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
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 $dW_j \sim \sqrt{\delta t} Z$.
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