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 $$\sigma >0$$
It is clear that $X_t$ is independent of $r$ and $\mu$ so that's why the case of $r >\mu$ doesn't effect us.
