Thinning of Poisson processes

hbghlyj

A8LectureNotes_MT22_24Sep2022.pdf Theorem 7.6 (Thinning of a Poisson process).
If $A∼$ Poisson$(μ)$, and, conditional on $A = n$, $B∼$ Binomial$(n, p)$, then in fact $B∼$ Poisson$(pμ)$. This fact was proved in two different ways in the Prelims course. For example, it can be done using generating functions: let $B = X_1 + X_2 + \dotsb + X_A$ where $X_i$ are i.i.d. Bernoulli random variables; then $G_B (s) = G_A(G_X (s))$.
計算如下
$A∼$ Poisson$(μ)$, $G_A(s)=e^{(s-1)μ}$
$X∼$ Bernoulli$(p)$, $G_X(s)=\color{magenta}{1-p+ps}$
$B=X_1 + X_2 + \dotsb + X_A$
$$G_B (s) = G_A(G_X (s))=e^{(\color{magenta}{1-p+ps}-1)μ}=e^{(s-1)pμ}$$
所以$B∼$ Poisson$(pμ)$

hbghlyj

Theorem 7.6 (Thinning of a Poisson process).
But if $A \sim \operatorname{Poisson}(\mu)$, and, conditional on $A=n, B \sim \operatorname{Binomial}(n, p)$, then in fact $B \sim \operatorname{Poisson}(p \mu)$. This fact was proved in two different ways in the Prelims course. For example, it can be done using generating functions: let $B=X_1+X_2+\cdots+X_A$ where $X_i$ are i.i.d. Bernoulli random variables; then $G_B(s)=G_A\left(G_X(s)\right)$. Alternatively, by direct calculation:
$$
\begin{aligned}
\mathbb{P}(B=k)= & \sum_{n \geq 0} \mathbb{P}(B=k \mid A=n) \mathbb{P}(A=n) \\
= & \sum_{n \geq k} \frac{e^{-\mu} \mu^n}{n !}\left(\begin{array}{l}
n \\
k
\end{array}\right) p^k(1-p)^{n-k} \\
& \vdots \\
= & \frac{e^{-p \mu}(p \mu)^k}{k !} .
\end{aligned}
$$
Hence indeed we have here that $M(s, t] \sim \operatorname{Poisson}(p \lambda(t-s))$. So indeed property (iii) holds as desired, and $M$ is a Poisson process of rate $p \lambda$.

hbghlyj

$B=X_1 + X_2 + \dotsb + X_A$
$\Rightarrow G_B (s) = G_A(G_X (s))$

这里使用了page 22的结论:
We can also treat the sum of a random number of random variables. Let $X_1, X_2, \ldots$ be i.i.d. random variables (taking non-negative integer values), and let $N$ be another random variable, also taking non-negative integer values, independent of the sequence $X_i$. Define $S=X_1+\cdots+X_N$. Then we can write the generating function of $S$ in terms of the common generating function of the $X_i$ and the generating function of $N$ by the law of total probability for expectations
\begin{aligned}
G_S(z) &=\mathbb{E}\left(z^S\right) \\
&=\mathbb{E}\left(z^{X_1+\cdots+X_N}\right) \\
&=\sum_{n=0}^{\infty} \mathbb{E}\left(z^{X_1+\cdots+X_n} \mid N=n\right) \mathbb{P}(N=n) \\
&=\sum_{n=0}^{\infty} \mathbb{E}\left(z^{X_1+\cdots+X_n}\right) \mathbb{P}(N=n) \quad \text { (by independence) } \\
&=\sum_{n=0}^{\infty}\left(\mathbb{E}\left(z^{X_1}\right)\right)^n \mathbb{P}(N=n) \quad \text { (by independence) } \\
&=\mathbb{E}\left(\left(G_X(z)\right)^N\right) \\
&=G_N\left(G_X(z)\right)
\end{aligned}

Czhang271828

其实统计意义很简单. 例如一天内进入超市的人数呈 Poisson 分布(记均值=方差=$\lambda$), 再假定城市中男性占人口比例的 $p$, 那么一天内进入超市的男性顾客数量也呈 Poisson 分布(相应地, 均值=方差=$p\lambda$). 此处忘却掉女性顾客即可.

hbghlyj

MSE
独立随机变量 $X,Y$ 服从参数为 $a,b$ 的泊松分布.
\begin{eqnarray*}
  \Pr(X+Y=k) &=& \sum_{p=0}^\infty \sum_{q=0}^\infty \mathrm{e}^{-a-b} \frac{a^p}{p!} \frac{b^q}{q!} \mathbf{1}_{p+q=k} = \mathrm{e}^{-a-b}\sum_{p=0}^{k}  \frac{a^p}{p!} \frac{b^{k-p}}{(k-p)!} \\ &=& \mathrm{e}^{-a-b} \frac{1}{k!} \cdot \sum_{p=0}^{k}  \binom{k}{p} a^p  b^{k-p} = \mathrm{e}^{-a-b} \frac{1}{k!} \left(a+b\right)^k
\end{eqnarray*}(最后一个等式使用二项式定理) 故 $X+Y∼\text{Poi}(a+b)$.

