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Categories: Mathematics, Statistics, Stochastic analysis.

Kolmogorov equations

Consider the following general Itō diffusion $X_t \in \mathbb{R}$, which is assumed to satisfy the conditions for unique existence on the entire time axis:

$$\begin{aligned} \dd{X}_t = f(X_t, t) \dd{t} + g(X_t, t) \dd{B_t} \end{aligned}$$

Let $\mathcal{F}_t$ be the filtration to which $X_t$ is adapted, then we define $Y_s$ as shown below, namely as the conditional expectation of $h(X_t)$, for an arbitrary bounded function $h(x)$, given the information $\mathcal{F}_s$ available at time $s \le t$. Because $X_t$ is a Markov process, $Y_s$ must be $X_s$-measurable, so it is a function $k$ of $X_s$ and $s$:

$$\begin{aligned} Y_s \equiv \mathbf{E}[h(X_t) | \mathcal{F}_s] = \mathbf{E}[h(X_t) | X_s] = k(X_s, s) \end{aligned}$$

Consequently, we can apply Itō’s lemma to find $\dd{Y_s}$ in terms of $k$, $f$ and $g$:

$$\begin{aligned} \dd{Y_s} &= \bigg( \pdv{k}{s} + \pdv{k}{x} f + \frac{1}{2} \pdvn{2}{k}{x} g^2 \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s} \\ &= \bigg( \pdv{k}{s} + \hat{L} k \bigg) \dd{s} + \pdv{k}{x} g \dd{B_s} \end{aligned}$$

Where we have defined the linear operator $\hat{L}$ to have the following action on $k$:

$$\begin{aligned} \hat{L} k \equiv \pdv{k}{x} f + \frac{1}{2} \pdvn{2}{k}{x} g^2 \end{aligned}$$

At this point, we need to realize that $Y_s$ is a martingale with respect to $\mathcal{F}_s$, since $Y_s$ is $\mathcal{F}_s$-adapted and finite, and it satisfies the martingale property, for $r \le s \le t$:

$$\begin{aligned} \mathbf{E}[Y_s | \mathcal{F}_r] = \mathbf{E}\Big[ \mathbf{E}[h(X_t) | \mathcal{F}_s] \Big| \mathcal{F}_r \Big] = \mathbf{E}\big[ h(X_t) \big| \mathcal{F}_r \big] = Y_r \end{aligned}$$

Where we used the tower property of conditional expectations, because $\mathcal{F}_r \subset \mathcal{F}_s$. However, an Itō diffusion can only be a martingale if its drift term (the one containing $\dd{s}$) vanishes, so, looking at $\dd{Y_s}$, we must demand that:

$$\begin{aligned} \pdv{k}{s} + \hat{L} k = 0 \end{aligned}$$

Because $k(X_s, s)$ is a Markov process, we can write it with a transition density $p(s, X_s; t, X_t)$, where in this case $s$ and $X_s$ are given initial conditions, $t$ is a parameter, and the terminal state $X_t$ is a random variable. We thus have:

$$\begin{aligned} k(x, s) = \int_{-\infty}^\infty p(s, x; t, y) \: h(y) \dd{y} \end{aligned}$$

We insert this into the equation that we just derived for $k$, yielding:

$$\begin{aligned} 0 = \int_{-\infty}^\infty \!\! \Big( \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y) \Big) h(y) \dd{y} \end{aligned}$$

Because $h$ is arbitrary, and this must be satisfied for all $h$, the transition density $p$ fulfills:

$$\begin{aligned} 0 = \pdv{}{s}p(s, x; t, y) + \hat{L} p(s, x; t, y) \end{aligned}$$

Here, $t$ is a known parameter and $y$ is a “known” integration variable, leaving only $s$ and $x$ as free variables for us to choose. We therefore define the likelihood function $\psi(s, x)$, which gives the likelihood of an initial condition $(s, x)$ given that the terminal condition is $(t, y)$:

$$\begin{aligned} \boxed{ \psi(s, x) \equiv p(s, x; t, y) } \end{aligned}$$

And from the above derivation, we conclude that $\psi$ satisfies the following PDE, known as the backward Kolmogorov equation:

$$\begin{aligned} \boxed{ - \pdv{\psi}{s} = \hat{L} \psi = f \pdv{\psi}{x} + \frac{1}{2} g^2 \pdvn{2}{\psi}{x} } \end{aligned}$$

