Exercise 6.8 (Kolmogorov backward equation). Consider the stochastic differential equation
$$
d X(u)=\beta(u, X(u)) d u+\gamma(u, X(u)) d W(u) .
$$
We assume that, just as with a geometric Brownian motion, if we begin a process at an arbitrary initial positive value $X(t)=x$ at an arbitrary initial time $t$ and evolve it forward using this equation, its value at each time $T>t$ could be any positive number but cannot be less than or equal to zero. For $0 \leq t<T$, let $p(t, T, x, y)$ be the transition density for the solution to this equation (i.e., if we solve the equation with the initial condition $X(t)=x$, then the random variable $X(T)$ has density $p(t, T, x, y)$ in the $y$ variable). We are assuming that $p(t, T, x, y)=0$ for $0 \leq t<T$ and $y \leq 0$.
Show that $p(t, T ; x, y)$ satisfies the Kolmogorov backward equation
$$
-p_t(t, T, x, y)=\beta(t, x) p_x(t, T, x, y)+\frac{1}{2} \gamma^2(t, x) p_{x x}(t, T, x, y) .
$$
(Hint: We know from the Feynman-Kac Theorem, Theorem 6.4.1, that, for any function $h(y)$, the function
$$
g(t, x)=\mathbb{E}^{t, x} h(X(T))=\int_0^{\infty} h(y) p(t, T, x, y) d y
$$
satisfies the partial differential equation
$$
g_t(t, x)+\beta(t, x) g_x(t, x)+\frac{1}{2} \gamma^2(t, x) g_{x x}(t, x)=0 .
$$
Use (6.9.44) to compute $g_t, g_x$, and $g_{x x}$, and then argue that the only way (6.9.45) can hold regardless of the choice of the function $h(y)$ is for $p(t, T, x, y)$ to satisfy the Kolmogorov backward equation.)