Given: $\Lambda$ is a poisson process on $\mathbb{R}$. Each point $x_i$ in $\Lambda$ underoges some markging process' resulting in $N=\{(x_1,U_1), (x_2,U_2) \dots (x_n, U_n)\}$ where $U_i$ are iid generated from some process witg density $\mu$
This essentially is a generalisation of the Colouring theorem. Let the joint 'marked' distribution $\Lambda'$ be denoted by: $f(x,y)$ where $x\ \in \ \Lambda$ and $y\ \in U$ and $(x,y) \in \Lambda'$ which is in $\mathbb{R}^2$ Consider $F = \sum_{x_i\ \in\ \Lambda}f(x_i,y_i)$
Fact 1: $\Lambda' = \{(X,U)|X\in \Lambda\}$ is an independent process. (This is only true conditional on $X$)
Thus, $E[e^{-sF}|X \in \Lambda] = \Pi_{X_i \in \Lambda} f(X_i, U_i) = \Pi_{X_i \in \Lambda}\int_{U}e^{-sf(X_i,U_i)}p(X_i,U_i)dU_i$
Fact 2: In any region $\Lambda_i$, the number of points $N(\Lambda_i) \sim Poisson(\mu_i)$ where $\mu_i = \int_{\Lambda_i} \mu(x)dx$ and hence, consider:
$F = \sum_{i=1}^k f(X_i) = \sum_{i=1}f_iN_i^k$ being $k$ disjoint sets where $f$ is any measurable function. then
$$ \begin{align} E[e^{sF}] &= \Pi_{i=1}^k E[e^{sf_iN_i}]\\ &= \Pi(e^{\lambda_i(e^{fs}-1)})\ \text{where} \lambda_i = \int_{A_i}\lambda(x)dx\\ &= e^{\sum_{i=1}^k \int(e^{fs}-1)\lambda(x)dx}\\ &= e^{\sum_{i=1}^k \int(e^{fs}-1)\lambda(x)dx}\\ &= e^{ \int_{\Lambda}(e^{fs}-1)\lambda(x)dx}\\ \end{align} $$or $E[e^{-sF}] = exp(-\int_{\Lambda}(1-e^{fs})\lambda(x)dx)$
Now, consider $E[e^{-sF'}]$ where $F'$ is the sum of independent random variable $f(X_i, U_i)$ Conditional on $\Lambda$:
$E[e^{-sF'}|\Lambda] = \Pi_{X_i\in \Lambda}E[e^{-sf(X_i,U_i)}]$
Consider the measurable function $f'(x) = -log(\int_U e^{-f(X,U)}p(X,U)du)$, then:
[Couldn't take it further from here, but the idea should be to define $F^* = \sum f(X,U)$. I was not able to come up with a generating function for this. The idea was to show that it has generating function that for a poisson]
Aliter
[Source Grimmett and Stirzker, 3rd Edition 6.13.5]
Consider $f: R \longrightarrow R^2$ denoting the distribution of points in $\Lambda'$ then for any set $B\subset R^2$, the number of points of f(\Lambda') in B is given by $N_B = |\Lambda' \cup f^{-1}B| \sim Poisson(B)$ Given project disjoint sets $B_i$ in $R^2$, there pre-images in $R$ are also disjoint and
$\Lambda(A) \sim Poisson(\int_A \lambda(x)dx)$
$\Lambda'(B) \sim \Lambda(f^{-1}B) = Poisson(\int_{f^{-1}B}\lambda(x)dx) = Poisson(\int_B \lambda(x)dx\mu(y)dy$
Given Raindrops fall as a $PPP(\lambda drops/cm^2)$ on $R^2 \times [0, \infty)$ and each drop scatters as in a radius of circle $r \sim exp(1/cm)$. To find: Probability density of first drop touching the origin.
