SABR and smile asymptotics, financial engineering for quants lesson

Two-year caplets on a forward rate of 4 percent are quoted at several strikes, and Black’s formula needs a different volatility at each strike to reproduce the prices. A model for such a row of volatilities has to let the volatility itself move, and its option prices then come from a pricing problem in two state variables with no closed form in general. This lesson specifies the SABR model, derives its generator and a coordinate that removes the elasticity from the diffusion, states the Hagan formula as the asymptotic result it is, and checks that formula at the forward, in the constant volatility case and in price space.

Key terms

SABR model

Definition Let $F_t$ be the simple forward rate of a period that fixes at the expiry $T$ and pays at $T_p \ge T$ , and let $\mathbb{Q}^{T_p}$ be the $T_p$ -forward measure, the measure under which the forward rate of that period is a martingale. Time is measured from the valuation date $0$ , so the time to expiry is $T$ . The SABR model specifies

\begin{gathered} dF_t = a_t F_t^{\beta}\, dW_t, \\[0.4em] da_t = \nu\, a_t\, dZ_t, \\[0.4em] d[W, Z]_t = \rho\, dt, \end{gathered}

where $W$ and $Z$ are $\mathbb{Q}^{T_p}$ -Brownian motions, $a_t$ is the volatility scale with $a_0 > 0$ , the elasticity parameter satisfies $0 \le \beta \le 1$ , the volatility of volatility satisfies $\nu \ge 0$ , and the correlation satisfies $-1 \le \rho \le 1$ , with $-1 < \rho < 1$ needed for the smile formula below. For a swap rate the same equations are written under the annuity measure $\mathbb{Q}^{A}$ . The name comes from stochastic alpha, beta and rho, where alpha is the scale written here as $a$ . A complete specification also fixes where $F$ lives. For $\beta = 1$ the forward stays positive, for $0 < \beta < 1$ it can reach zero and a rule at zero, usually absorption, is part of the model, and for $\beta = 0$ it can become negative. With no drift term $F$ is a local martingale. When it stays nonnegative it is a supermartingale, so $\mathbb{E}^{\mathbb{Q}^{T_p}}[F_T] \le F_0$ . A consistent model needs equality, because the forward rate is a true martingale under its own forward measure, and whether equality holds depends on $\beta$ , $\rho$ and the rule at zero, which this lesson does not settle. A caplet with notional $N$ , accrual fraction $\delta$ and strike $K$ is worth

\begin{aligned} &N\delta\, P(0,T_p) \\[0.4em] &\quad \times \mathbb{E}^{\mathbb{Q}^{T_p}}\bigl[(F_T - K)^{+}\bigr], \end{aligned}

and the parameters, fitted to such prices, describe the dynamics under $\mathbb{Q}^{T_p}$ , not under $\mathbb{P}$ .

In plain words SABR lets the volatility of a forward rate wander as a process of its own, and it lets the size of the forward’s moves depend on the level of the forward through the power $\beta$ . Four numbers, $a_0$ , $\beta$ , $\nu$ and $\rho$ , describe the smile of one expiry.

Example The constructed example of this lesson uses $F_0 = 0.04$ , $a_0 = 0.2$ , $\beta = 1$ , $\nu = 0.4$ , $\rho = -0.3$ and $T = 2$ years. With $\beta = 1$ the instantaneous lognormal volatility of the forward is $a_t$ itself, $0.2$ per square root of a year at the start, and its instantaneous normal volatility is $a_0 F_0 = 0.008$ , or $80$ bp per square root of a year. The variance rates at the start are $a_0^2 F_0^2 = 0.000064$ for the forward and $\nu^2 a_0^2 = 0.0064$ for the scale, and the covariance rate is $\rho\nu a_0^2 F_0 = -0.000192$ , which is $-0.3$ times the product $0.008 \cdot 0.08$ of the two volatilities. The negative correlation means that, under $\mathbb{Q}^{T_p}$ , a fall in the forward tends to come with a rise in its volatility scale.

