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Diffstat (limited to 'source/know/concept/detailed-balance')
| -rw-r--r-- | source/know/concept/detailed-balance/index.md | 54 |
1 files changed, 27 insertions, 27 deletions
diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md index 9745959..b89d5da 100644 --- a/source/know/concept/detailed-balance/index.md +++ b/source/know/concept/detailed-balance/index.md @@ -24,9 +24,9 @@ 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 +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): @@ -37,7 +37,7 @@ $$\begin{aligned} We start by demanding **stationarity**, which is a weaker condition than detailed balance. -We want the probability $P$ of being in an arbitrary state volume $V$ +We want the probability $$P$$ of being in an arbitrary state volume $$V$$ to be constant in time: $$\begin{aligned} @@ -56,11 +56,11 @@ $$\begin{aligned} = - \oint_{\partial V} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}} \end{aligned}$$ -In other words, the "flow" of probability *into* the volume $V$ -is equal to the flow *out of* $V$. +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: +it is called a **stationary distribution** $$\phi(t, x) = \pi(x)$$. +Because $$V$$ was arbitrary, $$\pi$$ can be found by solving: $$\begin{aligned} 0 @@ -70,7 +70,7 @@ $$\begin{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 +the probability flux through an arbitrary surface $$S$$ must vanish (compare to closed surface integral above): $$\begin{aligned} @@ -78,7 +78,7 @@ $$\begin{aligned} = - \int_{S} \big( \vb{u} \phi - D \nabla \phi \big) \cdot \dd{\vb{S}} \end{aligned}$$ -And since $S$ is arbitrary, this is only satisfied if the flux is trivially zero +And since $$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} @@ -93,7 +93,7 @@ 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 $k(x)$, +Formally, given two arbitrary functions $$h(x)$$ and $$k(x)$$, we have the property: $$\begin{aligned} @@ -109,9 +109,9 @@ $$\begin{aligned} <div class="hidden" markdown="1"> <label for="proof-reversibility">Proof.</label> Consider the following weighted inner product, -whose weight function is a stationary distribution $\pi$ +whose weight function is a stationary distribution $$\pi$$ satisfying detailed balance, -where $\hat{L}$ is the Kolmogorov operator: +where $$\hat{L}$$ is the Kolmogorov operator: $$\begin{aligned} \inprod{\hat{L} h}{k}_\pi @@ -128,8 +128,8 @@ $$\begin{aligned} = -\nabla \cdot (\vb{u} \pi k - D k \nabla \pi - D \pi \nabla k) \end{aligned}$$ -Since $\pi$ is stationary by definition, -we know that $\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$, +Since $$\pi$$ is stationary by definition, +we know that $$\nabla \cdot (\vb{u} \pi - D \nabla \pi) = 0$$, meaning: $$\begin{aligned} @@ -138,7 +138,7 @@ $$\begin{aligned} = \nabla \pi \cdot (D \nabla k) + \pi \nabla \cdot (D \nabla k) \end{aligned}$$ -Detailed balance demands that $\vb{u} \pi = D \nabla \pi$, +Detailed balance demands that $$\vb{u} \pi = D \nabla \pi$$, leading to the following: $$\begin{aligned} @@ -150,9 +150,9 @@ $$\begin{aligned} = \pi \hat{L}\{k\} \end{aligned}$$ -Where we recognized the definition of $\hat{L}$ +Where we recognized the definition of $$\hat{L}$$ from the backward Kolmogorov equation. -Now that we have established that $\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$, +Now that we have established that $$\hat{L}{}^\dagger\{\pi k\} = \pi \hat{L}\{k\}$$, we return to the inner product: $$\begin{aligned} @@ -170,7 +170,7 @@ $$\begin{aligned} Now, consider the time evolution of the [conditional expectation](/know/concept/conditional-expectation/) -$\mathbf{E}\big[ k(X_t) | X_0 \big]$: +$$\mathbf{E}\big[ k(X_t) | X_0 \big]$$: $$\begin{aligned} \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big] @@ -184,10 +184,10 @@ $$\begin{aligned} Where we used the forward Kolmogorov equation and the definition of an adjoint operator. -Therefore, since the expectation $\mathbf{E}$ -does not explicitly depend on $t$ (only implicitly via $X_t$), +Therefore, since the expectation $$\mathbf{E}$$ +does not explicitly depend on $$t$$ (only implicitly via $$X_t$$), we can naively move the differentiation inside -(only valid within $\mathbf{E}$): +(only valid within $$\mathbf{E}$$): $$\begin{aligned} \pdv{}{t}\mathbf{E}\big[ k(X_t) | X_0 \big] @@ -195,9 +195,9 @@ $$\begin{aligned} = \mathbf{E}\bigg[ \hat{L}\{k(X_0)\} \bigg| X_0 \bigg] \end{aligned}$$ -A differential equation of the form $\ipdv{k}{t} = \hat{L}\{k(t, x)\}$ -for a time-independent operator $\hat{L}$ -has a general solution $k(t, x) = \exp(t \hat{L})\{k(0,x)\}$, +A differential equation of the form $$\ipdv{k}{t} = \hat{L}\{k(t, x)\}$$ +for a time-independent operator $$\hat{L}$$ +has a general solution $$k(t, x) = \exp(t \hat{L})\{k(0,x)\}$$, therefore: $$\begin{aligned} @@ -220,8 +220,8 @@ $$\begin{aligned} = \mathbf{E}\big[ h(X_t) \: k(X_0) \big] \end{aligned}$$ -Where the integral gave the expectation value at $X_0$, -since $\pi$ does not change in time. +Where the integral gave the expectation value at $$X_0$$, +since $$\pi$$ does not change in time. </div> </div> |