Diffusion Processes

If you’re not not accustomed to the language of probability now it’s the time to read some lecture notes on measure theory. The following is not an easy read, but it contains all the necessary definitions and results we will need throughout the course.

Filtrations

Recall that a \(\sigma\)-algebra represents available information.

In a dynamical setting, information changes with time. We therefore consider a family of \(\sigma\)-algebras

\[ \{\mathcal{F}_t\}_{t\geq 0}. \]

A filtration is an increasing family of sub-\(\sigma\)-algebras of \(\mathcal{F}\):

\[ \mathcal{F}_s \subseteq \mathcal{F}_t \subseteq \mathcal{F}, \qquad s\leq t. \]

Hence \(\mathcal{F}_t\) contains all information available up to time \(t\). Imagine yourself reading a book: the information you have at time \(t\) is the information contained in the first \(t\) pages, and your information set is increasing as you read more pages.

The natural filtration

Given a stochastic process \(\{X_t\}_{t\geq0}\), its natural filtration is

\[ \mathcal{F}_t^X = \sigma(X_s:0\leq s\leq t). \]

It is the smallest \(\sigma\)-algebra that makes all observations

\[ X_s, \qquad s\leq t, \]

measurable. I like to think of this as the least information that should be contained in the first \(t\) pages of a book in order to understand the story up to that point.

Adapted processes

A stochastic process \(\{X_t\}\) is adapted to a filtration \(\{\mathcal{F}_t\}\) if

\[ X_t \quad\text{is }\mathcal{F}_t\text{-measurable} \]

for every \(t\).

Informally, the value \(X_t\) can be determined using information available at time \(t\).

An adapted process therefore cannot depend on future information.

Every process is automatically adapted to its own natural filtration:

\[ X_t \text{ is } \mathcal{F}_t^X\text{-measurable}. \]

To end the book analogy, in our book we are able to understand the story up to page \(t\) without having to peek ahead.

Markov Process

Informally, a Markov process is a stochastic process that retains no memory.

We now set

\[ \mathcal F_t^X:=\sigma\{X_s:s\leq t\}. \]

Then a process \(\{X_t,\ t\in\mathbb R_+\}\) is a Markov process if

\[ \mathbb P(X_t\in\Gamma\mid\mathcal F_s^X) = \mathbb P(X_t\in\Gamma\mid X_s), \qquad t\geq s. \]

which is, the history of the process up to time \(s\) is irrelevant for predicting the future at time \(t\). Only the present matters.

Transition function

Define the transition function

\[ P(\Gamma,t\mid x,s) := \mathbb P(X_t\in\Gamma\mid X_s=x). \]

It is a probability measure on \(\mathbb R^d\) and satisfies the Chapman–Kolmogorov equation

\[ P(\Gamma,t\mid x,s) = \int_{\mathbb R^d} P(\Gamma,t\mid y,u)\, P(dy,u\mid x,s), \]

where \(\Gamma\in\mathcal B(\mathbb R^d)\), \(x\in\mathbb R^d\), and \(s\leq u\leq t\).

This formula merely states that the probability of going from \(x\) at time \(s\) to \(\Gamma\) at time \(t\) is the sum of the probabilities of going from \(x\) at time \(s\) to \(y\) at time \(u\), and then from \(y\) at time \(u\) to \(\Gamma\) at time \(t\), integrated over all possible intermediate states \(y\). Iterating the argument can be useful to see why it works. The proof can also be useful:

