---
title: "Dynkin's formula"
date: 2021-11-28
categories:
- Mathematics
- Stochastic analysis
layout: "concept"
---

Given an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
with a time-independent drift $f$ and intensity $g$
such that the diffusion uniquely exists on the $t$-axis.
We define the **infinitesimal generator** $\hat{A}$
as an operator with the following action on a given function $h(x)$,
where $\mathbf{E}$ is a
[conditional expectation](/know/concept/conditional-expectation/):

$$\begin{aligned}
    \boxed{
        \hat{A}\{h(X_0)\}
        \equiv \lim_{t \to 0^+} \bigg[ \frac{1}{t} \mathbf{E}\Big[ h(X_t) - h(X_0) \Big| X_0 \Big] \bigg]
    }
\end{aligned}$$

Which only makes sense for $h$ where this limit exists.
The assumption that $X_t$ does not have any explicit time-dependence
means that $X_0$ need not be the true initial condition;
it can also be the state $X_s$ at any $s$ infinitesimally smaller than $t$.

Conveniently, for a sufficiently well-behaved $h$,
the generator $\hat{A}$ is identical to the Kolmogorov operator $\hat{L}$
found in the [backward Kolmogorov equation](/know/concept/kolmogorov-equations/):

$$\begin{aligned}
    \boxed{
        \hat{A}\{h(x)\}
        = \hat{L}\{h(x)\}
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-kolmogorov"/>
<label for="proof-kolmogorov">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-kolmogorov">Proof.</label>
We define a new process $Y_t \equiv h(X_t)$, and then apply Itō's lemma, leading to:

$$\begin{aligned}
    \dd{Y_t}
    &= \bigg( \pdv{h}{x} f(X_t) + \frac{1}{2} \pdvn{2}{h}{x} g^2(X_t) \bigg) \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
    \\
    &= \hat{L}\{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
\end{aligned}$$

Where we have recognized the definition of $\hat{L}$.
Integrating the above equation yields:

$$\begin{aligned}
    Y_t
    = Y_0 + \int_0^t \hat{L}\{h(X_s)\} \dd{s} + \int_0^\tau \pdv{h}{x} g(X_s) \dd{B_s}
\end{aligned}$$

As always, the latter [Itō integral](/know/concept/ito-integral/)
is a [martingale](/know/concept/martingale/), so it vanishes
when we take the expectation conditioned on the "initial" state $X_0$, leaving:

$$\begin{aligned}
    \mathbf{E}[Y_t | X_0]
    = Y_0 + \mathbf{E}\bigg[ \int_0^t \hat{L}\{h(X_s)\} \dd{s} \bigg| X_0 \bigg]
\end{aligned}$$

For suffiently small $t$, the integral can be replaced by its first-order approximation:

$$\begin{aligned}
    \mathbf{E}[Y_t | X_0]
    \approx Y_0 + \hat{L}\{h(X_0)\} \: t
\end{aligned}$$

Rearranging this gives the following,
to be understood in the limit $t \to 0^+$:

$$\begin{aligned}
    \hat{L}\{h(X_0)\}
    \approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
\end{aligned}$$
</div>
</div>

The general definition of resembles that of a classical derivative,
and indeed, the generator $\hat{A}$ can be thought of as a differential operator.
In that case, we would like an analogue of the classical
fundamental theorem of calculus to relate it to integration.

Such an analogue is provided by **Dynkin's formula**:
for a stopping time $\tau$ with a finite expected value $\mathbf{E}[\tau|X_0] < \infty$,
it states that:

$$\begin{aligned}
    \boxed{
        \mathbf{E}\big[ h(X_\tau) | X_0 \big]
        = h(X_0) + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
    }
\end{aligned}$$

<div class="accordion">
<input type="checkbox" id="proof-dynkin"/>
<label for="proof-dynkin">Proof</label>
<div class="hidden" markdown="1">
<label for="proof-dynkin">Proof.</label>
The proof is similar to the one above.
Define $Y_t = h(X_t)$ and use Itō’s lemma:

$$\begin{aligned}
    \dd{Y_t}
    &= \bigg( \pdv{h}{x} f(X_t) + \frac{1}{2} \pdvn{2}{h}{x} g^2(X_t) \bigg) \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
    \\
    &= \hat{L} \{h(X_t)\} \dd{t} + \pdv{h}{x} g(X_t) \dd{B_t}
\end{aligned}$$

And then integrate this from $t = 0$ to the provided stopping time $t = \tau$:

$$\begin{aligned}
    Y_\tau
    = Y_0 + \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} + \int_0^\tau \pdv{h}{x} g(X_t) \dd{B_t}
\end{aligned}$$

All [Itō integrals](/know/concept/ito-integral/)
are [martingales](/know/concept/martingale/),
so the latter integral's conditional expectation is zero for the "initial" condition $X_0$.
The rest of the above equality is also a martingale:

$$\begin{aligned}
    0
    = \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
\end{aligned}$$

Isolating this equation for $\mathbf{E}[Y_\tau | X_0]$ then gives Dynkin's formula.
</div>
</div>

A common application of Dynkin's formula is predicting
when the stopping time $\tau$ occurs, and in what state $X_\tau$ this happens.
Consider an example:
for a region $\Omega$ of state space with $X_0 \in \Omega$,
we define the exit time $\tau \equiv \inf\{ t : X_t \notin \Omega \}$,
provided that $\mathbf{E}[\tau | X_0] < \infty$.

To get information about when and where $X_t$ exits $\Omega$,
we define the *general reward* $\Gamma$ as follows,
consisting of a *running reward* $R$ for $X_t$ inside $\Omega$,
and a *terminal reward* $T$ on the boundary $\partial \Omega$ where we stop at $X_\tau$:

$$\begin{aligned}
    \Gamma
    = \int_0^\tau R(X_t) \dd{t} + \: T(X_\tau)
\end{aligned}$$

For example, for $R = 1$ and $T = 0$, this becomes $\Gamma = \tau$,
and if $R = 0$, then $T(X_\tau)$ can tell us the exit point.
Let us now define $h(X_0) = \mathbf{E}[\Gamma | X_0]$,
and apply Dynkin's formula:

$$\begin{aligned}
    \mathbf{E}\big[ h(X_\tau) | X_0 \big]
    &= \mathbf{E}\big[ \Gamma \big| X_0 \big] + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
    \\
    &= \mathbf{E}\big[ T(X_\tau) | X_0 \big] + \mathbf{E}\bigg[ \int_0^\tau \hat{L}\{h(X_t)\} + R(X_t) \dd{t} \bigg| X_0 \bigg]
\end{aligned}$$

The two leftmost terms depend on the exit point $X_\tau$,
but not directly on $X_t$ for $t < \tau$,
while the rightmost depends on the whole trajectory $X_t$.
Therefore, the above formula is fulfilled
if $h(x)$ satisfies the following equation and boundary conditions:

$$\begin{aligned}
    \boxed{
        \begin{cases}
            \hat{L}\{h(x)\} + R(x) = 0 & \mathrm{for}\; x \in \Omega \\
            h(x) = T(x) & \mathrm{for}\; x \notin \Omega
        \end{cases}
    }
\end{aligned}$$

In other words, we have just turned a difficult question about a stochastic trajectory $X_t$
into a classical differential boundary value problem for $h(x)$.



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