Why Stochastic Volatility Can Produce an Implied Volatility Smile
Summary
The answer explains a route from stochastic future volatility to strike-dependent implied volatility. It writes a vanilla option value as an expectation over possible future realized volatilities, or equivalently as an average of Black–Scholes prices evaluated at those volatility outcomes. Since a Black–Scholes option price is monotonic in volatility, the answer defines an implied volatility for each strike that matches this averaged price. If volatility is deterministic and does not vary by strike, the implied volatility is flat, so the model produces no smile in that case.
The argument is intended to address the general relationship between stochastic volatility and smiles, rather than derive SABR specifically. The author notes that nonzero correlation can be incorporated with more work and adds that the reasoning assumes asset dynamics are restricted to pure stochastic-volatility models. Other model features can also generate smiles, so the exposition is not a universal claim that stochastic volatility is the only source. It gives intuition, but no calibration or empirical evidence.
Key ideas
- A vanilla option price can be expressed as an average of Black–Scholes prices over possible future volatility outcomes.
- Matching that averaged price with Black–Scholes defines an implied volatility for each strike.
- Deterministic volatility leads to flat implied volatility across strikes under the stated setup.
- The response limits its argument to pure stochastic-volatility dynamics and notes that other models can also generate smiles.
- Nonzero asset-volatility correlation is acknowledged but not worked through in detail.
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Full text
# Do all stochastic volatility models capture volatility smile?
# Do all stochastic volatility models capture volatility smile?
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.
## Answer by user34971 (score 3, accepted)
https://quant.stackexchange.com/a/53967
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.