# Proving $X_t = W_t - \int_0^t \frac{X_s}{1 - s} \, ds$ is the Brownian Bridge

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?

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 $$t and $$t>s$$ shows that this equals (1).