Local volatility surface corresponding to the implied volatility surface

In Derman/Kani/Zou paper about local vol they rebuilt a local vol surface from an implied vol surface. Each implied volatility depicted in the surface of the "implied Vol" is the Black-Scholes implied volatility. Bascially the volatility you have to enter into the Black-Scholes formula to have its theoretical option value match the option’s market price.

Derman Paper

Now, in the local vol model, they extract the market’s consensus for future local volatilities σ(S,t), as a function of future index level S and time t, from the spectrum of available options prices as quoted by their implied Black-Scholes volatilities. The model fits a consistent implied tree to these quoted option prices, and then allows the calculation of the fair values and exposures of all (standard and exotic) options, consistent with all the initial liquid options prices.

Question: If I compare the graphs in the paper of the implied vol surface and the local vol surface why is it so different? The local vol should be consistent with the liquid option prices. i.e. Term 1.0, 550 level: implied surface 13.5% vol, local vol surface 18% vol. If there is a liquid strike in the market they should have the same vol, am i right?

Many thanks

You should not expect the local vol to be equal to the implied vol except in the trivial case where both are constant (Black-Scholes model). I haven't read the Derman articles but it is quite clear using Dupire's formula (see Gatheral's book for example).

Local volatility can be computed in terms of call prices using Dupire's formula $$\sigma^2(T,K) = \frac{\frac{\partial C}{\partial T} + (r - q)K \frac{\partial C}{\partial K} + qC}{ \frac{1}{2} K^2 \frac{\partial^2C}{\partial K^2}}$$ To get the relationship with implied volatility, it is better to think in terms of the log-moneyness forward $y = \ln(K/F_0^T)$ rather than strike. Writing $w(T,y) = T\Sigma^2(T,y)$ for the total implied variance, the Black-Scholes formula reads $$C(T,K) = C_{BS}(T,K,\Sigma(T,\ln(K/F)),r,q) = S_0 \left( N(-\frac{y}{\sqrt{w}} + \frac{1}{2}\sqrt{w}) - e^y N(-\frac{y}{\sqrt{w}} - \frac{1}{2}\sqrt{w}) \right)$$ Plugging it into the Dupire formula, one gets
$$\sigma_{\mathrm{Dup}}(T,K)^2 = \frac{ \frac{\partial w}{\partial T} }{1 - \frac{y}{w} \frac{\partial w}{\partial y}+ \frac{1}{4}\left( - \frac{1}{4} + \frac{1}{w} + \frac{y^2}{w^2} \right) \left(\frac{\partial w}{\partial y}\right)^2 + \frac{1}{2}\frac{\partial^2 w}{\partial y^2} }$$

This general formula can be simplified in limit cases:

No skew: in this case, $\sigma_{\mathrm{Dup}}(T)^2 = \frac{\partial w}{\partial T} = \Sigma(T)^2 + 2T\Sigma\frac{\partial \Sigma}{\partial T}$. Local vol is already different from implied vol unless they are both constant (Black-Scholes model).

Short maturities: When $T\to 0$, and the derivatives of the implied vol stay bounded, one can check that
$$\Sigma(0,y) = \frac{1}{\int_0^1 \frac{dt}{\sigma(0,ty)}}$$ so that
$$\frac{\partial \Sigma}{\partial y} (0,0) = \frac{1}{2}\frac{\partial \sigma_{Dup}}{\partial y} (0,0)$$ in other words, at the money, for very short maturities, the implied volatility skew is half the local volatility skew.

Also note that if you start with a twice differentiable implied vol surface, the local vol will only be continuous.

• so if i price an exotic option such as a Down in put i.e., i would take the local vol model which would give me the future price of the option. I would then hedge with a vanilla, which is priced with Black-Scholes? – Catchitup Mar 9 '15 at 0:24
• Your hedging instruments should always be liquid assets. So their price is not given by a model, it is directly observed on the market. You can use the BS formula to imply a volatility from these but you do not use the BS model to price them. Pricing models are for exotics derivatives, hedging instruments prices are inputs given by the market. – AFK May 1 '15 at 21:31

Loosely speaking:

Local volatility is the instantaneous volatility after time T if the spot is S at that time.

Implied volatility is the expected integrated volatility from today up to time T if the spot ends up at S at that time.

• Let's say you have 5 paths leading to given strike and time. Path 1 local vol 11%, Path 2 local vol 12%, path 3 local vol 12.5%, path 4 local vol 13%, path 5 local vol 14%. The implied vol for 0 - 5th path is the avergage, therefore 12.5%. Right? So using 12.5% implied vol would give me the market price of the vanilla for a given strike and time. What if i now price the same vanilla with local vol? – Catchitup Mar 30 '15 at 17:05

Adding as answer as I don't have enough reputation to comment -- there is a typo in AFK's local vol formula, it should be: $$\sigma_{\mathrm{Dup}}(T,K)^2 = \frac{ \frac{\partial w}{\partial T} }{1 - \frac{y}{w} \frac{\partial w}{\partial y}+ \frac{1}{4}\left( - \frac{1}{4} - \frac{1}{w} + \frac{y^2}{w^2} \right) \left(\frac{\partial w}{\partial y}\right)^2 + \frac{1}{2}\frac{\partial^2 w}{\partial y^2} }$$ see Gatheral's "The Volatility Surface", p.13, eq. (1.10) (i.e., the sign of $\frac{1}{w}$ in the denominator is wrong).

• Why is local vol a function of K? Implied vol is a function of S, K and T. Shouldnt local vol a function purely of T and S? – user1559897 May 7 at 15:22
• Within the local volatility model, there is a duality between S and K. The Dupire equation (which is equivalent to the above) is formulated in terms of K, whereas the model dynamics of the local vol model are of course given in terms of S. Hence, in the above formula to calculate the local vol, you put K, but when you simulate using the local vol, you put S. – Philipp Dörsek Aug 19 at 20:36