ファインマン・カッツの公式

最近確率的計算法を学んでいます。一つな計算式はPDEの問いをブラウン運動に接続する。 the Feynman-Kac formula is problematic in that it seems to be stated in different forms depending on the source (in some cases u satisfies a PDE like that in Black Scholes, in others the heat equation), the notation for the conditional expectation is unclear, and the proofs are inconsistent, it seems, in their use of different Ito lemmas.
英語版のウィキベディアの解はu(x,t) = E(Q)(INTEGRAL(t,T)(exp(-INTEGRAL(t,r)(V(Xtau,tau)dtau))*f(Xr,r)dr) + exp(-INTEGRAL(t,T)(V(Xtau,tau)dtau))psi(XT)|Xt=x)
This basically indicates that the solution function u is a conditional expectation of psi's value at the terminal time T and the V and f values at times in the interim.
しかしシンガポールの南洋理工大学のLeon教授の説明は理解できました。
In this case, the solution to the PDE, g, was simply E(Q,x,t)(h(X(T))), or the expectation over all t and x of the value of some measurable function h at time T. 証明の由来も分かりました。　Basically, just start with a stochastic X(t) such that
(here, x = X(t))
dX(t) = b(t,x)dt + c(t,x)dW(t). Consider that g is a function of X and t, and apply Ito's lemma to find dg (drop arguments for clarity).
dg = (dg/dt + b*dg/dx + (1/2)*c^2*d^2g/dx^2)dt + c*(dg/dx)dW
However, since g is a Martingale (proven based on definition of g as conditional expectation of h), the dt term must be 0, meaning that (dg/dt + b*dg/dx + (1/2)c^2*d^2g/dx^2) = 0. In addition, substituting T for t into the g expression,
g(T,X(T)) = E(Q,t,x)(h(X(T))) = h(X(T)) = h(x).
The final two statements are the content of Feynman-Kac．
しかし利子の拡散方程式を表すハル・ワイト公式の偏微分方程式をファインマン・カッツで由来すればこの方程式の解の源を知りません。