I am reading a proof about the existence of the uniqueness of a SDE. There arises the following problem.
Let $X_t$ and $Y_t$ be two solutions of the SDE $dX_t=b(t,X_t)dt+\sigma(t,X_t)dW_t, X_0=\Xi$, where W is an r-dimensional Brownian motion.
Now by Ito, $$\Vert X_t-Y_t\Vert^2=2\int_0^t \langle b(s,X_s)-b(s,Y_s),X_s-Y_s\rangle ds +2\sum_{i=1}^d\sum_{j=1}^r\int_0^t (X_s^i-Y_s^i)(\sigma_{i,j}(s,X_s)-\sigma_{i,j}(s,Y_s))dW_s^j+\int_0^t \Vert\sigma(s,X_s)-\sigma(s,Y_s) \Vert^2 ds$$
How does this follow? I tried to apply Ito formula to $f(X_t-Y_t)$ with $f(x):=\Vert x\Vert^2$, but this does not work out.
Could someone help me here and show how to apply Ito correclty here.
Best Answer
Let $f:\mathbb R^d \to \mathbb R, (t, x) \mapsto \|x\|^2$. Then $f$ is smooth and time-independent. The derivatives of $f$ are as follows:
$$\frac{\partial}{\partial x_i} f(x) = 2x_i, \quad \frac{\partial^2}{\partial x_i\partial x_j} f(x) = 2\delta_{ij}$$
By Itô's formula: \begin{align} \mathrm d f(X_t-Y_t) &= \sum_{i=1}^d \frac{\partial f}{\partial x_i} (X_t-Y_t) \mathrm d(X^i_t-Y^i_t)\\ &+ \frac 12 \sum_{i,k=1}^d \frac{\partial^2 f}{\partial x_i\partial x_k} (X_t-Y_t) \sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t))(\sigma_{kj}(t,X_t)-\sigma_{kj}(t,Y_t)) \mathrm dt\\ &= \sum_{i=1}^d 2(X^i_t-Y^i_t) (b^i(t,X_t)-b^i(t,Y_t)) \mathrm dt\\ &+ \sum_{i=1}^d 2(X^i_t-Y^i_t) \sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t)) \mathrm dW^j_t\\ &+ \frac 12 \sum_{i=1}^d 2\sum_{j=1}^r (\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t))^2 \mathrm dt\\ &= 2\langle X_t-Y_t, b(t,X_t)-b(t,Y_t)\rangle \mathrm dt\\ &+ 2\sum_{i=1}^d \sum_{j=1}^r (X^i_t-Y^i_t)(\sigma_{ij}(t,X_t)-\sigma_{ij}(t,Y_t)) \mathrm dW^j_t\\ &+ \|\sigma(t,X_t)-\sigma(t,Y_t))\|^2 \mathrm dt \end{align} Here, $\|\cdot\|$ denotes the Frobenius norm. Furthermore: $$f(X_0-Y_0) = f(\Xi-\Xi) = f(0) = 0$$ Integrating $\mathrm df(X_t-Y_t)$ from $0$ to $t$ now yields the following identity: \begin{align} \|X_t-Y_t\|^2 &= 2\int_0^t \langle X_t-Y_s, b(s,X_s)-b(s,Y_s)\rangle \mathrm ds\\ &+ 2\sum_{i=1}^d \sum_{j=1}^r \int_0^t (X^i_s-Y^i_s)(\sigma_{ij}(s,X_s)-\sigma_{ij}(s,Y_s)) \mathrm dW^j_s\\ &+ \int_0^t \|\sigma(s,X_s)-\sigma(s,Y_s))\|^2 \mathrm ds \end{align}