5
$\begingroup$

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.

$\endgroup$

2 Answers 2

1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.