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As explained in this Wikipedia page, the infinitesimal generator of the standard Brownian motion is $\frac{1}{2}\Delta$ and for the Brownian motion it has an extra $\partial_t f$ term. Can anybody please explain how one can reach to $\frac{1}{2}\Delta$ from $Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(X_t))-f(x)}{t}$ in the standard Brownian motion, for a person who is not that much familiar with the stochastic math?

In the first answer here, the Brownian motion was studied. I would appreciate if i can have the proof (not deep stochastic math) for the standard Brownian motion.

And two points which might be covered in the answer of the question above:

  • Unlike the Brownian case, in the standard case we have $\mathbb{E}^x(B_t-x)=-x$.

  • In the equation (1) in that answer, how "one can show $\frac{d}{dt} P_t f(x) = A P_tf(x)$ " ?

Thanks in advance.

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    $\begingroup$ $\mathbb{E}^x[B_t - x] = x$ implies that $\mathbb{E}^x[B_t] = 2x$. I think you've made a sign error. Regardless, it makes little sense to talk about the generator of a standard Brownian motion where "standard" includes the requirement that the process starts at $0$. The generator captures behaviour of the transition semigroup which requires you to be able to start your process at different points. (i.e. by definition $\mathbb{E}^x$ is the expectation for the law of Markov process in question started at point $x$). $\endgroup$
    – Rhys Steele
    Jun 2, 2020 at 22:21
  • $\begingroup$ Thanks. And yes you are right, it should be $-x$. $\endgroup$
    – Denis
    Jun 2, 2020 at 22:27
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    $\begingroup$ If you've understood the need to consider the family of processes $(B_t^x)_{x \in \mathbb{R}^d}$ where $B_t^x$ is a Brownian motion started at $x$ when finding the generator then the accepted answer at the other question shows you how to show that the generator is given by the Laplacian in the one dimensional case. You get the multivariate version in pretty much the same way, using the multivariate taylor expansion. $\endgroup$
    – Rhys Steele
    Jun 2, 2020 at 22:30
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    $\begingroup$ Also btw $\mathbb{E}^x$ refers to the expectation with respect to the law of the process started from $x$ and hence if $B_t$ is a Brownian motion then $\mathbb{E}^x[B_t - x] = 0 \neq - x$; you can't postulate this to have a different value. $\endgroup$
    – Rhys Steele
    Jun 2, 2020 at 22:39
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    $\begingroup$ I think the definition of standard used there doesnt include the assumption that $B_0 = 0$ and instead refers to the other properties (such as the form of the covariance) that determine the properties of the transition functions associated to $B$. $\endgroup$
    – Rhys Steele
    Jun 3, 2020 at 9:11

1 Answer 1

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I'm in the process of learning this myself, and I did the following.

So in the 1D case, we want to show that $Af(x) := \lim_{t\rightarrow 0} \frac{\mathbb{E}^x[f(B_t)] - f(x)}{t} = \frac{1}{2}f''(x)$.

A Taylor expansion about $f(x)$ gives \begin{align} \mathbb{E}^x[f(B_t)] &\approx \mathbb{E}^x[ f(x) + (B_t-x) f'(x) + \frac{1}{2}(B_t-x)^2 f''(x)]\\ &= f(x) + \mathbb{E}^x[B_t-x] f'(x) + \frac{1}{2} \mathbb{E}^x[(B_t-x)^2] f''(x)\\ &= f(x) + \frac{t}{2} f''(x) \end{align}

where the last equality comes from the properties of a Brownian motion with $B_0=x$, $\mathbb{E}^x[B_t-x] = 0$ and $\mathbb{E}^x[(B_t-x)^2] = t$.

Plugging back into $Af(x)$, the $f(x)$'s cancel and we get $$ Af(x) = \lim_{t\rightarrow 0} \frac{\frac{t}{2}f''(x)}{t} = \frac{1}{2}f''(x)$$
(The t's trivially go away: $\lim_{t\rightarrow 0} \frac{t}{t} = \lim_{t\rightarrow 0} 1 = 1$)

You can extend this reasoning to the multidimensional case $x= (x_1, \dots, x_n)$, where the Taylor expansion becomes $$\mathbb{E}^x[f(B_t)] = f(x) + \frac{t}{2} \sum_{i=1}^n \frac{\partial}{\partial x_i} f(x).$$

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