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I am trying to calculate the conditional distribution of $$\biggl( \int_s^t W_u du \; \biggl| \; W_s = x, W_t= y\biggl) $$ where $W$ is a Standard Brownian Motion and $s\leq u \leq t$.

Any help would be greatly appreciated :)

My approach is the following: Using this helpful answer I can show that $$ \mathcal{L}\biggl( W_u \; \biggl| \; W_s = x, W_t= y\biggl) \sim \mathcal{N}\biggl( \frac{t-u}{t-s}x+\frac{u-s}{t-s}y, \frac{(t-u)(u-s)}{t-s} \biggl) $$ and thus

\begin{align} \mathbb{E}\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl] &= \int_s^t \mathbb{E}[W_u \; | \; W_s = x, W_t= y] du \\ &= \int_s^t \biggl(\frac{t-u}{t-s}x+\frac{u-s}{t-s}y\biggl) du \\ &= \frac{t-s}{2}x+ \frac{t-s}{2}y \end{align}

However, I struggle with calculating the variance... Is the following correct so far? \begin{align} Var\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl] &= \mathbb{E}\biggl[\biggl(\int_s^t W_u du \biggl)^2\; \biggl| \; W_s = x, W_t= y \biggl] - \biggl(\mathbb{E}\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl]\biggl)^2 \\ &= \mathbb{E}\biggl[\int_s^t \int_s^t W_v W_u du dv\; \biggl| \; W_s = x, W_t= y \biggl] - \biggl(\frac{t-s}{2}x+ \frac{t-s}{2}y\biggl)^2 \\ &= \int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv - \biggl(\frac{t-s}{2}x+ \frac{t-s}{2}y\biggl)^2 \end{align}

Emmy
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    Hint: Let's denote the process $W_u | W_s = x, W_t = y$ by $X_u$. Notice that this is a Brownian bridge that starts at $x$ and ends at $y$, so it admits the following representation: $X_u = x + B(u-s) - \frac{u-s}{t-s} \left( B(t-s) + x-y \right)$, where the equality is in law and $B$ is another Brownian motion. Then your integrand becomes $E(X_v X_u)$ and you can use this representation to compute it. – Jose Avilez May 13 '21 at 17:23
  • @JoseAvilez: Thank you very much for your hint! I can verify that the expectation of $X_u$ and $W_u | W_s=x, W_t=y$ is the same. Can I assume that $B=B_u$? – Emmy May 13 '21 at 20:51
  • Additionally, is there a simpler way to calculate $\mathbb{E}[X_uX_v]$ than solving the product $X_uX_v$ and using the properties of the expectation / Brownian motion (because I tried it and its a painful calculation)? – Emmy May 13 '21 at 20:52
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    The variance actually is independent of $x$ and $y$, so you can simplify the computation by setting $x=y=0$. (This is btw a general property of Gaussian distribution: if $(X,Y)$ is a Gaussian element, then the conditional variance of $Y$ given $X=x$ is independent of $X$, it depends only on the covariance structure.) – zhoraster May 14 '21 at 08:03
  • @zhoraster: Thanks a lot for your help! Unfortunately, I am unsure how to apply your hint. If I set $x=y=0$, I end up with $X_u = B(u-s)-\frac{u-s}{t-s}B(t-s)=0$. – Emmy May 14 '21 at 10:08

1 Answers1

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As mentioned in the comments, the variance is independent of $x$ and $y$, so we can assume $x=y=0$. Also set for simlicity $s=0, t=1$. Then, conditionally on $W_0=W_1=0$, $W$ is a standard Brownian bridge, so $$ \mathrm{Var}\left(\int_0^1 W_u du\,\middle|\, W_0=W_1=0 \right) = \mathrm{E}\left(\int_0^1\int_0^1 W_z W_u du\,dz\,\middle|\, W_0=W_1=0 \right) \\ =2 \int_0^1\int_0^z u (1-z) du\,dz = \int_0^1 z^2(1-z)dz = \frac1{12}. $$

Thanks to self-similarity and homogeneity of increments, for general $s,t$ the answer is ${(t-s)^3}/{12}$.

zhoraster
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  • Thank you so much for you answer! I can understand your approach and think its correct. The only open question I cannot answer myself is that the variance can get negative for large values of x and y using your solution in my calculation. Can you see my mistake? Again, thank you so much! – Emmy May 16 '21 at 08:06
  • @Emmy, no, it can't. Why the variance does not depend on $x,y$, the expectation $\mathrm{E}\left(\int_0^1\int_0^1 W_z W_u du,dz,\middle|, W_0=x, W_1=y \right)$ does. The extra contribution will be exactly equal to the squared expected value. – zhoraster May 16 '21 at 08:45
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    Ah my mistake! Thought for some reason it would be independent too... Thank you for clarifying and cannot thank you enough for answering the question!! :) – Emmy May 16 '21 at 10:23
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    My apologies for asking again as I already accepted the answer but is it possible that the correct answer is $(t-s)^3/12$? – Emmy Jul 05 '21 at 20:01
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    Yep, sorry, my bad. Indeed, by self-similarity, $\int_0^t W_s ds = t\int_0^1 W_{ts} ds \overset{d}{=} t^{3/2} \int_0^1 W_s ds$, so the variance is proportional to the cube of length of the interval. – zhoraster Jul 06 '21 at 05:16