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Grzelak Poisson Paths

Background

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Grzelak Poisson Paths

Educational script demonstrating grzelak poisson paths concepts.

Code

```python """

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Grzelak Poisson Paths

Educational script demonstrating grzelak poisson paths concepts. """

This file was placed here in error — Poisson process content belongs in ch07.

See ch07/codes/grzelak_poisson_process_paths.py for the correct placement.

This file can be safely deleted.

if name == "main": pass ```

Exercises

Exercise 1. Let \(N(t)\) be a Poisson process with intensity \(\lambda = 3\). Compute \(E[N(2)]\) and \(\mathrm{Var}(N(2))\).

Solution to Exercise 1

For a Poisson process with intensity \(\lambda\), \(N(t) \sim \mathrm{Poisson}(\lambda t)\). Therefore

\[ E[N(2)] = \lambda \cdot 2 = 3 \cdot 2 = 6 \]
\[ \mathrm{Var}(N(2)) = \lambda \cdot 2 = 6 \]

For a Poisson distribution, the mean and variance are both equal to \(\lambda t\).

Exercise 2. Define the compensated Poisson process \(\tilde{N}(t) = N(t) - \lambda t\). Show that \(\tilde{N}(t)\) is a martingale.

Solution to Exercise 2

We need to show \(E[\tilde{N}(t) \mid \mathcal{F}_s] = \tilde{N}(s)\) for \(s \le t\).

\[ E[\tilde{N}(t) \mid \mathcal{F}_s] = E[N(t) - \lambda t \mid \mathcal{F}_s] = E[N(t) \mid \mathcal{F}_s] - \lambda t \]

By the independent increments property, \(E[N(t) \mid \mathcal{F}_s] = N(s) + \lambda(t - s)\). Therefore

\[ E[\tilde{N}(t) \mid \mathcal{F}_s] = N(s) + \lambda(t-s) - \lambda t = N(s) - \lambda s = \tilde{N}(s) \]

which confirms the martingale property.

Exercise 3. Compute the probability that a Poisson process with \(\lambda = 5\) has exactly 3 jumps in the interval \([0, 1]\).

Solution to Exercise 3

The number of jumps in \([0,1]\) follows \(\mathrm{Poisson}(\lambda \cdot 1) = \mathrm{Poisson}(5)\). Therefore

\[ P(N(1) = 3) = \frac{e^{-5} \cdot 5^3}{3!} = \frac{e^{-5} \cdot 125}{6} = \frac{125}{6} \cdot e^{-5} \approx 0.1404 \]

Exercise 4. Describe how to simulate a Poisson process on \([0, T]\) using the inter-arrival time method. If the inter-arrival times are \(\tau_1, \tau_2, \ldots\), what is the distribution of each \(\tau_k\)?

Solution to Exercise 4

Each inter-arrival time \(\tau_k\) is independently distributed as \(\mathrm{Exponential}(\lambda)\) with density \(f(\tau) = \lambda e^{-\lambda \tau}\) for \(\tau \ge 0\).

The simulation algorithm is:

  1. Set \(t = 0\) and \(n = 0\).
  2. Generate \(\tau \sim \mathrm{Exponential}(\lambda)\) (e.g., \(\tau = -\ln(U)/\lambda\) where \(U \sim \mathrm{Uniform}(0,1)\)).
  3. Set \(t = t + \tau\).
  4. If \(t > T\), stop; otherwise set \(n = n + 1\) and record the jump time \(t_n = t\).
  5. Repeat from step 2.

The resulting process \(N(t) = \max\{n : t_n \le t\}\) is a Poisson process with intensity \(\lambda\).