This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
world:kolmogorov [2023/11/14 13:12] rdouc [Forward Kolmogorov equation] |
world:kolmogorov [2023/11/15 00:23] (current) rdouc [Forward Kolmogorov equation] |
||
---|---|---|---|
Line 12: | Line 12: | ||
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 ===== | ||
Line 20: | Line 19: | ||
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}} |
$$ | $$ | ||
Line 26: | Line 25: | ||
- | 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 | ||
Line 32: | Line 31: | ||
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}} |
$$ | $$ | ||
Line 47: | Line 46: | ||
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}{s} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{x}{s} | + | -\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} |
$$ | $$ | ||
Line 56: | Line 55: | ||
\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, | ||
Line 72: | Line 71: | ||
\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}{s} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{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. |