Elasticity parameter

Definition The elasticity parameter $\beta$ sets how the diffusion coefficient $aF^{\beta}$ of the forward depends on its level. Its elasticity with respect to $F$ is

\frac{d \ln (aF^{\beta})}{d \ln F} = \beta,

so the instantaneous lognormal volatility of $F$ is $aF^{\beta - 1}$ and its instantaneous normal volatility is $aF^{\beta}$ . With $\beta = 1$ the scale $a$ is a proportional volatility per square root of a year. With $\beta = 0$ it is a normal volatility in rate units per square root of a year. For $0 < \beta < 1$ it is measured in rate units to the power $1 - \beta$ per square root of a year, so fitted values of $a_0$ for different $\beta$ have different units and cannot be compared directly. Rates that can be negative are handled by a shifted SABR model, which applies the dynamics to $F + s$ for a fixed shift $s > 0$ and evaluates the formulas at $K + s$ . For $0 \le \beta < 1$ and $F > 0$ the natural coordinate is the antiderivative of $F^{-\beta}$ ,

y(F) = \frac{F^{1 - \beta}}{1 - \beta}.

Its derivatives are $y'(F) = F^{-\beta}$ and $y''(F) = -\beta F^{-\beta - 1}$ , and Itô’s formula with $d[F]_t = a_t^2 F_t^{2\beta}\, dt$ gives, while $F_t > 0$ ,

\begin{aligned} dy_t &= a_t\, dW_t \\[0.4em] &\quad - \tfrac12 \beta\, a_t^2 F_t^{\beta - 1}\, dt. \end{aligned}

For $\beta = 1$ the coordinate is $y = \ln F$ , and $dy_t = a_t\, dW_t - \tfrac12 a_t^2\, dt$ . The change of variables removes the level from the diffusion coefficient and leaves a drift from the curvature of $y$ . It explains the geometry behind the smile formula. It does not solve the pricing problem in the two variables $F$ and $a$ .

In plain words The parameter $\beta$ decides what stays fixed when the forward moves. With $\beta = 1$ the percentage volatility stays fixed, with $\beta = 0$ the volatility in basis points stays fixed, and values in between sit between these two cases. In the coordinate $y$ the forward moves with volatility $a$ alone.

Example For three choices of $\beta$ , with values of $a$ that give the same lognormal volatility $0.2$ at $F = 0.04$ , the table lists the instantaneous lognormal volatility $aF^{\beta - 1}$ and normal volatility $aF^{\beta}$ at $F = 0.04$ and then at $F = 0.08$ .

$\beta$	$a$	Lognormal	Normal
$0$	$0.008$	$0.2$ and $0.1$	$0.008$ and $0.008$
$0.5$	$0.04$	$0.2$ and $0.141421$	$0.008$ and $0.011314$
$1$	$0.2$	$0.2$ and $0.2$	$0.008$ and $0.016$

For $\beta = 0.5$ and $a = 0.04$ the coordinate is $y = 2\sqrt{F}$ , so $y_0 = 0.4$ , and the initial drift of $y$ is $-\tfrac14 \cdot 0.0016 \cdot 5 = -0.002$ per year. For $\beta = 0$ and $\nu = 0$ the forward is Gaussian, and with $F_0 = 0.04$ , $a = 0.008$ and $T = 2$ it ends below zero with probability $\Phi(-0.04/(0.008\sqrt{2})) = \Phi(-3.5355) \approx 0.0002$ under $\mathbb{Q}^{T_p}$ .

