Dynkin's Formula

Proof.

The diffusion is an Itô process whose coefficients are \(\mathbf{b}\) and \(\sigma\) read along the path. Its membership in that class, and the boundedness of the coefficients over the support of \(f\), were both verified when the generator was computed, the latter from continuity on a compact set. 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.

Proof.

The stopped path stays in the closure.
For \(s \lt \tau( \omega )\) the definition of the first exit time gives \(X_s( \omega ) \in U\). The time \(\tau( \omega )\) is positive, because the path starts in the open set \(U\), and it is finite almost surely, because \(E^x[ \tau ]\) is. At \(s = \tau( \omega )\) continuity of the path therefore 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.

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 \(\mathcal{F}_t^{(n)}\). 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 \(| a + \mathbf{w}_s | \lt R\), and continuity of the path gives \(| a + \mathbf{w}_\tau | \leq R\) on \(\{ \tau \lt \infty \}\). Since \(\theta_k \leq \tau\), every position visited up to time \(\theta_k\) satisfies \(| a + \mathbf{w}_s | \leq R\). 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

\[ \begin{align*} E^a \bigl[ | a + \mathbf{w}_{\theta_k} |^2 \bigr] &= | a |^2 + E^a \Bigl[ \int_0^{\theta_k} n \, ds \Bigr] \\\\ &= | a |^2 + n\, E^a[ \theta_k ] . \end{align*} \]

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 \(a + \mathbf{w}_\tau\). That position satisfies \(| a + \mathbf{w}_\tau | = R\): 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 \[ R^2 = | a |^2 + n\, E^a[ \tau ] , \] which rearranges to the value stated in the theorem.

Proof.

The annulus.
The annulus \(U\) is open, bounded, and contains \(a\). A path leaves \(U\) exactly when it enters \(\{ | x | \leq r \}\) or \(\{ | x | \geq R \}\), so the first exit time from \(U\) is the smaller of the two entrance times, \(\tau_U = T_r \wedge \tau_R\). A point outside the open ball of radius \(R\) lies outside \(U\), so \(\tau_U \leq \tau_R\) at every \(\omega\), and the mean exit time from a ball gives

\[ E^a[ \tau_U ] \leq E^a[ \tau_R ] = \frac{ R^2 - | a |^2 }{ n } . \]

Hence \(\tau_U\) has finite mean and is almost surely finite.

Where the path leaves.
On \(\{ \tau_U \lt \infty \}\) the path lies in \(U\) before \(\tau_U\), so continuity gives \(r \leq | a + \mathbf{w}_{\tau_U} | \leq R\). If \(T_r \lt \tau_R\), then \(\tau_U = T_r\), and the exit position from \(\{ | x | \gt r \}\) has radius at most \(r\) by the first exit theorem, hence exactly \(r\). If \(\tau_R \leq T_r\), then \(\tau_U = \tau_R\), and the exit position from the open ball has radius at least \(R\), hence exactly \(R\). The two times never coincide on \(\{ \tau_U \lt \infty \}\), since no point has radius both at most \(r\) and at least \(R\). Almost surely, therefore, \(| a + \mathbf{w}_{\tau_U} | = r\) on \(\{ T_r \lt \tau_R \}\) and \(| a + \mathbf{w}_{\tau_U} | = R\) off it.

A test function harmonic off the origin.
For \(x \neq 0\) put \(g( x ) = \varphi_n( | x | )\). The radius is smooth away from the origin, so \(g\) is smooth on \(\mathbb{R}^n \setminus \{ 0 \}\), with \(\partial_i g = \varphi_n'( | x | )\, x_i / | x |\). Differentiating once more and summing over \(i\),

\[ \begin{align*} \partial_{ii} g &= \varphi_n''( | x | ) \frac{ x_i^2 }{ | x |^2 } + \varphi_n'( | x | ) \Bigl( \frac{ 1 }{ | x | } - \frac{ x_i^2 }{ | x |^3 } \Bigr) , \\\\ \Delta g &= \varphi_n''( | x | ) + \frac{ n - 1 }{ | x | }\, \varphi_n'( | x | ) . \end{align*} \]

The last expression vanishes in every case. For \(n = 1\) both terms are zero. For \(n = 2\) it is \(- \rho^{-2} + \rho^{-1} \cdot \rho^{-1} = 0\), and for \(n \geq 3\) it is \(( 2 - n )( 1 - n ) \rho^{-n} + ( n - 1 )( 2 - n ) \rho^{-n} = 0\), with \(\rho = | x |\).

The function \(g\) is undefined at the origin, so it is cut off there. Let \(\psi\) be the smooth bump function with radii \(r/4\) and \(r/2\), and put \(f = ( 1 - \psi )\, g\) away from the origin and \(f( 0 ) = 0\). On the open ball of radius \(r/4\) the factor \(1 - \psi\) vanishes, so \(f\) is identically zero there. Away from the origin \(f\) is a product of smooth functions. Hence \(f \in C^2( \mathbb{R}^n )\). Where \(| x | \geq r/2\) the factor \(\psi\) vanishes, so \(f = g\) on the open set \(\{ | x | \gt r/2 \}\), which contains the closed annulus \(\overline{U} = \{ r \leq | x | \leq R \}\). With drift zero and noise coefficient the identity, the differential operator of the diffusion is \(L = \tfrac{1}{2} \Delta\), so \(L f = \tfrac{1}{2} \Delta g = 0\) throughout \(\overline{U}\).

