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====== Forward and Backward Kolmogorov equations ======

Consider the following SDE:

$$
dX_s = \mu(X_s,s)ds + \sigma(X_s,s)dW_s
$$

We provide only the ideas of the proofs. Additional assumptions are necessary to justify the use of all the tools.

Let $y\mapsto \trans{t}{x}{y}$ 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 $X_{s}=x$.

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

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

</WRAP>

Set $ Y_s=h(X_s) $ where $h$ is for example $C^2$ with bounded second derivatives. By Itô's Formula, 
$$
\rmd Y_s=h'(X_s) \rmd X_s + \frac 1 2 h''(X_s) \rmd\cro{X}_s=\lrb{h'(X_s) \mu(X_s,s)+\frac 1 2 h''(X_s) \sigma^2(X_s,s)} \rmd s + h'(X_s) \sigma(X_s,s) \rmd W_s
$$
Hence, 
\begin{align*}
    \int_\rset h(y) \trans{t}{x}{y} \rmd y - h(x) &= \PE_x[h(X_t)]-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 } \\ 
    & =\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_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 
\end{align*}
where the last equality is obtained from integration by parts. Differentiating both sides of the equation wrt $t$ yields
$$
\partial_t \trans{t}{x}{y} = - \partial_y \lrb{{\mu(y,t) \trans{t}{x}{y}}} +\frac 1 2 \partial^2_{yy} \lrb{\sigma^2(y,t) \trans{t}{x}{y}}
$$

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

<WRAP center round tip 90%>
The Backward Kolmogorov equation writes
$$
-\partial_s \trans{t-s}{x}{y}= \mu(x,s) \partial_x \trans{t-s}{x}{s} + \frac 1 2 \sigma^2(x,s) \partial^2_{xx} \trans{t-s}{x}{s}=0
$$

</WRAP>

Recall that 
$$
dX_s=\mu(X_s,s) \rmd s +  \sigma(X_s,s) \rmd W_s
$$
Now, define for $ s\leq t $, $u(x,s)=\left. \PE[h(X_t)|X_s] \right|_{X_s=x}=\int h(y) \trans{t-s}{x}{y} \rmd y$.

Set $Y_v=u(X_v,v)$. By Itô's Formula, 
\begin{align*}
    \rmd Y_v&=\partial_v 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 \\ 
    & = \lrb{\partial_v 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 
\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_v 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}
\end{align*}
Dividing by $t-s$ and letting $t\to s$, we get 
$$
\lr{\partial_s u + \mu \partial_x u  + \frac 1 2 \sigma^2 \partial^2_{xx}u}(x,s)=0
$$
Since $u(x,s)=\int h(y) \trans{t-s}{x}{y} \rmd y$, we finally obtain
$$
\partial_s \trans{t-s}{x}{y}+ \mu(x,s) \partial_x \trans{t-s}{x}{s} + \frac 1 2 \sigma^2(x,s) \partial^2_{xx} \trans{t-s}{x}{s}=0
$$
which completes the proof.

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