\[ \begin{aligned} P(\Gamma,t\mid x,s) &= \mathbb P(X_t\in\Gamma\mid X_s=x)\\ &= \mathbb P(X_t\in\Gamma\mid\mathcal F_s^X)\\ &= \mathbb E[\mathbf 1_\Gamma(X_t)\mid\mathcal F_s^X]\\ &= \mathbb E\left[ \mathbb E[ \mathbf 1_\Gamma(X_t) \mid\mathcal F_s^X ] \mid\mathcal F_u^X \right]\\ &= \mathbb E\left[ \mathbb E[ \mathbf 1_\Gamma(X_t) \mid\mathcal F_u^X ] \mid\mathcal F_s^X \right]\\ &= \mathbb E\left[ \mathbb E[ \mathbf 1_\Gamma(X_t) \mid X_u ] \mid\mathcal F_s^X \right]\\ &= \mathbb E\left[ \mathbb P(X_t\in\Gamma\mid X_u=y) \mid X_s=x \right]\\ &= \int_{\mathbb R^d} P(\Gamma,t\mid y,u)\, P(dy,u\mid x,s). \end{aligned} \]

Transition probability density

Now rewrite the transition function as

\[ P(\Gamma,t\mid x,s) = \int_\Gamma p(y,t\mid x,s)\,dy, \]

where \(p(\cdot\mid\cdot)\) is the transition probability density.

Chapman–Kolmogorov becomes

\[ \int_\Gamma p(y,t\mid x,s)\,dy = \int_{\mathbb R^d} P(\Gamma,t\mid z,u) P(dz,u\mid x,s), \]

hence

\[ \int_\Gamma p(y,t\mid x,s)\,dy = \int_{\mathbb R^d} \int_\Gamma p(y,t\mid z,u) p(z,u\mid x,s) \,dy\,dz. \]

It follows that

\[ p(y,t\mid x,s) = \int_{\mathbb R^d} p(y,t\mid z,u) p(z,u\mid x,s) \,dz. \]

as before.

Consider now a time-homogeneous Markov process,

\[ P(\Gamma,t\mid x,s) = P(\Gamma,t-s\mid x,0) =:P(t-s,x,\Gamma). \]

Then

\[ P(t+s,x,\Gamma) = \int_{\mathbb R^d} P(t,z,\Gamma)\, P(s,x,dz). \]

Or, in terms of the transition density,

\[ p(t+s,x,y) = \int_{\mathbb R^d} p(t,z,y) p(s,x,z)\,dz. \]

Which makes sense if we think of restarting the clock at every intermediate step.

Notice that the transition function and the initial distribution are sufficient to determine the law of a Markov process.

If the process starts from an arbitrary point \(x_0\in\mathbb R^d\), its initial distribution is the Dirac delta

\[ \nu(dx)=\delta_{x_0}(dx), \]

where \(\delta_{x_0}\) is the probability measure concentrated at \(x_0\), defined by

\[ \int_{\mathbb R^d} f(x)\,\delta_{x_0}(dx)=f(x_0). \]

Hence,

\[ \mathbb P_{x_0}(X_t\in\Gamma) = \int_{\mathbb R^d} P(x,t,\Gamma)\,\delta_{x_0}(dx) = P(x_0,t,\Gamma). \]

More generally, if the initial condition is distributed according to \(\nu\),

\[ \mathbb P_\nu(X_t\in\Gamma) = \int_{\mathbb R^d} P(x,t,\Gamma)\,\nu(dx). \]

So the general formula simply averages the transition probabilities over all possible starting points.

Similarly, we can calculate the expectation for an observable \(f\):

\[ \mathbb E_\nu[f(X_t)] = \int_{\mathbb R^d} f(x) \underbrace{ \int_{\mathbb R^d} \nu(dx_0)\, P(x_0,t,dx) }_{\mu_t(dx)}. \]

Semigroup

To describe time-homogeneous Markov processes, introduce a single-parameter family of linear operators such that

\[ P_0=I, \]

and

\[ P_{t+s}=P_tP_s, \qquad \forall s,t\geq0. \]

Define

\[ P_t f(x) := \mathbb E[f(X_t)\mid X_0=x] = \int_{\mathbb R^d} f(y)P(t,x,dy). \]

As expected,

\[ P_0f(x) = \mathbb E[f(X_0)\mid X_0=x] = f(x). \]

