Expected value of exponential martingale

Question

$(B_t)_{t\ge 0}$ is a Brownian motion starting from $0$. For $x>0$ define $\sigma_x=\inf\lbrace t\ge 0:B_t\le t-x\rbrace$ and $W_t=\exp(\lambda B_t-\frac{\lambda^2}{2}t)$. I am trying to verify that

$$E(W_{\sigma_x})=1 \iff \lambda \le 1. $$

I don't see how the condition $\lambda \le 1$ applies. Since $\sigma_x$ is a stopping time, by the martingale stopping theorem I have $E(W_{\sigma _x})=E(\exp(\lambda B_{\sigma_x}-\frac{\lambda ^2}{2}\sigma_x))=E(\exp(\lambda B_0 - \frac{\lambda}{2} 0))=E(\exp(0))=1$.

How do I obtain the constraint $\lambda \le 1$?

Answer 1

Define $\tilde{B}_t$ and $\tilde{\sigma}_x$ by

$$ \tilde{B}_t = B_t - \lambda t, \qquad \tilde{\sigma}_x = \inf \{ t \geq 0 : B_t \leq (1-\lambda)t - x \}.$$

For each fixed $T$, define the probability measure $\Bbb{Q}$ by $\Bbb{Q}( \cdot ) = \Bbb{E}[W_T \mathbf{1}_{(\cdot)}]$, where $\Bbb{E}$ denotes the expectation w.r.t. $\Bbb{P}$. Then by the Girsanov's theorem, $(\tilde{B}_t : 0 \leq t \leq T)$ is a standard Brownian motion w.r.t. $\Bbb{Q}$ and thus

$$ \Bbb{P}(\tilde{\sigma}_x > T) = \Bbb{Q}(\sigma_x > T) = \Bbb{E}[ W_T \mathbf{1}_{\{ \sigma_x > T \}} ].$$

Here, we utilized the fact that $\sigma_x$ can be written as $ \sigma_x = \inf \{ t \geq 0 : \tilde{B}_t \leq (1-\lambda)t - x \}$. Now by the optional stopping theorem applied to the bounded stopping time $T \wedge \sigma_x$, we have

\begin{align*} 1 = \Bbb{E}[W_{T\wedge \sigma_x}] &= \Bbb{E}[ W_{\sigma_x} \mathbf{1}_{\{ \sigma_x \leq T \}} ] + \Bbb{E}[ W_T \mathbf{1}_{\{ \sigma_x > T \}} ] \\ &= \Bbb{E}[ W_{\sigma_x} \mathbf{1}_{\{ \sigma_x \leq T \}} ] + \Bbb{P}(\tilde{\sigma}_x > T). \end{align*}

Letting $T \to \infty$, monotone convergence theorem shows that

$$ \Bbb{E}[W_{\sigma_x}] = \Bbb{P}(\tilde{\sigma}_x < \infty). $$

This becomes $1$ exactly when $\lambda \leq 1$.