PoissonProcess - Maple Help

Finance

 PoissonProcess
 create new Poisson process

 Calling Sequence PoissonProcess(lambda) PoissonProcess(lambda, X)

Parameters

 lambda - algebraic expression; intensity parameter X - algebraic expression; jump size distribution

Description

 • A Poisson process with intensity parameter $0<\mathrm{\lambda }\left(t\right)$, where $\mathrm{\lambda }\left(t\right)$ is a deterministic function of time, is a stochastic process $N$ with independent increments such that $N\left(0\right)=0$ and

$\mathrm{Pr}\left(N\left(t+h\right)-N\left(t\right)=1\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}N\left(t\right)\right)=\mathrm{lambda}\left(t\right)h+o\left(h\right)$

 for all $0\le t$. If the intensity parameter $\mathrm{\lambda }\left(t\right)$ itself is stochastic, the corresponding process is called a doubly stochastic Poisson process or Cox process.
 • A compound Poisson process is a stochastic process $J\left(t\right)$ of the form $J\left(t\right)={\sum }_{i=1}^{N\left(t\right)}{Y}_{i}$, where $N\left(t\right)$ is a standard Poisson process and ${Y}_{i}$ are independent and identically distributed random variables. A compound Cox process is defined in a similar way.
 • The parameter lambda is the intensity. It can be constant or time-dependent. It can also be a function of other stochastic variables, in which case the so-called doubly stochastic Poisson process (or Cox process) will be created.
 • The parameter X is the jump size distribution. The value of X can be a distribution, a random variable or any algebraic expression involving random variables.
 • If called with one parameter, the PoissonProcess command creates a standard Poisson or Cox process with the specified intensity parameter.

Examples

 > $\mathrm{with}\left(\mathrm{Finance}\right):$
 > $J≔\mathrm{PoissonProcess}\left(1.0\right):$
 > $\mathrm{PathPlot}\left(J\left(t\right),t=0..3,\mathrm{timesteps}=50,\mathrm{replications}=20,\mathrm{thickness}=3,\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true},\mathrm{markers}=\mathrm{false}\right)$

Create a subordinated Wiener process with $J$ as a subordinator.

 > $W≔\mathrm{WienerProcess}\left(J\right):$
 > $\mathrm{PathPlot}\left(W\left(t\right),t=0..3,\mathrm{timesteps}=20,\mathrm{replications}=10,\mathrm{markers}=\mathrm{false},\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{thickness}=3,\mathrm{gridlines}=\mathrm{true},\mathrm{axes}=\mathrm{BOXED}\right)$

Next define a compound Poisson process.

 > $Y≔\mathrm{Statistics}\left[\mathrm{RandomVariable}\right]\left(\mathrm{Normal}\left(0.3,0.5\right)\right):$
 > $\mathrm{\lambda }≔0.5$
 ${\mathrm{\lambda }}{≔}{0.5}$ (1)
 > $X≔\mathrm{PoissonProcess}\left(\mathrm{\lambda },Y\right):$
 > $\mathrm{PathPlot}\left(X\left(t\right),t=0..3,\mathrm{timesteps}=20,\mathrm{replications}=10,\mathrm{markers}=\mathrm{false},\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{thickness}=3,\mathrm{gridlines}=\mathrm{true},\mathrm{axes}=\mathrm{BOXED}\right)$

Compute the expected value of $X\left(T\right)$ for $T=3$ and verify that this is approximately equal to $\mathrm{\lambda }T$ times the expected value of $Y$.

 > $T≔3$
 ${T}{≔}{3}$ (2)
 > $\mathrm{ExpectedValue}\left(X\left(T\right),\mathrm{replications}={10}^{4},\mathrm{timesteps}=100\right)$
 $\left[{\mathrm{value}}{=}{0.4435146732}{,}{\mathrm{standarderror}}{=}{0.007164725012}\right]$ (3)
 > $\mathrm{\lambda }T\mathrm{Statistics}\left[\mathrm{ExpectedValue}\right]\left(Y\right)$
 ${0.45}$ (4)

Here is an example of a doubly stochastic Poisson process for which the intensity parameter evolves as a square-root diffusion.

 > $\mathrm{\kappa }≔0.354201$
 ${\mathrm{\kappa }}{≔}{0.354201}$ (5)
 > $\mathrm{\mu }≔1.21853$
 ${\mathrm{\mu }}{≔}{1.21853}$ (6)
 > $\mathrm{\nu }≔0.538186$
 ${\mathrm{\nu }}{≔}{0.538186}$ (7)
 > $\mathrm{y0}≔1.81$
 ${\mathrm{y0}}{≔}{1.81}$ (8)
 > $y≔\mathrm{SquareRootDiffusion}\left(\mathrm{y0},\mathrm{\kappa },\mathrm{\mu },\mathrm{\nu }\right):$
 > $J≔\mathrm{PoissonProcess}\left(y\left(t\right)\right):$
 > $\mathrm{PathPlot}\left(y\left(t\right),t=0..3,\mathrm{timesteps}=100,\mathrm{replications}=10,\mathrm{thickness}=3,\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)$
 > $\mathrm{PathPlot}\left(J\left(t\right),t=0..3,\mathrm{timesteps}=100,\mathrm{replications}=10,\mathrm{thickness}=3,\mathrm{color}=\mathrm{red}..\mathrm{blue},\mathrm{axes}=\mathrm{BOXED},\mathrm{gridlines}=\mathrm{true}\right)$

References

 Glasserman, P., Monte Carlo Methods in Financial Engineering. New York: Springer-Verlag, 2004.

Compatibility

 • The Finance[PoissonProcess] command was introduced in Maple 15.