Volatility of volatility

Definition The volatility of volatility $\nu \ge 0$ is the lognormal volatility of the scale, per square root of a year. The scale equation has the exact solution

a_t = a_0 \exp\bigl(\nu Z_t - \tfrac12 \nu^2 t\bigr),

so $a_t > 0$ , the logarithm $\ln a_t$ is normal with mean $\ln a_0 - \tfrac12 \nu^2 t$ and variance $\nu^2 t$ under $\mathbb{Q}^{T_p}$ , and $a$ is a true martingale with $\mathbb{E}^{\mathbb{Q}^{T_p}}[a_t] = a_0$ . This alone does not give the joint law of $F_T$ and $a_T$ . Together with $\rho$ the parameter enters the generator of the pair. The quadratic variations are $d[F]_t = a^2 F^{2\beta}\, dt$ , $d[a]_t = \nu^2 a^2\, dt$ and $d[F, a]_t = \rho\nu a^2 F^{\beta}\, dt$ . For a function $u(t, F, a)$ that is once continuously differentiable in $t$ and twice in $(F, a)$ , Itô’s formula gives

\begin{aligned} du &= (\partial_t u + \mathcal{L}u)\, dt \\[0.4em] &\quad + u_F\, aF^{\beta}\, dW \\[0.4em] &\quad + u_a\, \nu a\, dZ, \end{aligned}

where

\begin{aligned} \mathcal{L}u &= \tfrac12 a^2 F^{2\beta} u_{FF} \\[0.4em] &\quad + \rho\nu a^2 F^{\beta} u_{Fa} \\[0.4em] &\quad + \tfrac12 \nu^2 a^2 u_{aa}. \end{aligned}

The two mixed second-order terms of Itô’s formula combine into the single covariance term. If $u$ solves $\partial_t u + \mathcal{L}u = 0$ with $u(T, F, a) = g(F)$ and the local martingale $u(t, F_t, a_t)$ is a true martingale, then $u(t, F_t, a_t) = \mathbb{E}^{\mathbb{Q}^{T_p}}[g(F_T) \mid \mathcal{F}_t]$ , the undiscounted price of the payoff $g(F_T)$ . With $\nu = 0$ the scale is constant, and SABR reduces to the constant elasticity of variance model, or to Black’s model when $\beta = 1$ .

In plain words The volatility of volatility says how fast the volatility scale itself moves, and $\rho$ says whether it tends to rise when the forward falls. The covariance term is what makes the two move together, and dropping it gives a different model.

Example With $a_0 = 0.2$ , $\nu = 0.4$ and $T = 2$ , the logarithm $\ln a_2$ has standard deviation $0.4\sqrt{2} \approx 0.565685$ . The mean of $a_2$ stays $0.2$ , but its median is $0.2e^{-0.16} \approx 0.170429$ , and its central $90$ percent interval runs from about $0.0672$ to $0.4322$ . The probability that the scale ends above its start is

\Phi\Bigl(-\frac{0.16}{0.565685}\Bigr) \approx 0.388649.

Most paths end with a smaller scale, and a minority of large values keeps the mean at $0.2$ . These are probabilities under $\mathbb{Q}^{T_p}$ , which are pricing weights and not forecasts.

Implied volatility smile

Definition Fix the expiry $T > 0$ and the forward $F_0 > 0$ . For a strike $K > 0$ and $\sigma > 0$ the undiscounted Black call price is

\begin{aligned} c_B(K, \sigma) &= F_0\Phi(d_1) \\[0.4em] &\quad - K\Phi(d_2), \end{aligned}

where

\begin{aligned} d_1 &= \frac{\ln(F_0/K)}{\sigma\sqrt{T}} + \tfrac12 \sigma\sqrt{T}, \\[0.4em] d_2 &= d_1 - \sigma\sqrt{T}, \end{aligned}

and a caplet is quoted as $N\delta P(0,T_p)\, c_B(K, \sigma)$ . For fixed $K$ the map $\sigma \mapsto c_B(K, \sigma)$ is continuous and strictly increasing, with limit $(F_0 - K)^{+}$ as $\sigma \downarrow 0$ and limit $F_0$ as $\sigma \to \infty$ . So every undiscounted price $c(K)$ with $(F_0 - K)^{+} < c(K) < F_0$ has exactly one Black implied volatility $\sigma_B(K) > 0$ with $c_B(K, \sigma_B(K)) = c(K)$ . The implied volatility smile at expiry $T$ is the function $K \mapsto \sigma_B(K)$ on the strikes where it is defined. It is flat when one Black volatility prices every strike, and it is called a skew when it mainly falls or rises with the strike. A normal implied volatility, found by inverting the Bachelier formula instead, is another coordinate for the same prices.

