Winning money from random walks?

Question

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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?

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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?

Answer 1

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 $T:=\inf\{t\ge0\colon W_t\notin(-1,0.1)\}$.

Alternatively and more specifically, one can use Wald's lemma for the Brownian motion -- see e.g. THM 29.12 in the linked paper.