# 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.

You state $$X(t_j) - X(t_{j-1}) \backsim \mathcal{N}(0, \frac{t}{n})$$. Thus: $$$$X(t_j) - X(t_{j-1}) = \sqrt{\frac{t}{n}} Z ,$$$$ where $$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$$.