= Solution
Write $B=\beta$ for the driving <Brownian motion> throughout; the two symbols in the displayed formulas refer to the same process. For $t<1$ the deterministic integrand is square-integrable on $[0,t]$, so
$$
I_t=\int_0^t\frac{dB_s}{1-s},\qquad X_t=(1-t)I_t
$$
is well defined. The <Itô product rule>, with zero <quadratic covariation> against the deterministic factor, gives
$$
dX_t=-I_t\,dt+(1-t)\frac{dB_t}{1-t}
=dB_t-\frac{X_t}{1-t}\,dt,\qquad X_0=0.
$$
For the other form, apply the same <Itô product rule> to $B_t/(1-t)$:
$$
\frac{B_t}{1-t}=\int_0^t\frac{dB_s}{1-s}+\int_0^t\frac{B_s}{(1-s)^2}\,ds.
$$
Multiplying by $1-t$ gives the <stochastic integral representation of a Brownian bridge>
$$
\boxed{X_t=B_t-(1-t)\int_0^t\frac{B_s}{(1-s)^2}\,ds},\qquad t<1.
$$
The drift coefficient is singular only at the terminal endpoint, so this calculation is performed on compact subintervals of $[0,1)$.
Back to article page
