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In the following formula

$$Af(x) := \lim_{t \downarrow 0} \frac{\mathbb{E}^x(f(B_t))-f(x)}{t} $$

  1. If $B_t$ is the Brownian motion, what are $f(B_t)$ and $f(x)$? here some explanation was given. Can anybody please provide with more intuitive explanation?
  2. In the way to prove that $A=\frac{1}{2}\Delta$ here, and after $P_t f(x) = \mathbb{E}^x[f(X_t)]$, he defines $u(t,x) := P_t f(x)$. Which interpretations/properties of $P_t$ led to the latter definition? In other words, why the expected value of f(x) is a deterministic function depending on $x$ and $t$?
Kevin
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1 Answers1

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(1) If $f$ is a function then $f(B_t)$ and $f(x)$ are simply the evaluation of that function at the points $B_t$ and $x$ respectively. Here, since $x$ is simply the starting point of your Brownian motion $f(x)$ needs no further interpretation. For fixed $t$, $f(B_t)$ is itself a random variable since for each $\omega \in \Omega$ (the underlying probability space), you should set $(f(B_t))(\omega) = f(B_t(\omega))$ where the right hand side is just evaluating the function $f$ at the number $B_t(\omega)$.

(2) Note that the claim is not that $\mathbb{E}^x[f(x)]$ is a function that depends on $x$ and $t$ but rather that $$u(t,x) := \mathbb{E}^x[f(X_t)]$$ is a function depending on $x$ and $t$. If you are specifically interested in Brownian motion, you should replace the general Markov process $X_t$ with $B_t$. This is a function of $t$ since the expectation depends on the time at which we look at the process (e.g. our Brownian motion). It is also a function of a space point $x$ because we have to decide which point our Brownian motion should start from; in the definition of $u$ this is denoted by writing $\mathbb{E}^x$ which is the expectation with respect to the law of the Brownian motion started from the point $x$. Finally, this is a deterministic function since we took an expectation so there is no $\omega$ dependence (randomness) left on the right hand side of the definition.

Rhys Steele
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  • Thank you!
    1. I have difficulty in undestanding what function $f$ is, and what is its importance. If $B_t$ is the position of the brownian motion at time t, I cannot imagine what $f(B_t)$ is?
    2. How can we evaluate $\mathbb{E}^x[f(B_t)]$ using the integral formula (the general case $E^{x}[f(X_t)]=\int_{\mathbb{R}^n}f(y)p_t(x,y)dy$). Please let me know if should open a new question for the last question.
     – Kevin Nov 21 '23 at 16:25
  • And another reason for my difficulty with $f(B_t)$. If it is random, how the taylor formula (and not ito lemma) can be applied to that as $\mathbb{E}^x(f(B_t)) \approx \mathbb{E}^x \left[f(x)+f'(x)(B_t-x)+\frac{1}{2} f''(x)(B_t-x)^2 \right] $ – Kevin Nov 21 '23 at 16:46
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    (1) I'm not sure what you mean by "what the function $f$ is". $f$ is an arbitrary element of some space of functions (usually something like the space of continuous functions that vanish at infinity). The reason to study objects like $f(X_t)$ when $X$ is a Markov process is that the knowledge of all the $f(X_t)$ determine e.g. the law of the process (this would be immediate if you were allowed to take $f = 1_A$ for $A$ a measurable set. Since it is convenient to work with smooth enough functions, you can proceed for this fact by approximating the indicator functions by smooth enough functions) – Rhys Steele Nov 21 '23 at 16:49
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    (2) for completely arbitrary $f$, I wouldn't expect a nice explicit formula for the integral. Why would you want to compute it that way?

    (3) Because there you are applying Taylor's formula to the function $x \mapsto f(x)$. You write $f(y) = f(x) + f'(x)(y-x) + \frac12 f''(x) (y-x)^2 + o([y-x]^3)$ and then choose $y = B_t(\omega)$.

     – Rhys Steele Nov 21 '23 at 16:51