Let $X(t)=(1-t)\int_{0}^{t}\frac{dB(s)}{1-s}$ I want find $dX(t)$

Question

Let $X(t)=(1-t)\int_{0}^{t}\frac{dB(s)}{1-s}$, where $0\le t < 1$.Find $dX(t)$.

thanks for help.

Answer 1

Let $Y(t) = \int_0^t \frac{dB(s)}{1-s}$ so that $X(t) = (1-t) Y(t)$. You can now use Ito formula to get $$dX(t) = -Y(t) dt + (1-t) dY(t) = -Y(t) dt + (1-t) \frac{dB(t)}{1-t} = -\frac{X(t)}{1-t} dt + dB(t)$$ that is the required differential equation for $X(t)$. Notice that $X(t)$ is the Brownian bridge with initial point $a=0$ and final point $b=0$ in the time interval $(0,1)$.

Answer 2

HINT : $dB(s) = B'(s)ds$ so let $$f(s) = \frac{B'(s)}{1-s}$$ Therefore you can write the integral as :$$\int^t_0f(s)ds = F(t) - F(0)$$

Differentiate to find $dX(t)$.

Note that using my notation, $dF(t) = f(t)dt$.