4
$\begingroup$

In Lorenzo Bergomi, Stochastic Volatility Modeling, Chapter 5 Appendix A.1, Equation (5.64), as shown below, seems to assume $\hat\sigma$ to be constant. If that is the case, why do we bother to invoke the Feynman-Kac formula? We can simply use the Black-Scholes formula.

enter image description here

$\endgroup$
6
  • $\begingroup$ Hi Hans, would like to try to help you but don't have Bergomi's book with me (and don't have an electronic copy). Can you give more background / post a screenshot perhaps? $\endgroup$
    – Frido
    Mar 14 at 6:55
  • $\begingroup$ @Frido: I added the screenshot of the cited derivation. $\endgroup$
    – Hans
    Mar 14 at 7:11
  • $\begingroup$ I think Bergomi uses Feynman Kac to show that dollar gamma and dollar delta are martingales. But as you say, he is only showing this under Black-Scholes so could have just used BS PDE. Before I quit QFSE (and joined again) I had answered ((under user34971) a similar questions but for more general processes. In that case using expectations = Feynman Kac might have more added value: quant.stackexchange.com/a/45474/65759 Does this answer your question? $\endgroup$
    – Frido
    Mar 14 at 7:16
  • 1
    $\begingroup$ In other words, you're right, for just showing vega-gamma relationship in BS world you don't need Feynman-Kac $\endgroup$
    – Frido
    Mar 14 at 7:26
  • 1
    $\begingroup$ @Frido: See my answer below. $\endgroup$
    – Hans
    May 26 at 21:29

2 Answers 2

4
$\begingroup$

Just want to add the observation that the pricing PDE solution can be formally written as $$ C(\tau) = e^{\tau \mathcal H} C(0) \quad (*) $$ where $\tau$ is time to maturity and $\mathcal H$ is a differential operator. For example, in the BS world with zero interest rate it is $$ \mathcal H = \tfrac12 \sigma^2 S^2 \frac{\partial^2}{\partial S^2} $$ Thus $U(\tau) = e^{\tau \mathcal H}$ is an 'evolution operator'.

In the BS case $\sigma$ is not a variable but a parameter. So you can differentiate both sides of equation (*) to very quickly obtain the vega gamma relation by noting that the operator $U(\tau)$ depends on the parameter $\sigma$. You could reinsert dividends and rates to also obtain sensitivities to $r$ and $q$ in the same manner.

In stochastic volatility models you can similarly show that the sensitivity of the option price to the correlation parameter is the stochastic volatility model vanna.

$\endgroup$
3
  • $\begingroup$ Very nice. The operator expression which is not just formal but can be well defined via spectral operator integral i.e. Dunford integral. It relies on the $C(\tau)$ being a one-parameter analytic semigroup and the spectral theorem of a parabolic equation though, which are involved. It would be great to give more technical details and references so as to make the answer more substantial. $\endgroup$
    – Hans
    May 28 at 6:14
  • $\begingroup$ @Hans Agree that it could be more technical. If I could I would, but this is the theory of semigroups of linear operators which is somewhat beyond my knowledge at the moment. I 'just' apply it. However here is the canonical reference: link.springer.com/book/10.1007/978-1-4612-5561-1 $\endgroup$
    – Frido
    May 28 at 7:34
  • 1
    $\begingroup$ OK, you echoed my point about operator semigroup and fair enough. +1 and accepted. $\endgroup$
    – Hans
    May 28 at 7:44
2
$\begingroup$

Even though it is true that the volatility is constant in this setting, the relationship is valid for all terminal condition or pay-off function -- beyond the typical $(\pm(S-K))_+$ -- so long as the pay-off function is independent of the volatility. We can certainly write out the integral expressions of vega and gamma (of arbitrary pay-off functions) and find their relationship. But it seems simpler dealing with the PDE directly. Moreover, this methodology can be used to find other high order partial derivatives.

$\endgroup$

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.