Furthermore,

\[ \begin{aligned} P_{t+s}f(x) &= \int_{\mathbb R^d} f(y)P(t+s,x,dy)\\ &= \int_{\mathbb R^d} \int_{\mathbb R^d} f(y) P(t,z,dy) P(s,x,dz)\\ &= \int_{\mathbb R^d} P_tf(z) P(s,x,dz)\\ &= P_s\circ P_tf(x). \end{aligned} \]

Infinitesimal generator

Now define \(D(\mathcal L)\) as the set of all \(f\in C_b(E)\) such that

\[ \mathcal Lf := \lim_{t\to0} \frac{P_tf-f}{t} \]

exists.

Then the operator \(\mathcal L\) is the infinitesimal generator of \(P_t\), or the generator of the Markov process.

Now,

\[ \mathcal L = \lim_{h\to0} \frac{P_h-I}{h}, \]

which reminds us of the definition of the derivative. Notice:

\[ P_h\approx I+h\mathcal L. \]

Then

\[ P_t = \left(P_{t/n}\right)^n \approx \left( I+\frac{t}{n}\mathcal L \right)^n. \]

(verify the first equality using the semigroup property). Taking the limit,

\[ P_t = \lim_{n\to\infty} \left( I+\frac{t}{n}\mathcal L \right)^n = e^{t\mathcal L}. \]

Now define

\[ u(x,t):=P_tf(x) \]

and say we want to find the equation governing \(P_t\), i.e. determining the evolution of the conditional expectation of the observable \(f\).

Then

\[ \begin{aligned} \frac{\partial}{\partial t}u(x,t) &= \lim_{h\to0} \frac{ P_{t+h}f(x)-P_tf(x) }{h}\\ &= P_t\mathcal Lf(x)\\ &= \mathcal L P_tf(x). \end{aligned} \]

Therefore

\[ \frac{\partial u}{\partial t} = \mathcal Lu. \]

Since \(P_0=I\), we have

\[ u(x,0)=f(x). \]

Hence \(u(x,t)\) satisfies the initial value problem

\[ \begin{cases} \dfrac{\partial u}{\partial t} = \mathcal Lu,\\[6pt] u(x,0)=f(x). \end{cases} \]

This is the backward Kolmogorov equation.

Adjoint semigroup

While \(P_t\) acts on continuous bounded functions, we can define \(P_t^*\), which acts on probability measures:

\[ P_t^*\mu(\Gamma) = \int_{\mathbb R^d} P(t,x,\Gamma)\,d\mu(x). \]

Formally, we have

\[ \int_{\mathbb R^d} P_tf(x)\,d\mu(x) = \int_{\mathbb R^d} f(x)\,d(P_t^*\mu)(x). \]

Which makes sense if we think of \(P_t^*\) as the \(L^2\) adjoint of \(P_t\). We can also define, as before, the generator \(\mathcal L^*\) of the semigroup \(P_t^*\), and we have

\[ P_t^* = e^{t\mathcal L^*}. \]

Also,

\[ \int (\mathcal Lf)\,d\mu = \int f(\mathcal L^*\mu)\,dx. \]

Now define the law of the Markov process as

\[ \mu_t = P_t^*\mu, \]

which evolves according to

\[ \begin{cases} \dfrac{\partial\mu_t}{\partial t} = \mathcal L^*\mu_t,\\[6pt] \mu_0=\mu. \end{cases} \]

Assuming \(\mu\) and \(\mu_t\) have densities, denote them by \(\rho_0(\cdot)\) and \(\rho(\cdot,t)\).

Then

\[ \begin{cases} \dfrac{\partial\rho}{\partial t} = \mathcal L^*\rho,\\[6pt] \rho(x,0)=\rho_0(x). \end{cases} \]

Is the Fokker–Planck, or forward Kolmogorov, equation.

To conclude this introduction, the backward and forward Kolmogorov equations describe the evolution of the same Markov process from two complementary perspectives.

