# Winning money from random walks?

The same question was posted on StackExchange.

# Informal problem description

Assume that we have a stock whose price behaves exactly like a Wiener process. (There are multiple reasons why this is not the case in real life, but bear with me). I invest $$1$$ dollar into the stock. Then I wait until the stock rises to $$1.1$$ dollars. If the stock price hits $$0$$ dollars before it hits $$1.1$$ dollars, I go bankrupt and lose my investment.

Is this a viable strategy? And if yes, why is it possible to make money off of a random walk?

# Formal Problem setup

Assume we are given a standard Wiener process $$(W_t)_{t\in[0,\infty[}$$, i.e. a family of real random variables on a probability space $$(\Omega,\mathcal A, \mathsf P)$$ (event space, $$\sigma$$-algebra and the probability measure, respectively) such that for every $$t\geq0$$, we have

• $$W_0=0$$,
• for every $$t>0$$ and $$s\geq0$$, we have that $$W_{t+s}-W_t$$ is independent of every $$W_u$$ for $$u\le t$$,
• for every $$t\geq0$$ and $$s\geq0$$, the difference $$W_{t+s}-W_t$$ is normally distributed with expected value $$0$$ and variance $$s$$,
• for every $$\omega\in\Omega$$, the function $$\begin{split}W(\omega): [0,\infty[&\to\mathbb R,\\ t&\mapsto W_t(\omega)\end{split}$$ is continuous.

# Question

My strategy, I will call it $$S$$, is to wait until I make $$0.1$$ dollars in profit, i.e. until the Wiener process hits $$0.1$$. If the process hits $$-1$$ before hitting $$0.1$$, I loose all my money.

So we have $$S =0.1$$ if there exists a $$t\geq0$$ such that $$W_t=0.1$$ and $$W_s>-1$$ for all $$s\le t$$ and we have $$S=-1$$ otherwise.

What is the expected value of $$S$$? It think it is $$0$$. However, I don't know how to prove this.

# Some attempts

It is well-known$${}^1$$ that the running maximum $$M_t\overset{\text{Def.}}=\max_{0\le s\le t} W_s$$ has the cumulative distribution function $$\mathsf P(M_t\le m)=\begin{cases}\operatorname{erf}\left(\frac m{\sqrt{2t}}\right), &\text{if }m \geq0\\0, &\text{if }m\le0\end{cases}.$$

So for a fixed $$t_0\geq0$$, we know that $$\mathsf P(W_s>-1\text{ for all }s\le t_0) = \mathsf P(M_{t_0}<1)=\operatorname{erf}\left(\frac{1}{\sqrt{2t}}\right).$$

I don't see how to compute $$\mathsf E(S)$$ from that, though.

# Another idea

Maybe one can try to solve a discretized problem as was done in this great answer and then use Donsker's Theorem to conclude?

The expectation of $$S$$ is indeed $$0$$. This follows by the optional stopping theorem; see e.g. THM 29.11 with $$X=W$$, $$S=0$$, and $$T:=\inf\{t\ge0\colon W_t\notin(-1,0.1)\}$$.