• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathbb F=(\mathcal F)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
  • $B=(B_t)_{t\ge 0}$ be an $\mathbb F$-adapted Brownian motion with respect to $\mathbb F$

Let $H=(H_t)_{t\ge 0}$ be $\mathbb F$-adapted, locally bounded and of the form $$H_t(\omega)=\sum_{i=1}^nH_{t_{i-1}}(\omega)1_{(t_{i-1},t_i]}(t)\;\;\;\text{for all }\Omega\times[0,\infty)\;$$ for some $0=t_0<\ldots<t_n$. I absolutely don't get why $$\operatorname E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right]=\operatorname E\left[H_{t_{i-1}}^2\right](t_i-t_{i-1})\;.\tag{1}$$ Clearly, $$\operatorname E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2\right]=t_i-t_{i-1}\tag{2}\;,$$ but that seems to imply that we need to have $$\operatorname E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right]=\operatorname E\left[H_{t_{i-1}}^2\right]\operatorname E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2\right]\tag{3}\;.$$ Why does $(3)$ hold?


$$E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2\right] = E\left[E\left[H_{t_{i-1}}^2\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] = E\left[H_{t_{i-1}}^2E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \right] \right] = E\left[H_{t_{i-1}}^2(t_i-t_{i-1}) \right]$$

Justification for the equalities (from left to right): iterated conditioning, adaptedness of $H$, independence of increments of BM and the variance of increments of BM.

  • $\begingroup$ Independence of increments is the key. I'd done the same as you, but wasn't sure what I could do with $$\operatorname E\left[ H_{t_{i-1}}^2\operatorname E\left[\left(B_{t_i}-B_{t_{i-1}}\right)^2 \mid \mathbb{F}_{t_{i-1}}\right] \right] $$ $\endgroup$ – 0xbadf00d Jul 7 '15 at 12:38

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