This problem was motivated by my bitcoin trading and recalling some of my math education back in the day. I thought I'd ask people who know this much better than I...
Suppose there is a continuous, log-normally distributed random walk -- Brownian Motion basically. (As an aside, it's a martingale, right? But not necessarily a Markov process because the variance / volatility may depend on the past slightly while the mean does not.)
Anyway, my question is, suppose the X[t] is currently = c. What is the probability, given positive numbers x and y, that X[s] = x for some s while X[u] != y for any u between t and s. That is, X hits x before y?
I am asking because this seems to me like a winning strategy:
- Buy and hold 10 bitcoins on http://bitfinex.com
- Take a short position (borrow and sell) 5-6 bitcoins on margin
- Since the value of the bitcoins can go arbitrarily low before finally coming back up and giving me profit in a finite time, I can close my margin position when the price has sufficiently dropped and wait until my actual coins go back to what they were worth, so I don't have to worry about not making a profit eventually, assuming actual log-normal distribution.
- If the value of the bitcoins goes up, however, my bitcoins supporting the margin also go up in value. So it can keep covering the margin for a very long time. I only get margin-called if the value of the bitcoins shoots up 10x or so and doesn't come back down.
- So here's the dilemma. What are the chances that the value of the bitcoins goes up 10x and NEVER goes below the point at which I borrowed+sold them? Or similarly, if it goes down 10x and NEVER goes back up to the starting point?
According to the efficient market hypothesis, any strategy with martingales won't yield a positive expected value, so if the process is really a martingale, then this should all even out and in the long run I will make the same as if I just held the bitcoins and didn't do anything, which if it's a martingale is 0 profit, right?
I think you may need to define your process and your question more precisely. If you're talking about a discrete-time process (or one that changes only by discrete jumps) with a continuous distribution, the probability that it ever hits (exactly) $x$ is $0$. If you're talking about a martingale that has continuous sample paths, if $c$ is between $x$ and $y$, the probability $p$ of hitting $x$ before $y$ is exactly what you need to have $0$ expected gain by stopping the first time you hit $x$ or $y$: thus $c = p x + (1-p) y$, or $p = \dfrac{c - y}{x - y}$.
As for the probability of never coming back to the starting point, that may be $0$ in typical models... but "never" is a very long time. And any particular model is not going to reflect reality forever.