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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.

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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.

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