Small Noise and Large Deviations¶

Concept Definition¶

Consider the small-noise diffusion on \([0,T]\):

\[ \mathrm{d}X_t^{\varepsilon} = b(X_t^{\varepsilon})\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma(X_t^{\varepsilon})\,\mathrm{d}W_t, \qquad X_0^{\varepsilon} = x_0, \]

where \(\varepsilon > 0\) is a small parameter. As \(\varepsilon \downarrow 0\), \(X^{\varepsilon}\) concentrates around the deterministic skeleton (the ODE solution):

\[ \dot{\bar{x}}(t) = b(\bar{x}(t)), \qquad \bar{x}(0) = x_0. \]

Large deviations theory (Freidlin–Wentzell) makes precise how exponentially unlikely it is for \(X^{\varepsilon}\) to deviate from \(\bar{x}\).

Definition: Rate Function (Freidlin–Wentzell Action)

For \(\phi \in C([0,T]; \mathbb{R}^d)\), define the rate function

\[ I_{0,T}(\phi) := \begin{cases} \dfrac{1}{2}\displaystyle\int_0^T \bigl\|\sigma(\phi(t))^{-1}\bigl(\dot{\phi}(t) - b(\phi(t))\bigr)\bigr\|^2\,\mathrm{d}t & \text{if } \phi \text{ is abs.\ continuous and } \sigma(\phi(t)) \text{ invertible a.e.}, \\[6pt] +\infty & \text{otherwise.} \end{cases} \]

Equivalently, using the control representation: if \(\dot{\phi}(t) = b(\phi(t)) + \sigma(\phi(t))\,u(t)\) for some \(u \in L^2([0,T]; \mathbb{R}^m)\), then

\[ I_{0,T}(\phi) = \frac{1}{2}\int_0^T \|u(t)\|^2\,\mathrm{d}t. \]

This is the minimum energy (in \(L^2\) of the control) needed to force the system along the path \(\phi\).

Explanation¶

The Freidlin–Wentzell LDP¶

Theorem (Freidlin–Wentzell, 1970)

Assume \(b\) and \(\sigma\) are bounded and Lipschitz, and \(\sigma(x)\) is uniformly invertible. The family \((X^{\varepsilon})_{\varepsilon > 0}\), viewed as random variables in \(C([0,T]; \mathbb{R}^d)\), satisfies a Large Deviation Principle (LDP) with speed \(\varepsilon\) and good rate function \(I_{0,T}\):

(Upper bound) For every closed set \(F \subset C([0,T];\mathbb{R}^d)\):

\[ \limsup_{\varepsilon \downarrow 0}\;\varepsilon \log \mathbb{P}(X^{\varepsilon} \in F) \le -\inf_{\phi \in F} I_{0,T}(\phi). \]

(Lower bound) For every open set \(G \subset C([0,T];\mathbb{R}^d)\):

\[ \liminf_{\varepsilon \downarrow 0}\;\varepsilon \log \mathbb{P}(X^{\varepsilon} \in G) \ge -\inf_{\phi \in G} I_{0,T}(\phi). \]

On the Invertibility Assumption

The formula for \(I_{0,T}(\phi)\) requires \(\sigma(\phi(t))\) to be invertible along the path. For degenerate diffusions (where \(a(x) = \sigma(x)\sigma(x)^\top\) is not strictly positive definite), the rate function must be defined differently (as a variational infimum over controls). The above theorem applies in the uniformly elliptic case.

Informal Interpretation¶

Combining upper and lower bounds: for a set \(A\) satisfying \(\inf_{\phi \in \mathrm{int}(A)} I_{0,T}(\phi) = \inf_{\phi \in \bar{A}} I_{0,T}(\phi)\) (a continuity set of \(I_{0,T}\), meaning the infimum is the same over the interior and the closure),

\[ \mathbb{P}(X^{\varepsilon} \in A) \approx \exp\!\left(-\frac{1}{\varepsilon}\inf_{\phi \in A} I_{0,T}(\phi)\right) \quad (\varepsilon \downarrow 0). \]

The most likely atypical path (the instanton or optimal fluctuation path) is the minimiser of \(I_{0,T}\) over the constrained set \(A\). All other paths in \(A\) are exponentially less likely.

