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world:kolmogorov [2023/11/13 00:05] rdouc [Forward and Backward Kolmogorov equations] |
world:kolmogorov [2023/11/15 00:23] (current) rdouc [Forward Kolmogorov equation] |
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$$ | $$ | ||
- | dX_s = \mu(X_s,s)\rmd s + \sigma(X_s,s)\rmd W_s | + | \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. | We provide only the ideas of the proofs. Additional assumptions are necessary to justify the use of all the tools. | ||
- | Let $y\mapsto \trans{t}{x}{y}$ be the density of $X_t$ starting from $X_0=x$. Since $(X_t)$ is a homogeneous Markov process, we also have that $y\mapsto \trans{t}{x}{y}$ is the density of $X_{s+t}$ knowing $X_s$ at $X_{s}=x$. | + | 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 ===== | ===== Forward Kolmogorov equation ===== | ||
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The Forward Kolmogorov equation writes | The Forward Kolmogorov equation writes | ||
$$ | $$ | ||
- | \partial_t \trans{t}{x}{y} = - \partial_y \lrb{{\mu(y,t) \trans{t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma^2(y,t) \trans{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_s=h(X_s) $ 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_s=h'(X_s) \rmd X_s + \frac 1 2 h''(X_s) \rmd\cro{X}_s=\lrb{h'(X_s) \mu(X_s,s)+\frac 1 2 h''(X_s) \sigma^2(X_s,s)} \rmd s + h'(X_s) \sigma(X_s,s) \rmd W_s | + | \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, | Hence, | ||
\begin{align*} | \begin{align*} | ||
- | \int_\rset h(y) \trans{t}{x}{y} \rmd y - h(x) &= \PE_x[h(X_t)]-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_s) \mu(X_s,s)+\frac 1 2 h''(X_s) \sigma^2(X_s,s) \rmd s } \\ | + | & =\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(y,s) \trans{s}{x}{y}\rmd y +\frac 1 2 \int_\rset h''(y) \sigma^2(y,s) \trans{s}{x}{y} \rmd y}\rmd s \\ | + | & =\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(y,s) \trans{s}{x}{y}}}\rmd y +\frac 1 2 \int_\rset h(y) \partial^2_{yy} \lrb{\sigma^2(y,s) \trans{s}{x}{y}} \rmd y}\rmd s | + | & = \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{t}{x}{y} = - \partial_y \lrb{{\mu(y,t) \trans{t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma^2(y,t) \trans{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{t-s}{x}{y}= \mu(x,s) \partial_x \trans{t-s}{x}{s} + \frac 1 2 \sigma^2(x,s) \partial^2_{xx} \trans{t-s}{x}{s}=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} |
$$ | $$ | ||
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Recall that | Recall that | ||
$$ | $$ | ||
- | dX_s=\mu(X_s,s) \rmd s + \sigma(X_s,s) \rmd W_s | + | \rmd X_v=\mu_v(X_v) \rmd v + \sigma_v(X_v) \rmd W_v |
$$ | $$ | ||
- | Now, define for $ s\leq t $, $u(x,s)=\left. \PE[h(X_t)|X_s] \right|_{X_s=x}=\int h(y) \trans{t-s}{x}{y} \rmd y$. | + | 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 $Y_v=u(X_v,v)$. By Itô's Formula, | + | Set $Y_v=u_v(X_v)$. By Itô's Formula, |
\begin{align*} | \begin{align*} | ||
- | \rmd Y_v&=\partial_s u(X_v,v) \rmd s + \partial_x u(X_v,v) \rmd X_v + \frac 1 2 \partial^2_{xx}u(X_v,v) \rmd\cro{X}_v \\ | + | \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 + \mu \partial_x u + \frac 1 2 \sigma^2 \partial^2_{xx}u}(X_v,v) \rmd s + \lrb{\sigma \partial_{x}u} (X_v,v) \rmd W_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*} | \end{align*} | ||
Note that $Y_t=h(X_t)$ and $Y_s=\left. \PE[h(X_t)|X_s]\right|_{X_s=x}$. Hence, | Note that $Y_t=h(X_t)$ and $Y_s=\left. \PE[h(X_t)|X_s]\right|_{X_s=x}$. Hence, | ||
\begin{align*} | \begin{align*} | ||
- | 0= \PE[Y_t-Y_s| X_s]|_{X_s=x}= \left. \PE \lrb{\left. \int_s^t \lrb{\partial_s u + \mu \partial_x u + \frac 1 2 \sigma^2 \partial^2_{xx}u}(X_v,v) \rmd v \right| X_s} \right|_{X_s=x} | + | 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*} | \end{align*} | ||
Dividing by $t-s$ and letting $t\to s$, we get | Dividing by $t-s$ and letting $t\to s$, we get | ||
$$ | $$ | ||
- | \lr{\partial_s u + \mu \partial_x u + \frac 1 2 \sigma^2 \partial^2_{xx}u}(x,s)=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 $u(x,s)=\int h(y) \trans{t-s}{x}{y} \rmd y$, we finally obtain | + | Since $u_s(x)=\int h(y) \condtrans{t|s}{y}{x} \rmd y$, we finally obtain |
$$ | $$ | ||
- | \partial_s \trans{t-s}{x}{y}+ \mu(x,s) \partial_x \trans{t-s}{x}{s} + \frac 1 2 \sigma^2(x,s) \partial^2_{xx} \trans{t-s}{x}{s}=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. |