))) \\
&=& E(N(t)^2) + E(N(t))\cdot E(N(2t)-N(t))\qquad\text{by independence of separate intervals} \\
&=& E(N(t)^2) + E(N(t))^2 \\
&=& Var(N(t)) + 2E(N(t))^2 \\
&=& \lambda t + 2(\lambda t)^2.
\end{eqnarray*}
(e) $E(N(t)\mid N(2t)=5) = \dfrac{5}{2}$ since the rate of arrivals is constant for a homogeneous Poisson process.
