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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?

Emily
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  • PS: Sorry if this is a silly question! I feel like this should essentially be an unwinding of the definition of conditional expectation, but I'm a bit confused about it, and would really appreciate any help with this. – Emily May 25 '23 at 02:40
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    I think maybe it's best not to denote a stochastic process $W:\Omega\times[0,\infty)\to\Bbb R$ and a probability measure $W$ on $\mathcal C([0,t], \Bbb R)$ with the same symbol $W$. I also think that $\Bbb E^x[f(W_t)]$ does not make sense in case you're using the definition of $W$ that requires $W_0 = 0$. I think it would be more correct to write $\Bbb E[f(W_t + x)]$ which would then make everything trivial. – SBF May 25 '23 at 06:40
  • @SBF Thank you, these are both very good points! I think it's best not to change the notation now since it also appears in the answer, but I'll definitely keep this in mind when writing things in the future. Again, thank you very much! – Emily May 25 '23 at 06:47
  • Where did you take $(1)$ from? – SBF May 25 '23 at 06:48
  • I took it from Lemma 8.1/Table 8.1 from Schilling's Brownian Motion: A Guide to Random Processes and Stochastic Calculus. Here's the single page in question: Link – Emily May 25 '23 at 06:50
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    Given that they use generators and semi groups for Brownian motion there means that they treat it as a Markov process, which does not have a fixed initial point. The wiener measure you use in later equalities apparently requires a fixed point. That adds to the confusion of using the same symbol for the process and for the measure. – SBF May 25 '23 at 09:01
  • @SBF Oooh I see! Thank you so much for explaining this to me, I was indeed quite confused about it – Emily May 26 '23 at 14:57

1 Answers1

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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.)