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Subject: Expand knowledge base

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+---
+title: "Detailed balance"
+firstLetter: "D"
+publishDate: 2021-11-27
+categories:
+- Physics
+- Mathematics
+- Stochastic analysis
+
+date: 2021-11-25T20:42:35+01:00
+draft: false
+markup: pandoc
+---
+
+# Detailed balance
+
+Consider a system that can be regarded as a
+[Markov process](/know/concept/markov-process/),
+which means that its components (e.g. particles) are transitioning
+between a known set of states,
+with no history-dependence and no appreciable influence from interactions.
+
+At equilibrium, the principle of **detailed balance** then says that
+for all states, the rate of leaving that state is exactly equal to
+the rate of entering it, for every possible transition.
+In effect, such a system looks "frozen" to an outside observer,
+since all net transition rates are zero.
+
+We will focus on the case where both time and the state space are continuous.
+Given some initial conditions,
+assume that a component's trajectory can be described
+as an [Itō diffusion](/know/concept/ito-calculus/) $X_t$
+with a time-independent drift $f$ and intensity $g$,
+and with a probability density $\phi(t, x)$ governed by the
+[forward Kolmogorov equation](/know/concept/kolmogorov-equations/)
+(in 3D):
+
+$$\begin{aligned}
+    \pdv{\phi}{t}
+    = - \nabla \cdot \big( \vb{u} \phi - D \nabla \phi \big)
+\end{aligned}$$
+
+We start by demanding **stationarity**,
+which is a weaker condition than detailed balance.
+We want the probability $P$ of being in an arbitrary state volume $V$
+to be constant in time:
+
+$$\begin{aligned}
+    0
+    = \pdv{t} P(X_t \in V)
+    = \pdv{t} \int_V \phi \dd{V}
+    = \int_V \pdv{\phi}{t} \dd{V}
+\end{aligned}$$
+
+We substitute the forward Kolmogorov equation,
+and apply the divergence theorem:
+
+$$\begin{aligned}
+    0
+    = - \int_V \nabla \cdot \big( \vb{u} \phi - D \nabla \phi \big) \dd{V}
+    = - \oint_{\partial V} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
+\end{aligned}$$
+
+In other words, the "flow" of probability *into* the volume $V$
+is equal to the flow *out of* $V$.
+If such a probability density exists,
+it is called a **stationary distribution** $\phi(t, x) = \pi(x)$.
+Because $V$ was arbitrary, $\pi$ can be found by solving:
+
+$$\begin{aligned}
+    0
+    = - \nabla \cdot \big( \vb{u} \pi - D \nabla \pi \big)
+\end{aligned}$$
+
+Therefore, stationarity means that the state transition rates are constant.
+To get detailed balance, however, we demand that
+the transition rates are zero everywhere:
+the probability flux through an arbitrary surface $S$ must vanish
+(compare to closed surface integral above):
+
+$$\begin{aligned}
+    0
+    = - \int_{S} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}}
+\end{aligned}$$
+
+And since $S$ is arbitrary, this is only satisfied if the flux is trivially zero
+(the above justification can easily be repeated in 1D, 2D, 4D, etc.):
+
+$$\begin{aligned}
+    \boxed{
+        0 = \vb{u} \phi - D \nabla \phi
+    }
+\end{aligned}$$
+
+This is a stronger condition that stationarity,
+but fortunately often satisfied in practice.
+
+The fact that a system in detailed balance appears "frozen"
+implies it is **time-reversible**,
+meaning its statistics are the same for both directions of time.
+Formally, given two arbitrary functions $h(x)$ and $k(x)$,
+we have the property:
+
+$$\begin{aligned}
+    \boxed{
+        \mathbf{E}\big[ h(X_0) \: k(X_t) \big]
+        = \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-reversibility"/>
+<label for="proof-reversibility">Proof</label>
+<div class="hidden">
+<label for="proof-reversibility">Proof.</label>
+Consider the following weighted inner product,
+whose weight function is a stationary distribution $\pi$
+satisfying detailed balance,
+where $\hat{L}$ is the Kolmogorov operator:
+
+$$\begin{aligned}
+    \braket*{\hat{L} h}{k}_\pi
+    \equiv \int_{-\infty}^\infty \hat{L}\{h(x)\} \: \pi(x) \: k(x) \dd{x}
+    = \int_{-\infty}^\infty h(x) \: \hat{L}{}^\dagger\{\pi(x) k(x)\} \dd{x}
+\end{aligned}$$
+
+Where we have used the definition of an adjoint operator.
