The same question was posted on MathOverflow.

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 $1$ $ into the stock. Then I wait until the stock rises to $1.1$$. If the stock price hits $0$$ before it hits $1.1$$, 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 $(W_t)_{tin[0,infty[}$, 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 $tgeq0$, we have

• $W_0=0$,
• for every $t>0$ and $sgeq0$, we have that $W_{t+s}-W_t$ is independent of every $W_u$ for $ule t$,
• for every $tgeq0$ and $sgeq0$, the difference $W_{t+s}-W_t$ is normally distributed with expected value $0$ and variance $s$,
• for every $omegainOmega$, the function begin{split}W(omega): [0,infty[&tomathbb 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$$ 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 $tgeq0$ such that $W_t=0.1$ and $W_s>-1$ for all $sle t$ 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 $M_toverset{text{Def.}}=max_{0le sle t} W_s$ has the cumulative distribution function $$mathsf P(M_tle m)=begin{cases}operatorname{erf}left(frac m{sqrt{2t}}right), &text{if }m geq0\0, &text{if }mle0end{cases}.$$

So for a fixed $t_0geq0$, we know that $$mathsf P(W_s>-1text{ for all }sle t_0) = mathsf P(M_{t_0}<1)=operatorname{erf}left(frac{1}{sqrt{2t}}right).$$

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?