Business Simulation Exam
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PointProcesses.pptxADM 3305 Business Simulation Analytics
Simulation of Stochastic Processes
Point Processes
© 2020 Antoine Sauré
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Today’s Lecture
Point Processes:
Basis definitions
Poisson Process
Simulation of Point Processes
Learning goal:
By the end of this lecture, you will have a good feel for how Point Processes work and an idea of the things you can do with them
Point Processes
Motivation
Modeling the occurrence of a (financial) event
One can observe that there are some events that can repeatedly occur in financial markets. For example, there have been market crashes every few years, from the one in 2008 crisis to Black Monday in 1987 and back to the ones of the 1930s
How likely is that the next crash will occur within three years from now?
In the coming five years, how likely is it to see the first crash occurring within a year from now and the second crash occurring at least a year after?
Motivation
Modeling the losses resulting from the occurrences of events
Every crisis may result in different degrees of losses
In the coming five years, how likely is it to lose more than 1 million due to the crises?
Point Processes
The notion of when a sequence of events happen up to a given time T can be modeled by Point Processes
A simple point process ψ = {tn: n ≥ 1} is a sequence of strictly increasing points 0 < t1 < t2 < …
A counting process N(t) for ψ denotes the number of points falling in the interval (0,t], i.e., N(t) = max{n: tn ≤ t}
If tn are random variables, then ψ is called a random point process. Xn = tn − tn−1 is called the nth inter-arrival time
Renewal Process
A renewal process it is a random point process ψ for which the inter-arrival time {Xn} form an i.i.d. sequence Xn ∼ F
The rate of the renewal process is defined as λ = 1/E(X)
Implementation:
1. t = 0, N = 0
2. Generate an X from F
3. t = t + X. If t > T , then Stop
4. Set N = N + 1 and set tN = t
5. Go back to Step 2
Poisson Process
A Poisson process is a stochastic process N(t) such that
N(0) = 0
N(t) increments by +1 after a time, T, that is randomly distributed according to an exponential distribution with parameter l
In other words, an arrival process is said Poisson if the inter-arrival times are exponentially distributed
Poisson Process
A Poisson process at rate λ is a renewal point process in which the inter-arrival time X follows an exponential distribution with rate λ
Namely, {Xn: n ≥ 1} are i.i.d. with common distribution F(x) = P(X ≤ x) = 1 − e−λx, x ≥ 0; E(X) = 1/λ
Implementation:
1. t = 0, N = 0
2. Generate an X from F (x), i.e., X = −(1/λ)ln(1 - U)
3. t = t + X. If t > T, then Stop
4. Set N = N + 1 and set tN = t
5. Go back to Step 2
Properties of a Poisson Process
A Poisson process is a CTMC
N(t) has a Poisson distribution with parameter lt
E[N(t)] = λt, Var[N(t)] = λt
Properties of a Poisson Process
The counting process for a Poisson process is highly “regular”, namely the increments are stationary and independent
In other words, the increments depends only on the length of the time interval
E[N(s + t) − N(s)] = λt, Var[N(s + t) − N(s)] = λt
An Alternative Implementation
A more efficient (yet less intuitive) way to simulate a Poisson process:
1. Generate N distributed as Poisson with mean λT. If N = 0 Stop
2. Set n = N. Generate n i.i.d. Uniform(0,1) values, U1,...,Un and reset Ui = TUi so that Ui is Uniform (0,T)
3. Place Ui in ascending order to obtain U(1) < U(2) < · · · < U(n)
4. Set ti = U(i)
Properties of Poisson Processes
The superposition of n Poisson processes of parameter li is a Poisson process of parameter Sili
Assume that a Poisson process is split into n processes with probabilities pi. These n processes are independent Poisson process with parameter lpi
Partition of a Poisson Process
Suppose that a Poisson process (e.g., number of financial crises) consists of two independent Poisson processes (e.g., number of US crises and number of non-US crises)
The simulation can be done by simulating a combined one and then partitioning it
Let λ1, λ2 be the rates for the two sub-processes
1. t = 0, t1 = 0, t2 = 0, N1 = 0, N2 = 0. Set λ = λ1 + λ2 and p = λ1/λ
2. Generate U
3. t = t + [−(1/λ)ln(1 - U)]. If t > T then Stop
4. Generate U. If U ≤ p then set N1 =N1 + 1 and set tN1,1 = t; otherwise N2 = N2 + 1 and set tN2,2 = t
5. Go to Step 2
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Compound Poisson Process
In the case that the arrival might come in a batch, we have:
where N(t) is the counting process for the Poisson process. E(X(t)) = E(N(t))E(B) = λtE(B). Suppose that Bn ∼ G
Implementation:
1. t = 0, N = 0, X = 0
2. Generate a U
3. t = t + [−(1/λ)ln(1 - U)]. If t > T then Stop
4. Generate B from the distribution G
5. Set N = N + 1 and X = X + B. Go Step 2
Example: Financial Crises
Suppose that the average inter-arrival time between crises is 3 years and the losses resulting from the crises follow a normal distribution with mean half of a million and standard deviation 2 million
In the coming five years, how likely is it to see the first crash occurring within a year from now and resulting in less than one million loss, and the second crash occurring at least a year after and resulting in more than one million loss?
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