The backward Kolmogorov equation governs the evolution of an observable \(f\), as it tells us how the expected future value of \(f(X_t)\) depends on the initial state \(x\):

\[ \partial_t u=\mathcal L u. \]

The forward Kolmogorov equation, instead, governs the evolution of the law of the process. If \(\rho(x,t)\) is the probability density of \(X_t\), then

\[ \partial_t\rho=\mathcal L^*\rho. \]

Hence, \(\mathcal L\) propagates observables, while its adjoint \(\mathcal L^*\) propagates probability distributions.

Diffusion process

A Markov process consists of three parts:

a drift;
a random part;
a jump process.

A diffusion process is a Markov process with continuous paths, hence no jumps.

In one dimension, we need the following.

Continuity

For all \(\varepsilon>0\) and all \(x\),

\[ \int_{|x-y|>\varepsilon} P(dy,t\mid x,s) = o(t-s). \]

Drift

There exists \(b(x,s)\) such that, for all \(\varepsilon>0\) and all \(x\),

\[ \int_{|y-x|\leq\varepsilon} (y-x) P(dy,t\mid x,s) = b(x,s)(t-s) + o(t-s). \]

Diffusion

There exists \(\Sigma(x,s)\) such that, for all \(\varepsilon>0\) and all \(x\),

\[ \int_{|y-x|\leq\varepsilon} (y-x)^2 P(dy,t\mid x,s) = \Sigma(x,s)(t-s) + o(t-s). \]

Assuming that the first two moments exist, we can write, even more intuitively,

\[ b(x,s) = \lim_{t\to s} \mathbb E \left[ \frac{X_t-X_s}{t-s} \mid X_s=x \right], \]

and

\[ \Sigma(x,s) = \lim_{t\to s} \mathbb E \left[ \frac{(X_t-X_s)^2}{t-s} \mid X_s=x \right]. \]

Kolmogorov equations

From the definition of a diffusion process, we can derive explicit formulas for the forward and backward Kolmogorov equations.

Backward Kolmogorov

Let

\[ u(x,s) := \mathbb E[f(X_t)\mid X_s=x] = \int_{\mathbb R^d} f(y)P(dy,t\mid x,s). \]

Then \(u(x,s)\) solves the final value problem for a PDE of parabolic type:

\[ \begin{cases} -\dfrac{\partial u}{\partial s} = b(x,s) \dfrac{\partial u}{\partial x} + \dfrac{1}{2} \Sigma(x,s) \dfrac{\partial^2u}{\partial x^2}, \\[8pt] u(x,t)=f(x). \end{cases} \]

Parts of the proof are very interesting, so I recommend reading it at least once.

Proof:

The continuity assumption for the diffusion process, together with the boundedness of \(f\), implies that

\[ \begin{aligned} u(x,s) &= \int_{\mathbb{R}} f(y)P(dy,t\mid x,s) \\ &= \int_{|y-x|\leq\varepsilon} f(y)P(dy,t\mid x,s) + \int_{|y-x|>\varepsilon} f(y)P(dy,t\mid x,s) \\ &\leq \int_{|y-x|\leq\varepsilon} f(y)P(dy,t\mid x,s) + \lVert f\rVert_{L^\infty} \int_{|y-x|>\varepsilon} P(dy,t\mid x,s) \\ &= \int_{|y-x|\leq\varepsilon} f(y)P(dy,t\mid x,s) + o(t-s). \end{aligned} \]

We add and subtract the final condition \(f(x)\) and use the previous calculation to obtain

\[ \begin{aligned} u(x,s) &= \int_{\mathbb{R}} f(y)P(dy,t\mid x,s) \\ &= f(x) + \int_{\mathbb{R}} \bigl(f(y)-f(x)\bigr)P(dy,t\mid x,s) \\ &= f(x) + \int_{|y-x|\leq\varepsilon} \bigl(f(y)-f(x)\bigr)P(dy,t\mid x,s) \\ &\qquad + \int_{|y-x|>\varepsilon} \bigl(f(y)-f(x)\bigr)P(dy,t\mid x,s) \\ &= f(x) + \int_{|y-x|\leq\varepsilon} \bigl(f(y)-f(x)\bigr)P(dy,t\mid x,s) + o(t-s). \end{aligned} \]

