Dynkin's Formula
Two results stand ready and have not yet been put in the same sentence.
The
stopped expectation identity
holds for an arbitrary Itô process and an arbitrary stopping time of
finite mean, and writes the average of a function at the stopping time
as its initial value plus an accumulated second-order expression. The
generator theorem
identifies that expression, for a diffusion and a compactly supported
\(C^2\) function, as the value of the
generator.
Putting the second inside the first is the formula of this section.
The gain is not computational. Nothing in the resulting statement is
harder or easier to evaluate than what went into it. The gain is that
the coefficients disappear. The identity as first proved speaks of
\(\mathbf{u}\), \(\mathbf{v}\), and the entries of
\(\mathbf{v} \mathbf{v}^\top\), which are features of a particular
equation; the formula below speaks only of \(A\), which is defined from
the process alone with no equation mentioned. Two equations with the
same generator therefore obey the same formula, and a question about
averages can be posed and answered without ever exhibiting the
coefficients that produced the process.
Proof.
The diffusion is an Itô process whose coefficients are
\(\mathbf{b}\) and \(\sigma\) read along the path, and the
boundedness demanded over the support of \(f\) follows from their
continuity on a compact set, by the verification already made when
the generator was
computed. With \(f \in C_0^2( \mathbb{R}^m )\)
and \(E^x[ \tau ] \lt \infty\) assumed, every requirement of the
stopped expectation identity is met, and it gives
\[
E^x \bigl[ f( X_\tau ) \bigr]
= f( x )
+ E^x \Bigl[ \int_0^\tau
\Bigl( \sum_{i} b_i\, \partial_i f
+ \tfrac{1}{2} \sum_{i, l}
\bigl( \sigma \sigma^\top \bigr)_{il}\, \partial_{il} f
\Bigr)( X_s )\, ds \Bigr] .
\]
The generator theorem states that for \(f \in C_0^2\) the bracketed
expression equals \(A f\) at every point of \(\mathbb{R}^m\), so the
two integrands agree at every \(( s, \omega )\), and the displays
are the same statement written twice.
The hypotheses are worth separating by the work they do. Compact support
is what makes the identity applicable at all: it bounds the integrands
and secures the class membership that Itô's formula demands of
them. Finiteness
of \(E^x[ \tau ]\) is what makes the stopping time usable: it forces
\(\tau\) to be finite so that \(X_\tau\) exists, and it supplies the
dominating function that carries the truncated statement to the limit.
The first of these is a restriction on \(f\) and the second a
restriction on \(\tau\). The next section trades the one for the
other, dropping compact support at the cost of specializing the
stopping time to an exit time.
Considerably more general versions of the formula exist, in which
\(f\) is required only to be sufficiently regular near the range of the
stopped path and the stopping time is controlled by weaker means. They
belong to a theory of Markov processes broader than the one this track
develops, and nothing here will need them.
Beyond Compact Support
Compact support is an awkward hypothesis in practice, because the
functions one wants to feed to the formula are rarely compactly
supported. The squared distance to the origin is not; neither is a
logarithm, nor a power of the radius, and these are exactly the
functions whose behavior under a diffusion one wants to know. The
hypothesis is also visibly stronger than the situation requires. If the
stopping time is the moment the path first leaves a bounded region,
then the path never visits anything outside that region before being
stopped, and the values of \(f\) far away cannot possibly matter.
Turning that observation into a proof costs one construction. Replace
\(f\) by a compactly supported function agreeing with it near the
region, apply the formula already proved, and observe that neither side
can tell the two functions apart. The operator appearing in the
conclusion is then the
differential operator of the diffusion,
which is defined for every twice differentiable \(f\) and agrees with
\(A\) on \(C_0^2( \mathbb{R}^m )\). Stating the conclusion with \(L\)
rather than \(A\) is not fussiness: for a general \(f \in C^2\) the
limit defining \(A f\) need not exist, since the averages
\(E^x[ f( X_t ) ]\) need not even be finite.
