I'm trying to find

$$E \left[\int_3^4 B_t \, dt \mid \int_1^2 B_t \, dt= c\right]$$

where $B_t$ is standard Brownian motion and integral is just a Riemann integral and constant $c$ is a known real number and bounderies of integrals represent times.

If we write integrals in a Riemann sum approximation, then I think the above expectation can be found by solving a similar expectation: $ E [\sum_1^N B_{t_j} \Delta t_j \mid \sum_1^M B_{t_k} \Delta t_k=c]. $

So for a simple example to find $E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c]$ for some real coefficients $a_1, a_2, a_3, a_4$, we can write \begin{align*} E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c]&= \\E[ a_4 (B_4-B_3) + (a_3 +a_4)B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] & = \\a_4 E[B_4 -B_3 \mid a_2 (B_2 -B_1) + (a_2+a_1) B_1=c]+ (a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] \end{align*}

Since $B_4- B_3$ is independent of $B_2-B_1$ and $B_1$ so is independent of linear combination of both and in particular $a_2 (B_2-B_1) + (a_2+a_1)B_1$. So we have \begin{align*} E[a_4 B_4 + a_3 B_3 \mid a_2 B_2+ a_1 B_1 =c] &= \\a_4 E[B_4-B_3] + (a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c]=(a_3+a_4) E[B_3 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c] =(a_3+a_4) E[B_3-B_2 +B_2 \mid a_2 (B_2-B_1) + (a_2+a_1)B_1=c]=(a_3+a_4) E[B_2 |a_2 (B_2-B_1) + (a_2+a_1)B_1=c] \end{align*} So according to these calculations I think we have

$$E \left[\int_3^4 B_t \,dt \mid \int_1^2 B_t \, dt= c\right]=E\left[B_2\mid \int_1^2 B_t \, dt= c \right]$$

Is it correct? if so, how can I finish this?

Thank you in advance!

  • 1
    $\begingroup$ Hint: Since $\int_a^b B_t\,dt$ is a linear functional of Brownian motion, which is a centered Gaussian process, the random variables $X = \int_1^2 B_t\,dt$ and $Y = \int_3^4 B_t\,dt$ have a joint centered Gaussian distribution. So all you need to do is find their variances and covariance, $E[X^2]$, $E[Y^2]$, $E[XY]$, which is a nice exercise in Fubini's theorem. $\endgroup$ Jan 21, 2019 at 23:41

1 Answer 1


I suspect something like the following can be justified: \begin{align} & \operatorname{cov}\left( \int_3^4 B_s\,ds, \int_1^2 B_t\,dt \right) \\[10pt] = {} & \int_3^4 \left( \int_1^2 \operatorname{cov}(B_s,B_t) \, dt \right) \, ds \\[10pt] = {} & \int_3^4 \int_1^2 t\, dt\, ds \quad \text{since } \operatorname{cov}(B_s,B_t) = t \text{ because } t<s. \end{align} With the variances, covariance, and marginal expected values (both $0$ in this case) of jointly normally distributed random variables, there is a standard formula for the conditional expected value of one of them given the other.


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