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world:kolmogorov
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$$ \newcommand{\arginf}{\mathrm{arginf}} \newcommand{\argmin}{\mathrm{argmin}} \newcommand{\argmax}{\mathrm{argmax}} \newcommand{\asconv}[1]{\stackrel{#1-a.s.}{\rightarrow}} \newcommand{\Aset}{\mathsf{A}} \newcommand{\b}[1]{{\mathbf{#1}}} \newcommand{\ball}[1]{\mathsf{B}(#1)} \newcommand{\bbQ}{{\mathbb Q}} \newcommand{\bproof}{\textbf{Proof :}\quad} \newcommand{\bmuf}[2]{b_{#1,#2}} \newcommand{\card}{\mathrm{card}} \newcommand{\chunk}[3]{{#1}_{#2:#3}} \newcommand{\condtrans}[3]{p_{#1}(#2|#3)} \newcommand{\convprob}[1]{\stackrel{#1-\text{prob}}{\rightarrow}} \newcommand{\Cov}{\mathbb{C}\mathrm{ov}} \newcommand{\cro}[1]{\langle #1 \rangle} \newcommand{\CPE}[2]{\PE\lr{#1| #2}} \renewcommand{\det}{\mathrm{det}} \newcommand{\dimlabel}{\mathsf{m}} \newcommand{\dimU}{\mathsf{q}} \newcommand{\dimX}{\mathsf{d}} \newcommand{\dimY}{\mathsf{p}} \newcommand{\dlim}{\Rightarrow} \newcommand{\e}[1]{{\left\lfloor #1 \right\rfloor}} \newcommand{\eproof}{\quad \Box} \newcommand{\eremark}{</WRAP>} \newcommand{\eqdef}{:=} \newcommand{\eqlaw}{\stackrel{\mathcal{L}}{=}} \newcommand{\eqsp}{\;} \newcommand{\Eset}{ {\mathsf E}} \newcommand{\esssup}{\mathrm{essup}} \newcommand{\fr}[1]{{\left\langle #1 \right\rangle}} \newcommand{\falph}{f} \renewcommand{\geq}{\geqslant} \newcommand{\hchi}{\hat \chi} \newcommand{\Hset}{\mathsf{H}} \newcommand{\Id}{\mathrm{Id}} \newcommand{\img}{\text{Im}} \newcommand{\indi}[1]{\mathbf{1}_{#1}} \newcommand{\indiacc}[1]{\mathbf{1}_{\{#1\}}} \newcommand{\indin}[1]{\mathbf{1}\{#1\}} \newcommand{\itemm}{\quad \quad \blacktriangleright \;} \newcommand{\jointtrans}[3]{p_{#1}(#2,#3)} \newcommand{\ker}{\text{Ker}} \newcommand{\klbck}[2]{\mathrm{K}\lr{#1||#2}} \newcommand{\law}{\mathcal{L}} \newcommand{\labelinit}{\pi} \newcommand{\labelkernel}{Q} \renewcommand{\leq}{\leqslant} \newcommand{\lone}{\mathsf{L}_1} \newcommand{\lrav}[1]{\left|#1 \right|} \newcommand{\lr}[1]{\left(#1 \right)} \newcommand{\lrb}[1]{\left[#1 \right]} \newcommand{\lrc}[1]{\left\{#1 \right\}} \newcommand{\lrcb}[1]{\left\{#1 \right\}} \newcommand{\ltwo}[1]{\PE^{1/2}\lrb{\lrcb{#1}^2}} \newcommand{\Ltwo}{\mathrm{L}^2} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mcbb}{\mathcal B} \newcommand{\mcf}{\mathcal{F}} \newcommand{\meas}[1]{\mathrm{M}_{#1}} \newcommand{\norm}[1]{\left\|#1\right\|} \newcommand{\normmat}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}} \newcommand{\nset}{\mathbb N} \newcommand{\N}{\mathcal{N}} \newcommand{\one}{\mathsf{1}} \newcommand{\PE}{\mathbb E} \newcommand{\pminfty}{_{-\infty}^\infty} \newcommand{\PP}{\mathbb P} \newcommand{\projorth}[1]{\mathsf{P}^\perp_{#1}} \newcommand{\Psif}{\Psi_f} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\pscal}[2]{\langle #1,#2\rangle} \newcommand{\psconv}{\stackrel{\PP-a.s.}{\rightarrow}} \newcommand{\qset}{\mathbb Q} \newcommand{\revcondtrans}[3]{q_{#1}(#2|#3)} \newcommand{\rmd}{\mathrm d} \newcommand{\rme}{\mathrm e} \newcommand{\rmi}{\mathrm i} \newcommand{\Rset}{\mathbb{R}} \newcommand{\rset}{\mathbb{R}} \newcommand{\rti}{\sigma} \newcommand{\section}[1]{==== #1 ====} \newcommand{\seq}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\set}[2]{\lrc{#1\eqsp: \eqsp #2}} \newcommand{\sg}{\mathrm{sgn}} \newcommand{\supnorm}[1]{\left\|#1\right\|_{\infty}} \newcommand{\thv}{{\theta_\star}} \newcommand{\tmu}{ {\tilde{\mu}}} \newcommand{\Tset}{ {\mathsf{T}}} \newcommand{\Tsigma}{ {\mathcal{T}}} \newcommand{\ttheta}{{\tilde \theta}} \newcommand{\tv}[1]{\left\|#1\right\|_{\mathrm{TV}}} \newcommand{\unif}{\mathrm{Unif}} \newcommand{\weaklim}[1]{\stackrel{\mathcal{L}_{#1}}{\rightsquigarrow}} \newcommand{\Xset}{{\mathsf X}} \newcommand{\Xsigma}{\mathcal X} \newcommand{\Yset}{{\mathsf Y}} \newcommand{\Ysigma}{\mathcal Y} \newcommand{\Var}{\mathbb{V}\mathrm{ar}} \newcommand{\zset}{\mathbb{Z}} \newcommand{\Zset}{\mathsf{Z}} $$
Forward and Backward Kolmogorov equations
Consider the following SDE:
$$ \rmd X_s = \mu_s(X_s)\rmd s + \sigma_s(X_s)\rmd W_s $$
We provide only the ideas of the proofs. Additional assumptions are necessary to justify the use of all the tools.
