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Assume the stockprice as in the Black-Scholes model (Geometric Brownian Motion):

$$S_t=S_0e^{(\mu-\sigma^2/2)\cdot t+\sigma W_t}$$

Wouldn't there be an immediate arbitrage opportunity, to just buy the stock and wait until it reaches level above the riskfree asset (then sell stock to repay loan and gain remainder as profit)?

As we know, the Black-Scholes model is assumed to be arbitrage-free with unlimited debt and time horizon.

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What would you choose the T in the definition of an arbitrage to be, to ensure that you will not loose money? :) –  Henrik Aug 7 '14 at 18:50
@Henrik $T:=\inf t>0:S_t>S_0e^{rt}$ –  emcor Aug 7 '14 at 19:44
Is that T a valid choice for the no arbitrage definition? –  Henrik Aug 8 '14 at 6:52
@Henrik I am not sure, that is probably the question... $T$ would be a stopping time here. –  emcor Aug 8 '14 at 7:05
I like the question. But to 'make money' you should change your definition a bit. Due to the continuity of paths the $\inf\{t:S_t > S_0\ e^{rt}\} = \inf\{t:S_t = S_0\ e^{rt}\}$. Therefore you just wait to 'break even'. Perhaps you should change your stopping time slightly to $\inf\{t:S_t = S_0\ e^{rt} + \varepsilon\}$, to make $\varepsilon$ at every iteration. –  Kiwiakos Aug 8 '14 at 11:07

3 Answers 3

up vote 2 down vote accepted

It does not seem you feel the question is answered so I will try to elaborate over what I think seems to bother you.

Let $S_t = e^{(\mu -\sigma^2/2) t + \sigma W_t}$ be the stock price process and $B_t=e^{rt}$ be the risk free. The arbitrage you describe is then choosing a nice $\varepsilon >0$ and setting $\tilde{T}=\inf \{t>0 : (\mu -r -\sigma^2/2)t +\sigma W_t> \varepsilon\}$. Then one would have an "arbitrage" at $\tilde{T}$, as you say this will eventually happen which is true. In fact one even know the distribution of when your "arbitrage" will occur see

which is unfortunately also the problem. Since the inverse Gaussian distribution has mass over entire $\mathbb{R}^+$ you will not be able to choose $T\in \mathbb{R}$ such that $P(T \geq \tilde{T})=1$, ergo you can not in this way find an arbitrage.

It is in fact very easy to find an equivalent martingale measure in this model formally implying that the model is arbitrage free.

It is a different issue whether your "mini arbitrage strategy" is a attractive feature of a model. It is as you say simply a consequence of having a model where volatility accumulates without bound together with no possibility of default.

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Thanks, can you please elaborate a bit what you mean by $P(T\geq\tilde{T})=1$. Does it mean $\tilde{T}$ goes to infinite? It may be that the expected time of $\tilde{T}$ is finite? –  emcor Aug 10 '14 at 9:03
Ah I see, its finite $E(\tilde{T})=\mu<\infty$, but its not certain on $\mathbb{R}$. –  emcor Aug 10 '14 at 9:26
It seems you are still a little confused. Maybe this will clarify: Let $X\sim N(0,1)$ and i ask you to find a number c such that $c>X$ (a.s.). X does not go to infinity (whatever that would mean), nor is it relevant that the mean is finite - only because X does not have bounded support it is impossible. –  Henrik Aug 12 '14 at 7:54

No this is not a risk free arbitrage. What you are talking about is modeling a stock price with GBM and it has nothing to do with Black-Scholes. Black-Scholes is an option pricing formula that assumes that stocks follow GBM (which is a bad assumption to begin with but we won't get into that). What you are talking about doing is taking on leverage.

$ E[S_T]= S_0e^{ut} $ where $u$ is the growth rate of the stock. So if you take a loan out at time 0 for $S_0$ then at time T you will owe back $S_0e^{rt}$ where r is the risk free rate.

Now if $u > r$ it is true that you would EXPECT to make money. This is not arbitrage. Arbitrage is when you are guaranteed to make money with no risk. In the situation you are describing there will be times when you lose money and will not be able to pay back your loan.

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Hi Matthew, as you are rightly saying, the Black-Scholes model assumes this GBM for the underlying, so Black-Scholes does have alot to do with it. Any arbitrage strategy requires a short position, i.e. debt, to start an investment with $0$ initial capital. As I wrote, I would only sell the stock at the time when it crosses the loan value, so it is guaranteed profit in this sense? –  emcor Aug 8 '14 at 7:12
I agree you were correct if $T$ was a fixed time. –  emcor Aug 8 '14 at 7:52
No the stock has no guarantee to ever grow in such a way that its value is greater than that of the loan. It may be true that expected value for any time t is greater but it make take an infinite amount of time for that to happen. –  meh Aug 9 '14 at 19:06
Again, arbitrage does not require guaranteed profit, it just requires $V_T\geq0$ and $P(V_T>0)>0$. I do not care about expected value, the thing is that Brownian Motion is continuous distributed on $\mathbb{R}$, so there is a certain probability to make profit by just waiting for it. You cannot say just because waiting takes long, it would be infinite. Or you may show here $E(T^*)\rightarrow \infty$ e.g. by Laplace Transform of stopping time. –  emcor Aug 9 '14 at 20:29
Ok, so I'm posting here this definition for you: You should also understand, that $P(V_t>0)=1$ is not neccessary for arbitrage, since if you get riskfree profit with 1% probability (and 0 profit else), you would still always do it (arbitrage). Also, you should understand that question was about Black-Scholes world (not real world, where anything can happen). Gamblers Ruin cannot occur because BS-world allows unlimited debt (frictionless markets). –  emcor Aug 10 '14 at 16:50

This is not arbitrage. Your construction of the portfolio is zero but there is no guarantee that your stock will cover your loan. The GBM is a stochastic process, you have a probability that it'll not cover up your loan.

By the way, the question is invalid because you misunderstood the BS model. Under BS, GBM is simply an assumption. The BS model is used to price and hedge options not the underlying asset.

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But the GBM can never go to zero because $S_0>0$ and $e^x>0$. And there is some probability that $W_t$ will cross any level say $c=S_t$ at some time. –  emcor Aug 7 '14 at 14:28
Even it won't goto zero, there's still no guarantee that you can cover the loan. Remember the longer you wait, the more interest you'll have to pay. –  Student T Aug 7 '14 at 14:29
Arbitrage requires no guarantee, it just requires $V_T\geq0$ with $P(V_T>0)>0$ (some infinitesimal probability). The Black-Scholes loan is modeled as zero-bond without coupon and frictionless markets (so unlimited borrowing). –  emcor Aug 7 '14 at 14:30
Arbitrage requires you can always generate profit in all scenarios. If that's no guarantee, it's not an arbitrage, it's an investment. –  Student T Aug 7 '14 at 14:39
No, that is incorrect. –  emcor Aug 7 '14 at 14:41

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