I started reading SABR model recently. In Wiki page, it states that the SABR model can capture volatility smile in derivative market. However, I do not see how it does so.

  • $\begingroup$ What exactly don't you see? $\endgroup$ – Frido Rolloos May 7 '20 at 8:31
  • $\begingroup$ How does SABR model capture volatility smile? In other words, which part of the model shows that it captures volatility smile? $\endgroup$ – Idonknow May 7 '20 at 8:36
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    $\begingroup$ Take a look at the implied volatility expansion for the SABR model in the same Wiki link you posted. For $\alpha = 0, \beta \neq 1$ you don't have a stochastic volatility model but a CEV model which is a local volatility model. That already gives a smile. For $\beta = 1, \alpha = 0$ you will have no smile since it is then the Black-Scholes model. For $\beta = 1, \alpha \neq 0$ you have a stochastic volatility model and therefore also a smile. $\endgroup$ – Frido Rolloos May 7 '20 at 8:45
  • $\begingroup$ @ilovevolatility Perhaps I should rephrase my question: Does all stochastic volatility models capture volatility smile? $\endgroup$ – Idonknow May 7 '20 at 8:53
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    $\begingroup$ Yes, and I suppose your next question will be why, so I am thinking how to best answer that. I'll let you know when I have a simple answer for that. $\endgroup$ – Frido Rolloos May 7 '20 at 9:01

I am going to try to answer your more general question "Do all SV models generate a smile?" which you put in one of the comments. (Maybe edit also the title of your question if you want, if my answer is satisfactory.)

I will take zero correlation between asset and the volatility process to start with. The generalisation to non-zero correlation is straightforward (but more tedious).

Let $\bar{\sigma}$ denote future realised volatility. If volatility is stochastic, it will have a distribution. The price of a vanilla option is $$ C(S,K) = E[(S_T - K)_+] $$ By conditioning we can write \begin{align} C(S,K) &= E[(S_T - K)_+] \\ &= E[E[(S_T - K)_+] | \bar{\sigma}] \\ &= E[C^{BS}(S,K,\bar{\sigma})] \end{align}

Since the Black-Scholes vanilla option price is monotonic in volatility, we can always find a parameter, call it $\Sigma$, such that $$ C(S,K) = C^{BS}(S,K,\Sigma(K)) $$ whatever the value of $C(S,K)$ may be. Hence, $$ C^{BS}(S,K,\Sigma(K)) = E[C^{BS}(S,K,\bar{\sigma})],\quad \forall K $$ Thus, if volatility is not stochastic, then $$ C^{BS}(S,K,\Sigma(K)) = C^{BS}(S,K,\bar{\sigma}),\quad \forall K $$ But since the Black-Scholes price formula is monotonic in volatility, and $\bar{\sigma}$ does not depend on $K$ this must mean that, $$ \Sigma(K) = \bar{\sigma} \,\, \forall K \Rightarrow \frac{\partial \Sigma}{\partial K} =0 $$ So, if volatility is not stochastic then no smile. Hence not(no smile) implies not(not stochastic).

Hope this makes sense.

EDIT: I should have added one maybe two assumptions to make this "proof" completely airtight since other non SV models can also give a smile, but under the assumption that the assets can only follow pure SV models (potentially with zero volatility of volatility) then the proof is OK.

  • $\begingroup$ Where is the zero correlation between asset and volatility used in your proof? $\endgroup$ – Idonknow May 8 '20 at 12:03
  • $\begingroup$ In the conditioning part. $\endgroup$ – Frido Rolloos May 8 '20 at 12:05

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