Can someone explain me what this notation $\langle X \rangle_t$ means and why it is used in the context of SDEs ? For example reading a book I have found this two example:
- Given $ \frac{dS_t}{S_t}=\mu dt+\sigma dW_t$ then \begin{align*}dlog(S_t) = & \frac{d(log(S_t))}{d S_t} dS_t+\frac{1}{2}\frac{d^2(log(S_t))}{dS^2_t} \langle S \rangle_t \\ = & \frac{1}{S_t} (\mu S_tdt + \sigma S_t dW_t) + \frac{1}{2} \biggl (\frac{-1}{S^2_t} \biggr) \sigma^2 S^2_t dt \end{align*}
- Given $ \frac{dP}{P} = rdt $ and $ \frac{d \xi_t}{\xi_t} = f(\xi_t,S_t)dt+g(\xi_t,S_t)dW_t $ then \begin{align*} d(\xi_t,P_t)& = \xi_tdP_t+P_td\xi_t+d \langle \xi,P \rangle_t \\ & = \xi_t(rP_tdt)+P_t[f(\xi_t,S_t)\xi_tdt+g(\xi_T,S_t)\xi_tdW_t] \end{align*} I have tried to make sense of this notation. The first example implies $d\langle S \rangle_t = \sigma^2 S^2_t dt $ and I have found in another answer on this site that $\langle X_t \rangle = \int_0^t X_s^2 ds $ and so \begin{align*} d \langle S \rangle_t & = d \int_0^t S^2_s ds \\ & = \int_0^t dS^2_s ds\\ & = \int_0^t (\mu^2 ds^2 S^2_s + 2 \mu \sigma ds dW_t S_s + \sigma^2S_s^2dW^2) ds \\ & = \int_0^t (\sigma^2S_s^2dW^2) ds \\ & = \int_0^t (\sigma^2S_s^2ds) ds \end{align*} but at this point, despite some similarity, I do not know how to proceed since $ds^2=0$. ( I do not know if it is ok taking the differential inside the integral and if $d(S_t^2)=(dS_t)^2$; I have reported my trial only because it seems to me too much similar to the result to be completely wrong, I am sorry if it is not the case). Thank you for your answer