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Brownian bridge. Let $B$ be a $d$-dimensional Euclidean Brownian motion. Then the process $t \mapsto X_t = B_t - tB_1$ is called a Brownian bridge. Let $G_t = \sigma \{B_s, s \leq t; B_1\}$. Prove the following facts as an exercise:

$X$ is a semimartingale with respect to $G^*$, and there is a $G^*$-adapted Brownian motion W such that $X$ is the solution of $X_t = W_t - \int_0^t \frac{X_s}{1 - s} \, ds\,.$

This equation can be solved explicitly:
$X_t = (1 - t) \int_0^t \frac{1}{1 - s}\,dW_s\,.$

I tried to solve the equation $X_t = W_t - \int_0^t \frac{X_s}{1 - s} \, ds$ to obtain the Brownian bridge in the following way: $dX_s=dW_s-\frac{X_s}{1 - s}\,ds$.

Multiplying by $\frac{1}{1 - s}$:
$\frac{dX_s}{1-s} +\frac{X_s}{(1 - s)^2}\,ds =\frac{dW_s}{1-s}$ (By Ito) $\implies d\left(\frac{X_s}{1-s}\right)=\frac{dW_s}{1-s}$
$\implies \int_{0}^t d\left(\frac{X_s}{1-s}\right)=\int_0^t\frac{dW_s}{1-s}\implies \frac{X_t}{1-t}-X_0=\int_0^t\frac{dW_s}{1-s}$.

Now I am stuck with the integral $\int_0^t\frac{dW_s}{1-s}$.

I thought I could consider something as $W_{\frac{s}{1-s}}$, but it does not lead me anywhere.

Question: How do I solve the equation to obtain the Brownian Bridge?

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1 Answer 1

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To show that $X$ is a Brownian bridge on $[0,1]$ it suffices to show that it is centered and Gaussian with $$\tag{1}{\rm Cov}[X_t,X_s]=\min(t,s)-ts\,.$$ (see [1] p. 35). By $$ X_t=(1-t)\int_0^t\frac{dW_u}{1-u} $$ we see that $X$ is centered and Gaussian and \begin{align} {\rm Cov}[X_t,X_s]=(1-t)(1-s)\int_0^{\min(t,s)}\frac{1}{(1-u)^2}\,du=(1-t)(1-s)\frac{-\min(t,s)}{\min(t,s)-1}\,. \end{align} Checking the cases $t<s$ and $t>s$ shows that this equals (1).

[1] D. Revuz, M. Yor, Continuous Martingales and Brownian Motion. 1991.

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  • $\begingroup$ Thanks for your answer! If you prove the mean and the covariance is the same. How would that entail the law would be the same? How would this imply all the Brownian Bridge properties hold? I took a look at the Levy Brownian motion characterisation and it seems there would be needed a specialized form of that. Even so, the covariance would not be enough? $\endgroup$ Sep 24, 2023 at 13:17
  • $\begingroup$ I think those questions are legit. It seems there are various definitions of Brownian bridge. The one I took is from Revuz & Yor. George Lowther's very good website gives a better overview. Please research yourself. I'd like to move on. $\endgroup$
    – Kurt G.
    Sep 24, 2023 at 14:20

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