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I'm working through Le Gall's Brownian Motion, Martingales, and Stochastic Calculus, and I'm struggling on an exercise. The question concerns computing the finite-dimensional marginal distributions of a Brownian bridge. In particular, let $B_{t}$ be a Brownian motion on $[0,1]$ or $\mathbb{R}^{+}$ (doesn't matter which), and for $t\in [0,1]$ define the Brownian Bridge to be $W_t = B_t - t B_1$.

I've shown that $W_t$ is a centered Gaussian process with covariance function $K(s,t) = \min\{s,t\}- st$.

I'm now asked to prove that for $0<t_1<\cdots<t_p<1$, the law of $(W_{t_1}, \dots, W_{t_p})$ has density $$ g(w_1, \dots, w_p) = \sqrt{2\pi} \,p_{t_1}(w_1)\,p_{t_2 - t_1}(w_2 - w_1)\,\cdots\, p_{t_p - t_{p-1}}(w_p-w_{p-1})\,p_{1-t_{p}}(-w_p), $$ where $$ p_{t}(w)= \frac{1}{\sqrt{2\pi t}}\exp(-w^2/2t). $$

Solution Progress:

Attempt 1:

The density is factored into a bunch of products of Gaussian densities, where the variance of each is $t_{i}-t_{i-1}$. This makes me want to relate the vector of Brownian Bridge terms $(W_{t_1}, \dots, W_{t_p})$ to either the vector of Brownian motion $(B_{t_1}, \dots, B_{t_p})$ or to the independent increments $(B_{t_1}-B_{0}, B_{t_2}-B_{t_1},\dots ,B_{t_p}-B_{t_{p-1}})$. Both of these vectors have densities which are a product of individual gaussians.

We note that $$ \begin{pmatrix} W_{t_p}\\ \vdots\\ W_{t_1} \end{pmatrix}= \begin{pmatrix} B_{t_p}-t_{p}B_{1}\\ \vdots\\ B_{t_1}-t_{1}B_{1} \end{pmatrix}= \begin{pmatrix} -t_p&1&{}&{}&{}\\ -t_{p-1}&{}&1&{}&{}\\ \vdots&{}&{}&\ddots&{}\\ -t_1&{}&{}&{}&1 \end{pmatrix} \begin{pmatrix} B_{1}\\ B_{t_p}\\ \vdots\\ B_{t_1} \end{pmatrix} $$

I would love to use a change of variables, but the issue is that I'm required to bring in the additional $B_1$ term, and so the linear transformation above maps $p+1$-dimensional space into $p$ dimensional space. Thus, the determinant isn't defined. I'm not sure if I'm just being stupid, or this is really a problem?

Attempt 2:

Another approach I've thought of is that since the density $g(w_1, \dots, w_p)$ factors into densities of differences, let's first focus on the density of $(W_1 - W_{t_p}, \dots, W_{t_2} - W_{t_1}, W_{t_1})$. We have \begin{align*} \begin{pmatrix} W_{1} - W_{t_p}\\ W_{t_p}-W_{t_{p-1}}\\ \vdots\\ W_{t_2}-W_{t_1}\\ W_{t_1} \end{pmatrix}&= \begin{pmatrix} (B_{1}-B_{t_p})-(1-t_p)B_{1}\\ (B_{t_p}-B_{t_{p-1}})-(t_{p}-t_{p-1})B_{1}\\ \vdots\\ (B_{t_2}-B_{t_1})-(t_2-t_1)B_{1}\\ B_{t_1} - t_{1} B_{1} \end{pmatrix}\\ &= \bigg\{ \begin{pmatrix} 1&{}&{}\\ {}&\ddots&{}\\ {}&{}&1 \end{pmatrix}- \begin{pmatrix} (1-t_p)&\cdots&(1-t_p)\\ (t_p -t_{p-1})&\cdots&(t_p - t_{p-1})\\ \vdots&{}&\vdots\\ t_{1}&\cdots&t_1 \end{pmatrix} \bigg\} \begin{pmatrix} B_{1}-B_{t_p}\\ B_{t_p}-B_{t_{p-1}}\\ \vdots\\ B_{t_2}-B_{t_1}\\ B_{t_1} \end{pmatrix}. \end{align*} The matrix in curly braces is equal to $$ \begin{pmatrix} 1&{}&{}\\ {}&\ddots &{}\\ {}&{}&1 \end{pmatrix} - \begin{pmatrix} 1-t_p\\ \vdots\\ t_1 \end{pmatrix} \begin{pmatrix}1&\cdots&1\end{pmatrix}. $$ This is a rank-p $(p+1)\times (p+1)$ matrix. And thus, change of variables is not possible.

Attempt 3:

Let us first consider the density of the increments $W_{t_1}, W_{t_2}-W_{t_1}, \dots, W_{t_p}-W_{t_{p-1}}, W_{1}-W_{t_p}$, and factor it by successively conditioning $$ g(W_{t_1}, W_{t_2}-W_{t_1},\dots, W_{1}-W_{t_p})= g(W_{1}-W_{t_p}|W_{t_{p-1}}-W_{t_{p-2}},\dots, W_{t_1})\cdots g(W_{t_1}). $$

One can compute that \begin{align*} g(W_{t_1})&= \sqrt{2\pi} p_{t_1}(W_{t_1})p_{1-t_1}(-W_{t_1})\\ &= \frac{\sqrt{2\pi}}{\sqrt{2\pi t_1}\sqrt{2\pi (1-t_1)}}\exp\bigg( -\frac{W_{t_1}^{2}}{2t_1}\bigg) \exp\bigg(-\frac{W_{t_1}^{2}}{2(1-t_1)}\bigg) \end{align*} This is promising, as we have the desired form for a product of distributions. The next step is computing the conditional distributions $g(W_{t_2}-W_{t_1}|W_{t_1})$ onwards.

