# Discretization of Wiener process

The Wiener process $$(W_t)$$ is a continuous stochastic process that satisfies the following there conditions:

1. $$W_0 = 0$$,
2. the increments $$\mathrm{d}W_t = W_{t + \mathrm{d}t} - W_t$$ are normally distributed with mean $$0$$ and variance $$\mathrm{d}t$$,
3. 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.

• I dont know the derivation of the continuation stochastic process but I would venture that it came from this process in reverse. That is, a discrete model was developed as you stated and then the limit was evaluated as delta t goes to zero. – Attack68 Nov 22 '19 at 20:47

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

• Can you please explain a little more what you meant because I still not get it! – luca dibo Nov 23 '19 at 0:45
• thanks. Have added further detail, hope it is clearer now, but please do let me know if any part of the explanation is not clear! – Magic is in the chain Nov 23 '19 at 12:06
• thanks for having added further details! What you added is clear. It's certainly true that dW_j is a normal random variable with mean 0 and variance dt. What is not clear to me why you can write the recursive algorithm given the proprieties 2 and 3, so for example, where the indepedence of the increments enter in the statement? – luca dibo Nov 23 '19 at 22:51
• Properties 2 and 3 are needed for the central limit theorem to assert normality: "The central limit theorem states that the sum of a number of independent and identically distributed random variables with finite variances will tend to a normal distribution as the number of variables grows. " from en.wikipedia.org/wiki/Central_limit_theorem – Mats Lind Nov 25 '19 at 13:53
• If the increments were not independent, then the increments in the j-th interval could depend on the increments in the other intervals, which is going to complicate things, but only slightly - please check out the Ornstein Uhlenbeck process as an example. – Magic is in the chain Nov 25 '19 at 20:41