The formula.
The set \(U\) is open and bounded, it contains \(a\), and \(E^a[ \tau_U ] \lt \infty\), so Dynkin's formula on a bounded region applies to \(f\). Up to time \(\tau_U\) the path stays in \(\overline{U}\), where \(L f = 0\), so the integral term vanishes and

\[ E^a \bigl[ f( a + \mathbf{w}_{\tau_U} ) \bigr] = f( a ) = \varphi_n( | a | ) . \]

By the previous steps \(f( a + \mathbf{w}_{\tau_U} )\) equals \(\varphi_n( r )\) on \(\{ T_r \lt \tau_R \}\) and \(\varphi_n( R )\) off it, almost surely. Writing \(p = P^a[ T_r \lt \tau_R ]\), the left side is

\[ p\, \varphi_n( r ) + ( 1 - p )\, \varphi_n( R ) = \varphi_n( | a | ) . \]

The function \(\varphi_n\) is strictly increasing for \(n \leq 2\) and strictly decreasing for \(n \geq 3\), so \(\varphi_n( r ) \neq \varphi_n( R )\), and solving for \(p\) gives the stated value.

Proof.

The outer sphere recedes.
Fix \(0 \lt r \lt | a |\) and let \(R\) run through the integers exceeding \(| a |\). A larger radius is reached no earlier than a smaller one, so \(\tau_R\) is nondecreasing in \(R\). It also tends to infinity at every \(\omega\). For fixed \(t\) the continuous path is bounded on \([ 0, t ]\), so \(\tau_R \gt t\) once \(R\) exceeds that bound. The events \(\{ T_r \lt \tau_R \}\) therefore increase to \(\{ T_r \lt \infty \}\), and monotone convergence applied to their indicators gives \(P^a[ T_r \lt \infty ] = \lim_{R \to \infty} P^a[ T_r \lt \tau_R ]\). By the theorem, this limit is

\[ \begin{align*} \lim_{R \to \infty} \frac{ R - | a | }{ R - r } &= 1 , \quad n = 1 , \\\\ \lim_{R \to \infty} \frac{ \log R - \log | a | }{ \log R - \log r } &= 1 , \quad n = 2 , \\\\ \lim_{R \to \infty} \frac{ | a |^{2 - n} - R^{2 - n} }{ r^{2 - n} - R^{2 - n} } &= \Bigl( \frac{ r }{ | a | } \Bigr)^{n - 2} , \quad n \geq 3 . \end{align*} \]

This proves the claims about \(T_r\) in parts (2) and (3).

The inner sphere shrinks (\(n \geq 2\)).
Fix an integer \(R \gt | a |\). A path that visits the origin has radius \(0 \leq r\) at that moment, so \(T_r \leq T_0\) for every \(0 \lt r \lt | a |\), and \(\{ T_0 \lt \tau_R \} \subseteq \{ T_r \lt \tau_R \}\). The theorem therefore bounds \(P^a[ T_0 \lt \tau_R ]\) by

\[ \begin{align*} \frac{ \log R - \log | a | }{ \log R - \log r } &\longrightarrow 0 , \quad n = 2 , \\\\ \frac{ | a |^{2 - n} - R^{2 - n} }{ r^{2 - n} - R^{2 - n} } &\longrightarrow 0 , \quad n \geq 3 , \end{align*} \]

as \(r \downarrow 0\), because \(\log r \to - \infty\) and \(r^{2 - n} \to \infty\). Hence \(P^a[ T_0 \lt \tau_R ] = 0\) for every \(R\). As in the first step, the events \(\{ T_0 \lt \tau_R \}\) increase to \(\{ T_0 \lt \infty \}\), so \(P^a[ T_0 \lt \infty ] = 0\).

The line (\(n = 1\)).
Here the inner sphere \(\{ -r, r \}\) never contains the origin, so the start is shifted instead. Let \(s = a / | a | \in \{ -1, 1 \}\), fix \(r \gt 0\), and put \(a' = a + r s\), so that \(| a' | = | a | + r \gt r\). Both \(P^a\) and \(P^{a'}\) are probabilities of events of the one driving motion \(\mathbf{w}\). By the first step, applied at \(a'\), almost every path of \(a' + \mathbf{w}\) enters \([ -r, r ]\) at some time \(t\). At that time

\[ s\, ( a + \mathbf{w}_t ) = s\, ( a' + \mathbf{w}_t ) - r \leq 0 , \]

while \(s\, ( a + \mathbf{w}_0 ) = | a | \gt 0\). The intermediate value theorem, applied to the continuous function \(u \mapsto s\, ( a + \mathbf{w}_u )\) on \([ 0, t ]\), produces a time at which \(a + \mathbf{w}_u = 0\). Hence \(P^a[ T_0 \lt \infty ] = 1\).