---

\begin{eqnarray*}
    \Pr\left(X=k \mid X+Y=n\right) &=& \frac{\Pr\left(X=k, Y=n-k\right)}{\Pr(X+Y=n)} \\ &=& \frac{\exp(-a) a^k/k! \cdot \exp(-b) b^{n-k}/(n-k)!}{\exp(-a-b) \cdot (a+b)^n/n!} \\
&=& \binom{n}{k} \left(\frac{a}{a+b}\right)^k \left(1-\frac{a}{a+b}\right)^{n-k}
\end{eqnarray*}故 $(X \mid X+Y=n )\sim \operatorname{Bin}\left(n, \frac{a}{a+b} \right)$.

---

$X-Y$的分布称为Skellam distribution.
\begin{eqnarray*}
  \Pr(X-Y=n) &=& \sum_{p=0}^\infty \sum_{q=0}^\infty \mathrm{e}^{-a-b} \frac{a^p}{p!} \frac{b^q}{q!} \mathbf{1}_{p-q=n} = \mathrm{e}^{-a-b}\sum_{q=\max(0,-n)}^{\infty}  \frac{a^{n+q}}{(n+q)!} \frac{b^{q}}{q!} \\ &=& \begin{cases} a^n \left(a b\right)^{-n/2} I_n\left(2 \sqrt{a b}\right) & n \geqslant 0 \cr b^{-n} \left(a b\right)^{n/2} I_{-n}\left(2 \sqrt{a b}\right) & n < 0\end{cases} = \left(\frac{a}{b}\right)^{n/2} I_{\vert n\vert} \left(2 \sqrt{a b}\right)
\end{eqnarray*}where $I_n(x)$ is the modified Bessel function of the first kind, 如下图
import graph;
import gsl;
unitsize(1cm);
real xmin = 0;
real xmax = 3;

real f0(real x) { return I(0, x); }
real f1(real x) { return I(1, x); }

path g0 = graph(f0, xmin, xmax);
path g1 = graph(f1, xmin, xmax);

xaxis(Label("$x$",align=2E),Ticks("%f",NoZero,step=1),xmax=3.5);
yaxis(Label("$y$",align=2N),Ticks("%f",NoZero,step=1));

draw(g0, blue);
draw(g1, red);

label("$I_0(x)$", (xmax, f0(xmax)), E, blue);
label("$I_1(x)$", (xmax, f1(xmax)), E, red);

hbghlyj

SB3.1 Applied Probability
Theorem 2.12 (Thinning of Poisson processes). Let $Z$ be a Poisson process of rate $\lambda$ and let $p \in[0,1]$. Mark each point of the process independently with probability $p$. Let $X$ be the counting process of marked points, and let $Y$ be the counting process of unmarked points. Then $X$ is a Poisson process of rate $\lambda p, Y$ is a Poisson process of rate $\lambda(1-p)$ and $X$ and $Y$ are independent.

hbghlyj

Czhang271828 发表于 2023-4-14 11:26
其实统计意义很简单. 例如一天内进入超市的人数呈 Poisson 分布(记均值=方差=$\lambda$), 再假定城市中男性 ...

此外，男顾客数和女顾客数是独立的随机变量

A8LectureNotes_MT22_24Sep2022.pdf page 70
Remark 7.7. In fact, it is not too hard to prove something stronger. If $L$ is the process of unmarked points, then $L$ is a Poisson process of rate $(1 - p)λ$, and the processes $L$ and $M$ are independent.

以下网站有证明：
14.5: Thinning and Superpositon
For \(t \ge 0\), \(M_t\) has the Poisson distribution with parameter \(p r\), \(W_t\) has the Poisson distribution with parameter \((1 - p) r\), and \(M_t\) and \(W_t\) are independent.
Proof. The important observation is that the conditional distribution of \(M_t\) given \(N_t = n\) is binomial with parameters \(n\) and \(p\). Thus for \(j \inN\) and \(k \inN\),
\begin{align*} \Pr(M_t = j, W_t = k) & = \Pr(M_t = j, N_t = j + k) = \Pr(N_t = j + k) \Pr(M_t = j \mid N_t = j + k) \\ & = e^{-r t} \frac{(r t)^{j + k}}{(j + k)!} \frac{(j + k)!}{j! k!} p^j (1 - p)^k \\ & = e^{-p r t} \frac{(p r t)^j}{j!} e^{-(1 - p) r t} \frac{\left[(1 - p) r t\right]^k}{k!}=\Pr(M_t = j)\Pr(W_t = k)  \end{align*}