Now we show that \(u(s,x)\) solves the backward Kolmogorov equation. We use the Chapman–Kolmogorov equation:

\[ \begin{aligned} u(x,\sigma) &= \int_{\mathbb{R}} f(z)P(dz,t\mid x,\sigma) \\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(z)P(dz,t\mid y,\rho) P(dy,\rho\mid x,\sigma) \\ &= \int_{\mathbb{R}} u(y,\rho)P(dy,\rho\mid x,\sigma). \end{aligned} \]

Where we are integrating over future paths. We use Taylor’s theorem to obtain

\[ u(z,\rho)-u(x,\rho) = \frac{\partial u(x,\rho)}{\partial x}(z-x) + \frac{1}{2} \frac{\partial^2u(x,\rho)}{\partial x^2} (z-x)^2(1+\alpha_\varepsilon), \qquad |z-x|\leq\varepsilon, \]

where

\[ \lim_{\varepsilon\to0}\alpha_\varepsilon=0. \]

Now we will use the last two results:

\[ \begin{aligned} \frac{u(x,s)-u(x,s+h)}{h} &= \frac{1}{h} \left( \int_{\mathbb{R}} P(dy,s+h\mid x,s)u(y,s+h) - u(x,s+h) \right) \\ &= \frac{1}{h} \int_{\mathbb{R}} P(dy,s+h\mid x,s) \bigl(u(y,s+h)-u(x,s+h)\bigr) \\ &= \frac{1}{h} \int_{|x-y|<\varepsilon} P(dy,s+h\mid x,s) \bigl(u(y,s+h)-u(x,s+h)\bigr) + o(1) \\ &= \frac{\partial u}{\partial x}(x,s+h) \frac{1}{h} \int_{|x-y|<\varepsilon} (y-x)P(dy,s+h\mid x,s) \\ &\qquad + \frac{1}{2} \frac{\partial^2u}{\partial x^2}(x,s+h) \frac{1}{h} \int_{|x-y|<\varepsilon} (y-x)^2P(dy,s+h\mid x,s) (1+\alpha_\varepsilon) + o(1) \\ &= b(x,s) \frac{\partial u}{\partial x}(x,s+h) + \frac{1}{2} \Sigma(x,s) \frac{\partial^2u}{\partial x^2}(x,s+h) (1+\alpha_\varepsilon) + o(1). \end{aligned} \]

where we have used the definition of the drift and diffusion coefficients in the last step. The result follows by taking the limit \(h\to0\) and \(\varepsilon\to0\).

If \(b=b(x)\) and \(\Sigma=\Sigma(x)\), we can rewrite this as an initial value problem by letting

\[ v=u(x,t-s). \]

Then

\[ \begin{cases} \dfrac{\partial v}{\partial(t-s)} = b(x)\dfrac{\partial v}{\partial x} + \dfrac{1}{2} \Sigma(x) \dfrac{\partial^2v}{\partial x^2}, \\[8pt] v(x,0)=f(x). \end{cases} \]

In the time-homogeneous case, we can finally set \(s=0\) to write

\[ \begin{cases} \dfrac{\partial u}{\partial t} = b(x)\dfrac{\partial u}{\partial x} + \dfrac{1}{2} \Sigma(x) \dfrac{\partial^2u}{\partial x^2}, \\[8pt] u(x,0)=f(x), \end{cases} \]

where

\[ \mathcal L = b(x)\partial_x + \frac{1}{2} \Sigma(x)\partial_{xx}. \]

Forward Kolmogorov (Fokker–Planck)

We have

\[ P(dy,t\mid x,s) = p(y,t\mid x,s)\,dy. \]

Under suitable conditions, the transition probability density is the solution to the initial value problem

\[ \begin{cases} \dfrac{\partial p}{\partial t} = -\dfrac{\partial}{\partial y} \left( b(t,y)p \right) + \dfrac{1}{2} \dfrac{\partial^2}{\partial y^2} \left( \Sigma(t,y)p \right), \\[10pt] p(y,s\mid x,s) = \delta(x-y). \end{cases} \]

The proof is less interesting than the backward case, but it is presented for completeness.

