{{page>:defs}}
====== 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.