Theorem: Dynkin's Formula on a Bounded Region
Let \(X\) be an Itô diffusion in \(\mathbb{R}^m\), let
\(U \subseteq \mathbb{R}^m\) be open and bounded, let
\(x \in U\), and let \(\tau\) be the
first exit time
of \(X\) from \(U\), assumed to satisfy
\(E^x[ \tau ] \lt \infty\). Then for every
\(f \in C^2( \mathbb{R}^m )\),
\[
E^x \bigl[ f( X_\tau ) \bigr]
= f( x )
+ E^x \Bigl[ \int_0^\tau L f( X_s )\, ds \Bigr] .
\]
Proof.
The stopped path stays in the closure.
For \(s \lt \tau( \omega )\) the definition of the first exit time
gives \(X_s( \omega ) \in U\). At \(s = \tau( \omega )\), which is
finite almost surely because \(E^x[ \tau ]\) is, continuity of the
path makes \(X_\tau\) a limit of points of \(U\), so
\(X_\tau \in \overline{U}\). Hence the whole stopped path lies in
\(\overline{U}\), a compact set since \(U\) is bounded.
A compactly supported twin.
Since \(U\) is bounded, there is an \(R\) with \(\overline{U}\)
contained in the open ball \(B_R\) of radius \(R\) about the
origin. The
smooth bump function
with radii \(R\) and \(R + 1\) is a smooth
\(\psi : \mathbb{R}^m \to \mathbb{R}\) equal to \(1\) on
\(\overline{B_R}\) and vanishing outside \(B_{R + 1}\). Put
\(g = \psi f\). A product of a smooth function with a
\(C^2\) one is \(C^2\), and \(g\) vanishes wherever \(\psi\)
does, so its support lies in the compact set
\(\overline{B_{R + 1}}\) however \(f\) behaves far away; hence
\(g \in C_0^2( \mathbb{R}^m )\). On the open ball \(B_R\) the
factor \(\psi\) is identically \(1\), so \(g\) and \(f\)
coincide there together with all their first and second
derivatives, and therefore \(L g = L f\) throughout \(B_R\).
Transfer.
The first exit time is a stopping time by the
first exit theorem,
and \(E^x[ \tau ] \lt \infty\) by hypothesis, so
Dynkin's formula
applies to \(g\):
\[
E^x \bigl[ g( X_\tau ) \bigr]
= g( x )
+ E^x \Bigl[ \int_0^\tau A g( X_s )\, ds \Bigr] .
\]
Every quantity in this display is unchanged when \(g\) is replaced
by \(f\). The starting point \(x\) lies in \(U \subseteq B_R\), so
\(g(x) = f(x)\). The stopped position lies in
\(\overline{U} \subseteq B_R\), so
\(g( X_\tau ) = f( X_\tau )\) pointwise in \(\omega\). For
\(s \leq \tau\) the position \(X_s\) lies in \(B_R\), where
\(A g = L g = L f\), the first equality by the
generator theorem
applied to \(g \in C_0^2\) and the second by the previous step.
Substituting gives the display of the theorem.
The exchange is a fair one. What has been given up is generality in the
stopping time, which is now required to be an exit time from a bounded
region rather than an arbitrary integrable stopping time. What has been
gained is that \(f\) may be any twice differentiable function
whatsoever. For the questions this track will ask, that is exactly the
right trade. The duration of a stay inside a region, and the position
at which it ends, are questions whose stopping time is an exit time by
construction and whose test functions are not compactly supported.
One hypothesis remains unexamined. Nothing so far guarantees that
\(E^x[ \tau ] \lt \infty\) for the exit time from a bounded region, and
it has been assumed rather than proved at every appearance. The next
section shows, in the case that matters most, that the formula proves
it as a by-product rather than requiring it as an input.
Exit Times from a Ball
Here is a question the track has been able to pose since stopping times
were introduced and unable to answer. Start Brownian motion at a point
inside a ball. How long, on average, before it leaves? The quantity is
manifestly well defined and there has been no procedure for computing
it. Dynkin's formula supplies one, and the computation is short enough
to be worth following in full.
Take the driving motion and the state to have the same number of
components, so that the diffusion started at \(a\) is Brownian motion
\(a + w\), and fix \(R \gt 0\) with \(| a | \lt R\). Let
\[
K = \{ x \in \mathbb{R}^n : | x | \lt R \} ,
\quad
\tau = \inf \{ t \geq 0 : a + w_t \notin K \} .
\]
Theorem: Mean Exit Time from a Ball
With \(K\), \(\tau\) and \(a\) as above, \(\tau\) is almost
surely finite and
\[
E^a[ \tau ] = \frac{ R^2 - | a |^2 }{ n } .
