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.
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 $$\mathbb{E}^x[f(B_t)] = f(x) + \frac{t}{2} \sum_{i=1}^n \frac{\partial}{\partial x_i} f(x).$$