+We would like to rewrite this:
+
+$$\begin{aligned}
+    \hat{L}{}^\dagger \{\pi k\}
+    = -\nabla \cdot \big( \vb{u} \pi k - D \nabla(\pi k) \big)
+    = -\nabla \cdot (\vb{u} \pi k - D k \nabla \pi - D \pi \nabla k)
+\end{aligned}$$
+
+Since $\pi$ is stationary by definition,
+we know that $\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$,
+meaning:
+
+$$\begin{aligned}
+    \hat{L}{}^\dagger \{\pi k\}
+    = \nabla \cdot (D \pi \nabla k)
+    = \nabla \pi \cdot (D \nabla k) + \pi \nabla \cdot (D \nabla k)
+\end{aligned}$$
+
+Detailed balance demands that $\vb{u} \pi = D \nabla \pi$,
+leading to the following:
+
+$$\begin{aligned}
+    \hat{L}{}^\dagger \{\pi k\}
+    &= D \nabla \pi \cdot \nabla k + \pi \nabla \cdot (D \nabla k)
+    = \pi \vb{u} \cdot \nabla k + \pi \nabla \cdot (D \nabla k)
+    \\
+    &= \pi \big( \vb{u} \cdot \nabla k + \nabla \cdot (D \nabla k) \big)
+    = \pi \hat{L}\{k\}
+\end{aligned}$$
+
+Where we recognized the definition of $\hat{L}$
+from the backward Kolmogorov equation.
+Now that we have established that $\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$,
+we return to the inner product:
+
+$$\begin{aligned}
+    \braket*{\hat{L} h}{k}_\pi
+    = \int_{-\infty}^\infty h(x) \: \pi(x) \: \hat{L}\{k(x)\} \dd{x}
+    = \braket*{h}{\hat{L} k}_\pi
+\end{aligned}$$
+
+Consequently, the following weighted inner products must also be equivalent:
+
+$$\begin{aligned}
+    \braket{\exp\!(t \hat{L}) h}{k}_\pi
+    = \braket{h}{\exp\!(t \hat{L}) k}_\pi
+\end{aligned}$$
+
+Now, consider the time evolution of the
+[conditional expectation](/know/concept/conditional-expectation/)
+$\mathbf{E}\big[ k(X_t) | X_0 \big]$:
+
+$$\begin{aligned}
+    \pdv{t} \mathbf{E}\big[ k(X_t) | X_0 \big]
+    &= \pdv{t} \int_{-\infty}^\infty k(x) \: \phi(t, x) \dd{x}
+    = \int_{-\infty}^\infty k \pdv{\phi}{t} \dd{x}
+    \\
+    &= \int_{-\infty}^\infty k \: \hat{L}{}^\dagger\{\phi\} \dd{x}
+    = \int_{-\infty}^\infty \hat{L}\{k\} \: \phi \dd{x}
+    = \mathbf{E}\big[ \hat{L}\{k(X_t)\} | X_0 \big]
+\end{aligned}$$
+
+Where we used the forward Kolmogorov equation
+and the definition of an adjoint operator.
+Therefore, since the expectation $\mathbf{E}$
+does not explicitly depend on $t$ (only implicitly via $X_t$),
+we can naively move the differentiation inside
+(only valid within $\mathbf{E}$):
+
+$$\begin{aligned}
+    \pdv{t} \mathbf{E}\big[ k(X_t) | X_0 \big]
+    = \mathbf{E}\bigg[ \pdv{k(X_t)}{t} \bigg| X_0 \bigg]
+    = \mathbf{E}\bigg[ \hat{L}\{k(X_0)\} \bigg| X_0 \bigg]
+\end{aligned}$$
+
+A differential equation of the form $\pdv*{k}{t} = \hat{L}\{k(t, x)\}$
+for a time-independent operator $\hat{L}$
+has a general solution $k(t, x) = \exp\!(t \hat{L})\{k(0,x)\}$,
+therefore:
+
+$$\begin{aligned}
+    \mathbf{E}\big[ k(X_t) \big| X_0 \big]
+    = \mathbf{E}\big[ \exp\!(t \hat{L})\{k(X_0)\} \big| X_0 \big]
+    = \exp\!(t \hat{L})\{k(X_0)\}
+\end{aligned}$$
+
+With this, we can evaluate the two weighted inner products from earlier,
+which we know are equal to each other.
+Using the *tower property* of the conditional expectation:
+
+$$\begin{aligned}
+    \braket{h}{\exp\!(t \hat{L}) k}_\pi
+    &= \mathbf{E}\big[ h(X_0) \: \mathbf{E}[k(X_t) | X_0] \big]
+    = \mathbf{E}\big[ h(X_0) \: k(X_t) \big]
+    \\
+    = \braket{\exp\!(t \hat{L}) h}{k}_\pi
+    &= \mathbf{E}\big[ \mathbf{E}[h(X_t) | X_0] \: k(X_0) \big]
+    = \mathbf{E}\big[ h(X_t) \: k(X_0) \big]
+\end{aligned}$$
+
+Where the integral gave the expectation value at $X_0$,
+since $\pi$ does not change in time.
+</div>
+</div>
+
+
+
+## References
+1.  U.H. Thygesen,
+    *Lecture notes on diffusions and stochastic differential equations*,
+    2021, Polyteknisk Kompendie.
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