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Suppose $X_u$ satisfies the SDE $$ dX_u = a(u, X_u) du + b(u,X_u) dW_u\;. $$ Fix $T > 0$ and let $X_t = x$ for $0 \leq t \leq T$ be given. For any measurable function $h$, define $$ g(t,x) = E^{t,x}(h(X_T))\;. $$ Then $g$ satisfies the Feynman-Kac equation $$ g_t(t,x) + a(t, x) g_x(t,x) + \frac{1}{2}b^2(t,x)g_{xx}(t,x) = 0 $$ with terminal condition $$ g(T,x) = h(x) \;. $$ Fair enough, but now consider the Kolmogorov backward equation. Let $p(t,x; T, y)$ be the transition density of $X_u$; i.e., if $X_t = x$, then $X_T$ has density $p(t,x; T,y)$ in the $y$ variable. Then for each $T$ and $y$, $p$ satisfies the Kolmogorov backward equation $$ p_t(t,x; T,y) + a(t,x)p_x(t,x;T,y) + \frac{1}{2}b^2(t,x)p_{xx}(t,x;, T,y) = 0 $$ with terminal condition $$ p(T,x; T,y) = u_T(x) $$ for some function $u_T(x)$

As far as I can tell, these are the same equations; the only difference between the solutions being the terminal conditions. Is this a correct understanding? Is there some greater insight as to why these aren't both called the same equation? I see that $g$ and $p$ are related by $$ g(t,x) = \int_{-\infty}^\infty h(y) p(x,t;T,y)\,dy $$ but I'm not seeing anything insightful from this relationship, and am afraid I'm missing something. I've read the answer here but am still a bit unsure of the difference.

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  • if $X_t = x$, then $X_T$ has density $p(t,x; T,y)$ in the $y$ variable -- by so $p$ is conditional density, not transition density – athos Jan 04 '22 at 18:32

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I hate to resurrect a four year old question with an only-partially-helpful response, but this paper shows that Feynman-Kac is the same as a Kolmogorov backward equation. I haven't read it in detail, so I can't summarise and do it any justice, but hopefully it helps you. In case the link doesn't work, paper is: VERSIONS OF THE FEYNMAN-KAC FORMULA by A N Borodin.

Great question!