Why the Rate Function Has This Form¶

The Itô SDE says \(\dot{X}^{\varepsilon} \approx b(X^{\varepsilon}) + \sqrt{\varepsilon}\,\sigma(X^{\varepsilon})\,\dot{W}\). To force the path along \(\phi \ne \bar{x}\), one must effectively supply a "fake" Brownian increment \(\sqrt{\varepsilon}\,\sigma\,u\,\mathrm{d}t\) at each time. The Radon–Nikodym density of the original measure \(\mathbb{P}\) with respect to the shifted measure \(\mathbb{Q}^u\) is

\[ \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}^u}\bigg|_{\mathcal{F}_T} = \exp\!\left(-\frac{1}{2\varepsilon}\int_0^T \|u(t)\|^2\,\mathrm{d}t\right) = \exp\!\left(-\frac{I_{0,T}(\phi)}{\varepsilon}\right), \]

so the \(\mathbb{P}\)-probability of the path \(\phi\) is suppressed by a factor \(\exp(-I_{0,T}(\phi)/\varepsilon)\) relative to the shifted measure — directly giving the exponential rate. (Since \(u\) is a deterministic \(L^2\) function, Novikov's condition \(\mathbb{E}\exp(\frac{1}{2\varepsilon}\int\|u\|^2\,\mathrm{d}t) < \infty\) holds automatically, ensuring \(\mathbb{Q}^u\) is a true probability measure.)

Diagram / Example¶

Example: Escape from a Potential Well¶

Let \(d = 1\), \(b(x) = -V'(x)\) (gradient drift), \(\sigma = 1\):

\[ \mathrm{d}X_t^{\varepsilon} = -V'(X_t^{\varepsilon})\,\mathrm{d}t + \sqrt{\varepsilon}\,\mathrm{d}W_t. \]

Suppose \(V\) has a local minimum at \(0\) and a saddle point at \(x^* > 0\) with \(V(x^*) > V(0)\). The quasipotential for escape from the well is

\[ \Delta V := V(x^*) - V(0) = I_{0,\infty}(\phi^*), \]

where \(\phi^*\) is the minimising path (the instanton). By the LDP,

\[ \mathbb{P}(X^{\varepsilon} \text{ exits the well}) \approx \exp\!\left(-\frac{\Delta V}{\varepsilon}\right). \]

This recovers the Kramers escape rate from statistical physics.

Hamilton–Jacobi Connection¶

Define the Hamiltonian associated to the large deviations:

\[ H(x, p) = b(x) \cdot p + \frac{1}{2}\,p^\top a(x)\,p, \qquad a = \sigma\sigma^\top. \]

The quasipotential \(U(x) := \inf\{I_{0,\infty}(\phi) : \phi(0) = x_0,\, \phi(\infty) = x\}\) satisfies the stationary Hamilton–Jacobi equation:

\[ H(x, \nabla U(x)) = 0. \]

This equation has a meaningful solution when \(x_0\) is a stable equilibrium of the ODE \(\dot{x} = b(x)\) (i.e. \(b(x_0) = 0\) and the linearisation at \(x_0\) is stable); the quasipotential then measures the cost of reaching \(x\) from the basin of attraction of \(x_0\).

The time-dependent analogue: the value function \(V(t, x) = -\varepsilon \log \mathbb{E}^x[e^{-g(X_T^{\varepsilon})/\varepsilon}]\) satisfies, as \(\varepsilon \to 0\), the nonlinear Hamilton–Jacobi PDE

\[ \partial_t V + H(x, \nabla_x V) = 0, \]

in contrast to the linear Kolmogorov backward equation satisfied by \(\mathbb{E}^x[g(X_T^{\varepsilon})]\). Large deviations thus convert linear PDE problems into nonlinear Hamilton–Jacobi problems.

Proof / Derivation¶

We sketch the Girsanov-based derivation of the rate function.

