By Morters P., Peres Y.

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**Example text**

Moreover, the process is a supermartingale. Indeed, E Xk Fk−1 ≤ M (4k−1 ) + E max 0≤t≤4k −4k−1 B(t) − 2k+1 . 4 The martingale property of Brownian motion 61 using that E[ max B(t)] = E|B(1)| ≤ 1 by the reflection principle and a simple estimate. 0≤t≤1 Now let t = 4 and use the supermartingale property for τ ∧ to get E M (4τ ∧ t) = E Xτ ∧ + E 2τ ∧ +1 ≤ E[X0 ] + 2 E 2τ . Note that X0 = M (1) − 2, which has finite expectation and, by our assumption on the moments of T , we have E[2τ ] < ∞. Thus, by monotone convergence, E M (4τ ) = lim M (4τ ∧ t) < ∞ , t↑∞ which completes the proof of the theorem.

8. 3 Markov processes derived from Brownian motion In this section, we define the concept of a Markov process. Our motivation is that various processes derived from Brownian motion are Markov processes. Among the examples are the reflection of Brownian motion in zero, and the process {Ta : a ≥ 0} of times Ta when a Brownian motion reaches level a for the first time. We assume that the reader is familiar with the notion of conditional expectation given a σ-algebra, see [Wi91] for a reference. 28.

These results identify the first and second moments of the value of Brownian motion at well-behaved stopping times. 41 (Wald’s lemma for Brownian motion) Let {B(t) : t ≥ 0} be a standard linear Brownian motion, and T be a stopping time such that either (i) E[T ] < ∞, or (ii) B(t ∧ T ) : t ≥ 0 is L1 -bounded . Then we have E[B(T )] = 0. 42 The proof of Wald’s lemma is based on an optional stopping argument. 6. 47 for an optimal criterion. Proof. We first show that a stopping time satisfying condition (i), also satisfies condition (ii).