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In order to show that the infinitesimal generator of the Brownian motion is $\frac{1}{2}\Delta$, in this answer, first he writes equation $$ \frac{d}{dt} P_t f(x) = A P_tf(x), \tag{1} $$ then he derives the following approximation : $$ \mathbb{E}^x(f(B_t)) \approx f(x)+ \frac{t}{2} f''(x) $$ Then it is argued that " From (1) we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation"

As discussed here, we cannot simply replace the approximation into the heat equation. If so,

  1. Why did the author of that post made this approximation? how did he use this approximation for the proof? if he didn't use it,
  2. Can anybody explain more his argument that : " From (1) we see that $u(t,x) := \mathbb{E}^x(f(B_t))$ is the (unique) solution of the heat equation..."?
Arctic Char
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Denis
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1 Answers1

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Your confusion might arise because you think somehow the approximation serves to construct a solution to the heat equation. What is happening is that you start with a solution to some partial differential equation (PDE), and the approximation serves to identify this PDE as the heat equation. No proofs were given in either of the posts you linked. They are just formal arguments to help develop intuition.

Start with your second question. Equation (1) is $$\frac{d}{dt}P_t f(x) = A P_t f(x).$$ By definition, $P_t f(x) = \mathbb{E}^x [f(B_t)].$ Setting $u(t,x) = P_t f(x)$ in equation (1), we have $$\frac{d}{dt} u(t,x) = A u(t, x). \tag{$\spadesuit$}$$ Here, $A$ is a differential operator, so $u(t,x)$ solves some differential equation with some initial conditions. Which differential equation is it?

To guess which differential equation it is, the approximation $u(t,x) \approx f(x) + t f''(x)/2$ is used. Putting this directly into the left-hand side of ($\spadesuit$), you find $$\frac{d}{dt} u(t,x) = \frac{d}{dt}\left(f(x) + t\frac{f''(x)}{2}\right)=\frac{f''(x)}{2}=Au(t,x).$$ Based on this relation, can you guess what $A$ is?

snar
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  • Very helpful answer, thanks. Sorry if it is easy to be understood. But i can only "feel" that the operator A acting on u gives the $\frac{1}{2}\Delta$. If we had $u(t,x)=f(x)$, then i could understand that. Am i missing something? – Denis Aug 16 '20 at 17:23
  • @Denis Sorry, I don't understand your question. – snar Aug 16 '20 at 17:38
  • How can we conclude from $\frac{f''(x)}{2}=Au(t,x)$ that operator $A$ is the Laplacian operator? – Denis Aug 16 '20 at 17:51
  • I assume my word 'conclude' is not a correct word, as you wrote about guessing. But then, why the person in the linked answer, argued "we SEE from 1 that, it is the UNIQUE solution..". I mean, if it was only a guess, why he wrote like that? – Denis Aug 17 '20 at 12:54
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    Because they recognize that the necessary conditions for a heat equation to have a unique solution are satisfied. If I give you the differential equation $y'(x) = y(x)$ with $y(0)=1$ and you tell your friend "We see that the unique solution is $y(x) = e^x$", did you "see" or "guess"? I don't want to speculate further about what another member wrote, and I'm not sure this discussion is productive anymore. – snar Aug 17 '20 at 13:53