Stationary Solutions for Homogeneous Fokker-Plank Equation
If FP equation is given by
\[\frac{\partial p_s}{\partial t}(x) = -\frac{\partial}{\partial x} \left( A(x)p_s(x) \right) + \frac{1}{2} \frac{\partial^2}{\partial x^2} \left( B(x)p_s(x) \right)\], then a homogeneous solutions of FP is same to the following:
\[\frac{d}{d x} \left( A(x)p_s(x) \right) - \frac{1}{2} \frac{d^2}{d x^2} \left( B(x)p_s(x) \right) = 0\]It means that
\[A(x)p_s(x) - \frac{1}{2} \frac{d}{d x} \left( B(x)p_s(x) \right) = 0.\]It is a simple oridinary differential equation. For convenience, we set a function $C(x)$ such that $C(x) = B(x)p_s(x)$, so that we evaluate the equation,
\[\begin{aligned} A(x)p_s(x) &= \frac{1}{2} \frac{d}{d x} \left( B(x)p_s(x) \right) \\ 2 \cdot C(x)\frac{A(x)}{B(x)} &= \frac{d}{d x} C(x) \\ 2 \cdot \frac{A(x)}{B(x)} dx &= \frac{1}{C(x)} dC(x) \\ C_0 \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) &= C(x) \\ p_s(x) &= \frac{C_0}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) \end{aligned} \label{eq01:ST} \tag{S1}\]where $C_0$ is a normaliztion constant such that $\int_0^b dx p_s(x) = 1$, thus
\[\begin{aligned} 1 = \int_R p_s(x) dx &= \int_R \frac{C_0}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) dx\\ &= C_0 \int_R \frac{1}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) dx \\ \Rightarrow C_0 &= \frac{1}{\int_R \frac{1}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) dx} \end{aligned}\]Potential dynamics
The foward equation
\[\frac{\partial p}{\partial t}(z, t) = - \sum_{i}\frac{\partial p}{\partial z_i} A_i(z, t)p(z, t) + \frac{1}{2} \sum_{i,j} \frac{\partial^2}{\partial z_i \partial z_j} B_{ij} (z, t) p(z, t) \label{eq02:PD} \tag{S2}\]is equivalent to
\[\frac{\partial p}{\partial t}(z, t) + \sum_{i} \frac{\partial}{\partial z_i} J_i(z, t) = 0 \label{eq03:PD} \tag{S3}\]where we define the probablity current
\[J_i (z, t) = A_i (z, t) p(z, t) - \frac{1}{2} \sum_j \frac{\partial }{\partial z_j} B_{ij}(z, t)p(z, t).\]Consider some REgion $R$ with a boundatry $S$ and define
\[P(R, t) = \int_R p(z, t) dz = \int_R dz p(z, t)\]then $\eqref{eq03:PD}$ is equivalent to
\[\frac{\partial P(R, t)}{\partial t} = - \int_S n \cdot J(z,t) dS = = - \int_S dS n \cdot J(z,t) \label{eq04:PD} \tag{S4}\]where $n$ is the outward pointning normal to $S$.
Therefore, $\eqref{eq04:PD}$ indicates that the total loss of probablity is given by the surface integral of $J$ over the boundary of $R$.
- 그러므로 Homogeneous Fokker Plank 방정식의 경우 $\eqref{eq02:PD}$ 에서
인 경우이며, 이 경우는 정통적인 1-st order Necessity Condition이 된다.
Stationary Solution of Gradient Descent
If the learning equation based on a stochastic gradient rule is given by
\[dX_t = - \nabla U(X_t) dt + \sqrt{T(t)} dW_t \label{eq05:GD} \tag{S5}\], then the Fokker-Plank equation of $\eqref{eq05:GD}$ is
\[\frac{\partial p}{\partial t}(x, t) = \frac{\partial}{\partial x} \left( \nabla U(x) p(x, t) \right) + \frac{1}{2} T(t) \frac{\partial^2 p}{\partial x^2}(x, t) \label{eq06:GD} \tag{S6}\], and the stationary solution of the $\eqref{eq06:GD}$ is
\[\begin{aligned} p(x, t) &= \frac{C_0}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) \\ &= \frac{C_0}{T(t)} \exp \left( 2 \int_0^x \frac{-\nabla U (\bar{x})}{T(t)} d\bar{x} \right) \\ &= \frac{C_0}{T(t)} \exp \left( -2 \frac{1}{T(t)} \int_0^x \nabla U (\bar{x})d\bar{x} \right) \\ &= \frac{C_0}{T(t)} \exp \left( -2 \frac{U (x)}{T(t)}\right) \end{aligned} \label{eq07:GD} \tag{S7}\]By definition, $C_0$ is a normalization parameter such that
\[\begin{aligned} C_0 &= \frac{1}{\int_R \frac{1}{B(x)} \exp \left( 2 \int_0^x \frac{A(\bar{x})}{B(\bar{x})} d\bar{x} \right) dx} \\ &= \frac{1}{\frac{1}{T(t)}\int_R \exp \left( -2 \frac{U (x)}{T(t)}\right) dx} \\ &= \frac{T(t)}{\int_R \exp \left( -2 \frac{U (x)}{T(t)}\right) dx}. \end{aligned}\]Therefore, the stationary solution of gradient descent is
\[p(x, t) = \frac{1}{Z(x)} \exp \left( -2 \frac{U (x)}{T(t)}\right)\]where $Z(x) = \int_R \exp \left( -2 \frac{U (x)}{T(t)}\right) dx$.
The stationary solution of the Fokker Plank equation to the stochastic gradient rule $\eqref{eq05:GD}$ is Gibb’s distribution or Boltzmann distribution.
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