# Is the local volatility linear if smile is linear?

Assume $dS = S_t\sigma(S_t,t)dW$.

Given a implied volatility smile which is linear in, say, $(K - S_0)$, (we know its intercept and slope), we wish to calibrate $\sigma(S_t, t)$ to it. Will it too be linear? If so, with what intercept and slope?

If not, what will it be?

• AFK's answer here might be useful: quant.stackexchange.com/questions/16898/…
– bcf
Mar 28, 2017 at 1:43
• See @Quantuple's answer: quant.stackexchange.com/questions/33263/…. It appears as though his approximation for the volatility smile, $\Sigma$, is assumed to be linear function of $ln(S_0/K)$ over the integrated volatility since the integrated volatiity is a single quantity. However, there is no assumption that local volalities are linear functions of $ln(S_0/K)$. Mar 28, 2017 at 19:15

To simplify the problem, let us consider normal local volatilities $$\sigma \left ( S_t, t \right)$$ and implied volatilities $$\sigma_i \left ( K, T \right)$$ such that the model is:

$$dS_t = \sigma \left ( S_t, t \right) dW$$

(no rate, repo, dividends, etc.) and $$\sigma_i \left ( K, T \right)$$ is the normal volatility input into Bachelier's formula to give the call price $$C \left ( K, T \right)$$ (Bachelier is the normal version of Black-Scholes with $$dS_t = \sigma_i dW$$, see for instance the formula in my lecture notes, slide 81. A normal context considerably simplifies reasoning about volatility compared to log-normal Black-Scholes, and what follows also applies to log-normal volatilities with minor adjustements).

Dupire's well known formula expresses the local volatility $$\sigma$$ as a function of call prices $$C$$ (see slide 193 for a refresher):

$$\sigma^2 \left( x, t \right) = \frac{2C_t \left( x, t \right)}{C_{xx} \left( x, t \right)}$$

(where the subscripts denote derivatives). Since $$C \left ( K, T \right)$$ is given by Bachelier's formula applied in $$\sigma_i \left ( K, T \right)$$, we can compute the partial derivatives $$C_T$$ and $$C_{KK}$$ (with particular care that the implied volatility is itself a function of the strike and maturity) to find the (perhaps less well known) equivalent Dupire's formula expressing local volaitlities directly as a function of implied volatilities. The result is (see slide 195 for the log-normal version):

$$\sigma^2 \left( x,t \right) = \frac { \sigma_i^2 \left( x,t \right) + 2 t \sigma_i \left( x,t \right) \frac{\partial \sigma_i}{\partial t} \left( x,t \right)} { \left( 1 + \frac{s-x}{\sigma_i \left( x,t \right)} \frac{\partial \sigma_i}{\partial x} \left( x,t \right) \right)^2 + t \sigma_i \left( x,t \right) \frac{\partial ^2 \sigma_i}{\partial x^2} \left( x,t \right)}$$

Now I can answer your question precisely, assuming that implied volatilities are stationary (same for all expiries $$T$$) and linear, of the form $$\sigma_i \left( K,T \right) = \sigma_i \left( K \right) = \alpha + \beta \left( K - S\right)$$ ($$\alpha$$ is the ATM volatility and $$\beta$$ is its slope, commonly called 'skew'), then the formula simplifies, since the first derivative of the implied volatility to expiry, and its second derivative to strike both disappear, and we find: $$\sigma \left( x \right) = \frac{ \sigma_i \left( x \right) }{ 1 + \beta \frac{s-x}{\sigma_i\left(x\right)}} = \frac{ \alpha + \beta \left( x - s\right) }{ 1 - \beta \frac{x-s}{\alpha + \beta \left( x - s\right)}} = \frac{\left( \alpha + \beta \left( x - s \right) \right)^2 }{\alpha} = \frac{\sigma_i^2\left(x\right)}{\sigma_i\left(s\right)}$$

This formula, which I emphasize is exact, clearly shows that local volatility is quadratic when implied volatility is linear. See the chart below with ATM = 15, skew = -1 vol (in normal terms) every 10 points in strike. Note that the local vol is ~approximately linear after all, although we know for a fact that it is, in reality, quadratic:

In addition, it illustrates some enlightening phenomena:

• $$\sigma\left(s\right) = \sigma_i\left(s\right) = \alpha$$ the local and implied volatilities coincide at the money.

• The slope of the local volatility is $$\sigma'\left(x\right) = 2 \frac{\sigma_i\left(x\right)}{\sigma_i\left(s\right)} \sigma_i'\left(x\right) = 2 \frac{ \alpha + \beta \left( x - s \right) }{\alpha} \beta$$, in particular $$\sigma'\left(s\right) = 2 \sigma_i'\left(s\right) = 2 \beta$$: at the money, the slope of local volatility is twice the slope of implied volatility (twice the skew).

• The second derivative of the local volatility $$\sigma''\left(x\right) = \frac{2 \beta^2}{\alpha}$$. Note that it is constant, since local volatility is exactly quadratic. This is also typically a small number. In our example with ATM = 15 and skew = -0.1, the convexity is around 0.0013, which explains why the local volatility is ~almost linear, in our example, and typically, when implied volatility is linear.

• The local volatility is also stationary (same for any time $$t$$), although this is no longer true when implied volatility is not stationary (obviously) or not linear (as can be seen from the term in the second derivative on the general formula, which has a $$t$$).

Finally, here is the answer to the question, expressed in its clearest form:

$${\sigma _i}\left( K \right) = ATM + \beta \left( {K - {S_0}} \right) \Rightarrow \sigma \left( S \right) = ATM + 2\beta \left( {S - {S_0}} \right) + \frac{{{\beta ^2}{{\left( {S - {S_0}} \right)}^2}}}{{ATM}}$$

This result is exact in normal volatility terms and ~mostly exact with log-normal volatilities a la Black-Scholes (in which case we must disregard the convexity bias in the terms $$d_1$$ and $$d_2$$). It helps to do the maths in a normal context and rely on what the intuitions apply, although approximately, in a log-normal context too. Although academic papers tend to be written in log-normal terms (as I did in my lecture notes), practitioners tend to reason in normal terms. Note that Bruno Dupire originally derived his famous formula in normal terms too.

Thanks to Brian Huge for meaningful insights.

• Hi Antoine, it seems you have another account: quant.stackexchange.com/users/17606/antoine-conze Could you merge them here? Jan 24, 2019 at 20:22
• No this is not mine. Jan 24, 2019 at 22:11
• Yes, I concur :) Jan 25, 2019 at 13:03
• Then it’s all good :) Jan 25, 2019 at 21:10
• Excellent answer Jan 27, 2019 at 14:21