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world:kolmogorov [2023/11/14 13:13]
rdouc [Backward Kolmogorov equation]
world:kolmogorov [2023/11/15 00:23] (current)
rdouc [Forward Kolmogorov equation]
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 We provide only the ideas of the proofs. Additional assumptions are necessary to justify the use of all the tools. 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 $s\leq t$.  +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 $X_s=x$. ​
-Let $y\mapsto \trans{s,t}{x}{y}$ be the density of $X_t$ starting from $X_s=x$. ​+
  
 ===== Forward Kolmogorov equation ===== ===== Forward Kolmogorov equation =====
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 The Forward Kolmogorov equation writes The Forward Kolmogorov equation writes
 $$ $$
-\partial_t \trans{s,t}{x}{y} = - \partial_y \lrb{{\mu_t(y) \trans{s,t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \trans{s,t}{x}{y}}+\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}}
 $$ $$
  
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-Set $ Y_u=h(X_u) $ where $h$ is $C^\infty$ with bounded support. By Itô's Formula, ​+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 \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
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 Hence, ​ Hence, ​
 \begin{align*} \begin{align*}
-    \int_\rset h(y) \trans{s,t}{x}{y} \rmd y - h(x) &= \PE[h(X_t)|X_s]|_{X_s=x}-h(x) ​  \\  +    \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_0^t h'​(X_u) \mu_u(X_u)+\frac 1 2 h''​(X_u) \sigma_u^2(X_u) \rmd u } \\  +    & =\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_0^t  \lr{\int_\rset h'(y) \mu_u(y) \trans{s,u}{x}{y}\rmd y +\frac 1 2 \int_\rset h''​(y) \sigma^2_u(y) \trans{s,u}{x}{y} \rmd y}\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_0^t  \lr{-\int_\rset h(y) \partial_y \lrb{{\mu_u(y) \trans{s,u}{x}{y}}}\rmd y +\frac 1 2 \int_\rset h(y) \partial^2_{yy} \lrb{\sigma^2_u(y) \trans{s,u}{x}{y}} \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*} \end{align*}
 where the last equality is obtained from integration by parts. Differentiating both sides of the equation wrt $t$ yields where the last equality is obtained from integration by parts. Differentiating both sides of the equation wrt $t$ yields
 $$ $$
-\partial_t \trans{s,t}{x}{y} = - \partial_y \lrb{{\mu_t(y) \trans{s,t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \trans{s,t}{x}{y}}+\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}}
 $$ $$
  
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 The Backward Kolmogorov equation writes The Backward Kolmogorov equation writes
 $$ $$
--\partial_s \trans{s,t}{x}{y}= \mu_s(x) \partial_x \trans{s,t}{x}{y} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{x}{y}+-\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}
 $$ $$
  
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 \rmd X_v=\mu_v(X_v) \rmd v +  \sigma_v(X_v) \rmd W_v \rmd X_v=\mu_v(X_v) \rmd v +  \sigma_v(X_v) \rmd W_v
 $$ $$
-Now, define for $ s\leq t $, $u_s(x)=\left. \PE[h(X_t)|X_s] \right|_{X_s=x}=\int h(y) \trans{s,t}{x}{y} \rmd y$. +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, ​ Set $Y_v=u_v(X_v)$. By Itô's Formula, ​
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 \lr{\partial_s u_s + \mu_s \partial_x u_s  + \frac 1 2 \sigma_s^2 \partial^2_{xx}u_s}(x)=0 \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 $u_s(x)=\int h(y) \trans{s,t}{x}{y} \rmd y$, we finally obtain+Since $u_s(x)=\int h(y) \condtrans{t|s}{y}{x} \rmd y$, we finally obtain
 $$ $$
-\partial_s \trans{s,t}{x}{y}+ \mu_s(x) \partial_x \trans{s,t}{x}{y} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{x}{y}=0+\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. ​ which completes the proof. ​
world/kolmogorov.1699964027.txt.gz · Last modified: 2023/11/14 13:13 by rdouc