I'm currently self-learning stochastic process and trying to understand the theorem related sampling Poisson process from this resource
I'm trying to work out the steps of the following theorem from the resource
Theorem 4.2. Consider a homogeneous Poisson process N¯(t) with intensity function λ¯. Let t¯1,t¯2, . . . ,t¯N¯(T) be the points of the process in the interval (0, T]. Suppose that for 0 ≤ t ≤ T, 0 ≤ λ(t) ≤ λ¯. For k = 1, 2, . . . , N¯(T), delete the point t¯k with probability 1 − λ(t¯k)/λ¯; then the remaining points form a point process N(t)
One of the steps in the proof is as follows -
$p_{n} = \int_{a}^{b} \int_{t1}^{b} ... \int_{t_{n-1}}^{b} \prod_{k=1}^{n} \left ( 1 - \frac{\lambda (t_{k})}{\bar{\lambda }} \right ) dt_{1} dt_{2}...dt_{n}$
$ = \frac{e-^{\bar{\lambda}(b-a)}}{n!}\int_{a}^{b} \int_{a}^{b}...\int_{a}^{b} \prod_{k=1}^{n}(\bar{\lambda } - \lambda (t_{k})) ds_{1} ds_{2}...ds_{n} $
where $a \leqslant t1 < ... < t_{n} \leq b$ and unordered $\left \{ s_{1},...,s_{n} \right\} = \left \{ t_{1},...,t_{n} \right\}$ and $s_{k}$ takes value in $\left [ a,b \right ]$
Can someone help me understand the change of variable from t to s, please?
My intuitive understanding is that since the limit of t is hard to integrate, we construct a variable s which spans the whole space [a,b]. Since there are n! ways of constructing s from t we divide by n!
Yuanda Chen's paper contains a few typos, but the biggest issue here is that you haven't quoted the statement from Yuanda Chen's paper incorrectly. The correct claim is as follows: let $$p_n=\int_a^b{\int_{t_1}^b{\dots\int_{t_{n-1}}^b{\overline{\lambda}^ne^{-\overline{\lambda}(b-a)}\prod_{k=1}^n{\left(1-\frac{\lambda(t_k)}{\overline{\lambda}}\right)}\,dt_n}\,\dots\,dt_2}\,dt_1}\text{;}$$ then $$p_n=\frac{e^{-\overline{\lambda}(b-a)}}{n!}\int_a^b{\int_a^b{\dots\int_a^b{\prod_{k=1}^n{\left(\overline{\lambda}-\lambda(s_k)\right)}\,ds_n}\,\dots\,ds_2}\,ds_1}$$
With this correction, we can see that most of the change is due to Chen rearranging constants: he is factoring $e^{-\overline{\lambda}(b-a)}$ out of the integrals, and canceling the factors of $\frac{1}{\overline{\lambda}}$ in the denominators of the product. We thus have, without a change of variable, $$p_n=e^{-\overline{\lambda}(b-a)}\int_a^b{\int_{t_1}^b{\dots\int_{t_{n-1}}^b{\prod_{k=1}^n{\left(\overline{\lambda}-\lambda(t_k)\right)}\,dt_n}\,\dots\,dt_2}\,dt_1}$$
You are correct that the remaining factor of $\frac{1}{n!}$ comes from the ways to make $\{s_k\}_k$ from the $\{t_k\}_k$. The easiest way I see to write it is to note that we never use the ordering of $\{t_k\}_k$ in the latter expression for $p_n$. So let $S_n$ act on $[a,b]^n$ in the natural way; then, for all $\sigma\in S_n$, $\sigma$ has constant Jacobian $1$. Thus \begin{align*} p_n&=e^{-\overline{\lambda}(b-a)}\int_a^b{\int_{t_{\sigma(1)}}^b{\dots\int_{t_{\sigma(n-1)}}^b{\prod_{k=1}^n{\left(\overline{\lambda}-\lambda\left(t_{\sigma(k)}\right)\right)}\,dt_{\sigma(n)}}\,\dots\,dt_{\sigma(2)}}\,dt_{\sigma(1)}} \end{align*} As the $S_n$-orbit of the space of all $\{t_k\}_k$ (the so-called Weyl chamber) is precisely $[a,b]^n$, we obtain \begin{align*} n!\cdot p_n&=\|S_n\|p_n \\ &=e^{-\overline{\lambda}(b-a)}\sum_{\sigma\in S_n}{\int_a^b{\int_{t_{\sigma(1)}}^b{\dots\int_{t_{\sigma(n-1)}}^b{\prod_{k=1}^n{\left(\overline{\lambda}-\lambda\left(t_{\sigma(k)}\right)\right)}\,dt_{\sigma(n)}}\,\dots\,dt_{\sigma(2)}}\,dt_{\sigma(1)}}} \\ &=e^{-\overline{\lambda}(b-a)}\int_a^b{\int_a^b{\dots\int_a^b{\prod_{k=1}^n{\left(\overline{\lambda}-\lambda(s_k)\right)}\,ds_n}\,\dots\,ds_2}\,ds_1} \end{align*}
Dividing by $n!$ gives the claimed result.