Step 1: Girsanov change of measure. For a deterministic control \(u \in L^2([0,T];\mathbb{R}^m)\), define the controlled process \(\phi\) by \(\dot{\phi} = b(\phi) + \sigma(\phi)\,u\), \(\phi(0) = x_0\). In the small-noise SDE \(\mathrm{d}X^\varepsilon = b\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma\,\mathrm{d}W\), forcing \(X^\varepsilon \approx \phi\) requires shifting the standard BM \(W\) by \(u/\sqrt{\varepsilon}\) — that is, replacing \(\mathrm{d}W_t\) by \(\mathrm{d}W_t - (u(t)/\sqrt{\varepsilon})\,\mathrm{d}t\). By Girsanov's theorem, the change-of-measure density for this shift is

\[ \frac{\mathrm{d}\mathbb{Q}^u}{\mathrm{d}\mathbb{P}}\bigg|_{\mathcal{F}_T} = \exp\!\left( \frac{1}{\sqrt{\varepsilon}}\int_0^T u(t)^\top\,\mathrm{d}W_t - \frac{1}{2\varepsilon}\int_0^T \|u(t)\|^2\,\mathrm{d}t \right). \]

Under \(\mathbb{Q}^u\), the process \(\widetilde{W}_t := W_t - \frac{1}{\sqrt{\varepsilon}}\int_0^t u(s)\,\mathrm{d}s\) is a standard BM, and the SDE becomes \(\mathrm{d}X^\varepsilon = (b + \sigma u)\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma\,\mathrm{d}\widetilde{W}\), which has \(\phi\) as its leading-order (\(\varepsilon \to 0\)) solution. The log-likelihood of the shift is \(-\frac{1}{2\varepsilon}\int\|u\|^2\,\mathrm{d}t = -I_{0,T}(\phi)/\varepsilon\), directly giving the exponential rate.

Step 2: Exponential tightness. One shows that the family \((X^{\varepsilon})\) is exponentially tight in \(C([0,T];\mathbb{R}^d)\) under Lipschitz assumptions, ruling out escape to infinity.

Step 3: Identification of the rate function. Using Varadhan's lemma and the duality formula for the log-Laplace functional, the rate function is identified as the Legendre transform of the log-moment generating functional, yielding \(I_{0,T}(\phi) = \frac{1}{2}\int_0^T \|u(t)\|^2\,\mathrm{d}t\) for the optimal control \(u\) realising path \(\phi\). \(\square\)

What to Remember¶

Small-noise diffusions satisfy a path-space LDP with speed \(\varepsilon\) and rate function \(I_{0,T}(\phi)\).
The rate function measures the minimum \(L^2\)-energy of the control needed to force the path \(\phi\).
Rare events occur along instanton paths (minimisers of \(I_{0,T}\) subject to the constraint).
Large deviations connect stochastic dynamics to nonlinear Hamilton–Jacobi PDEs, replacing the linear Kolmogorov equations.
The Freidlin–Wentzell LDP applies under uniform ellipticity; degenerate cases require separate treatment.

Exercises¶

Exercise 1. Consider the small-noise SDE \(\mathrm{d}X_t^\varepsilon = -X_t^\varepsilon\,\mathrm{d}t + \sqrt{\varepsilon}\,\mathrm{d}W_t\) with \(X_0^\varepsilon = 1\). Write down the deterministic skeleton \(\bar{x}(t)\) (the ODE solution with \(\varepsilon = 0\)). For the path \(\phi(t) = e^{-t} + \delta\sin(\pi t/T)\) on \([0, T]\), compute the control \(u(t)\) such that \(\dot{\phi} = b(\phi) + u\) and express the rate function \(I_{0,T}(\phi)\) as an integral involving \(u\).

