Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $h:[0,\infty) \to \mathbb{R}$ be a measurable, square integrable function on $[0,t]$, for all $t \geq 0$.

I want to show that if $H_t = \int_0^t h(s)\;dB_s$, where $(B_t)_{t\geq0}$ is a standard Brownian Motion, then $$\operatorname{Cov}(H_s,H_t)=\int_0^sh^2(u)\;du,\text{ for }s<t$$

Using a characteristic funtion of $\mathcal{N}$ and Ito's lemma I have shown that

$$H_t \sim \mathcal{N(0,\int_0^th^2(u)\;du)}$$

So I am only left to show that $\int_s^t h(u)\;dB_u$ is independent of $\mathcal{F}_s$. ($\mathcal{F}_{t\geq 0}$ is the natural filtration of $B_t$)

The problem here is that $\int_s^t h(u)\;dB_u$ is a limit of the functions of a form $\int_0^{\infty}\displaystyle\sum_{k=1}^{N_k}a_k\mathbf{1}_{(s_k,t_k]}dB_u$ in $\displaystyle \sup_{t\geq 0}||\cdot||_2$ norm, where $a_k's$ are deterministic constants in this case and $s_1 > s$

The stochastic integrals of simple processes (as above) are definitely independent of $\mathcal{F}_s$, but how can I deduce that the independence is preserved after taking the limits in the $\displaystyle \sup_{t\geq 0}||\cdot||_2$ norm?

share|cite|improve this question
Notice the inequality $\| E[ X | \mathcal{G} ] \|_p \leq \| X \|_p$ for $p \geq 1$. Thus if $X_n, X \in L^p$ and $X_n \stackrel{L^p}{\longrightarrow} X$, then $E[X_n|\mathcal{G}] \stackrel{L^p}{\longrightarrow} E[X|\mathcal{G}]$. Thus if $(\phi_n)$ is a sequence of simple processes approximating $h$, then $$0 = E \left[ \int_{s}^{t} \phi_n \; dB \Bigg| \mathcal{F}_{s} \right] \stackrel{L^p}{\longrightarrow} E \left[ \int_{s}^{t} h \; dB \Bigg| \mathcal{F}_{s} \right].$$ – Sangchul Lee Feb 22 '12 at 17:29
This computes the covariance beautifully and shows that $E \left[ \int_{s}^{t} h \; dB \Bigg| \mathcal{F}_{s} \right]=0$ a.s. Thank you a lot. Still can't deduce the independence directly though. I could invoke now that I have two normal random variables with 0 covariance ($\int_0^sh(u)\;dB_u$, $\int_s^th(u)\;dB_u$), and so must be independent. But $\mathcal{F}_s$ could be bigger than $\sigma(\int_0^vh(u)\;dB_u,\;v \leq s)$ – Tom Artiom Fiodorov Feb 22 '12 at 20:39

Just use the integration by parts formula with some cleverly chosen processes: Integration by parts says $$ X_tY_t = X_0Y_0 + \int_0^t X_s dY_s + \int_0^t Y_s dX_s + [X,Y]_s $$

For $s < t$ we define the processes $X_t = \int_0^t h(u) dB_u = H_t$ and $Y_t = \int_0^t 1_{[0,s]}(u) h(u) dB_u = H_s$ then applying integration by parts. Note that $dX_u = h(u) dB_u$ and $dY_u = h(u)1_{[0,s]}(u) dB_u$

$$ H_tH_s = 0 + \int_0^t \left[\int_0^u h(v) dB_u \right] h(u)1_{[0,s]}(u) dB_u + \int_0^t \ldots dB_v + \int_0^t h(u)^2 1_{[0,s]}(u) du $$

I didn't develop the interior of the stochastic integrals since it won't matter afterwards. The $h$ function is square integrable and so the stochastic integrals are martingales, and so their expectations are equal to the expectation at time 0 which is exactly zero!. For the same reason $E(H_t) = 0$ for all t.

And so the covariance is given by $$ E(H_tH_s) - E(H_t)E(H_s) = E(\int_0^t h(u)^2 1_{[0,s]}(u) du) - 0 = E(\int_0^s h(u)^2 du)$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.