Understanding specific step in Wald Lemma's proof

65 Views Asked by At

A friend and I are having a hard time getting our head around a specific passage in proving Wald's Lemma for Brownian motion. The proof comes specifically from "Brownian Motion" by Mörters and Peres.

The theorem and proofs are as follows.

Let $\{B(t):t\geq0\}$ be a standard Brownian motion, and let $T$ be a stopping time such that either

(i) $E(T) < \infty$

(ii) $\{B(t \wedge T):t\geq 0\}$ is dominated by and integrable random variable.

Then we have that $E[B(T)] = 0$.

Proof:

We first show that if a stopping time satisfies $(i)$ then it also satisfies $(ii)$.

Suppose $(i)$ is true. Define

$$M_k = \max_{0\leq t \leq 1}\mid B(t+k)- B(k) \mid \qquad \text{and} \qquad M= \sum_{k=1}^{\lceil T \rceil}M_k$$

then we have that

$$E(M) = E\left[ \sum_{k=1}^{\lceil T \rceil}M_k\right] = \sum_{k=0}^\infty E[I_{\{T>k-1\}}M_k] = \\=E(M_0)\sum_{k=0}^\infty P(T>k-1) = E(M_0)E(T+1) < \infty$$

It is clear to us why this product is almost surely finite, and the rest of the proof is also clear. However, what we really don't understand is how $\sum_{k=0}^\infty P(T>k-1) = E(T+1)$. This is the only passage that is obscure to us and we would appreciate any help.