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\title{Reformulation without f}
\author{Shakil Rafi}
\begin{document}
\maketitle  

\begin{lemma}
	Let $T \in (0,\infty)$, let $\lp \Omega, \mathcal{F}, \mathbb{P}\rp $ be a probability space, let $\alpha_d \in \R^d \rightarrow \R$, $d \in \N$, be infinitely often differentiable functions, let $u_d \in C^{1,2} \lp \lb 0,T \rb \times, \R^d, \R \rp$, $d \in \N$, satisfy for all $d \in \N$, $t \in \lb 0,T\rb$, $x \in \R^d$ that:
	\begin{align}
		\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \lp \Delta_x u_d \rp \lp t,x \rp + \alpha_x \lp x \rp u_d \lp t,x \rp = 0 
	\end{align}
	Let $\mathcal{W}^d: \lb 0,T \rb \times \Omega \rightarrow \R^d$, $d \in \N$ be standard Brownian motions, and let $\mathcal{X}^{d,t,x}: \lb t,T\rb \times \Omega \rightarrow \R^d$, $d \in \N$, $t\in \lb 0,T \rb$, $s \in \lb t,T\rb$, $x\in \R^d$ we have $\mathbb{P}$-a.s. that:
	\begin{align}
		\mathcal{X}^{d,t,x}_s = x+ \int^t_s \sqrt{2} d\mathcal{W}^d_r
	\end{align}
	Then for all $d \in \N$, $t\in \lb 0,T\rb$, $x\in \R^d$ it holds that:
	\begin{align}
		u_d\lp t,x \rp + \mathbb{E} \lb \exp \lp \int^T_t \alpha_x \lp \mathcal{X}^{d,t,x}_r \rp dr \rp u_d \lp T, \mathcal{X}^{d,t,x}_T \rp \rb  
	\end{align}
\end{lemma}

\begin{proof}
	Let $T \in [0,\infty)$, and let $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$ be a probability space. For all $d \in \N$, let $V \in C^{1,1} \lp  \R^d \times [0,T],\R \rp $ be  $V(x,t) = \alpha_d(x)$, let $\sigma_d : \R^d \rightarrow \R^{d\times d}$ be given by $\sigma_d(x) = \diag_d \lp \sqrt{2} \rp$, let $\mu_d: \R^d \rightarrow \R^d$ be given by $\mu_d(x) = \mymathbb{0}_d$, and finally let $f(t,x) = 0$. By Feynman-Kac and substituting the above, the following expression:
	\begin{align}\label{0.0.4}
		\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \frac{1}{2}\Trace\lp \sigma(t,x) \lb \sigma(t,x) \rb^* \lp \Hess_x(u_d \rp \lp t,x \rp \rp + \la \mu(t,x), \lp \nabla_x u_d \rp \lp t,x\rp \ra +V(t,x)u_d(t,x) \nonumber\\ +f(t,x) = 0 
	\end{align} 
	is rendered: 
	\begin{align}\label{0.0.5}
		\lp \frac{\partial}{\partial t} u_d \rp \lp t,x \rp + \lp \Delta_x u_d \rp \lp t,x\rp + \alpha_d(x)u_d(x) = 0
	\end{align}
	Note then that Feyman-Kac sates that the solution to (\ref{0.0.4}) can be written as:
	\begin{align}
		u(t,x) = \E \lb \int^T_t e^{\int^r_t V (\mathcal{X}_t,\tau) d\tau}f(\mathcal{X}_r, r )dr + e^{-\int^T_t V(\mathcal{X}_\tau,\tau)d\tau } u(\mathcal{X}_T,T) \rb 
	\end{align} 
	Where $\mathcal{X}$ is an $\lp \Omega, \mathcal{F}, \mathbb{P} \rp$-adapted stochastic process given by:
	\begin{align}
		\mathcal{X}_t = x + \int^t_s \mu_d \lp \mathcal{X} \rp dr+\int^t_s\sqrt{2}d\mathcal{W}_r^d  
	\end{align}
	Note then that the substitutions then yield that the solution to (\ref{0.0.5}) is given by:
	\begin{align}
		u(t,x) = \E \lb \exp \lp \int^T_t \alpha_d \lp \mathcal{X} \rp dr \rp u_d \lp T,\mathcal{X}^{d,t,x}_T \rp \rb 
	\end{align}
\end{proof}






























\end{document}