Consider $W_t$ is a standard Wiener process. Assume $\tau = \inf\{t : W_t = a\}$. Consider $X_t = W_t \mathbb{1}(\tau > t) + (2W_{\tau} – W_t) \mathbb{1}(\tau \le t)$. We want to show that $(X_t, \mathcal{F}_t)$ is a martingale, where $\{\mathcal{F}_t\}$ is natural filtration of the Wiener process.
I know that it's possible to show that the process is also Wiener, but it's much more complicated (since we use strong Markov property). Here I guess the situation is much easier.
It's not hard to show that $X_t$ is $\mathcal{F}_t$ measurable and $L_1$. The major problem is to show $\mathbb{E}(X_t | F_s) = X_s$ precisely.
I tried so: consider different cases
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$s < t < \tau$: $\mathbb{E}(W_t|F_s) = W_s$
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$s < \tau < t$: $\mathbb{E}(2W_\tau – W_t | F_s) = \mathbb{E}(2W_\tau – 2W_s + W_s – W_t + W_s | F_s) = W_s$
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$\tau < s < t$: $\mathbb{E}(2W_\tau – W_t| F_s) = \mathbb{E}(2W_\tau – W_s – (W_t – W_s)|F_s) = 2W_\tau – W_s$
But I guess it's not so fair to write so, since $\tau$ is random variable and for some $\omega \in \Omega$ it might be true and for others not. Any hints?
Best Answer
First of all the expression can be rewritten as: $$ W_t \mathbb{1}_{\tau >s} - W_t \mathbb{1}_{t \geq \tau >s}+ (2W_{\tau}-W_t)\mathbb{1}_{\tau \leq s}+(2W_{\tau}-W_t)\mathbb{1}_{s<\tau \leq t}$$
On one hand $$E(W_t \mathbb{1}_{\tau >s}+(2W_{\tau}-W_t)\mathbb{1}_{\tau \leq s} \vert \mathcal{F}_s)= W_s \mathbb{1}_{\tau >s} + (2W_{\tau}-W_s)\mathbb{1}_{\tau \leq s} $$
Therefore we need to prove $$E(- W_t \mathbb{1}_{s < \tau \leq t} +(2W_{\tau}-W_t)\mathbb{1}_{s<\tau \leq t} \vert \mathcal{F}_s)=0$$
$$ \Leftrightarrow E((2W_{\tau}-2W_t)\mathbb{1}_{s<\tau \leq t} \vert \mathcal{F}_s)=0$$ Which is true because $W_t-W_{\tau}$ is independant of $\mathcal{F}_s$ when $s<\tau \leq t$ and has $0$ expected value when $ \tau \leq t$