# Conditional expectation of integral of brownian motion

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...

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: $$$$\mathbb{E}\bigg[ \bigg(\int_0^T B_u du\bigg)^2 \bigg].$$$$

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}

• 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?
– Emmy
Commented May 13, 2021 at 14:12
• @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) Commented May 13, 2021 at 14:17
• 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!
– Emmy
Commented May 13, 2021 at 14:27
• 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. Commented May 13, 2021 at 14:45
• 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 :)
– Emmy
Commented May 13, 2021 at 15:40