1.0 The Poisson Process

Assuming the event to be exactly one asset jump arrival within the interval \(\left[0,T\right]\), how can we determine the probability of this jump:

We consider a Poisson process and we denote the time of the n-th jump by \(T_n\). The sequence {\(T_1,T_2,...T_n\)} is the sequence of arrival(or jump) times. The number of events in a time interval \(\Delta t\) is poisson distributed with mean \(\lambda \Delta t\) for all t \(\ge\) 0, with arrival intensity \(\lambda\):

\[ Prob\left(N_{t+\Delta t} - N_t = n\right) = e^{-\lambda \Delta \frac{T}{2}} \frac{\left(\lambda \Delta \frac{T}{2}\right)^n}{n!}, n = 0,1,... \]

Event A Let Event A be the probability of 1 asset jump in a given time interval \(\left[0,T\right]\), n = 1.

\[ Prob\left(N_{t+\Delta t} - N_t = 1\right) = e^{-\lambda T} \frac{\left(\lambda T\right)^1}{1!} \\ Prob\left(A\right) = \lambda Te^{-\lambda T} \]

The probability of Event A is \(\lambda Te^{-\lambda T}\).

2.0 Properties of the poisson process, and mathematical proof

How can we prove that the probability of the event of exactly one asset jump within the interval \(\left[0,T\right]\) is the same as:

1 asset jump in the interval \(\left[0,\frac{T}{2}\right)\), and no asset jump in the interval \(\left[\frac{T}{2},T\right]\).
No asset jumps in the interval \(\left[0,\frac{T}{2}\right)\), and 1 asset jump in the interval \(\left[\frac{T}{2},T\right]\).

\[ Prob\left(N_{t+\Delta t} - N_t = n\right) = e^{-\lambda \Delta \frac{T}{2}} \frac{\left(\lambda \Delta \frac{T}{2}\right)^n}{n!}, n = 0,1,... \]

Expectation (Mean) is given as: E\(\left[N_{t+\Delta t} - N_t\right] = \lambda \Delta t\)

We know that the Poisson process is:

Independent and Identically distributed
A discrete probability distribution of a sum of independent Bernoilli trials

This means that at any point in time, the Poisson process probabilistically restarts itself.

Event X

Let Event X be the probability of exactly 1 jump from interval \(\left[0,\frac{T}{2}\right)\) and 0 jump from interval \(\left[\frac{T}{2},T\right]\), where \(X_1\) = 1 and \(X_2\) = 0:

First, we find the probability of 1 jump occuring in the interval \(\left[0,\frac{T}{2}\right)\):

\[ Prob\left(N_t = 1\right) = Prob\left(X_1 = 1\right)= e^{-\lambda \frac{T}{2}}\frac{\left(\lambda \frac{T}{2}\right)^n}{n!} = e^{-\lambda \frac{T}{2}} \frac{\lambda T}{2} \]

For interval \(\left[\frac{T}{2},T\right]\), the interval length is the same as the first interval \(\left[0,\frac{T}{2}\right]\). We can find the probability of 0 jump occuring in this interval:

\[ Prob\left(N_t = 0\right) = Prob\left(X_2 = 0\right) = e^{-\lambda \frac{T}{2}} \]

We can find the probability of Event X occuring below:

Where Event X is Prob(\(X_2 = 0\)) conditional on Prob(\(X_1 = 1\)) for interval \(\left[0,\frac{T}{2}\right)\)

\[ Prob\left(X_1 = 1\right) \times Prob\left(X_2 = 0\right) = e^{-\lambda \frac{T}{2}} \frac{\lambda T}{2} \times e^{-\lambda \frac{T}{2}} \\ \\ Prob\left(X\right) = Prob\left(X_2 = 0 | X_1 = 1\right) = e^{-\lambda T} \left(\frac{\lambda T}{2}\right) \]

Event Y

Let Event Y be the probability of 0 jump from interval \(\left[0,\frac{T}{2}\right)\) and exactly 1 jump from interval \(\left[\frac{T}{2},T\right]\):

First, we find the probability of 0 jump occuring in the interval \(\left[0,\frac{T}{2}\right)\):

\[ Prob\left(N_t = 0\right) = Prob\left(Y_1 = 0\right) = e^{-\lambda \frac{T}{2}} \]

For interval \(\left[\frac{T}{2},T\right]\), the interval length is the same as the first interval \(\left[0,\frac{T}{2}\right]\). We can find the probability of 1 jump occuring in this interval:

\[ Prob\left(N_t = 1\right) = Prob\left(Y_2 = 1\right) = e^{-\lambda \frac{T}{2}}\frac{\left(\lambda \frac{T}{2}\right)^n}{n!} = e^{-\lambda \frac{T}{2}} \frac{\lambda T}{2} \]

We can find the probability of Event Y occuring below:

