I know the following ($S_t$ = stock, $B_t$ = bond price) $$ d(X_tY_t) = dX_tY_t + X_tdY_t + dY_tdX_t $$ And i have the following multiplication table: $$ \begin{array} \hline \times & dW_t & dt\\ dW_t & dt & 0 \\ dt & 0 & 0 \\ \end{array} $$
Now i was solving the following: $dB_t^* =d\left(\frac{B_t}{S_t}\right)$ $$ dB_t^* =d\left(\frac{B_t}{S_t}\right) = B_tdS^{-1}_t + B_t dS_t^{-1} +dS_t^{-1}dB_t $$ was my first try. However the answer sheet suggested that the final term cancels out. Could someone point me in the right direction as to why this is the case? I would think that it would not be the case in general. This is of course homework, but i am missing the explanation as to why my gut feeling is incorrect.
For reference: $$ \begin{align} S_t &= S_0 \exp([\mu-\frac{1}{2}\sigma^2]t + \sigma W_t)\\ B_t &= e^{rt} \end{align} $$
Use Ito's lemma. First, \begin{align} d(S_t^{-1}) &= \left[\mu S_t(-S_t^{-2}) + \frac{1}{2} \sigma^2S_t^2(2S_t^{-3}) \right]dt + \sigma S_t (-S_t^{-2})dW_t \\ &= \left[ -\mu S_t^{-1} +\sigma^2S_t^{-1} \right]dt - \frac{\sigma}{S_t}dW_t \end{align} Then: \begin{align*} d(S_t^{-1})dB_t &= \left( \left[ -\mu S_t^{-1} +\sigma^2S_t^{-1} \right] dt - \frac{\sigma}{S_t} dW_t \right) re^{rt} dt \\ &= \left( \left[ -\mu S_t^{-1} +\sigma^2S_t^{-1} \right] \underbrace{(dt)^2}_0 - \frac{\sigma}{S_t} \underbrace{dW_tdt }_0 \right)re^{rt}\\ &=0 \end{align*}
Alternatively, use Ito's lemma on the full $dB^*_t=d(S_t^{-1}B_t)$. \begin{align} dB^*_t &= d\left(\frac{B_t}{S_t}\right) \\ &= \left[ re^{rt}S_t^{-1} + \mu S_t e^{rt} (-1)S_t^{-2} + \frac{\sigma^2S_t^2}{2} 2 B_t S_t^{-1} \right]dt + \sigma S_t(-B_tS_t^{-2})dW_t \\ &= re^{rt}dtS_t^{-1} + e^{rt}\left( S_t^{-1}[\sigma^2 - \mu]dt - \frac{\sigma }{S_t}dW_t \right) \\ &= dB_t S_t^{-1} +B_t d(S_t \end{align}
Note that $S_t$ satisfies: $$ dS_t = \mu S_tdt + \sigma S_t dW_t = \tilde{\mu}(S_t)dt + \tilde{\sigma}(S_t)dW_t $$