Let $N(t)$ be a Poisson process, compute $P\{N(s)=1,N(t)=2\}$ for any $0\leq s<t$

Question

I got the answer as $\lambda^2e^{-\lambda t}s(t-s)$ using the properties like independent increment and stationary increments. But I can't seem to understand the steps in the solution of the book.

For reference, the definition followed in the book is, let $\eta_1,\eta_2,\ldots$ be a sequence of independent random variables,each having the same exponential distribution of rate $\lambda$.We put

$\xi_n=\eta_1+\ldots+\eta_n$

$N(t)$, where $t\geq0$, is a Poisson process if $N(t)=max\{n:t\geq\xi_n\}$

Solution: Using the fact that $\eta_1,\eta_2,\ldots$ are independent and exponentially distributed, we obtain

$\begin{array} {lcl} P\{N(s)=1,N(t)=2\} & = & P\{\xi_1\leq s<\xi_2\leq t<\xi_3\} \\ & = & P\{\eta_1\leq s<\eta_1+\eta_2\leq t<\eta_1+\eta_2+\eta_3\}\\ &=& \displaystyle\int_{0}^{s}P\{s<u+\eta_2\leq t<u+\eta_2+\eta_3\}\lambda e^{-\lambda u}du\\ &=& \displaystyle\int_{0}^{s}\left(\displaystyle\int_{s-u}^{t-u}P\{t<u+v+\eta_3\}\lambda e^{-\lambda v}dv \right)\lambda e^{-\lambda u}du\\ &=& \displaystyle\int_{0}^{s}\left(\displaystyle\int_{s-u}^{t-u}e^{-\lambda (t-u-v)}\lambda e^{-\lambda v}dv \right)\lambda e^{-\lambda u}du\\ &=& \lambda^2e^{-\lambda t}s(t-s)\end{array}$

I understood the first two lines.For the next step, I thought maybe they used the PDF of the exponential distribution of $\eta_1$ and took the PDF as function of $u$. So the replacing of $\eta_1$ by $u$. But can we just do that? Didn't it become a probability of an event by the second line? So how can we consider a single inequality in it?

I'm trying to self learn Stochastic Process.So my knowledge isn't the best in this. Any help would be appreciated.

Btw I solved the problem just like https://math.stackexchange.com/questions/4356860/let-nt-be-a-poisson-process-calculate-pns-neq-nt-and-pns-0-nt

Answer 1

It involves the knowledge of multiple integrals.

Define $D=\{(\eta_1,\eta_2,\eta_3): \eta_1\leq s<\eta_1+\eta_2<t<\eta_1+\eta_2+\eta_3\}$,

and $A(u)=\{(\eta_2,\eta_3): s<u+\eta_2<t<u+\eta_2+\eta_3\}$, where $u \leq s$.

Given that $\eta_1,\eta_2,...$ are mutually independent, the PDF $f(\eta_1,\eta_2,\eta_3)=P(\eta_1)P(\eta_2)P(\eta_3)$.

We have

$$\begin{aligned}P(\eta_1<s<\eta_1+\eta_2<t<\eta_1+\eta_2+\eta_3)&=\iiint_Df(\eta_1,\eta_2,\eta_3) d\eta_1d\eta_2d\eta_3\\ &=\iiint_DP(\eta_1)P(\eta_2)P(\eta_3) d\eta_1d\eta_2d\eta_3\\ &=\iiint_DP(u)P(\eta_2)P(\eta_3) dud\eta_2d\eta_3\\ &=\int_0^sP(u)(\iint_{A(u)}P(\eta_2)P(\eta_3)d\eta_2d\eta_3)du\\ &=\int_0^s\lambda e^{-\lambda u}(\iint_{A(u)}P(\eta_2)P(\eta_3)d\eta_2d\eta_3)du\\ &=\int_0^s\lambda e^{-\lambda u}P(s<u+\eta_2\leq t<u+\eta_2+\eta_3) du \end{aligned}$$

Above describes the deduction process from the second line to the third line. The same reasoning applies to deducing the result from the third line.