*The same question was posted on [StackExchange](https://math.stackexchange.com/q/3930845/631742).*


Informal problem description
============================


Assume that we have a stock whose price behaves exactly like a [Wiener process](https://en.wikipedia.org/wiki/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[${}^1$](https://en.wikipedia.org/wiki/Wiener_process#Running_maximum) 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](https://math.stackexchange.com/a/3410056/631742) and then use [Donsker's Theorem](https://en.wikipedia.org/wiki/Donsker%27s_theorem) to conclude?



Winning money from random walks?
The expectation of $S$ is indeed $0$. This follows by the optional stopping theorem; see e.g. [THM 29.11](https://www.math.wisc.edu/%7Eroch/grad-prob/gradprob-notes29.pdf) 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.