Solution to Exercise 1

The deterministic skeleton solves \(\dot{\bar{x}}(t) = -\bar{x}(t)\) with \(\bar{x}(0) = 1\), giving

\[ \bar{x}(t) = e^{-t} \]

For the path \(\phi(t) = e^{-t} + \delta\sin(\pi t/T)\), the SDE has \(b(x) = -x\) and \(\sigma = 1\), so the control \(u(t)\) satisfies \(\dot{\phi}(t) = b(\phi(t)) + u(t) = -\phi(t) + u(t)\). Therefore

\[ u(t) = \dot{\phi}(t) + \phi(t) \]

Computing \(\dot{\phi}(t) = -e^{-t} + \delta\frac{\pi}{T}\cos(\pi t/T)\):

\[ u(t) = -e^{-t} + \delta\frac{\pi}{T}\cos(\pi t/T) + e^{-t} + \delta\sin(\pi t/T) = \delta\left(\frac{\pi}{T}\cos(\pi t/T) + \sin(\pi t/T)\right) \]

The rate function is

\[ I_{0,T}(\phi) = \frac{1}{2}\int_0^T u(t)^2\,\mathrm{d}t = \frac{\delta^2}{2}\int_0^T \left(\frac{\pi}{T}\cos(\pi t/T) + \sin(\pi t/T)\right)^2\mathrm{d}t \]

Expanding the square and using orthogonality of \(\cos\) and \(\sin\) over \([0,T]\):

\[ I_{0,T}(\phi) = \frac{\delta^2}{2}\left(\frac{\pi^2}{T^2}\cdot\frac{T}{2} + \frac{T}{2}\right) = \frac{\delta^2}{4}\left(\frac{\pi^2}{T} + T\right) \]

Note that as \(\delta \to 0\), \(\phi \to \bar{x}\) and \(I_{0,T}(\phi) \to 0\), confirming that the skeleton has zero cost.

Exercise 2. Let \(b(x) = -V'(x)\) with \(V(x) = \frac{1}{2}x^2\) and \(\sigma = 1\). Compute the quasipotential \(U(y) = \inf\{I_{0,\infty}(\phi) : \phi(0) = 0,\, \phi(\infty) = y\}\) and verify that \(U(y) = V(y) - V(0) = \frac{1}{2}y^2\). Show that the Hamiltonian \(H(x, p) = -V'(x)\,p + \frac{1}{2}p^2\) satisfies \(H(x, \nabla U(x)) = 0\).

Solution to Exercise 2

With \(b(x) = -V'(x) = -x\) and \(\sigma = 1\) (so \(a = 1\)), the rate function for a path \(\phi\) from \(\phi(0) = 0\) to \(\phi(\infty) = y\) is

\[ I_{0,\infty}(\phi) = \frac{1}{2}\int_0^\infty (\dot{\phi}(t) + \phi(t))^2\,\mathrm{d}t \]

The quasipotential is \(U(y) = \inf_\phi I_{0,\infty}(\phi)\) over paths with \(\phi(0) = 0\) and \(\phi(t) \to y\) as \(t \to \infty\).

The Euler–Lagrange equation for the minimizing path gives \(\ddot{\phi} = \phi\) (the "instanton equation"), with boundary conditions \(\phi(0) = 0\) and \(\phi(\infty) = y\). However, it is easier to use the known result for gradient systems: when \(b = -\nabla V\) and \(a = I\), the quasipotential is

\[ U(y) = 2(V(y) - V(0)) = 2\left(\frac{1}{2}y^2 - 0\right) = y^2 \]

Wait — let us verify more carefully. For the gradient system \(\dot{\phi} = -V'(\phi) + u\), the optimal control that takes \(\phi\) from \(0\) to \(y\) uses the time-reversed deterministic flow: \(\dot{\phi} = +V'(\phi) = \phi\) with \(\phi(0) = 0\), arriving at \(y\) in infinite time (approaching \(y\) as \(t \to \infty\) via \(\phi(t) = y(1 - e^{-2t})\) approximately). The control along this path is \(u = \dot{\phi} - b(\phi) = \dot{\phi} + \phi = 2\phi\).

Actually, for gradient systems with \(a = I\), there is a standard identity: \(U(y) = 2(V(y) - V(0))\). But let us verify via the Hamilton–Jacobi equation.

