show that $a \mathbb{P}(\tau \leq t) = \mathbb{E}[M_t \mathbb{1}_{\{\tau \leq t\}}]$

Question

let $(M_t)_{t \geq 0}$ be a positive, continuous martingale starting from $0$

let $a > 0$ and $\tau$ be the stopping time defined as follows :

$$\tau =\{ t\geq0 \,| \, M_t \geq a\}$$

by use of Doob's optional stopping theorem and the dominated convergence theorem we have that :

$M_{\tau} = a, \, \mathbb{E}[M_{t \wedge\tau}] = 0,\, a \mathbb{P}(\tau \leq t) + \mathbb{E}[M_{t} |\{\tau > t\}]\mathbb{P}(\tau > t) = 0 $

so basically we have to prove that :

$- \mathbb{E}[M_{t} |\{\tau > t\}](\mathbb{P}(\tau > t)) = \mathbb{E}[M_t \mathbb{1}_{\{\tau \leq t\}}] $

on the one hand we have :

$\mathbb{E}[M_t \mathbb{1}_{\{\tau \leq t\}}] = \mathbb{E}[M_t | \{\tau \leq t\}]\mathbb{P}(\tau \leq t)$

so at the end it all comes down to proving that :

$$\mathbb{E}[M_t | \{\tau \leq t\}]\mathbb{P}(\tau \leq t) +\mathbb{E}[M_{t} |\{\tau > t\}](\mathbb{P}(\tau > t)) = 0 $$

but don't we have by the law of total probabilities that :

$$\mathbb{E}[M_t | \{\tau \leq t\}]\mathbb{P}(\tau \leq t) +\mathbb{E}[M_{t} |\{\tau > t\}](\mathbb{P}(\tau > t)) = \mathbb{E}[M_t]$$

I'm really confused to what I'm doing wrong.

Answer 1

$EM_t$ is independent of $t$ by definition of a martingale. Since $M_0=0$ it follows that $EM_t=0$ for all $t$ Hence there is no contradiction!

Answer 2

Another way of proving it would be to use the Doob's optional theorem as you did: \begin{equation} E[M_{t\wedge \tau}] = 0 \end{equation} But now write

\begin{align} E[M_{t\wedge \tau}] &= E[M_{\tau}1_{\tau \leq t}] + E[M_{t}1_{\tau \geq t}] \\ &= aP(\tau \leq t) + E[M_t(1-1_{\tau \leq t})]\\ &= aP(\tau \leq t) + \underbrace{E[M_t]}_0 - E[M_t1_{\tau \leq t}] = 0 \end{align} Hence $a \mathbb{P}(\tau \leq t) = \mathbb{E}[M_t \mathbb{1}_{\{\tau \leq t\}}]$ a.s.