In plain words The smile writes a row of option prices as one Black volatility per strike. If the forward really were lognormal with one fixed volatility under $\mathbb{Q}^{T_p}$ , the line would be flat, and every other distribution of $F_T$ bends or tilts it. The smile describes prices. It is not a forecast of realized volatility.

Example The normal model $dF_t = \sigma_N\, dW_t$ is SABR with $\beta = 0$ and $\nu = 0$ . Take $F_0 = 0.04$ , $\sigma_N = 0.008$ and $T = 2$ , the same normal volatility of $80$ bp per square root of a year as in the constructed example. Exact Bachelier prices, inverted with Black’s formula, give the following smile.

Strike $K$	Price $c(K)$	$\sigma_B(K)$
$0.03$	$0.0111702$	$0.231170$
$0.035$	$0.0074473$	$0.214469$
$0.04$	$0.0045135$	$0.200671$
$0.045$	$0.0024473$	$0.189014$
$0.05$	$0.0011702$	$0.178991$

The smile falls with the strike although the model has no stochastic volatility. A fixed volatility in basis points is close to a lognormal volatility of $\sigma_N/\sqrt{F_0 K}$ , which is larger at low strikes. With $\beta = 1$ and $\nu = 0$ , in contrast, the model is Black’s model with volatility $a_0$ , and its smile is flat at $a_0$ .

Hagan formula

Definition Assume the unshifted SABR model with $F = F_0 > 0$ , $K > 0$ , $T > 0$ , $a_0 > 0$ , $0 \le \beta \le 1$ , $\nu \ge 0$ and $-1 < \rho < 1$ . Set

\begin{gathered} \ell = \ln(F/K), \\[0.4em] m = (FK)^{(1 - \beta)/2}, \\[0.4em] z = \frac{\nu}{a_0}\, m\, \ell, \end{gathered}

and

\begin{gathered} s(z) = \sqrt{1 - 2\rho z + z^2}, \\[0.4em] x(z) = \ln \frac{s(z) + z - \rho}{1 - \rho}. \end{gathered}

The Hagan formula approximates the Black implied volatility of SABR prices by

\begin{aligned} \sigma_B(K) &\approx \frac{a_0}{m\, b(\ell)} \cdot \frac{z}{x(z)} \\[0.4em] &\quad \times (1 + HT), \end{aligned}

with

\begin{aligned} b(\ell) &= 1 + \frac{(1 - \beta)^2 \ell^2}{24} \\[0.4em] &\quad + \frac{(1 - \beta)^4 \ell^4}{1920} \end{aligned}

and

\begin{aligned} H &= \frac{(1 - \beta)^2 a_0^2}{24\, m^2} \\[0.4em] &\quad + \frac{\rho\beta\nu a_0}{4m} \\[0.4em] &\quad + \frac{(2 - 3\rho^2)\nu^2}{24}. \end{aligned}

The formula comes from a singular perturbation expansion that treats the volatility parameters as small and keeps the first correction in $T$ . This lesson states it as that asymptotic result and does not derive it. It is not an exact SABR price, and no bound on its error is given here. Because $(1 - 2\rho z + z^2) - (z - \rho)^2 = 1 - \rho^2 > 0$ , the root $s(z)$ exceeds $\lvert z - \rho \rvert$ , so the ratio inside the logarithm is positive for every real $z$ . Moreover $x(0) = 0$ and $x'(z) = 1/s(z) > 0$ , so $z/x(z) > 0$ for $z \ne 0$ , and the factor is read as $1$ at $z = 0$ , which covers $K = F$ and the case $\nu = 0$ . Every other factor except $1 + HT$ is positive, so the output is positive exactly when $1 + HT > 0$ . The output is a volatility to insert into Black’s formula, and the prices it produces need not satisfy the no-arbitrage conditions of call prices, such as convexity in the strike or consistency across expiries. Those conditions have to be checked in price space.