Proof:

Fix \(f(y)\in C_0^2(\mathbb{R})\). An argument similar to that used in the proof of the backward Kolmogorov equation gives

\[ \lim_{h\to0} \frac{1}{h} \left( \int_{\mathbb{R}} f(y)p(y,s+h\mid x,s)\,dy-f(x) \right) = b(x,s)\frac{df}{dx}(x) + \frac{1}{2}\Sigma(x,s)\frac{d^2f}{dx^2}(x). \]

On the other hand,

\[ \begin{aligned} \int_{\mathbb{R}} f(y)\frac{\partial}{\partial t}p(y,t\mid x,s)\,dy &= \frac{\partial}{\partial t} \int_{\mathbb{R}}f(y)p(y,t\mid x,s)\,dy \\ &= \lim_{h\to0}\frac{1}{h} \int_{\mathbb{R}} \bigl(p(y,t+h\mid x,s)-p(y,t\mid x,s)\bigr)f(y)\,dy \\ &= \lim_{h\to0}\frac{1}{h} \left( \int_{\mathbb{R}}p(y,t+h\mid x,s)f(y)\,dy - \int_{\mathbb{R}}p(z,t\mid x,s)f(z)\,dz \right) \\ &= \lim_{h\to0}\frac{1}{h} \left( \int_{\mathbb{R}}\int_{\mathbb{R}} p(y,t+h\mid z,t)p(z,t\mid x,s)f(y)\,dy\,dz - \int_{\mathbb{R}}p(z,t\mid x,s)f(z)\,dz \right) \\ &= \lim_{h\to0} \int_{\mathbb{R}} p(z,t\mid x,s) \frac{1}{h} \left( \int_{\mathbb{R}}p(y,t+h\mid z,t)f(y)\,dy-f(z) \right)dz \\ &= \int_{\mathbb{R}} p(z,t\mid x,s) \left( b(z,t)\frac{df}{dz}(z) + \frac{1}{2}\Sigma(z,t)\frac{d^2f}{dz^2}(z) \right)dz \\ &= \int_{\mathbb{R}} b(z,t)p(z,t\mid x,s)f'(z)\,dz + \frac{1}{2} \int_{\mathbb{R}} \Sigma(z,t)p(z,t\mid x,s)f''(z)\,dz \\ &= \left[ b(z,t)p(z,t\mid x,s)f(z) \right]_{-\infty}^{\infty} - \int_{\mathbb{R}} \frac{\partial}{\partial z} \bigl(b(z,t)p(z,t\mid x,s)\bigr)f(z)\,dz \\ &\quad + \frac{1}{2} \left[ \Sigma(z,t)p(z,t\mid x,s)f'(z) \right]_{-\infty}^{\infty} - \frac{1}{2} \int_{\mathbb{R}} \frac{\partial}{\partial z} \bigl(\Sigma(z,t)p(z,t\mid x,s)\bigr)f'(z)\,dz \\ &= - \int_{\mathbb{R}} \frac{\partial}{\partial z} \bigl(b(z,t)p(z,t\mid x,s)\bigr)f(z)\,dz \\ &\quad - \frac{1}{2} \left[ \frac{\partial}{\partial z} \bigl(\Sigma(z,t)p(z,t\mid x,s)\bigr)f(z) \right]_{-\infty}^{\infty} \\ &\quad + \frac{1}{2} \int_{\mathbb{R}} \frac{\partial^2}{\partial z^2} \bigl(\Sigma(z,t)p(z,t\mid x,s)\bigr)f(z)\,dz \\ &= \int_{\mathbb{R}} \left( - \frac{\partial}{\partial z} \bigl(b(z,t)p(z,t\mid x,s)\bigr) + \frac{1}{2} \frac{\partial^2}{\partial z^2} \bigl(\Sigma(z,t)p(z,t\mid x,s)\bigr) \right)f(z)\,dz. \end{aligned} \]

