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In Bergomi [Stochastic Vol Modelling] (Sec. 2.5.2), in the section on surface dynamics, the following definition of the "Skew Stickiness Ratio" (SSR) is made:

$$ SSR = \dfrac{1}{\mathcal{S}_T}\frac{d\hat{\sigma}_{F_TT}}{d\log(S_0)}$$

where $\mathcal{S}_T=\frac{d\hat{\sigma}_{KT}}{d\log{K}}\Bigr\rvert_{K=F_T}$.

Combining these equations we get,

$$ SSR = \dfrac{1}{\frac{d\hat{\sigma}_{KT}}{d\log{K}}\Bigr\rvert_{K=F_T}}\frac{d\hat{\sigma}_{F_TT}}{d\log(S_0)}$$

Then Bergomi says that $SSR=1 =>$ Sticky Strike and $SSR=0 =>$ Sticky Delta.

I would like to prove these statements mathematically, but I am having trouble.

If $SSR=1$, then we have that: $$ \frac{d\hat{\sigma}_{KT}}{d\log{K}}\Bigr\rvert_{K=F_T}= \frac{d\hat{\sigma}_{F_TT}}{d\log(S_0)}$$

then how is this statement equivalent to sticky strike? In other words, how is this statement equivalent to $\hat{\sigma}_{KT}(t,S_t)=\hat{\sigma}_{KT}(t+\epsilon, S_{t+\epsilon})$?

If $SSR=0$, then we have that: $$ \frac{d\hat{\sigma}_{F_TT}}{d\log(S_0)}=0$$

then how is this statement equivalent to sticky delta? In other words, how is this statement equivalent to $\hat{\sigma}_{xT}(t,S_t)=\hat{\sigma}_{xT}(t+\epsilon, S_{t+\epsilon})$, where $x=\log{\dfrac{S}{F_T}}$?

Bergomi gives intuitive explanations for these statements which don't make much sense to me. I suppose I am looking for a formal proof to ease my mind.

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Some Notations

It's easy to get lost so let's introduce some notations and let $$ \sigma : (t, S, K, \tau) \to \sigma(K,\tau; S, t) $$ denote the implied volatility smile prevailing at time $t$ when the spot price is $S_t=S$ for an option with strike level $K$ and time to expiry $\tau=T-t$. From here onward, we drop the $t$ argument to keep notations uncluttered (everything happens at fixed $t$).

Back to the notations used in the book, the ATMF vol for the current ($t=0$) spot value of $S_0$ can then be rewritten as $$ \hat{\sigma}_{F_T T}(S_0) := \sigma(f(S_0),T; S_0) $$ where the forward price verifies $$F_T := f(S_0) = S_0 \exp((r-q-u)T)$$

Using these notations, the second term involved in the definition of the SSR can be readily evaluated as $$ \frac{ d \hat{\sigma}_{F_T T}(S_0) }{d S_0} = \lim_{\epsilon \to 0} \frac{\color{blue}{\sigma(f(S_0+\epsilon),T; S_0+\epsilon)}- \color{green}{\sigma(f(S_0),T ; S_0)} }{ \epsilon } $$

The stickiness assumptions allow us to relate the smile after a spot move $\color{blue}{\sigma(\cdot,T;S_0+\epsilon)}$ to the original smile $\color{green}{\sigma(\cdot,T;S_0)}$. In particular

Sticky strike rule

For all $K>0$ we by definition need to have $$\color{blue}{\sigma(K,T;S_0+\epsilon)} = \color{green}{\sigma(K,T;S_0)}$$ such that we can successively write \begin{align} \frac{ d \hat{\sigma}_{F_T T}(S_0) }{d S_0} &= \lim_{\epsilon \to 0} \frac{\color{blue}{\sigma(f(S_0+\epsilon),T; S_0+\epsilon)} - \color{green}{\sigma(f(S_0),T ; S_0)} }{ \epsilon } \\ &= \lim_{\epsilon \to 0} \frac{\color{green}{\sigma(f(S_0+\epsilon),T; S_0)} - \color{green}{\sigma(f(S_0),T ; S_0)} }{ \epsilon } \\ &= \frac{\partial \sigma(K,T;S_0)}{\partial K}(f(S_0)) f'(S_0) \end{align} Since from the definition of $f(.)$ we have $$ f'(S_0) = f(S_0)/S_0 $$ then indeed $$ \frac{ d \hat{\sigma}_{F_T T}(S_0) }{d \ln(S_0)} = \frac{\partial \sigma(K,T;S_0)}{\partial \ln(K) }(f(S_0)) = \mathcal{S}_T $$

Sticky moneyness rule

By definition $$ \sigma(K^*,T;S_0+\epsilon) = \sigma(K,T;S_0) $$ if and only if $K^*/(S_0+\epsilon) = K/S_0$.

Hence the equivalent formulation of the rule, for all $K > 0$ $$ \color{blue}{\sigma(K,T,S_0+\epsilon)} = \color{green}{\sigma(K S_0/(S_0+\epsilon), T; S_0)} $$

Applying this now leads to \begin{align} \frac{ d \hat{\sigma}_{F_T T}(S_0) }{d S_0} &= \lim_{\epsilon \to 0} \frac{\color{blue}{\sigma(f(S_0+\epsilon),T; S_0+\epsilon)} - \color{green}{\sigma(f(S_0),T ; S_0)}}{ \epsilon } \\ &= \lim_{\epsilon \to 0} \frac{\color{green}{\sigma(f(S_0+\epsilon) S_0/(S_0+\epsilon),T; S_0)} - \color{green}{\sigma(f(S_0),T ; S_0)} }{ \epsilon } \\ &= \lim_{\epsilon \to 0} \frac{\color{green}{\sigma(f(S_0),T; S_0)} - \color{green}{\sigma(f(S_0),T ; S_0)} }{ \epsilon } = 0 \end{align}

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  • $\begingroup$ Very clear and rigorous $\endgroup$ – John Doe Apr 11 '18 at 17:33
  • $\begingroup$ Glad I could help $\endgroup$ – Quantuple Apr 11 '18 at 17:37
  • $\begingroup$ @Quantuple: Nice answer +1. Would you care to take a look at this question quant.stackexchange.com/q/39202/6686 ? Thanks. $\endgroup$ – Hans Apr 12 '18 at 20:50

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