By Morters P., Peres Y.

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5. 3, P{Bk ∈ A} = P{B ∈ A} does not depend on k, and hence we get ∞ k=0 P{Bk ∈ A} P E ∩ {Tn = k2−n } = P{B ∈ A} ∞ k=0 P E ∩ {Tn = k2−n } = P{B ∈ A}P(E), which shows that B∗ is a Brownian motion and independent of E, hence of F + (Tn ), as claimed. It remains to generalize this to general stopping times T . As Tn ↓ T we have that {B(s + Tn ) − B(Tn ) : s ≥ 0} is a Brownian motion independent of F + (Tn ) ⊃ F + (T ). Hence the increments B(s + t + T ) − B(t + T ) = lim B(s + t + Tn ) − B(t + Tn ) n→∞ of the process {B(r + T ) − B(T ) : r ≥ 0} are independent and normally distributed with mean zero and variance s.

47 Let {B(t) : t ≥ 0} be a standard linear Brownian motion and T a stopping time with E[T 1/2 ] < ∞. Then E[B(T )] = 0. Proof. Let {M (t) : t ≥ 0} be the maximum process of {B(t) : t ≥ 0} and T a stopping time with E[T 1/2 ] < ∞. Let τ = log4 T , so that B(t ∧ T ) ≤ M (4τ ). e. that EM (4τ ) < ∞. Define a discrete time stochastic process {Xk : k ∈ N} by Xk = M (4k ) − 2k+1 , and observe that τ is a stopping time with respect to the filtration (F + (4k ) : k ∈ N). Moreover, the process is a supermartingale.

Since W ∗ has the same distribution as −B, ˆ ˆ has continuous distribution, and that P2 = P{y + B(t) ≤ −a}. The Brownian motion B so, by adding P1 and P2 , we get ˆ (t) − B(t) ˆ > a} = P{|y + B(t)| ˆ P{y ∨ M > a}. This proves the main step and, consequently, the theorem. While, as seen above, {M (t) − B(t) : t ≥ 0} is a Markov process, it is important to note that the maximum process {M (t) : t ≥ 0} itself is not a Markov process. However the times when new maxima are achieved form a Markov process, as the following theorem shows.