Hamilton–Jacobi verification: The Hamiltonian is

\[ H(x, p) = b(x)p + \frac{1}{2}p^2 = -xp + \frac{1}{2}p^2 \]

The Hamilton–Jacobi equation \(H(x, U'(x)) = 0\) with \(U(x) = \frac{1}{2}x^2\) gives \(U'(x) = x\):

\[ H(x, x) = -x \cdot x + \frac{1}{2}x^2 = -x^2 + \frac{1}{2}x^2 = -\frac{1}{2}x^2 \ne 0 \]

This does not vanish, so \(U(y) = \frac{1}{2}y^2\) does not satisfy \(H(x, \nabla U) = 0\). Let us try \(U(y) = y^2\) so \(U'(x) = 2x\):

\[ H(x, 2x) = -x(2x) + \frac{1}{2}(2x)^2 = -2x^2 + 2x^2 = 0 \checkmark \]

So \(U(y) = y^2 = 2V(y) - 2V(0)\) and \(V(y) - V(0) = \frac{1}{2}y^2\), confirming

\[ U(y) = 2(V(y) - V(0)) = 2 \cdot \frac{1}{2}y^2 = y^2 \]

and \(H(x, \nabla U(x)) = 0\) is verified. \(\square\)

Exercise 3. Consider the double-well potential \(V(x) = \frac{1}{4}x^4 - \frac{1}{2}x^2\) with minima at \(x = \pm 1\) and a saddle at \(x = 0\). For the gradient diffusion \(\mathrm{d}X_t^\varepsilon = -V'(X_t^\varepsilon)\,\mathrm{d}t + \sqrt{\varepsilon}\,\mathrm{d}W_t\) started at \(X_0 = -1\), compute the potential barrier \(\Delta V = V(0) - V(-1)\) and give the exponential rate at which \(X^\varepsilon\) escapes from the left well to the right well as \(\varepsilon \to 0\).

Solution to Exercise 3

The potential is \(V(x) = \frac{1}{4}x^4 - \frac{1}{2}x^2\). The critical points satisfy \(V'(x) = x^3 - x = 0\), giving \(x = 0, \pm 1\).

\(V(-1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\) (local minimum)
\(V(0) = 0\) (local maximum, i.e., saddle of the potential)
\(V(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}\) (local minimum)

The potential barrier for escape from the left well (at \(x = -1\)) over the saddle (at \(x = 0\)) is

\[ \Delta V = V(0) - V(-1) = 0 - \left(-\frac{1}{4}\right) = \frac{1}{4} \]

By the Freidlin–Wentzell large deviation principle, for the gradient diffusion with \(a = 1\) the quasipotential for escape is \(2\Delta V\) (using the identity \(U = 2(V(\text{saddle}) - V(\text{minimum}))\) for gradient systems). The probability of escape satisfies

\[ \mathbb{P}(X^\varepsilon \text{ escapes left well}) \approx \exp\!\left(-\frac{2\Delta V}{\varepsilon}\right) = \exp\!\left(-\frac{1}{2\varepsilon}\right) \]

as \(\varepsilon \downarrow 0\). The exponential rate is \(2\Delta V = 1/2\), so rare transitions between wells become exponentially suppressed as the noise vanishes.

Exercise 4. Explain the Girsanov-based derivation of the rate function. For the small-noise SDE \(\mathrm{d}X_t^\varepsilon = b(X_t^\varepsilon)\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma(X_t^\varepsilon)\,\mathrm{d}W_t\), describe how shifting \(W_t\) by \(u(t)/\sqrt{\varepsilon}\) via Girsanov's theorem produces a change-of-measure density proportional to \(\exp(-I_{0,T}(\phi)/\varepsilon)\). Why does Novikov's condition hold for a deterministic control \(u \in L^2\)?

Solution to Exercise 4

Consider the small-noise SDE \(\mathrm{d}X_t^\varepsilon = b(X_t^\varepsilon)\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma(X_t^\varepsilon)\,\mathrm{d}W_t\). To force \(X^\varepsilon\) to follow a path \(\phi\) with \(\dot{\phi} = b(\phi) + \sigma(\phi)u\), we shift the Brownian motion by \(u(t)/\sqrt{\varepsilon}\).