In what follows, we consider $s\leq t$ and we let $y\mapsto \condtrans{t|s}{y}{x}$ be the density of $X_t$ starting from $X_s=x$.
Forward Kolmogorov equation
The Forward Kolmogorov equation writes $$ \partial_t \condtrans{t|s}{y}{x} = - \partial_y \lrb{{\mu_t(y) \condtrans{t|s}{y}{x}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \condtrans{t|s}{y}{x}} $$
Set $ Y_u=h(X_u) $ where $h$ is $C^2$ with bounded support. By Itô's Formula, $$ \rmd Y_u=h'(X_u) \rmd X_u + \frac 1 2 h''(X_u) \rmd\cro{X}_u=\lrb{h'(X_u) \mu_u(X_u)+\frac 1 2 h''(X_u) \sigma_u^2(X_u)} \rmd u + h'(X_u) \sigma_u(X_u) \rmd W_u $$ Hence, \begin{align*} \int_\rset h(y) \condtrans{t|s}{y}{x} \rmd y - h(x) &= \PE[h(X_t)|X_s]|_{X_s=x}-h(x) \\ & =\PE_x\lrb{\int_s^t h'(X_u) \mu_u(X_u)+\frac 1 2 h''(X_u) \sigma_u^2(X_u) \rmd u } \\ & =\int_s^t \lr{\int_\rset h'(y) \mu_u(y) \condtrans{u|s}{y}{x}\rmd y +\frac 1 2 \int_\rset h''(y) \sigma^2_u(y) \condtrans{u|s}{y}{x} \rmd y}\rmd u \\ & = \int_s^t \lr{-\int_\rset h(y) \partial_y \lrb{{\mu_u(y) \condtrans{u|s}{y}{x}}}\rmd y +\frac 1 2 \int_\rset h(y) \partial^2_{yy} \lrb{\sigma^2_u(y) \condtrans{u|s}{y}{x}} \rmd y}\rmd u \end{align*} where the last equality is obtained from integration by parts. Differentiating both sides of the equation wrt $t$ yields $$ \partial_t \condtrans{t|s}{y}{x} = - \partial_y \lrb{{\mu_t(y) \condtrans{t|s}{y}{x}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \condtrans{t|s}{y}{x}} $$
Backward Kolmogorov equation
The Backward Kolmogorov equation writes $$ -\partial_s \condtrans{t|s}{y}{x}= \mu_s(x) \partial_x \condtrans{t|s}{y}{x} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \condtrans{t|s}{y}{x} $$
Recall that $$ \rmd X_v=\mu_v(X_v) \rmd v + \sigma_v(X_v) \rmd W_v $$ Now, define for $ s\leq t $, $u_s(x)=\left. \PE[h(X_t)|X_s] \right|_{X_s=x}=\int h(y) \condtrans{t|s}{y}{x} \rmd y$.
Set $Y_v=u_v(X_v)$. By Itô's Formula, \begin{align*} \rmd Y_v&=\partial_s u_v(X_v) \rmd s + \partial_x u_v(X_v) \rmd X_v + \frac 1 2 \partial^2_{xx}u_v(X_v) \rmd\cro{X}_v \\ & = \lrb{\partial_s u_v + \mu_v \partial_x u_v + \frac 1 2 \sigma_v^2 \partial^2_{xx}u_v}(X_v) \rmd s + \lrb{\sigma_v \partial_{x}u_v} (X_v) \rmd W_v \end{align*}
Note that $Y_t=h(X_t)$ and $Y_s=\left. \PE[h(X_t)|X_s]\right|_{X_s=x}$. Hence, \begin{align*} 0= \PE[Y_t-Y_s| X_s]|_{X_s=x}= \left. \PE \lrb{\left. \int_s^t \lrb{\partial_s u_v + \mu_v \partial_x u_v + \frac 1 2 \sigma_v^2 \partial^2_{xx}u_v}(X_v) \rmd v \right| X_s} \right|_{X_s=x} \end{align*} Dividing by $t-s$ and letting $t\to s$, we get $$ \lr{\partial_s u_s + \mu_s \partial_x u_s + \frac 1 2 \sigma_s^2 \partial^2_{xx}u_s}(x)=0 $$ Since $u_s(x)=\int h(y) \condtrans{t|s}{y}{x} \rmd y$, we finally obtain $$ \partial_s \condtrans{t|s}{y}{x}+ \mu_s(x) \partial_x \condtrans{t|s}{y}{x} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \condtrans{t|s}{y}{x}=0 $$ which completes the proof.
world/kolmogorov.txt · Last modified: 2023/11/15 00:23 by rdouc
