I want to ask about a reasoning part of calculation of the expected first exit time of Brownian motion from a ball, in Example 7.4.2 of Stochastic Differential Equations: An Introduction with Applications (Bernt Øksendal).
Before describing the proof, I wrote down Dynkin's formula (Theorem 7.4.1) as a lemma:
Let $f \in C_0^2(\mathbb{R}^n)$. Suppose $\tau$ is a stopping time, $E^x[\tau] < \infty$ where $E^x[\cdot]$ express a conditional expectation with initial value $x$. Then
$$
E^x[f(X_\tau)] = f(x) + E^x \left[ \int_0^\tau Af(X_s) ds \right],
$$
where $A$ refers to an infinitesimal generator of time-homogeneous Ito diffusion process $\{X_t\}_{t \ge 0}$.
The, the proof is like the following:
Consider $n$-dimensional Brownian motion $B = (B_1, …, B_n)$ starting at $a = (a_1, …, a_n) \in \mathbb{R}^n (n \ge 1)$ and assume $|a| < R$. Then, we examine the expected value of the first exit time $\tau_K$ of B from the ball $K = \{x \in \mathbb{R}^n; |x| < R \}$.
To calculate this, choose an integer $k$ and apply Dynkin's formula with $X = B$, $\tau = \sigma_K = \min(k, \tau_K)$, and $f \in C_0^2$ s.t. $f(x) = |x|^2 \ \mathrm{for} \ |x| \le R$ (which satisfies the assumption of Dynkin's formula):
\begin{align}
E^a[f(B_{\sigma_k})]
&= f(a) + E^a \left[ \int_0^{\sigma_k} \frac{1}{2} \Delta f(B_s) ds \right] \\
&= f(a) + E^a \left[ \int_0^{\sigma_k} n \cdot ds \right] \\
&= |a|^2 + n \cdot E^a [\sigma_k].
\tag{1}\label{eq1}\end{align}
Hence
$$
E^a[\sigma_k] \le \frac{1}{n} (R^2 – |a|^2) \ \mathrm{for \ all} \ k.
\tag{2}\label{eq2}$$
So letting $k \to \infty$, we conclude that $\tau_K = \lim \sigma_k < \infty \ \mathrm{a.s.}$ and
$$
E^a[\tau_K] = \frac{1}{n} (R^2 – |a|^2).
$$
I understand up to the formula $E^a[\sigma_k] \le \frac{1}{n} (R^2 – |a|^2)$ for all $k$, but I am not clear the rest of the proof from there. Here are the points I'm wondering:
- I know $\tau_K = \lim \sigma_k$ from the definition of $\tau_K$, but why is it almost surely finite? Does $E^a[\sigma_k] < \infty$ derive $\sigma_k < \infty$?
If this question is resolved, I understand that $\lim \limits_{k \to \infty} E^a[\sigma_k] = E^a[\tau_K]$ from the bounded convergence theorem. However, I have an another question:
- From which equation and how was the last equation derived? I have two possible ways.
- The first way is taking the limit $k \to \infty$ in $\eqref{eq1}$. In this way, I'm not sure that $\lim \limits_{k \to \infty} E^a[f(B_{\sigma_k})] = R^2$. I can derive $\lim \limits_{k \to \infty} E^a[f(B_{\sigma_k})] = E^a \left[ \lim \limits_{k \to \infty} f(B_{\sigma_k}) \right]$ the way $f(B_{\sigma_k}) \le R^2$ and I apply the bounded convergence theorem to exchange the limit and expectation, but it's not clear to me that $\lim \limits_{k \to \infty} f(B_{\sigma_k}) = f(B_{\tau_K}) \ (= f(R) = R^2)$.
- The second is the derivation from $\eqref{eq2}$, but I'm not convinced that $E^a[\tau_K] = \frac{1}{n} (R^2 – |a|^2)$.
There are some questions about the same topic, but they doesn't solve my problem because their approaches are different from the above:
Best Answer
Yes. Generally from Markov $$P[X>t]\leq \frac{E[X]}{t}<\frac{c}{t}$$ and so by taking $t\to +\infty$, we get $P(X=\infty)=0$.
Indeed, you need to use the first way as you just did because only at that level you have equality.
You can use dominated convergence theorem because $f,B$ are continuous and we have the domination
$$E^a[f(B_{\sigma_k})]\leq R^{2}$$
because $\sigma_{k}\leq \tau_{K}$ and so $B_{\sigma_k}\in K$.