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

Question

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?

Answer 1

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.