$\lambda=1$
Define $U_i$ to be iid $Bernoulli(p(x_k,y_k)$ given by: $$ U_k = \begin{cases} 1 & if \sqrt{x_k^2+y_k^2} \leq r_k\\ 0 & otherwise \end{cases} $$
By coloring theorem $\implies$ $\Lambda'=\{(r_k,x_k,y_k\}\sim PPP(\lambda p(x_k,y_k))$ For the drop to splash the origin with radius '$r$':
Consider $\Lambda^1 = \{(t_i,x_i,y_i,r_i) \in \Lambda: U_k=1\}$ $\tau = min\{t_k: (t_k,x_k,y_k,r_k) \in \Lambda'\}$
Consider $P(t < \tau, R<r)$ = P[no points in $[0,r] \times [0, t] \times R^2$] = $P[\Lambda^1([0,r] \times [0, t] \times R^2)=0]$
$P[\Lambda^1([0,r] \times [0, 1] \times R^2)=0]=exp(-\int_{0}^r \int_{0}^{1} \int_{R^2}\lambda p(x,y)dxdydtdr) =exp(-\lambda 2\pi tr)$
Now, $P(R<r)=\int P(t,R)dt = exp(-2\pi\lambda r)$
Thus, $R \sim exponential({2\pi\lambda})$
$ER=\int_0^{\infty} re^{-2\pi\lambda r}dr=\frac{1}{2\pi} = 0.15$
In [1]:
### Simulation
%matplotlib inline
from __future__ import division
import numpy as np
import matplotlib.pyplot as plt
np.random.seed(1)
import math
N=1000
s=0
def R(x,y):
return math.sqrt(x*x+y*y)
for i in range(N):
r=-100
y=0
x=0
while R(x,y)>r:
S=np.random.uniform(size=2)
x=S[0]
y=S[1]
r=np.random.exponential(1)
s+=r
print 'Average radius: {}'.format(s/N)
The simulation results do not seem to be close to the expected results of 0.15
In order to simulate the continuous time MC, we make use of the instantaneous rate matrix $Q$ given by: $Q=\lambda(P-I)$ where $\lambda=10^6$ The coninuous time MC is approximated to occur in discreted time steps of $10^{-6}$ seconds
$Q_{aa}$ = Total jump rate out of state a
$Q_{ab}$ = Jum rate from $a \longrightarrow b$
Due to the original transition matrix $P$ having extremely small entries, most of the time is spent in the same state. By approximating the holding time to be poisson, we arrive at the $(e^{tQ})_{ab}$ approximation for $P(Y_t=b|Y_0=a)$
$\tau_\dagger = inf\{t \geq 0: Y_t=\dagger\}$ i.e $\tag_\dagger$ is the hitting time and $u(a) = E[\tau_\dagger|Y_0=a]$ defines the mean hitting time.
$u(\dagger)=0$
For $a \neq \dagger$
$u(a) = \text{Hold time in state a} + \sum_b \text{(fractional jump rate from $a$ to $b$)} \times u(b)$
Alternatively:
$u(a) = \frac{1}{-Q_{aa}} + \sum_{b \neq a}(\frac{Q_{ab}}{-Q_{aa}})u(b)$ $\implies$ $-Q_{aa}u(a) = 1 + \sum_{b \neq a} Q_{ab}u(b)$
We thus solve for:
$Q\vec{u}=-\vec{1}$
In [2]:
k_a=2e-6
k_b=2e-6
k_p=5e-6
k_d=1e-5
ll = 1e6
P = np.matrix([[1-k_a-k_b, k_a ,k_b, 0, 0, 0],
[k_a, 1-k_a-k_b, 0, k_b, 0, 0],
[k_b, 0, 1-k_a-k_b, k_a, 0, 0],
[0, k_b, k_a, 1-k_a-k_b-k_p, k_p, 0],
[0, 0, 0, 0, 1-k_d, k_d],
[0, 0, 0, k_d, 0, 1-k_d]], dtype=np.float64)
Q = ll*(P-np.eye(6))
print(Q)
In [3]:
Qd= Q[:-1,:-1]
Qi = np.linalg.pinv(Qd)
u=(np.sum(Qi, axis=1)*-1)
u=u.tolist()
In [4]:
def h(x):
s=0
ht=0
cc=0
for i in range(1,10000):
new_state=x
while new_state!=5:
old_state=new_state
probs = Q[old_state,:]/-Q[old_state,old_state]
probs=probs.tolist()[0]
probs[old_state]=0
qaa = np.random.exponential(-1/Q[old_state,old_state])
z=np.random.choice(6, 1, p=probs)
new_state = z[0] #states[z[0]]
s+=qaa
return s/10000
In [5]:
print('From calculation: {}\t From Simulation: {}'.format(u[0][0],h(0)))
In [6]:
print('From calculation: {}\t From Simulation: {}'.format(u[1][0],h(1)))
In [7]:
print('From calculation: {}\t From Simulation: {}'.format(u[2][0],h(2)))
In [8]:
print('From calculation: {}\t From Simulation: {}'.format(u[3][0],h(3)))
In [9]:
print('From calculation: {}\t From Simulation: {}'.format(u[4][0],h(4)))