# Prove that $X_t = W_t \mathbb{1}(\tau > t) + (2W_{\tau} - W_t) \mathbb{1}(\tau \le t)$ is a martingale.

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

1. $$s < t < \tau$$: $$\mathbb{E}(W_t|F_s) = W_s$$

2. $$s < \tau < t$$: $$\mathbb{E}(2W_\tau - W_t | F_s) = \mathbb{E}(2W_\tau - 2W_s + W_s - W_t + W_s | F_s) = W_s$$

3. $$\tau < s < t$$: $$\mathbb{E}(2W_\tau - W_t| F_s) = \mathbb{E}(2W_\tau - W_s - (W_t - W_s)|F_s) = 2W_\tau - W_s$$

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?

• Did you mean to write $X_t = W_t \mathbb{1}(\color{red}\tau > t) + (2W_{\tau} - W_t) \mathbb{1}(\color{red}\tau \le t)$ instead? Jan 31, 2022 at 12:49
• @JoseAvilez yes! My bad. Edited. Jan 31, 2022 at 13:08
• Hint: consider the equality $E(X_t \vert \mathcal{F}_s)= E(X_t \mathbb{1}_{\tau \leq s} \vert \mathcal{F}_s )+ E(X_t \mathbb{1}_{\tau > s } \vert \mathcal{F}_s)$
– FOE
Jan 31, 2022 at 13:44
• @FOE and then use that $\mathbb{1}(\tau \le s)$ is $\mathcal{F}_s$ measurable? But there are at least one problem with $\mathbb{1}(\tau > s)$, since it might be that $\mathbb{1}(s < \tau < t)$ Jan 31, 2022 at 16:08

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 $$\tau \leq t$$

• I'm not sure you're right about term $(2W_\tau - W_t)\mathbb{1}(\tau \le s) - (2W_\tau - W_t)\mathbb{1}(s < \tau \le t)$ and $W_t \mathbb{1}(\tau > s)$. Because $\{\tau > s\}$ implies $\tau < t$, hence it will be $2W_\tau - W_t$, instead of $W_s$. Jan 31, 2022 at 19:19
• It seems there should be: $W_t \mathbb{1}(t > s > \tau) + W_t \mathbb{1}(s < \tau < t) + (2W_\tau - W_t) \mathbb{1}(\tau < s < t)$ Jan 31, 2022 at 19:26
• There was a mistake in the signs I just edited, , however I can not understand your statement: $\lbrace \tau >s \rbrace$ does not implies $\tau <t$
– FOE
Jan 31, 2022 at 19:52
• Yes, now it seems to be clear. Jan 31, 2022 at 20:02
• If you want to be more formal, the final statement (which is true and maybe not hard to see) could be proved with the Strong Markov Property and studying the distribution of the increments of the Brownian Motion.
– FOE
Jan 31, 2022 at 20:12