Assume $$dS_t = -S_t dt + tdB_t$$, where $B_t$ is a standard Brownian motion.
I want to find a formula for $$X_t = \lim_{\Delta t \rightarrow 0} (S_{t_k+1}-S_{t_k})^2$$, where $t_k=k\Delta t$.
I got $X_t = 0$, but I think my solution is not correct.
This is the steps I got so far:
Without thinking about the fact that $S_t$ is the Lagevin Equation, we have:
$$S_{t_k+1} = S_{t_k} + -S_{t_k} \Delta t_k + t\Delta B_{t_k}$$ $$S_{t_k+1}-S_{t_k} = -S_{t_k} \Delta t_k + t\Delta B_{t_k}$$ $$X_t = \lim_{\Delta t \rightarrow 0} (-S_{t_k} \Delta t_k + t\Delta B_{t_k})^2$$ $$X_t = \lim_{\Delta t \rightarrow 0} (S_{t_k}^2 \Delta t_k^2 -2tS_{t_k}\Delta t_k \Delta B_{t_k} + t^2\Delta B_{t_k}^2)$$ $$X_t = \lim_{\Delta t \rightarrow 0} (t^2\Delta t_k)=\lim_{\Delta t \rightarrow 0} (t^2k\Delta t)=0$$ since $dB_t^2 = dt, dt^2=0, dtdb_t = 0$ by Ito's lemma
I heard that this should not be equal to zero from my other classmate. Did I do something wrong here?
I haven't used the solution of the Langevin Equation. Do I have to use it? I know the solution of the Langevin Equation would be
$$S_t=S_0e^{-t}+t\int^t_0 e^{t-s}dB_s$$, but I am not sure I have to use this to this problem
Thanks in advance!