# Product rule with stochastic differentials

I am encountering difficulty in seeing how this relationship holds:

with $S_T$ being stock price at time $T$, I want to find the sde for $S_t e^{-rt}$ $$dS_t = rS_tdt + \sigma S_t d\hat B_t$$

Where $d\hat B_t$ is brownian motion such that $$X_t = S_t e^{-rt} = X_0 e^{\sigma \hat B_t - \frac{1}{2} \sigma^2 t}$$

So now, If the product rule for SDE's is:

$$d(X_tY_t) = X_tdY_t + Y_tdX_t + dX_tdY_t$$ then, $$d(S_t e^{-rt}) = (S_t)d(e^{-rt}) + e^{-rt}dX_t + d(S_t)d(e^{-rt})$$ correct?

However, my book says the end result is:

$$d(S_t e^{-rt}) = e^{-rt}dS_t - re^{-rt}S_tdt$$ $$= e^{-rt}(\mu S_tdt + \sigma S_tdB_t - rS_tdt)$$ $$= \sigma S_t e^{-rt}(\frac{\mu - r}{\sigma}dt + dB_t)$$

How is this achieved, and what have I done wrong?

• What's the SDE for $S_t$?
– user223391
May 9, 2015 at 16:51
• Whoops, edited..! May 9, 2015 at 16:56

The point here is $d(S_t)d(e^{-rt})=0$.
The reason is $dt\cdot dW=0$ and $dt\cdot dt=0$.
Heuritically, $d(S_t)d(e^{-rt})=-re^{-rt}dt(dS_t)\sim O(dt)^{3/2}$. Thus, it is not considered in the SDE for $S_te^{-rt}$.
• $$d(e^{-rt}) == -re^{-rt}dt$$ where does this result come from? May 10, 2015 at 1:22