4
$\begingroup$

Can Black-Scholes option values be derived via the Capital Asset Pricing Model, without resort to the use of a risk-free portfolio being created from the option and a Delta determined quantity of the underlying instrument?

$\endgroup$
  • $\begingroup$ What is a "Black-Scholes option value" ? $\endgroup$ – will Feb 17 '18 at 12:45
  • $\begingroup$ Is there a good reason why you posted the exact same question (quant.stackexchange.com/questions/35931/…) which already had an answer (which you didn't care to accept)? This kind of behaviour is being frowned upon here. $\endgroup$ – vonjd Feb 17 '18 at 19:11
  • 1
    $\begingroup$ @vonjd you're right, kind of sounded familiar, I didn't recalled I had already answered it... $\endgroup$ – Daneel Olivaw Feb 18 '18 at 14:18
  • $\begingroup$ The question I asked previously (quant.stackexchange.com/questions/35931/…) is not a duplicate of this question. Yes I asked for a derivation via the Capital Asset Pricing Model in that question, but the question asked here is what I intended but failed to state correctly. In that previous question I failed to state that I did not want a derivation that was dependent on the risk-free portfolio. The derivation without resort to this device has been requested here. $\endgroup$ – Andrew Beaven Feb 19 '18 at 10:00
5
$\begingroup$

From Frequently Asked Questions in Quantitative Finance (2009) by Paul Wilmott, p. 416:

This derivation, originally due to Cox & Rubinstein (1985) starts from the Capital Asset Pricing Model in continuous time. In particular it uses the result that there is a linear relationship between the expected return on a financial instrument and the covariance of the asset with the market. The latter term can be thought of as compensation for taking risk.

But the asset and its option are perfectly correlated, so the compensation in excess of the risk-free rate for taking unit amount of risk must be the same for each.

For the stock, the expected return (dividing by $dt$) is $\mu$. Its risk is $\sigma$.

From Ito we have $$dV = \frac{\partial V}{\partial t}dt + \frac{1}{2}\sigma^2S^2\frac{\partial ^2V}{\partial S^2}dt + \frac{\partial V}{\partial S}dS$$ Therefore the expected return on the option is $$\frac{1}{V}\left( \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial ^2V}{\partial S^2} + \mu S \frac{\partial V}{\partial S}\right)$$ and the risk is $$\frac{1}{V} \sigma S \frac{\partial V}{\partial S}$$ Since both the underlying and the option must have the same compensation, in excess of the risk-free rate, for unit risk $$\frac{\mu-r}{\sigma}= \frac{\frac{1}{V}\left( \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial ^2V}{\partial S^2} + \mu S \frac{\partial V}{\partial S}\right)}{\frac{1}{V} \sigma S \frac{\partial V}{\partial S}}$$ Now rearrange this. The $\mu$ drops out and we are left with the Black–Scholes equation.

$\endgroup$
  • $\begingroup$ This is clever, but isn’t it circular? Let’s start with the Black-Scholes, do some rearranging, and then return the Black-Scholes. $\endgroup$ – David Addison Feb 17 '18 at 17:48
  • 2
    $\begingroup$ @DavidAddison: We don't start with Black-Scholes. The first equation just states how the payoff of an option $V$ evolves as a function of $S$ and $t$. Because of the stochasticity of $S$ we have to use Ito's lemma. $\endgroup$ – vonjd Feb 17 '18 at 18:11
  • 1
    $\begingroup$ I see now. Still, very clever. $\endgroup$ – David Addison Feb 18 '18 at 21:04
  • 1
    $\begingroup$ An elegant answer. I will consider the math carefully, I have been looking at monte carlo simulations over non-instantaneous time periods and the relationship between option returns and simulated risk premia (location of the terminal underlying distribution). To satisfy myself I will need to convert these instantaneous derivatives to finite differences. $\endgroup$ – Andrew Beaven Feb 19 '18 at 17:03
  • $\begingroup$ The ''"help"" section did not make it really clear how to accept an answer.....so can that be made clear and left here for a week or so before this comment is deleted as off topic $\endgroup$ – Andrew Beaven Feb 22 '18 at 15:20
2
$\begingroup$

Yes, see page 16 from the below paper:

"Four Derivations of the Black-Scholes Formula" (F. Rouah)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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