\]
Proof.
A bounded truncation.
The set \(K\) is open, so \(\tau\) is a
stopping time.
A constant time \(k\) is one as well, since \(\{ k \leq t \}\)
is \(\Omega\) when \(k \leq t\) and empty otherwise, and both
lie in every \(\mathcal{F}_t\). The
calculus of stopping times
then makes \(\theta_k = \tau \wedge k\) a stopping time for each
positive integer \(k\), and \(E^a[ \theta_k ] \leq k\) is
finite. Dynkin's formula may therefore be applied to \(\theta_k\)
even though nothing is yet known about \(\tau\).
A test function equal to the squared radius.
Let \(\psi\) be the
smooth bump function
with radii \(R\) and \(R + 1\), and set
\(f( x ) = \psi( x ) | x |^2\). Then \(f \in C_0^2\), and
\(f( x ) = | x |^2\) whenever \(| x | \leq R\). On the open ball
the derivatives of \(f\) up to second order are those of the
squared radius, and by continuity of those derivatives the same
holds on the closed ball. The
generator of
Brownian motion, half the Laplacian, therefore gives
\[
A f( x ) = \tfrac{1}{2} \Delta | x |^2 = \tfrac{1}{2} \cdot 2 n = n ,
\quad | x | \leq R .
\]
The path is confined until it stops.
For \(s \lt \tau\) the definition of \(\tau\) gives
\(| a + w_s | \lt R\), and continuity of the path gives
\(| a + w_\tau | \leq R\) on \(\{ \tau \lt \infty \}\).
Since \(\theta_k \leq \tau\), every position visited up to time
\(\theta_k\) satisfies \(| a + w_s | \leq R\). Hence \(A f\)
equals the constant \(n\) all along the stopped path, and \(f\)
equals the squared radius at its endpoints.
The formula, and a uniform bound.
Dynkin's formula
applied to \(f\) and \(\theta_k\) reads
\[
E^a \bigl[ | a + w_{\theta_k} |^2 \bigr]
= | a |^2 + E^a \Bigl[ \int_0^{\theta_k} n \, ds \Bigr]
= | a |^2 + n\, E^a[ \theta_k ] .
\]
The left side is at most \(R^2\), because the stopped position
lies in the closed ball. Therefore
\(E^a[ \theta_k ] \leq ( R^2 - | a |^2 ) / n\) for every \(k\),
a bound that does not depend on \(k\).
Releasing the truncation.
The times \(\theta_k\) rise to \(\tau\) at every \(\omega\), so
monotone convergence
gives \(E^a[ \tau ] = \lim_k E^a[ \theta_k ]\), which the uniform
bound keeps finite. In particular \(\tau\) is almost surely
finite, so at almost every \(\omega\) one has
\(\theta_k = \tau\) for all large \(k\), and the stopped
position converges almost surely to \(a + w_\tau\). That position
satisfies \(| a + w_\tau | = R\): at most \(R\) by the
confinement step, and at least \(R\) because the exit position
lies outside \(K\), the second half of the first exit theorem.
The left side of the display is bounded
by \(R^2\), so
dominated convergence
sends it to \(R^2\). Passing to the limit on both sides gives
\(R^2 = | a |^2 + n\, E^a[ \tau ]\).
Three features of the answer are worth reading off, and the
computation delivered them without being aimed at any of them. The
mean exit time depends on the starting point
only through its distance from the centre, which the rotational symmetry
of Brownian motion makes plausible but which nothing in the computation
assumed. It scales as the square of the radius, the signature of a
process whose displacement grows like the square root of elapsed time.
And it decreases as the dimension grows. A ball of fixed radius is left
sooner in higher dimensions, because every additional component
contributes its own independent wandering to the radius while the
radius to be reached stays where it was.
The hypothesis that would not go away has also been disposed of.
Both formulas above carried the requirement that the stopping time
have finite mean, and here that requirement was never assumed. It was
obtained instead, from a bound the formula itself produced on the
truncations, and the exact value came out of the same limit. The pattern
is worth keeping. Applied to a bounded truncation of a stopping time one
knows nothing about, Dynkin's formula frequently returns the
integrability that its own hypotheses demand.