I am working through an answer to the following question and I do not understand a statement given towards the end of the solution, specifically why $\tilde{W}(\sigma) = 1$.
(This question is related to my previous post Expectation of a Wiener process at a Stopping Time but I believe it may be viewed without reference to that post). \begin{align} \end{align}
$\textbf{Question:}$ Define the following stochastic process \begin{align} \tilde{W}(t) := W(t) + \int_{0}^{t} e^{W(s)}\mathbb{1}_{\{|W(s)\le 1\}}\,ds \quad \forall \, t \in [0,T] \end{align} For the stopping time $\sigma := \inf\{t \ge 0 : W(t) = 1 - \int_{0}^{t} e^{W(s)}\mathbb{1}_{\{|W(s)\le 1\}}\,ds \}$ compute \begin{align} \mathbb{E}_{\mathcal{Q}}\left[e^{-\sigma/2} \right] \end{align} \begin{align} \end{align} $\textbf{Solution:}$ Define $M(t) = \exp\left(\tilde{W}(t) - \tfrac{1}{2}t \right)$ as a martingale $(M(t) : t \ge 0)$.
Then from the $\textbf{optional}$ $\textbf{sampling}$ $\textbf{theorem}$, \begin{align} 1 = \mathbb{E}_{\mathcal{Q}}\left[M(0) \right] = \mathbb{E}_{\mathcal{Q}}\left[M(t\wedge \sigma) \right] \end{align}
Moreover, we have $M(t \wedge \sigma) \rightarrow M(\sigma)$ $\mathcal{Q}$-a.s. as $t \rightarrow \infty$ and it follows that \begin{align} &|M(t \wedge \sigma| \le \exp\left(\tilde{W}(t \wedge \sigma) - \tfrac{1}{2}(t\wedge \sigma)\right) \le \exp\left(\tilde{W}(t\wedge \sigma) \right) \le \exp(1),\\ &\mathbb{E}_{\mathcal{Q}}\left[\exp(1)\right] = \exp(1) < \infty \end{align}
By Lebesgue's theorem of dominated convergence, \begin{align} \mathbb{E}_{\mathcal{Q}}\left[M(\sigma)\right] = \mathbb{E}_{\mathcal{Q}}\left[\lim_{t \rightarrow \infty} M(t \wedge \sigma) \right] = \lim_{t \rightarrow \infty} \mathbb{E}_{\mathcal{Q}}\left[M(t\wedge \sigma) \right] = 1 \end{align} On the other hand, since $\tilde{W}(\sigma) = 1$ we have \begin{align} \mathbb{E}_{\mathcal{Q}}\left[M(\sigma) \right] &= \mathbb{E}_{\mathcal{Q}}\left[e^{1 - \tfrac{1}{2}\sigma} \right]\\ 1 &= \mathbb{E}_{\mathcal{Q}}\left[e^{1 - \tfrac{1}{2}\sigma} \right]\\ \end{align} so, \begin{align} \mathbb{E}_{\mathcal{Q}}\left[e^{-\tfrac{1}{2}\sigma} \right] = \frac{1}{e} \end{align}
$\textbf{My Question}:$ It is stated above that $\tilde{W}(\sigma) = 1$, unfortunately I cannot see why this is the case. Could someone please explain why this is?
All help is appreciated.
Many thanks,
John
The step you are asking about is just algebra:
$$W(\sigma)=1-\int_0^\sigma e^{W(s)} 1_{|W(s)| \leq 1} ds$$
Substitute that into the definition of $\tilde{W}$:
$$\tilde{W}(\sigma)=1-\int_0^\sigma e^{W(s)} 1_{\{ |W(s)| \leq 1\}} ds + \int_0^\sigma e^{W(s)} 1_{\{ |W(s)| \leq 1 \}} ds = 1.$$