Fourth Moments of Wiener Process Increments
Summary
The document explains how to calculate the fourth moment of a Wiener process increment by expressing the normally distributed increment as its standard deviation times a standard normal variable. It uses the standard normal moments, including a fourth moment of three, to derive the increment’s fourth moment. The same scaling approach also accounts for the zero odd moments and the second moment.
There is a discrepancy in the source: its opening statement gives 3t²/n, but the derivation gives 3t²/n². Given the stated increment variance of t/n, the derivation is consistent: scaling a standard normal by √(t/n) multiplies its fourth moment by (t/n)². The document is a focused moment calculation, not a broader treatment of stochastic processes or their use in pricing and risk models.
Key ideas
- A Wiener process increment with variance t/n can be represented as √(t/n) times a standard normal variable.
- The fourth moment of a standard normal variable is three.
- Scaling a random variable by a constant scales its fourth moment by the constant raised to the fourth power.
- For the stated variance, the resulting fourth moment is 3t²/n², so the document’s initial expression appears to omit a factor of n.
Tags
Full text
# How to Evaluate Expected Value powered 4 of a Wiener Process?
# How to Evaluate Expected Value powered 4 of a Wiener Process?
Since $X(t_j) - X(t_{j-1})$ is Normally distributed with mean zero and variance $t/n$ we have
$$ \operatorname{E} [(X(t_j) - X(t_{j-1}))^2 ] = \frac{t}{n} \tag{1}$$ and $$ \operatorname{E} [(X(t_j) - X(t_{j-1}))^4 ] = \frac{3t^2}{n} \tag{2}$$
I can't seem to understand how the second result (2) is obtained. This is from Quantitative Finance by Paul Wilmott.
## Answer by Colin T Bowers (score 1)
https://quant.stackexchange.com/a/51219
You state $X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$. Thus: \begin{equation} X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z , \end{equation} where $Z \backsim \mathcal{N}(0, 1)$. Note that: \begin{align} & \mathbb{E} \sqrt{\frac{t}{n}} Z = 0 \\ & \mathbb{E} \left( \sqrt{\frac{t}{n}} Z \right)^2 = \frac{t}{n} \mathbb{E} Z^2 = \frac{t}{n} \\ & \mathbb{E} \left( \sqrt{\frac{t}{n}} Z \right)^3 = \left( \frac{t}{n} \right)^{\frac{3}{2}} \mathbb{E} Z^3 = 0 \\ & \mathbb{E} \left( \sqrt{\frac{t}{n}} Z \right)^4 = \left( \frac{t}{n} \right)^2 \mathbb{E} Z^4 = \frac{3 t^2}{n^2} \end{align} The third line follows since $\mathbb{E} Z^3 = 0$ and the fourth line follows since $\mathbb{E} Z^4 = 3$.
This result for the fourth moment of the Standard Normal is in many textbooks, but a proof can be found here on Mathematics StackExchange.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.