Here the boundary terms vanish since \(f\in C_0^2(\mathbb{R})\). Hence

\[ \int_{\mathbb{R}} \left[ \frac{\partial p}{\partial t} + \frac{\partial}{\partial z}(bp) - \frac{1}{2}\frac{\partial^2}{\partial z^2}(\Sigma p) \right]f(z)\,dz =0 \]

for every \(f\in C_0^2(\mathbb{R})\). Therefore the integrand is zero (in the weak sense, and pointwise under the assumed smoothness), giving

\[ \boxed{ \frac{\partial p}{\partial t} = -\frac{\partial}{\partial z}(bp) + \frac{1}{2}\frac{\partial^2}{\partial z^2}(\Sigma p). } \]

Assume that the initial distribution of \(X_t\) is \(\rho_0(x)\) and set \(s=0\) above.

Define

\[ \rho(y,t) := \int p(y,t\mid x,0) \rho_0(x)\,dx. \]

Then we can obtain the probability density at y of the process \(X_t\), initially distributed according to \(\rho_0(x)\).

\[ \begin{cases} \dfrac{\partial\rho}{\partial t} = -\dfrac{\partial}{\partial y} \left( b \rho \right) + \dfrac{1}{2} \dfrac{\partial^2}{\partial y^2} \left( \Sigma \rho \right), \\[10pt] \rho(y,0)=\rho_0(y). \end{cases} \]

Diffusion processes in \(\mathbb R^d\)

For a process in \(\mathbb R^d\), we define the drift and diffusion coefficients

\[ b(x,s) = \lim_{t\to s} \frac{1}{t-s} \int_{|y-x|<\varepsilon} (y-x) P(dy,t\mid x,s), \]

and

\[ \Sigma(x,s) = \lim_{t\to s} \frac{1}{t-s} \int_{|y-x|<\varepsilon} (y-x)\otimes(y-x) P(dy,t\mid x,s). \]

As before, the generator is

\[ \mathcal L = b(x,s)\cdot\nabla + \frac{1}{2} \Sigma(x,s):\nabla\nabla, \]

where

\[ A:\nabla\nabla f = \sum_{i,j} A_{ij} \frac{\partial^2f} {\partial x_i\partial x_j}. \]

Backward Kolmogorov

The backward Kolmogorov equation is

\[ \begin{cases} -\dfrac{\partial u}{\partial s} = b(x,s)\cdot\nabla_xu + \dfrac{1}{2} \Sigma(x,s):\nabla_x\nabla_xu, \\[10pt] u(t,x)=f(x). \end{cases} \]

For a time-homogeneous process, this becomes

\[ \begin{cases} \dfrac{\partial u}{\partial t} = \mathcal Lu, \\[6pt] u(x,0)=f(x). \end{cases} \]

Forward Kolmogorov

The forward Kolmogorov equation is

\[ \begin{cases} \dfrac{\partial p}{\partial t} = \nabla_y\cdot \left( -b(t,y)p + \dfrac{1}{2} \nabla_y\cdot \left( \Sigma(t,y)p \right) \right), \\[10pt] p(y,s\mid x,s) = \delta(x-y). \end{cases} \]

For a time-homogeneous process,

\[ \begin{cases} \dfrac{\partial\rho}{\partial t} = \mathcal L^*\rho, \\[6pt] \rho(x,0)=\rho_0(x). \end{cases} \]