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    $\begingroup$ +1 : full credit for presentation of the question! Three attempts , each detailed, I will try to understand and provide feedback on all three attempts when I answer this question. $\endgroup$ Commented Apr 3, 2020 at 4:53
  • $\begingroup$ Thank you! looking forward to reading your comments. $\endgroup$ Commented Apr 3, 2020 at 5:01
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    $\begingroup$ @AlexLapanowski is this relevant math.stackexchange.com/questions/3348869/… ? $\endgroup$ Commented Apr 3, 2020 at 7:11
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    $\begingroup$ I did not want to leave you alone for a long time, so I thought I should tell you that I've been struggling too, and with the same problem of rank issues in the transformation matrix. Have you seen the formulation of the Brownian bridge as "Brownian motion conditioned on $B_1 = 0$"? Of course you cannot condition on a measure zero event, but this process is shown to be the weak limit of the Brownian motion conditioned on $|B_1| < \epsilon$ with $\epsilon$ dropping to zero. $\endgroup$ Commented Apr 5, 2020 at 4:30
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    $\begingroup$ Essentially, the fact that you are conditioning on a measure zero event creates the rank problem, I would think. That is why even when I searched online I could not find too many calculations involving this definition, because it is not amenable enough. $\endgroup$ Commented Apr 5, 2020 at 4:34

1 Answer 1

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The simplest approach is to use the fact that the Brownian bridge $\{W_t,t\in[0,1]\}$ has the same distribution as $\{B_t,t\in[0,1]\}$ conditioned to $B_1 = 0$, which immediately gives the required density. This, however, requires a priori knowledge of the said fact (which is not too hard to prove though).

I will use another approach, which in a sense is reverse to this one. First note that the Brownian bridge $\{W_t = B_t - t B_1\}$ is independent of $B_1$, which can be checked by computing the covariance: in the Gaussian case, zero correlation implies independence.

Therefore, $$ f_{W_{t_1},\dots, W_{t_p}, B_1}(w_1,\dots,w_p,x) = f_{W_{t_1},\dots,W_{t_p}}(w_1,\dots,w_p)f_{B_1}(x). $$ Consequently, $$ f_{W_{t_1},\dots,W_{t_p}}(w_1,\dots,w_p) = \frac{f_{W_{t_1},\dots, W_{t_p}, B_1}(w_1,\dots,w_p,x)}{f_{B_1}(x)}.\tag{1} $$ Denoting $w_0= t_0 = 0$, $$ f_{W_{t_1},\dots, W_{t_p}, B_1}(w_1,\dots,w_p,x) = f_{B_{t_1},\dots, B_{t_p}, B_1}(w_1 + t_1 x,\dots,w_p+t_p x,x)\\ = \prod_{i=1}^{p} \frac{1}{\sqrt{2\pi (t_{i} - t_{i-1})}}\exp \Bigl\{-\frac{\big(w_{i} - w_{i-1}+(t_{i}-t_{i-1}) x\big)^2}{2(t_{i}-t_{i-1}) }\Bigr\} \\\times \frac{1}{\sqrt{2\pi (1-t_p)}}\exp \Bigl\{-\frac{\big(x(1-t_p)-w_p\big)^2}{2(1-t_p)}\Bigr\} \\ = \prod_{i=1}^{p} \frac{1}{\sqrt{2\pi(t_{i}-t_{i-1}) }} \exp \Bigl\{- \frac{(w_i-w_{i-1})^2}{2(t_i-t_{i-1})^2} - x(w_i-w_{i-1}) - \frac{x^2(t_i-t_{i-1})}{2}\Bigr\}\\ \times \frac{1}{\sqrt{2\pi(1-t_{p}) }} \exp \Bigl\{-\frac{{w_p}^2}{2(1-t_p)} + w_p x - \frac{x^2(1-t_p)}{2}\Bigr\} \\ = \prod_{i=1}^{p}\frac{1}{\sqrt{2\pi(t_{i}-t_{i-1}) }}\exp \Bigl\{-\frac 12 \frac{(w_i-w_{i-1})^2}{(t_i-t_{i-1})^2}\Bigr\}\\ \times \frac{1}{\sqrt{2\pi(1-t_{p}) }} \exp \Bigl\{-\frac{{w_p}^2}{2(1-t_p)} - \frac{x^2}{2}\Bigr\}\\ = \prod_{i=1}^{p} p_{t_{i}-t_{i-1}}(w_i-w_{i-1}) \cdot p_{1-t_p}(w_p) e^{-x^2/2}. $$ Plugging into (1), $$ f_{W_1,\dots,W_1}(w_1,\dots,w_p) = \sqrt{2\pi}\prod_{i=1}^{p} p_{t_{i}-t_{i-1}}(w_i-w_{i-1}) \cdot p_{1-t_p}(w_p), $$ as required.


One could simplify the computation: since the ratio in (1) is independent of $x$ and since everything is continuous, one can take $x=0$. By not doing so we have actually shown the independence once more.

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  • $\begingroup$ Thank you so much for your help! I've given you the bounty and the check mark for accepted answer. I'm going to work through it for myself and leave some more detailed comments in a bit. $\endgroup$ Commented Apr 5, 2020 at 18:45
  • $\begingroup$ what's the relevance of continuity for your comment at the bottom? $\endgroup$ Commented Feb 1, 2021 at 13:00
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    $\begingroup$ @Timkinsella, I am speaking about densities, which are defined modulo sets of Lebesgue measure zero. So appealing to value in a particular point won't work unless we employ some additional ideas, like continuity here. $\endgroup$
    – zhoraster
    Commented Feb 1, 2021 at 14:43

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