Define \(\widetilde{W}_t = W_t - \frac{1}{\sqrt{\varepsilon}}\int_0^t u(s)\,\mathrm{d}s\). Under the original measure \(\mathbb{P}\), this shift changes the SDE to

\[ \mathrm{d}X_t^\varepsilon = \bigl(b(X_t^\varepsilon) + \sigma(X_t^\varepsilon)u(t)\bigr)\mathrm{d}t + \sqrt{\varepsilon}\,\sigma(X_t^\varepsilon)\,\mathrm{d}\widetilde{W}_t \]

By Girsanov's theorem, the Radon–Nikodym derivative for making \(\widetilde{W}\) a standard BM under a new measure \(\mathbb{Q}^u\) is

\[ \frac{\mathrm{d}\mathbb{Q}^u}{\mathrm{d}\mathbb{P}}\bigg|_{\mathcal{F}_T} = \exp\!\left(\frac{1}{\sqrt{\varepsilon}}\int_0^T u(t)^\top\,\mathrm{d}W_t - \frac{1}{2\varepsilon}\int_0^T \|u(t)\|^2\,\mathrm{d}t\right) \]

Under \(\mathbb{Q}^u\), the leading-order dynamics are \(\dot{\phi} = b(\phi) + \sigma(\phi)u\), so \(X^\varepsilon\) concentrates near \(\phi\). The \(\mathbb{P}\)-probability of paths near \(\phi\) is suppressed by

\[ \frac{\mathrm{d}\mathbb{P}}{\mathrm{d}\mathbb{Q}^u} = \exp\!\left(-\frac{1}{\sqrt{\varepsilon}}\int_0^T u^\top\,\mathrm{d}W_t + \frac{1}{2\varepsilon}\int_0^T \|u\|^2\,\mathrm{d}t - \frac{1}{2\varepsilon}\int_0^T\|u\|^2\,\mathrm{d}t\right) \]

Taking logarithms and expectations under \(\mathbb{Q}^u\), the dominant term is \(-\frac{1}{2\varepsilon}\int_0^T\|u\|^2\,\mathrm{d}t = -I_{0,T}(\phi)/\varepsilon\), giving the exponential rate.

Novikov's condition requires \(\mathbb{E}\exp\!\left(\frac{1}{2\varepsilon}\int_0^T\|u(t)\|^2\,\mathrm{d}t\right) < \infty\). Since \(u \in L^2([0,T];\mathbb{R}^m)\) is deterministic, \(\int_0^T\|u\|^2\,\mathrm{d}t\) is a finite constant (not random). Therefore the exponential is a finite constant, and Novikov's condition holds trivially.

Exercise 5. State the upper and lower bounds of the Freidlin–Wentzell LDP. For a continuity set \(A\) (where \(\inf_{\mathrm{int}(A)} I = \inf_{\bar{A}} I\)), show that the two bounds combine to give

\[ \lim_{\varepsilon \downarrow 0}\;\varepsilon\log\mathbb{P}(X^\varepsilon \in A) = -\inf_{\phi \in A} I_{0,T}(\phi). \]

Give a concrete example of a set \(A\) in path space that is a continuity set and one that is not.

Solution to Exercise 5

The Freidlin–Wentzell LDP states:

Upper bound: For every closed \(F\): \(\limsup_{\varepsilon \downarrow 0}\,\varepsilon\log\mathbb{P}(X^\varepsilon \in F) \le -\inf_{\phi \in F}I_{0,T}(\phi)\)
Lower bound: For every open \(G\): \(\liminf_{\varepsilon \downarrow 0}\,\varepsilon\log\mathbb{P}(X^\varepsilon \in G) \ge -\inf_{\phi \in G}I_{0,T}(\phi)\)

For a continuity set \(A\) with \(\inf_{\phi \in \mathrm{int}(A)}I = \inf_{\phi \in \bar{A}}I =: c\), apply:

Upper bound with \(F = \bar{A}\): \(\limsup\,\varepsilon\log\mathbb{P}(X^\varepsilon \in A) \le \limsup\,\varepsilon\log\mathbb{P}(X^\varepsilon \in \bar{A}) \le -\inf_{\bar{A}}I = -c\)
Lower bound with \(G = \mathrm{int}(A)\): \(\liminf\,\varepsilon\log\mathbb{P}(X^\varepsilon \in A) \ge \liminf\,\varepsilon\log\mathbb{P}(X^\varepsilon \in \mathrm{int}(A)) \ge -\inf_{\mathrm{int}(A)}I = -c\)

Combining: \(-c \le \liminf \le \limsup \le -c\), so

\[ \lim_{\varepsilon \downarrow 0}\,\varepsilon\log\mathbb{P}(X^\varepsilon \in A) = -c = -\inf_{\phi \in A}I_{0,T}(\phi) \]

Continuity set example: For the SDE \(\mathrm{d}X^\varepsilon = -X^\varepsilon\,\mathrm{d}t + \sqrt{\varepsilon}\,\mathrm{d}W\) with \(X_0 = 0\), consider \(A = \{\phi : \|\phi - \bar{x}\|_\infty > \delta\}\) for some \(\delta > 0\). This set has \(\mathrm{int}(A) = A\) (it is open) and \(\bar{A} = \{\phi : \|\phi - \bar{x}\|_\infty \ge \delta\}\). If the infimum of \(I\) over \(\bar{A}\) is not attained on the boundary \(\{\|\phi - \bar{x}\|_\infty = \delta\}\) exclusively (which generically it is), then \(\inf_{\mathrm{int}(A)}I = \inf_{\bar{A}}I\) and \(A\) is a continuity set.

Non-continuity set example: The singleton \(A = \{\bar{x}\}\) (the deterministic skeleton itself). Here \(\mathrm{int}(A) = \emptyset\), so \(\inf_{\mathrm{int}(A)}I = +\infty\), while \(\inf_{\bar{A}}I = I(\bar{x}) = 0\). Since \(+\infty \ne 0\), this is not a continuity set.

Exercise 6. For the small-noise diffusion \(\mathrm{d}X_t^\varepsilon = b(X_t^\varepsilon)\,\mathrm{d}t + \sqrt{\varepsilon}\,\sigma(X_t^\varepsilon)\,\mathrm{d}W_t\), define the Hamiltonian \(H(x, p) = b(x) \cdot p + \frac{1}{2}p^\top a(x)\,p\) with \(a = \sigma\sigma^\top\). Write down the Hamilton equations (the Euler–Lagrange equations for minimising \(I_{0,T}\)) and show that the instanton path \(\phi^*\) satisfies

\[ \dot{\phi}^* = b(\phi^*) + a(\phi^*)\,p^*, \qquad \dot{p}^* = -(\nabla_x b)^\top p^* - \frac{1}{2}\nabla_x(p^{*\top} a\,p^*), \]

where \(p^*\) is the conjugate momentum.

Solution to Exercise 6

The rate function is \(I_{0,T}(\phi) = \frac{1}{2}\int_0^T\|u(t)\|^2\,\mathrm{d}t\) where \(\dot{\phi} = b(\phi) + \sigma(\phi)u\), i.e., \(u = \sigma(\phi)^{-1}(\dot{\phi} - b(\phi))\). This is a calculus of variations problem. Define the Lagrangian

\[ L(\phi, \dot{\phi}) = \frac{1}{2}\|\sigma(\phi)^{-1}(\dot{\phi} - b(\phi))\|^2 = \frac{1}{2}(\dot{\phi} - b(\phi))^\top a(\phi)^{-1}(\dot{\phi} - b(\phi)) \]

The conjugate momentum is

\[ p = \frac{\partial L}{\partial \dot{\phi}} = a(\phi)^{-1}(\dot{\phi} - b(\phi)) \]

so \(\dot{\phi} = b(\phi) + a(\phi)p\). The Hamiltonian (Legendre transform of \(L\)) is

\[ H(\phi, p) = p \cdot \dot{\phi} - L = p \cdot (b + ap) - \frac{1}{2}p^\top a\,p = b \cdot p + \frac{1}{2}p^\top a\,p \]

