# Derivation of Backward Kolmogorov Equation

I'm following Kallianpur-Gopinath's textbook "Stochastic analysis and diffusion processes" to study Kolmogorov equations and I got stuck in a step of the derivation of the backward equation.

In particular the authors (guess it's a pretty standard construction actually) start from the solution of a d-dimensional SDE

$$dX_t = b(t,X_t)dt + \sigma(t,X_t)dB_t$$

and, after computing the Taylor expansion of the probability transition function, arrive to write something like

$$-\big(p(s,x;t,y) -p(s-h,x;t,y)\big) = \sum_{i=1}^d \frac{\partial}{\partial x_i}p(s,x;t,y) \int_{\mathbb{R}^d}(z_i-x_i)p(s-h,x;s,z)dz + \\ \frac{1}{2} \sum_{i=1}^d\sum_{j=1}^d \frac{\partial^2}{\partial x_i \partial x_j}p(s,x;t,y) \int_{\mathbb{R}^d}(z_i-x_i)(z_j-x_j)p(s-h,x;s,z)dz +\\ \int_{\mathbb{R}^d} o\big(|z-x|^3\big) p(s-h,x;s,z)dz\\$$

By dividing by $h$ and letting $h \to 0$ we should get the backward equation

$$-\frac{ \partial p(s,x;t,y) }{ \partial s } = b(s,x) \cdot \nabla p(s,x;t,y) + \frac{1}{2} (\sigma(s,x), \nabla^2 p(s,x;t,y) \sigma(s,x) )$$

what I don't understand is the step

$$\lim_{h\to 0} \frac{1}{h} \int_{\mathbb{R}^d} (z_i-x_i) p(s-h,x;s,z)dz \stackrel{?}{=} b_i(s,x)$$

I understand this should be related with the fact that the solutions of SDE are diffusion processes, but the definition of diffusion process requires

$$\lim_{h\to 0} \frac{1}{h} \int_{U_x} (z_i-x_i) p(s-h,x;s,z)dz = b_i(s,x)$$

for any bounded neighborhood $U_x$ of $x$, but instead we are considering an integral over $\mathbb{R}^d$. Even if I split the integral in two, one over a bounded set and one over its complement I don't see how to prove that the second goes to $0$. Any help is appreciated.

EDIT:

Quoting from the same book:

A Markov process $X_t$ with transition function $p(s,x;t,A)$ is called a diffusion process if it has continuous paths and if there exists an $\mathbb{R}^d$-valued function $b(s,x)$ and a $d \times d$-matrix-valued function $a(t,x)$ s.t. for every bounded open neighborhood $U_x$ of $x$

• $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x^c} p(t,x;t+h,dy) = 0$

• $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x} (y_i - x_i) p(t,x;t+h,dy) = b_i(t,x) \ \ \ \ i \in \{1...d\}$

• $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x} (y_i - x_i)(y_j - x_j) p(t,x;t+h,dy) = a_{i,j}(t,x) \ \ \ \ i,j \in \{1...d\}$

and immediately after there is a theorem that states that the solution of the SDE written at the top of this post, under the usual hypotheses of Lipschitzianity of $b$ and $\sigma$, is a diffusion process with drift $b$ and diffusion coefficient $\sigma \sigma^*$.

• Could you please add the definition of a diffusion process to your question? There are several ones and it is hard to answer your question without knowing the "proper" definition.
– saz
Jun 15, 2015 at 15:56
• @saz Sorry, didn't know there are different definitions for diffusions. I edited the post with the definition written on the book, thanks! Jun 15, 2015 at 20:13
• Because of the first condition $\lim_{h \downarrow 0} \frac{1}{h} \int_{U_x^c} p(t,x;t+h,dy) = 0$, the integration outside $U_x$ is 0. Jan 4 at 22:30