Stock price is a martingale if the riskless interest rate is zero?

I came across a question as such:

Suppose company IBC is trading at \$75 per share. What does it cost to construct a derivative security that pays exactly one dollar when IBC hits$100 for the first time? Ignore dividends, assume a riskless interest rate of zero, assume all assets are infinitely divisible, ignore any short sale restriction.

There is a solution using the no-arbitrage argument. But my intuition is to use the martingale. Since the interest rate is zero, if the stock follows geometric Brownian motion, then the drift term become zero, so the stock price becomes a martingale. If we use the martingale property $E[S_{0}] = E[S_{T}]$, and assume the upper bound of stock price is 100, lower bound is 0, then we can calculate the probability $\alpha$ of hitting \$100 at time T $$E[S_{T}] = \alpha\times\100 + (1-\alpha)\times\0 = E[S_{0}] = 75$$ we get$\alpha = 0.75$, so the expected pay off of the derivative is $$\0.75=0.75\times\1 + (1-0.75)\times\0$$ hense the price of the derivative should be \$0.75. I don't have much background in Probability or martingale theory, so is this a valid argument to solve this problem?

Your derivation is incorrect because the option does not have a fixed maturity $T$. Instead it ends whenever the stock reaches 100. Mathematically this means that we have a stopping time which is a random variable $\tau(\omega) = \inf \{t \ge 0 | S_t(\omega) = 100\}$.
It is usually assumed that the stock price eventually reaches 100 i.e. that $\tau$ is almost surely finite.
Under this assumption this case $S_\tau$ is well defined and is always equal to $1$. Note that $E[S_\tau] \neq E[S_0]$ even though $S$ is a martingale. This is because the stopping time $\tau$ is not bounded so the optional stopping theorem does not apply.
The solution to the exercise is to think in terms of delta-hedging. How much of the stock do you need to hold to hedge your risk? At time $\tau$, the stock is worth $S_\tau = \$100$. So you can replicate your payoff by buying$1\%$of the stock at time 0 and holding it until it reaches$\$100$ to pay the $\$1$. So the price of the derivative is$S_0/S_{\tau} = \$0.75$.
• If $\tau$ is almost surely finite, then doesn't mean it's bounded? Also I was looking at the random walk example of optional stopping theorem on Wiki and it says "the expected time at which the walk ends is finite (say, from Markov chain theory)". My intuition is the stock price movement is just like the random walk example, so why the optional stopping theorem doesn't apply? Thank you. – astr627 Apr 26 '15 at 16:09
• Bounded would mean that there exists a deterministic time $T_{max}$ such that $\tau \leq T_{max}$ almost surely. This is clearly not the case. Here we assume that the stock will always reach 100 but it could take 1 day, 1 year, 10 years etc.. we don't know. – AFK Apr 26 '15 at 16:44
• The example you cite is a bit different: the stopping time corresponds to reaching either an upper barrier $b$ or a lower barrier $a$. In this case, the theorem applies because the stopped martingale $X_{t\wedge \tau_{a,b}}$ is bounded (it stays in the corridor $[a,b]$). If you only consider the upper barrier, this is no longer the case: the random walk can go as low as you can imagine even if it will eventually come back up and hit the barrier. Anyway, the theorem cannot apply to your problem since, as I explained $S_\tau = 1$ by definition so $E[S_\tau] = 1$ and cannot be equal to $S_0$. – AFK Apr 26 '15 at 16:53