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I want to do the following exercise of Durrett - Brownian motion and Martingales in Analysis: enter image description here

Im pretty sure that we have to use the optional stopping theorem for the martingale $M_t= |B_t|^2-td$. If we could use the OST, we would have $$E^x[M_T] =E^x[M_0]$$ which would give us the claim. But to use the OST in that way we need that either $M_t$ is uniformly integrable or that $T$ is bounded. Well, $T$ is not bounded and I don't think that $M_t$ is uniformly integrable either. So what to do here?

Falrach
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Mimimi
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1 Answers1

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We claim that $$\Bbb E_x [M_T] = \Bbb E_x[M_0]$$ For every $t\geq 0$ the stopping time $t\wedge T$ is almost surely bounded by $t$ and therefore optimal stopping leads to the equation $$E_x [M_{t\wedge T}] = \Bbb E_x[M_0]$$ for every $t\geq 0$. Now note that since $t\wedge T \leq T$ almost surely we have $\vert B_{t\wedge T} \vert ^2 \leq r^2$ almost surely, and additionally $t \to t\wedge T$ is non-decreasing. Therefore by dominated convergence theorem and monotone convergence theorem we have $$ \lim_{t\to\infty} \Bbb E_x[ \vert B_{t\wedge T}\vert^2] = \Bbb E_x[ \lim_{t\to\infty} \vert B_{t\wedge T}\vert^2] = \Bbb E_x[ \vert B_{ T}\vert^2]$$ and $$\lim_{t\to\infty} \Bbb E_x [(t\wedge T )d] = \Bbb E_x[Td]$$ Consequently $$\lim_{t\to \infty} E_x[M_0] =\lim_{t\to \infty} E_x [M_{t\wedge T}] = \Bbb E_x[ \vert B_{ T}\vert^2] - \Bbb E_x[Td] = \Bbb E_x [M_T]$$

Falrach
  • 4,102