In a mathematical finance text by Ubbo F Wiersema, I came across the following
Say $\Delta t$ is very small. $\Delta B(t)$ denotes $\textit{brownian motion increment}$. Then $E[\Delta t\Delta B(t)]=0$. $Var[\Delta t\Delta B(t)]=(\Delta t)^2$.
It is also written that $Var[(\Delta B(t))^2]=2(\Delta t)^2$.
How was the calculations of the above two variances done?
First of all, I think the variance result is wrong, see below.
$\Delta B(t)$ is centered and normally distributed with variance $\Delta t$.
We can rewrite it as $$\Delta B(t)=\sqrt{\Delta t} Y$$ where $Y$ is a standard normal variable.
We know the moments of $Y$ , wich are $E(Y)=0, E(Y^2)=1$ and $E(Y^4)=3$
$$var(\Delta t\Delta B(t))=var(\Delta t\sqrt{\Delta t}Y)=(\Delta t\sqrt{\Delta t})^2*var(Y)=\Delta t^3$$
For the other one, same thing
$$var(\Delta B(t)^2)=var(\Delta t Y^2)=\Delta t^2var(Y^2)$$
WE know that $$var(Y^2)=E(Y^4)-E(Y^2)^2=3-1^2=2$$ Hence, $$var(\Delta B(t)^2)=2\Delta t^2$$