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+---
+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.