Discretizing a Wiener Process for Numerical Simulation
Summary
The document explains how to represent a Wiener process on a uniform time grid for simulation. Divide the interval into equal steps, then construct each sampled value by adding the next increment to the previous value, starting from zero. The increment over each step is normally distributed with mean zero and variance equal to the step length, and increments on separate steps are independent.
The answer clarifies that each increment can be generated by multiplying a standard normal draw by the square root of the time step. Scaling a standard normal this way gives the required zero mean and step-length variance. This describes the basic discrete sampling scheme, rather than a full stochastic differential equation solver; simulation accuracy for a particular model depends on how that model is discretized and on the chosen time grid.
Key ideas
- A Wiener process starts at zero and has independent, normally distributed increments.
- On a uniform grid, each process value is the preceding value plus the increment for that interval.
- Each increment has variance equal to the interval length.
- A standard normal draw scaled by the square root of the time step produces an increment with the required distribution.
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# Discretization of Wiener process
# Discretization of Wiener process
The Wiener process $(W_t)$ is a continuous stochastic process that satisfies the following there conditions:
- $W_0 = 0$,
- the increments $\mathrm{d}W_t = W_{t + \mathrm{d}t} - W_t$ are normally distributed with mean $0$ and variance $\mathrm{d}t$,
- the increments are mutually independents, i.e. $\mathrm{d}W_i$ is independent from $\mathrm{d}W_j$ for every $i$ different from $j$.
I now want to discretize the Wiener process in order to simulate it as described in the beginning of “An algorithmic introduction to numerical simulatiom of stochastic differential equation” by Higham (2001).
- First, I have to discretize the time interval $[0,T]$ in $N$ sub-intervals of equal length $\delta_t = \frac{T}{N}$. In this way, each of the $N+1$ time instants are given by $t_i = i \cdot \delta_t$.
- Then, at each time instant $t_i$, the discretized version of the Wiener process is \begin{align*} W_i = W_{i-1} + \mathrm{d}W_{i-1}, \end{align*} where $\mathrm{d} W_{i-1}\sim N(0,\delta_t)$ and $W_0=0$.
I would like to undestand how proprieties 2 and 3 imply the above iteration formula.
## Answer by Magic is in the chain (score 6, accepted)
https://quant.stackexchange.com/a/49870
I think what is meant is that the increment of the brownian between discrete points $(i-1)$ and $i$ is normally distributed with mean 0 and variance equal to $\delta t$ which represents the length of the interval between the two discrete points. You can then write the increment as $\sqrt{\delta t}$ times a standard normal random. So the equation is to be read in conjunction with the statement following the equation.
To see it more clearly, let's represent the standard normal by Z so $Z \sim N\left(0,1\right)$ and the claim is that $dW_j \sim \sqrt{\delta t} Z$. A linear transformation of a normal is normal, so $\sqrt{\delta t} Z$ is indeed normal, and its mean and variance are easy to calculate:
$E \left[\sqrt{\delta t} Z \right]=\sqrt{\delta t}E \left[Z\right]=0$
$V \left[\sqrt{\delta t} Z \right]=\delta t V \left[Z\right]=\delta t$
And because $dW_j$ has the same properties, we can say $dW_j \sim \sqrt{\delta t} Z$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.