Where Prob(\(Y_2 = 1\)) is conditional on Prob(\(Y_1 = 0\)) for interval \(\left[0,\frac{T}{2}\right)\)

\[ Prob\left(Y_1 = 0\right) \times Prob\left(Y_2 = 1\right) = e^{-\lambda \frac{T}{2}} \times e^{-\lambda \frac{T}{2}} \frac{\lambda T}{2} \\ \\ Prob\left(Y\right) = Prob\left(Y_2 = 1 | Y_1 = 0\right) = e^{-\lambda T} \left(\frac{\lambda T}{2}\right) \]

Probability Comparison

Event A:

\(Prob\left(A\right) = \lambda Te^{-\lambda T}\)

Event X or Event Y:

\(Prob\left(X \cup Y\right) = Prob\left(X\right) + Prob\left(Y\right) = e^{-\lambda T} \left(\frac{\lambda T}{2}\right) + e^{-\lambda T} \left(\frac{\lambda T}{2}\right) = \lambda Te^{-\lambda T}\)

3.0 Programming the mathematical proof

We can assign arbitrary numbers to the variables and conduct below checks:

Set \(\lambda\) = 0.5, T = 1, \(\mu\) = \(\lambda \frac{T}{2}\)

Check that probabilities of both Events X and Y are the same
Calculate probabilities and check against inbuilt Poisson PMF function from Scipy Library

import numpy as np

lamb = 0.5
T = 1
mu = lamb*(T/2)

# Event X probabilities
Prob_X1 = np.exp(-lamb*T/2)*(((lamb*T/2)**1)/1)
Prob_X2 = np.exp(-lamb*T/2)

# Event Y probabilities
Prob_Y1 = np.exp(-lamb*T/2)
Prob_Y2 = np.exp(-lamb*T/2)*(((lamb*T/2)**1)/1)

print('Probability of 1 asset jump in interval [0,T/2]: '+str(Prob_X1))

Probability of 1 asset jump in interval [0,T/2]: 0.19470019576785122

print('Probability of 0 asset jump in interval [T/2,0]: '+str(Prob_X2))

Probability of 0 asset jump in interval [T/2,0]: 0.7788007830714049

print('Probability of Event X: '+str(Prob_X1*Prob_X2))

Probability of Event X: 0.15163266492815836

print('Probability of 1 asset jump in interval [0,T/2]: '+str(Prob_Y1))

Probability of 1 asset jump in interval [0,T/2]: 0.7788007830714049

print('Probability of 0 asset jump in interval [T/2,0]: '+str(Prob_Y2))

Probability of 0 asset jump in interval [T/2,0]: 0.19470019576785122

print('Probability of Event Y: '+str(Prob_Y1*Prob_Y2))

Probability of Event Y: 0.15163266492815836

# Part A probability
Prob_A = np.exp(-lamb*T)*(lamb*T)

print('Probability of Event A: '+str(Prob_A))

Probability of Event A: 0.3032653298563167

print('Probability of Event X or Event Y: '+str(Prob_X1*Prob_X2 + Prob_Y1*Prob_Y2))

Probability of Event X or Event Y: 0.3032653298563167

We can use the Poisson Probability Mass Function from the Scipy library to calculate the individual event probability for each aportioned intervals:

from scipy.stats import poisson

# Event X
PMF_X1 = poisson.pmf(1, mu)
PMF_X2 = poisson.pmf(0, mu)

# Event Y
PMF_Y1 = poisson.pmf(0, mu)
PMF_Y2 = poisson.pmf(1, mu)

print('Probability of 1 asset jump in interval [0,T/2]: '+str(PMF_X1))

Probability of 1 asset jump in interval [0,T/2]: 0.19470019576785122

print('Probability of 0 asset jump in interval [T/2,0]: '+str(PMF_X2))

Probability of 0 asset jump in interval [T/2,0]: 0.7788007830714049

print('Probability of Event X: '+str(PMF_X1*PMF_X2))

Probability of Event X: 0.15163266492815836

print('Probability of 0 asset jump in interval [0,T/2]: '+str(PMF_Y1))

Probability of 0 asset jump in interval [0,T/2]: 0.7788007830714049

print('Probability of 1 asset jump in interval [T/2,0]: '+str(PMF_Y2))

Probability of 1 asset jump in interval [T/2,0]: 0.19470019576785122

print('Probability of Event Y: '+str(PMF_Y1*PMF_Y2))

Probability of Event Y: 0.15163266492815836

Finally:

# Event A vs (Event X or Event Y)
PMF_A = poisson.pmf(1, mu*2)
print('Probability of Event A: '+str(PMF_A))

Probability of Event A: 0.3032653298563167

print('Probability of Event X or Event Y in interval [0,T]: '+str(PMF_X1*PMF_X2 + PMF_Y1*PMF_Y2))

Probability of Event X or Event Y in interval [0,T]: 0.3032653298563167

Hence, we have proved that the event of 1 asset jump in interval \(\left[0,T\right]\) in Part A gives rise to the exact same probability as:

The event of 1 jump in the interval \(\left[0,\frac{T}{2}\right)\) and no jumps in interval \(\left[\frac{T}{2},T\right]\)

The event of 0 asset jump in interval \(\left[0,\frac{T}{2}\right)\) and 1 jump in interval \(\left[\frac{T}{2},T\right]\)