When the Black–Scholes Model Has an Equivalent Martingale Measure

# Why is the black-scholes model arbitrage free when σ>0?


# Why is the black-scholes model arbitrage free when σ>0?












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.

## Answer by q.t.f. (score 1, accepted)

https://quant.stackexchange.com/a/17879

If $\sigma=0$ there is no randomness: the spot follows a single deterministic path. That is, the measure consists of a point mass at that path. Any equivalent measure can again only give a point mass at that same path, with the same drift. So in this case we must have $\mu = r$ to have an equivalent martingale measure. This is arbitrage free, but there is no longer a market-price-of-risk giving different risk-neutral vs. real-world measures. This makes sense: there is no randomness so no risk!

## Answer by Jesse (score 0)

https://quant.stackexchange.com/a/43808

> I want to show that: if $\sigma$ is positive then there is no arbitrage in the model

You will not be able to show this, because it is not true. The Black-Scholes model does allow for arbitrage opportunities if one places no restrictions on the size of the allowable trading strategies, cf. Harrison-Pliska (1981).

What is true is that there are no arbitrage opportunities among the set of "tame" trading strategies (where a tame strategy is one whose corresponding value process is always non-negative). This does follow from the existence of an equivalent martingale measure (and an EMM always exists as long as $\sigma \neq 0$); cf Theorem 6.1.1 in Risk-Neutral Valuation by Bingham-Kiesel.

Perhaps what you really intend to ask is "why does an EMM not exist (in general) when $\sigma = 0$?". This question has been adequately addressed by q.t.f., but it's different from the question in the title...