Key words: Ito operator, Property of trace, Fokker-Planck equation, L2 adjoint, Hilbert space.
1. Ito operator
Let $X$ be the solution of the following SDE.
$d X_t = b(X_t)dt + \sigma(X_t)dW_t$
where $X_t \in \mathbb{R}^d$, and $W$ is an n-dimensional standard Brownian motion ($\sigma$ is a $d \times n$). Define $\mathcal{L}$ as the infinitesimal generator of X on a $C^2$-function $f$:
$\mathcal{L} f(X_{t}) = f_{x}(X_{t})^{T} d X_{t} + \frac{1}{2} dX_t^{T}f_{xx}(X_t) dX_t$
Note that the second term $\in \mathbb{R}$. Therefore, we can further rewrite the formula like
$\mathcal{L} f(X_{t}) = f_{x}(X_{t})^{T} d X_{t} + \frac{1}{2} tr\Big(dX_t^{T}f_{xx}(X_t) dX_t\Big) $
$\quad \quad \quad= f_{x}(X_aw{t})^{T} d X_{t} + \frac{1}{2} tr\Big(dX_tdX_t^{T}f_{xx}(X_t)\Big) $
$\quad \quad \quad= f_{x}(X_{t})^{T} d X_{t} + \frac{1}{2} tr\Big(\sigma(X_t)^Tf_{xx}(X_t)\sigma(X_t)\Big) dt$
Note that $tr(ABC) = tr(BCA) = tr(CAB)$ (However, $tr(ABC)$ may not equal to $tr(ACB)$). Therefore, L can be written as:
$L = \sum\limits_{i=1}^d b_i(t, x) \partial_{i} + \sum\limits_{i, j= 1}^d a_{i, j}(t, x)\partial_{i, j}$
, where $a_{i, j}$ is the $i$th row and $j$th columns of the matrix $\sigma(X_t)^T\sigma(X_t)$ (the transpose of $\sigma(X_t)\sigma(X_t)^T$). (Recall $tr(A^TB) = \sum_{i, j}a_{i, j}b_{i, j}$)
2 Fokker-Planck equation (Kolmogorov forward equation)
It is an intentional practice for the derivation of L2 adjoint.
Theorem:
Let H and K be Hilbert spaces, and let $A: H \xrightarrow{} K$ be a bounded, linear map. Then there exist a unique bounded linear map $A^*: K \xrightarrow{} H$ such that:
$\forall x \in H, \quad \forall y \in K, \quad \langle Ax, y \rangle = \langle x, A^*y \rangle$
By the definition of linear operator, Ito operator is a linear operator. By the definition of Ito integral, we can ensure that it is bounded. Assume $X_t$ has a density distribution $p(t, \cdot)$. Therefore, there exist a unique L2 adjoint such that:
$\frac{d}{dt} \mathbb{E}\Big(f(X_t)\Big) = \frac{d}{dt} \int_{x} f(x) p(t, x)dx = \int_{x} Lf(x) p(t, x) dx = \int_{x}f(x)L^{*}p(t, x) dx$
By definition, $L^{*}p(t, x)$ should be the Fokker-Planck equation.
$\int_{x} Lf(x) p(t, x) dx = \int_{x} \Big(f_{x}^Tb(x, t) + \frac{1}{2}tr(\sigma^T f_{xx} \sigma)\Big) p(t, x) dx$
$\int_{x} f_{x}^Tb(x, t) p(t, x)dx = \int_{x} \sum_{i}\Big(p(t, x)b(t, x_i)\frac{\partial}{\partial x_i}f(x, t)\Big) dx$
Note that
$ \int_{x} p(t, x)b(t, x_i)\frac{\partial}{\partial x_i}f(x) dx = \int_{x} p(t, x)b_i(t, x) d f dx_{-i}$
$ = \int_{x} p(t, x)b_i(t, x)f(x) |_{x_i = -\infty}^{x_i= + \infty} - f(x)d\Big(p(t, x)b_i(t, x)\Big) d x_{-i}$
$ = -\int_{x}f(x)d\Big(p(t, x)b_i(t, x)\Big) d x_{-i} = -\int_{x}f(x)p(t, x)\frac{\partial b_{i}(t, x)}{\partial x_i} d x -\int_{x}f(x)b_{i}(t, x)\frac{\partial p(t, x)}{\partial x_i} d x$
Therefore, $\int_{x} f_{x}^Tb(x, t) p(t, x)dx = -\int_{x} f(x) \sum\limits_{i}\frac{\partial}{\partial x_i} \Big(p(t, x) b_{i}(t, x)\Big) dx$
Similarly, the second term can be written as:
$\int_{x} \frac{1}{2}tr(\sigma^T f_{xx} \sigma) p(t, x) dx = \int_{x}\frac{1}{2} \sum_{i, j}^d \frac{\partial^2}{\partial x_i\partial x_j} a_{i, j}p(x, t) dx$.
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