1
$\begingroup$

Given a stochastic model for the evolution of St, with a given SDE for its volatility, how can you tell if the given model satisfy the sticky delta (or the sticky strike) property? Is it possible to prove analytically this property? Or the only way is to actually compute the prices?

$\endgroup$
  • $\begingroup$ You'll not be able to do it for $t_0$, as you'll need to bump your data, so you'll be forced to choose how you bump your smile (i.e. sticky delta or strike), i.e. you force the model to do what you want. You can however look at the conditional vol in the future, take a grid of points, say $S_t=90, S_t=91, \ldots, S_t=110$ at $t=0.5$ and for each point measure the conditional distribution at $t=1$ (with stoch vol you need to sample the vol too, so you'll have a 2d grid). You'll get a corresponding smile for each, and now you can observe how the model changes the smile for different spots. $\endgroup$ – will May 27 at 9:40
  • $\begingroup$ I asked explicitly for an analytic solution, I already know, given prices, how to do it, but that was not the question... $\endgroup$ – Giacomo Giannoni May 27 at 10:03
3
$\begingroup$

I agree with the comment made by will: for a given model, you can potentially compute a Delta according to any "stickiness rule" depending on which data you decide to bump vs. keep constant.

That being said, if you look at the following quantity $$ \Delta = \left. \frac{\partial V}{\partial S_0} \right\vert_{\Theta} $$ that we could call the in-model Delta as in "all parameters and state variables except the spot price are held constant" (e.g. $\Theta = (v_0,\theta,\kappa,\rho,\xi)$ in Heston), then you can say that:

For a (log-) space homogeneous diffusion model, $\Delta = \left. \frac{\partial V}{\partial S_0} \right\vert_{\Theta}$ will be a sticky-moneyness Delta.

A (log-) space homogeneous model is simply one where $$ \frac{dS_t}{S_t} = \mu_t dt + \sigma_t dW_t $$ where both the drift and diffusion coefficients cannot be direct functions of $S_t$ (e.g. no a local volatility model), such that after using Itô, you can directly integrate to obtain that $S_T/S_t$ does not depend on $S_t$ for any $T \geq t$.

As a result of this last property, European vanilla prices end up being homogeneous functions of degree 1 in space i.e. for a spot price $S_0$, strike and expiry $(K,T)$ $$ C(\xi S_0, \xi K, T; \Theta) = \xi C(S_0, K, T; \Theta), \,\,\forall \xi > 0 $$ such that (Euler's theorem, or just deriving the above wrt $\xi$ and setting $\xi = 1$ $$ C = \Delta S_0 + \frac{\partial C}{\partial K} K \tag{1} $$

Now, if you assume the model generates a volatility surface $\Sigma(S_0;K,T,\Theta)$ where $\Sigma$ is the function defined through $$ C(S_0,K,T;\Theta) := C_{BS}(S_0, K, T; \Sigma(S_0,K,T;\Theta)) $$ then, starting from $(1)$, using the chain-rule and the fact that BS model is (log)-space homogeneous, you will get that $$ \frac{\partial \Sigma}{\partial S_0}(S_0,K,T;\Theta) = -\frac{K}{S_0} \frac{\partial \Sigma}{\partial K}(S_0,K,T;\Theta) \tag{2} $$ which is indeed the definition of the sticky-moneyness rule.

Indeed, sticky moneyness suggests that $$ \Sigma(S_0+\delta S_0, K, T) = \Sigma(S_0, K^*, T) $$ provided, as the name indicates, that $$\frac{K^*}{S_0} = \frac{K}{S_0+\delta S_0} \iff K^* = K(1 + \delta S_0/S_0)^{-1}$$ Under such circumstances, \begin{align} \frac{\partial \Sigma}{\partial S_0}(S_0, K, T) &= \lim_{\delta S_0 \to 0} \frac{\Sigma(S_0+\delta S_0, K, T) - \Sigma(S_0, K, T)}{\delta S_0} \nonumber \\ &= \lim_{\delta S_0 \to 0} \frac{\Sigma\left(S_0, K(1 + \delta S_0/S_0)^{-1}, T\right) - \Sigma(S_0, K, T)}{\delta S_0} \nonumber \\ &= \lim_{\delta S_0 \to 0} \frac{\Sigma\left(S_0, K(1 - \delta S_0/S_0), T\right) - \Sigma(S_0, K, T)}{\delta S_0} \nonumber \\ &= \lim_{\delta K \to 0} \frac{\Sigma\left(S_0, K-\delta K, T\right) - \Sigma(S_0, K, T)}{\frac{S_0}{K}\delta K} \nonumber\\ &= -\frac{K}{S_0} \frac{\partial \Sigma}{\partial K}(S_0, K, T) \end{align}

$\endgroup$

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.