I am trying to calculate $$\mathbb{E}\biggl[\biggl(\int_s^t W_u du\biggl)^2 \biggl|W_s=x, W_t=y\biggl] $$ where $W$ is a Standard Brownian Motion and $s\leq u \leq t$. Any help or tips would be greatly appreciated :)

My approach is the following \begin{align} \mathbb{E}\biggl[\biggl(\int_s^t W_u du\biggl)^2 \biggl|W_s=x, W_t=y\biggl] &=\mathbb{E}\biggl[\int_s^t \int_s^t W_v W_u du dv\; \biggl| \; W_s = x, W_t= y \biggl]\\ &=\int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \end{align} For $v\leq u$ I can rewrite this to
\begin{align} \mathbb{E}[W_v W_u | \; W_s = x, W_t= y] &= \mathbb{E}[W_v ((W_u-W_v)+W_v) | \; W_s = x, W_t= y] \\ &=\underbrace{\mathbb{E}[W_v (W_u-W_v) | \; W_s = x, W_t= y]}_{=0}+\mathbb{E}[W_v^2 | \; W_s = x, W_t= y]\\ &=\mathbb{E}[W_v^2 | \; W_s = x, W_t= y]\\ &= \frac{(t-v)(v-s)}{t-s} - \biggl(\frac{t-v}{t-s}x+\frac{v-s}{t-s}y\biggl)^2 \end{align} Where I used in the last equation that $(W_v | \; W_s = x, W_t= y) \sim \mathcal{N}( \frac{t-v}{t-s}x+\frac{v-s}{t-s}y, \frac{(t-v)(v-s)}{t-s} )$. I end up with this aweful calculation \begin{align} &\int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \\ = &\int_s^t \int_s^u \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv + \int_s^t \int_u^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv \\ = &\int_s^t \int_s^u \frac{(t-v)(v-s)}{t-s} - \biggl(\frac{t-v}{t-s}x+\frac{v-s}{t-s}y\biggl)^2du dv + \int_s^t \int_u^t \frac{(t-u)(u-s)}{t-s} - \biggl(\frac{t-u}{t-s}x+\frac{u-s}{t-s}y\biggl)^2du dv \end{align} I am sure there must be a better solution than this endless calculation but I cannot think of one...


1 Answer 1


I found this to be a very interesting question, and I took a different approach to your working. Here's my attempt:

Instead of considering the integral $\int_s^t W_u du \rvert W_s=x, W_t=y$, we can consider the integral $\int_s^tB_u du$ where $B_u$ is a Brownian bridge process with $B_s = x$, $B_t = y$.

Furthermore, we can shift the limits of the integral from $[s, t]$ to $[0, T]$ where $T := t-s$. In this case, we define $B_0 = x$, $B_T = y$. So we want to find: \begin{equation} \mathbb{E}\bigg[ \bigg(\int_0^T B_u du\bigg)^2 \bigg]. \end{equation}

We can re-write our integral as follows \begin{align} \int_0^T B_u du &= \int_0^T(T-u)dB_u. \end{align}

Then, \begin{align} \mathbb{E}\bigg[ \bigg(\int_0^T(T-u)dB_u\bigg)^2 \bigg] &= \mathbb{E}\bigg[ \int_0^T (T-u)^2 d[B]_u \bigg] \\ &= \int_0^T (T-u)^2 du \\ &= \frac{(t-s)^3}{3} \end{align}

  • 1
    $\begingroup$ Thank you so much for your reply! I was not familar with Brownian Bridge processes but it is a very smart idea and way more elegant approach than mine! You used the Itô Isometry, so this is also true for Brownian Bridge processes? $\endgroup$
    – Emmy
    Commented May 13, 2021 at 14:12
  • $\begingroup$ @Emmy, Ito isometry can be applied to any semi-martingale in $L^2$ (see math.stackexchange.com/questions/1972607/…). Furthermore, I think Browinan bridge is a semi-martingale (not totally sure - I just did a quick google search) $\endgroup$
    – R. Rayl
    Commented May 13, 2021 at 14:17
  • $\begingroup$ Ah great! I will have a look at it :) And if you allow me one last question: The background of this question is that I am trying to calculate the variance of $\int_s^t W_u du | W_s=x, W_t=y$ (see this question). Using your solution in the calculation, the variance can get negative. Can you see my mistake? And again, thank you so much! $\endgroup$
    – Emmy
    Commented May 13, 2021 at 14:27
  • $\begingroup$ Yes, you are correct. The variance could indeed be negative for large $x$ or $y$, so I have probably made a mistake. Your calculations look correct to me (provided you are allowed change the expectation and the integral - I assume you use Fubini?). Also, where did you find this question? If you ever find an answer could you please share it? I'd be really interested. $\endgroup$
    – R. Rayl
    Commented May 13, 2021 at 14:45
  • $\begingroup$ Thanks a lot for taking a look! I really appreciate it :) And yes I was using Fubini to exchange the integrals. Regarding the source of the question: I am trying to replicate the Monte Carlo simulation that was done in this paper. They present 3 Monte Carlo schemes for pricing continous arithmetic Asian options and the last scheme requires the simulation of this conditional integral (see Remark 2.2.). And sure will do so :) $\endgroup$
    – Emmy
    Commented May 13, 2021 at 15:40

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