The killed transition operator is
$$
P_tf(x)=\mathbb E_x[f(X_t)\mathbf1_{\{t<\mathcal T\}}]
=\int_0^1p(t,x,y)f(y)dy.
$$
The <infinitesimal generator> of the diffusion is
$$
L=\frac12\frac{d^2}{dx^2}+\frac12v(x)\frac d{dx}.
$$
The <Markov property> and the <Chapman-Kolmogorov equation> give, for each fixed $y$,
$$
p(t+h,x,y)=\int_0^1p(h,x,z)p(t,z,y)dz
=P_h[p(t,\mathord\cdot,y)](x).
$$
Dividing by $h$, letting $h\downarrow0$, and using the generator definition together with the assumed $C^{1,2}$ regularity gives the pointwise <Kolmogorov backward equation>
$$
\partial_tp(t,x,y)
=\frac12\partial_x^2p(t,x,y)
+\frac12v(x)\partial_xp(t,x,y).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 5 / e / Solution

{c}
{wiki}

For a time-homogeneous Markov process with transition kernel $P_t(x,dy)$,
$$
P_{s+t}(x,A)=\int P_s(x,dy)P_t(y,A).
$$
When transition densities exist, this becomes
$$
p(s+t,x,z)=\int p(s,x,y)p(t,y,z)dy.
$$


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