2
$\begingroup$

In a paper titled Investing in Volatility published in 1998 by Emanuel Derman, Michael Kamal, Iraj Kani, John McClure, Cyrus Pirasteh, and Joseph Z. Zou, I found the following assertion (on page 9) that I am trying to clarify: The quantity $(1/2)\Gamma(\Delta S)^2$ is the gain from an instantaneous index move. The key principle of options valuation is that no free lunch can be obtained by using options. Therefore, if the index actually moves with a realized volatility identical to the implied volatility $\Sigma$ at which the option was purchased, the gain from small index moves must cancel the loss in option value due to the passage of time. Figure 4c shows that this loss due to "time decay"; its magnitude in an instant $\Delta t$ is given by $(1/2)\Gamma(\Sigma^2S^2\Delta t)$.

EDIT: The gain mentioned before is that of a delta-hedged option.

Is there an exact (or approximate relationship) that can be proven (preferably as rigorous as possible) between time decay (or theta) and gamma (as discussed above), in a model-free way?
I have seen some heuristic arguments based on binomial trees but they do not look very convincing. I was looking for something rather more general, for example an argument in continuous time using stochastic calculus.

EDIT: I am interested in finding a mathematically rigorous derivation for the assertions quoted above. Actually, even an good approximation would be ok, as long as I understand its limitations.

EDIT: Any reference would be very welcome.


In view of the answer below by Soumirai, I want to reformulate the question as follows: Knowing that $\Theta = \frac{1}{2}\sigma^2S^2 \Gamma$ holds in a B-S model without rates and dividends, is there an analogous formula that holds for more general stochastic volatility models (of even model-free) ?

$\endgroup$
3
$\begingroup$

The relationship between theta and gamma is the Black-Scholes PDE.

Let's take normal B-S dynamics with $r=0$: $dS_t = \sigma S_t dW_t$

The pricing PDE for a derivative $g(S_T)$ is (with terminal condition $g$):

$\frac{dp}{dt} = \frac{1}{2}\sigma^2S^2 \frac{d^2p}{dS^2}$

Or

$\Theta = \frac{1}{2}\sigma^2S^2 \Gamma$

This PDE has a solution (Feynman-Kac Theorem): $p(t,S_t) = \mathbb{E}(g(S_T))$, which is the derivative price. What the PDE tells us is that the value of the derivative changes with time ($\frac{dp}{dt}$), at a rate that is proportional to $\Gamma$ times some stuff (variance).

There are many sources that derive B-S PDE in different ways, and give explanations. For instance: https://www.frouah.com/finance%20notes/Black%20Scholes%20PDE.pdf

EDIT: For the generalized version, see https://en.wikipedia.org/wiki/Feynman%E2%80%93Kac_formula

$\endgroup$
3
  • $\begingroup$ Yes, this is indeed what we get under a B-S model. But I am interested to see if that relationship holds for a general (unspecified) model. Would that relation be difficult to prove for a very general/abstract SDE model ? $\endgroup$ – fwd_T Dec 30 '20 at 21:20
  • $\begingroup$ If your SDE is more general, your PDE will be more general as well. For instance if you add a second underlying and price an option on both underlyings, you will have something like $\Theta=c_1\Gamma_1 + c_2\Gamma_2 + c_3\Gamma_{12}$ where $\Gamma_{12}$ is the cross-derivative. Or if you add stochastic rates you will have $\Theta=c_1\Gamma_1 + c_2\Gamma_r + c_3\Gamma_{1r}$ where $\Gamma_{1r}$ is the cross derivative between rates and spot. Last if you add jumps, you will have some integral term in the PDE to account for the non-locality due to the jump. Have a look at Feynman-Kac and pricing $\endgroup$ – Soumirai Dec 30 '20 at 21:23
  • $\begingroup$ I was rather thinking about the situation when the volatility is stochastic (rather than a model with potentially multiple correlated underlyings). Is there a way to prove the formula you wrote for other, more general stochastic volatility models, where the stochastic volatility is not specified ? Otherwise said, how does $\Theta=\frac{1}{2}\sigma^2S^2\Gamma$ change when the volatility is not a constant parameter anymore, for example for $dS_t=\sigma_tS_tdW_t$ ? $\endgroup$ – fwd_T Dec 30 '20 at 21:28

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.