Let $(B_t)$ be a Brownian motion. Let $0<t<1$. What is the distribution of $X_t=\int_0^t \frac{B_s-sB_1}{1-s} ds + B_t – tB_1$ ?
The tricky part is of course the integral. I do not see how to exploit the increments of the Brownian motion like one usually does when faced with integrals of Brownian motion. My intuition tells me that $X_t \sim\mathcal N (0,t)$.
Best Answer
Using integration-by-parts $$ \int_0^t\frac{B_s}{1-s}\,ds=\int_0^t\log(1-s)\,dB_s-\log(1-t)B_t\, $$ and $$ B_1\int_0^t\frac{s}{1-s}\,ds $$ and writing $f(t)=\int_0^t\frac{s}{1-s}\,ds+t$ we get that
\begin{align} X_t&=\int_0^t1+\frac{\log(1-s)}{\log(1-t)}\,dB_s-B_1\int_0^t\frac{s}{1-s}\,ds-tB_1\\ &=\int_0^t1+\frac{\log(1-s)}{\log(1-t)}\,dB_s-f(t)B_1\\ &=\underbrace{\int_0^t1+\frac{\log(1-s)}{\log(1-t)}+f(t)\,dB_s}_{=:\,Y_t\text{ Gaussian }}-\underbrace{f(t)(B_1-B_t)}_{\text{Gaussian and independent of }Y_t}\,. \end{align}