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====== 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.

Let $y\mapsto \trans{s,t}{x}{y}$ be the density of $X_t$ starting from $X_s=x$.

===== Forward Kolmogorov equation =====

<WRAP center round tip 90%>
The Forward Kolmogorov equation writes
$$
\partial_t \trans{s,t}{x}{y} = - \partial_y \lrb{{\mu_t(y) \trans{s,t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \trans{s,t}{x}{y}}
$$

</WRAP>

Set $ Y_u=h(X_u) $ where $h$ is $C^\infty$ 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) \trans{s,t}{x}{y} \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 } \\ 
    & =\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_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 
\end{align*}
where the last equality is obtained from integration by parts. Differentiating both sides of the equation wrt $t$ yields
$$
\partial_t \trans{t}{x}{y} = - \partial_y \lrb{{\mu_t(y) \trans{t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma_t^2(y) \trans{t}{x}{y}}
$$

===== Backward Kolmogorov equation =====

<WRAP center round tip 90%>
The Backward Kolmogorov equation writes
$$
-\partial_s \trans{s,t}{x}{y}= \mu_s(x) \partial_x \trans{s,t}{x}{s} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{x}{s}
$$

</WRAP>

Recall that 
$$
\rmd X_v=\mu_v(X_v) \rmd v +  \sigma_v(X_v) \rmd W_v
$$
Now, define for $ s\leq t $, $u_s(x)=\left. \PE[h(X_t)|X_s] \right|_{X_s=x}=\int h(y) \trans{s,t}{x}{y} \rmd y$.

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 $u_s(x)=\int h(y) \trans{s,t}{x}{y} \rmd y$, we finally obtain
$$
\partial_s \trans{s,t}{x}{y}+ \mu_s(x) \partial_x \trans{s,t}{x}{s} + \frac 1 2 \sigma^2_s(x) \partial^2_{xx} \trans{s,t}{x}{s}=0
$$
which completes the proof.

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