Moving on, we can define the traditional probability density function $\phi(t, y)$ from the transition density $p$, by fixing the initial $(s, x)$ and leaving the terminal $(t, y)$ free:

$$\begin{aligned} \boxed{ \phi(t, y) \equiv p(s, x; t, y) } \end{aligned}$$

With this in mind, for $(s, x) = (0, X_0)$, the unconditional expectation $\mathbf{E}[Y_t]$ (i.e. the conditional expectation without information) will be constant in time, because $Y_t$ is a martingale:

$$\begin{aligned} \mathbf{E}[Y_t] = \mathbf{E}[k(X_t, t)] = \int_{-\infty}^\infty k(y, t) \: \phi(t, y) \dd{y} = \Inprod{k}{\phi} = \mathrm{const} \end{aligned}$$

This integral has the form of an inner product, so we switch to Dirac notation. We differentiate with respect to $t$, and use the backward equation $\ipdv{k}{t} + \hat{L} k = 0$:

$$\begin{aligned} 0 = \pdv{}{t}\Inprod{k}{\phi} = \Inprod{k}{\pdv{\phi}{t}} + \Inprod{\pdv{k}{t}}{\phi} = \Inprod{k}{\pdv{\phi}{t}} - \Inprod{\hat{L} k}{\phi} = \Inprod{k}{\pdv{\phi}{t} - \hat{L}{}^\dagger \phi} \end{aligned}$$

Where $\hat{L}{}^\dagger$ is by definition the adjoint operator of $\hat{L}$, which we calculate using partial integration, where all boundary terms vanish thanks to the existence of $X_t$; in other words, $X_t$ cannot reach infinity at any finite $t$, so the integrand must decay to zero for $|y| \to \infty$:

$$\begin{aligned} \Inprod{\hat{L} k}{\phi} &= \int_{-\infty}^\infty \pdv{k}{y} f \phi + \frac{1}{2} \pdvn{2}{k}{y} g^2 \phi \dd{y} \\ &= \bigg[ k f \phi + \frac{1}{2} \pdv{k}{y} g^2 \phi \bigg]_{-\infty}^\infty - \int_{-\infty}^\infty k \pdv{}{y}(f \phi) + \frac{1}{2} \pdv{k}{y} \pdv{}{y}(g^2 \phi) \dd{y} \\ &= \bigg[ -\frac{1}{2} k g^2 \phi \bigg]_{-\infty}^\infty + \int_{-\infty}^\infty - k \pdv{}{y}(f \phi) + \frac{1}{2} k \pdvn{2}{}{y}(g^2 \phi) \dd{y} \\ &= \int_{-\infty}^\infty k \: \big( \hat{L}{}^\dagger \phi \big) \dd{y} = \Inprod{k}{\hat{L}{}^\dagger \phi} \end{aligned}$$

Since $k$ is arbitrary, and $\ipdv{\Inprod{k}{\phi}}{t} = 0$ for all $k$, we thus arrive at the forward Kolmogorov equation, describing the evolution of the probability density $\phi(t, y)$:

$$\begin{aligned} \boxed{ \pdv{\phi}{t} = \hat{L}{}^\dagger \phi = - \pdv{}{y}(f \phi) + \frac{1}{2} \pdvn{2}{}{y}(g^2 \phi) } \end{aligned}$$

This can be rewritten in a way that highlights the connection between Itō diffusions and physical diffusion, if we define the diffusivity $D$, advection $u$, and probability flux $J$:

$$\begin{aligned} D \equiv \frac{1}{2} g^2 \qquad \quad u = f - \pdv{D}{x} \qquad \quad J \equiv u \phi - D \pdv{\phi}{x} \end{aligned}$$

Such that the forward Kolmogorov equation takes the following conservative form, so called because it looks like a physical continuity equation:

$$\begin{aligned} \boxed{ \pdv{\phi}{t} = - \pdv{J}{x} = - \pdv{}{x}\Big( u \phi - D \pdv{\phi}{x} \Big) } \end{aligned}$$

Note that if $u = 0$, then this reduces to Fick’s second law. The backward Kolmogorov equation can also be rewritten analogously, although it is less noteworthy:

$$\begin{aligned} \boxed{ - \pdv{\psi}{t} = u \pdv{\psi}{x} + \pdv{}{x}\Big( D \pdv{\psi}{x} \Big) } \end{aligned}$$

Notice that the diffusivity term looks the same in both the forward and backward equations; we say that diffusion is self-adjoint.

References

1. U.H. Thygesen, Lecture notes on diffusions and stochastic differential equations, 2021, Polyteknisk Kompendie.

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