# If we have a SDE $dX_t=b(X_t,t)dt+\sigma(X_t,t)dB_t$, why is $(dX_t)^2=\sigma^2(X_t)dt$?

Let $$(X_t)$$ be an Ito process with SDE $$dX_t=b(X_t,t)dt+\sigma(X_t,t)dB_t,$$ where $$(B_t)$$ is a Brownian Motion. Now Ito's formula says that $$df(X_t)=f'(X_t)dX_t+\frac{1}{2} f'(X_t)(dX_t)^2$$ is an Ito process too. I always read in books and papers, that $$(dX_t)^2=\sigma^2(X_t)dt$$ but it is never stated why. Is there a short proof or a reference that I can read?

• Take a look at this and the linked questions.
– saz
Jul 21, 2019 at 19:11

As mentioned in Wiener Process $dB^2=dt$

it's natural to define

$$\int_0^t g(B_s) \, dB_s^2 := \lim_{|\Pi| \to 0}\sum_{j=1}^n g(B_{t_{j-1}}) \cdot (B_{t_j}-B_{t_{j-1}})^2.$$

Consequently, using the result below we have

$$\int_0^t g(B_s) \, dB_s^2 = \int_0^t g(B_s) \, ds.$$

"Lecture 15. Quadratic variation of the Brownian motion paths

Let us denote

$$V_n=\sum_{k=1}^{n} \left( B_{t^n_k} -B_{t^n_{k-1}}\right)^2.$$

Thanks to the stationarity and the independence of Brownian increments, we have:

\begin{align} \mathbb{E} \left( (V_n-t)^2\right)=\mathbb{E} \left(V_n^2\right)-2t\mathbb{E} \left( V_n\right)+t^2 \\ =\sum_{j,k=1}^n\mathbb{E} \left( \left( B_{t^n_j} -B_{t^n_{j-1}}\right)^2\left( B_{t^n_k}-B_{t^n_{k-1}}\right)^2\right)-t^2 =\sum_{k=1}^n\mathbb{E} \left( \left( B_{t^n_j}-B_{t^n_{j-1}}\right)^4\right)+2\sum_{1\le j