Feynman–Kac formula: conditional expectation vs. Wiener integral

Question

The Feynman–Kac formula for the solution $u(t,x)$ of the one-dimensional heat equation \begin{align*} \partial_t u &= \frac{1}{2}\Delta_x u,\\ u(0,x) &= f(x) \end{align*} is given by \begin{equation} u(t,x)=\mathbb{E}^{x}[f(W_t)],\tag{1} \end{equation} with $W$ a Brownian motion and where $\mathbb{E}^{x}[f(W_t)]$ is the expectation of $f(W_t)$ conditioned on $W_0=x$. I've also seen this expressed as a Wiener integral \begin{equation} u(t,x)=\int_{\mathcal{C}([0,t],\mathbb{R})}f(\gamma(t)+x)\,\mathrm{d}W(\gamma).\tag{2} \end{equation} How can we show that $$\mathbb{E}^{x}[f(W_t)]=\int_{\mathcal{C}([0,t],\mathbb{R})}f(\gamma(t)+x)\,\mathrm{d}W(\gamma)$$ holds?

Answer 1

Yes, it's about understanding what it means for the expectation $\mathbb{E}^{x}(f(W_t))$ to be conditioned on $W_0 = x$. It means that the expectation is taken over paths that start at $x$, weighted according to the Wiener measure.

So the sample space is $\{ \gamma \in \mathcal{C}([0, t], \mathbb{R}) : \gamma(0) = x \}$, we're taking the expectation of $f(\gamma(t))$, and we're using the measure $dW(\gamma)$; we can set up the integral using the definition of expected value.

If we write $\gamma(t) = \gamma'(t) + x$, we can write the integral over $\{ \gamma' \in C([0, t], \mathbb{R}) : \gamma'(0) = 0 \}$, which will give what you've written. (There's an implicit condition that the integral is taken over $\gamma$ so $\gamma(0) = 0$ in your formula.)