Itô's formula application

This exercise is from M.Jeanblanc's Stochastic calculus exercises.

Let $$ X_t=(1-t)\int_{0}^{t}{\frac{dB_s}{1-s}}\hspace{0.6cm}, 0\leq t<1$$

We want to calculate $dX_t$.

My approach : If we consider $$f(X,Y)=(1-X)Y$$ and $$Y_t=\int_{0}^{t}{\frac{dB_s}{1-s}}\implies dY_t=\frac{dB_t}{1-t}$$ we'll get using the formula $$df=f_{X}dX + f_{Y}dY +\frac{1}{2}f_{XX}dX^2 +\frac{1}{2}f_{YY}dY^2 +f_{XY}dXdY$$ $$\begin{cases} \begin{aligned} &f_{X}dX=-YdX\\ &f_{Y}dY=(1-X)dY\\ &\frac{1}{2}f_{XX}dX^2=0\\ &\frac{1}{2}f_{YY}dY^2=0\\ &f_{XY}dXdY=-dXdY\\ \end{aligned} \end{cases} $$ By considering $X_t=f(t,Y_t)$ we get $$ dX_t=(-dt)\int_{0}^{t}{\frac{dB_s}{1-s}}+(1-t)\frac{dB_t}{1-t}-dt\frac{dB_t}{1-t}$$

The answer in the reference doesn't have the last term $$dt\frac{dB_t}{1-t}$$

Am I working with a wrong formula or did I miss something ?

Solution 1:

The expression $dtdB_t$ is short notation for the quadratic variation process. Check the definition and apply it to the processes $\{t\}$ and $\{B_t\}.$ You will see you get zero.

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