variance of square of brownian motion increment

Question

In other words,

$$\text{Var}\left\{ [W(t) - W(s)]^2 \right\} = \mathbb E \left\{ (W(t) - W(s))^4 \right\} - \left[ E\left\{ (W(t) - W(s))^2 \right\} \right]^2 $$

How is this equal to $(t-s)^2$ given that $\mathbb E[[W(t)-W(s)]^2] = t-s$?

Answer 1

Since for $Z\sim \mathcal (0, \sigma ^2)$, we have $\mathbb E \{Z^4\} = 3 \sigma ^4$ (http://en.wikipedia.org/wiki/Normal_distribution#Moments), we can deduce that

\begin{align}\text{Var}\left\{ (W(t)-W(s))^2 \right\} &= \mathbb E \left\{ (W(t)-W(s))^4 \right\} - E\left\{ (W(t)-W(s))^2 \right\} ^2 \\&=3 (t-s)^2 - (t-s)^2 = 2(t-s)^2 \end{align}

Answer 2

$$\text{Var}((W_t-W_s)^2) = \text{Var}(W_{t-s}^2) = \mathbb{E}\left[(W_{t-s}^2-(t-s))^2 \right] \stackrel{\ast}{=} (t-s)^2 \cdot \underbrace{\mathbb{E}((W_1^2-1)^2)}_{2} = 2(t-s)^2$$

In $(\ast)$ we used scaling property, i.e. $W_{t-s} \sim \sqrt{t-s} \cdot W_1$.