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I want to show that: if $σ$ is positive then there is no arbitrage in the model, even if $r > µ$. Whilst I have satisfied this for $ r > \mu$, I cannot see why the conditioning on $\sigma>0 $ is necessary.

Given: $S_0 = 1$, $B_t$ = Brownian motion and $S_t$ = stock price And our Black-scholes model: $$dS_t = \mu S_t dt + \sigma S_t dB_t$$

Then this model is arbitrage free if there is some Equivalent Martingale Measure $\mathbb{Q}$ such that $S_t e^{-rt}$ is a martingale.

So why do we require that $\sigma > 0$ for this to be arbitrage free?

The solution to BSM: $$S_T = S_0 e^{(r - \frac{1}{2}\sigma^2)T + B_T}$$

Now to discount it, let $X_t = S_T e^{-rT}$

So $$X_t = e^{(\mu - \frac{\sigma^2}{2} - r)T + \sigma(B_t)}$$ So we need to show $X_t$ is EMM under $\mathbb{Q}$

$$dX_t = \sigma S_t e^{-rt}( \frac{\mu - r}{\sigma}dt + dB_t)$$

then by Girsanov's theorem with $c = \frac{\mu - r}{\sigma}$, There's $\mathbb{Q}$ such that $ct + B_t = \hat B_t$ is a Brownian motion (is this called brownian motion with drift?)

Gives:

$$X_t = X_0 e^{\sigma \hat B_t - \frac{1}{2}\sigma^2t}$$

And this is an exponential martingale.

BUT, why does this rely on $$\sigma >0$$

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  • $\begingroup$ Since $\sigma$ is the square root volatility, this has to be positive. The only condition is that shifting measure is finite. But I could be wrong. $\endgroup$ – Chinny84 May 19 '15 at 6:54
  • $\begingroup$ The exercise I am doing alludes to the fact that under $\sigma \leq 0$, BSM is no longer arbitrage free. If I could rephrase, what is it about $\sigma >0$ that makes BSM arbitrage-free. $\endgroup$ – elbarto May 19 '15 at 9:41
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    $\begingroup$ Something here doesn't make sense. Even though I'm far from being an expert, as much as I understand $dB_t$ has a random sign, being a single Brownian motion step. So basically you can show a symmetry under flipping the sign of both $\sigma$ and $dB_t$, but since flipping the sign of $dB_t$ doesn't really change the situation I can't see a reason for the sign of $\sigma$ to be of any consequence. Of course if $\sigma=0$ we have a totally different situation. $\endgroup$ – Shahar Even-Dar Mandel May 20 '15 at 9:01

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