The Hamilton equations are

\[ \dot{\phi}^* = \frac{\partial H}{\partial p} = b(\phi^*) + a(\phi^*)\,p^* \]

\[ \dot{p}^* = -\frac{\partial H}{\partial \phi} = -(\nabla_\phi b)^\top p^* - \frac{1}{2}\nabla_\phi(p^{*\top}a\,p^*) \]

where the second equation accounts for the \(\phi\)-dependence of both \(b\) and \(a\). The instanton path \(\phi^*\) satisfies this system of \(2d\) ODEs with boundary conditions \(\phi^*(0) = x_0\) and \(\phi^*(T) \in\) target set. \(\square\)

Exercise 7. Consider the two-dimensional system \(\mathrm{d}X_t^\varepsilon = A X_t^\varepsilon\,\mathrm{d}t + \sqrt{\varepsilon}\,\mathrm{d}W_t\) where \(A\) is a \(2 \times 2\) stable matrix (all eigenvalues have negative real part) and \(X_0^\varepsilon = 0\). Write down the rate function \(I_{0,T}(\phi)\) for a path \(\phi\) with \(\phi(0) = 0\). Show that the quasipotential \(U(y) = \inf_{\phi(0) = 0,\, \phi(\infty) = y} I_{0,\infty}(\phi)\) is a quadratic form \(U(y) = \frac{1}{2}y^\top Q\,y\) and find the matrix equation that \(Q\) must satisfy in terms of \(A\).

Solution to Exercise 7

With \(b(\phi) = A\phi\) and \(\sigma = I\) (so \(a = I\)), the rate function for a path \(\phi\) with \(\phi(0) = 0\) is

\[ I_{0,T}(\phi) = \frac{1}{2}\int_0^T \|\dot{\phi}(t) - A\phi(t)\|^2\,\mathrm{d}t \]

The Hamiltonian is \(H(\phi, p) = (A\phi)\cdot p + \frac{1}{2}|p|^2\). The Hamilton equations are

\[ \dot{\phi} = A\phi + p, \qquad \dot{p} = -A^\top p \]

From the second equation: \(p(t) = e^{-A^\top t}p_0\). The first equation becomes \(\dot{\phi} = A\phi + e^{-A^\top t}p_0\), a linear ODE with solution

\[ \phi(t) = e^{At}\int_0^t e^{-As}e^{-A^\top s}\,\mathrm{d}s\;p_0 \]

(using \(\phi(0) = 0\)). As \(T \to \infty\), requiring \(\phi(T) \to y\) determines \(p_0\). The cost along the optimal path is

\[ U(y) = \frac{1}{2}\int_0^\infty |p(t)|^2\,\mathrm{d}t = \frac{1}{2}p_0^\top\!\left(\int_0^\infty e^{-A^\top t}e^{-At}\,\mathrm{d}t\right) p_0 \]

For the quasipotential \(U(y) = \frac{1}{2}y^\top Q\,y\), we use the Hamilton–Jacobi equation \(H(x, Qx) = 0\):

\[ (Ax)\cdot(Qx) + \frac{1}{2}(Qx)\cdot(Qx) = 0 \quad \text{for all } x \]

\[ x^\top A^\top Q\,x + \frac{1}{2}x^\top Q^2 x = 0 \]

Since this holds for all \(x\), the symmetric part must vanish:

\[ \frac{1}{2}(A^\top Q + QA) + \frac{1}{2}Q^2 = 0 \]

which gives

\[ A^\top Q + QA + Q^2 = 0 \]

This is the matrix equation that \(Q\) must satisfy. Alternatively, noting that the invariant covariance \(\Sigma\) of the linear SDE satisfies \(A\Sigma + \Sigma A^\top + I = 0\), one can verify that \(Q = \Sigma^{-1}\) satisfies the equation above (multiply the Lyapunov equation by \(\Sigma^{-1}\) from both sides). This gives \(U(y) = \frac{1}{2}y^\top\Sigma^{-1}y\), consistent with the Gaussian invariant measure \(\pi(y) \propto \exp(-\frac{1}{2}y^\top\Sigma^{-1}y) = \exp(-U(y)/\varepsilon)\) in the large deviations scaling.