In plain words The formula is a shortcut from four parameters to a whole smile without solving the pricing equation. It approximates the model, so when its output misbehaves, the approximation has failed, not the model.

Example In the constructed example $\beta = 1$ , so $m = 1$ , $b(\ell) = 1$ and $H = \rho\nu a_0/4 + (2 - 3\rho^2)\nu^2/24 = -0.006 + 0.0115333 = 0.0055333$ . At $K = 0.05$ we get $\ell = \ln 0.8 \approx -0.223144$ , $z = 2\ell \approx -0.446287$ , $s(z) \approx 0.965091$ and $x(z) \approx -0.462275$ , so

\begin{aligned} \sigma_B(0.05) &\approx 0.2 \cdot 0.965414 \\[0.4em] &\quad \times 1.011067 \\[0.4em] &\approx 0.195220. \end{aligned}

For the normal model of the previous card, the formula with $\beta = 0$ , $\nu = 0$ and $a_0 = 0.008$ gives $0.2006667$ at the forward, against the exact $0.2006714$ . Now keep every constructed input but take $T = 20$ . The formula gives about $0.662710$ , $0.584839$ and $0.537875$ at the strikes $0.001$ , $0.002$ and $0.003$ , and Black’s formula turns these into undiscounted call prices of about $0.0394657$ , $0.0388457$ and $0.0381976$ . The butterfly that buys the two outer calls and sells two middle calls has a nonnegative payoff that is positive when $0.001 < F_T < 0.003$ , yet these prices give it the value

\begin{aligned} &c(0.001) - 2c(0.002) \\[0.4em] &\quad + c(0.003) \approx -0.0000281, \end{aligned}

which is negative. Checking second differences of the formula’s call prices at strikes spaced $0.0002$ apart finds negative values at every strike from $0.0002$ to $0.0106$ when $T = 20$ , and at none of the strikes up to about $0.2$ when $T = 2$ .

At-the-money limit

Definition As $K \to F$ , both $\ell$ and $z$ tend to $0$ , so $z$ and $x(z)$ both vanish. Since $x(0) = 0$ and $x'(0) = 1/s(0) = 1$ ,

\lim_{z \to 0} \frac{z}{x(z)} = 1,

and the apparent singularity at the forward is removable. With $b(0) = 1$ and $m = F^{1 - \beta}$ at $K = F$ , the at-the-money limit of the Hagan formula is

\sigma_{\mathrm{ATM}} \approx \frac{a_0}{F^{1 - \beta}}\,(1 + H_0 T),

where

\begin{aligned} H_0 &= \frac{(1 - \beta)^2 a_0^2}{24 F^{2 - 2\beta}} \\[0.4em] &\quad + \frac{\rho\beta\nu a_0}{4 F^{1 - \beta}} \\[0.4em] &\quad + \frac{(2 - 3\rho^2)\nu^2}{24}. \end{aligned}

For $\beta = 1$ this is $\sigma_{\mathrm{ATM}} \approx a_0(1 + H_0 T)$ with $H_0 = \rho\nu a_0/4 + (2 - 3\rho^2)\nu^2/24$ , which does not depend on $F$ . With $\nu = 0$ and $\beta = 1$ the formula returns $a_0$ at every strike, and this is exact, because the model is then Black’s model with volatility $a_0$ . That limit checks both the removable quotient and the time correction. Near $z = 0$ the quotient has the Taylor expansion

\begin{aligned} \frac{z}{x(z)} &= 1 - \frac{\rho z}{2} \\[0.4em] &\quad + \frac{(2 - 3\rho^2) z^2}{12} + O(z^3). \end{aligned}

For $\beta = 1$ , where $z = (\nu/a_0)\ln(F/K)$ , the formula therefore has slope $\rho\nu(1 + H_0 T)/2$ with respect to $\ln K$ at the forward, and second derivative $(2 - 3\rho^2)\nu^2(1 + H_0 T)/(6a_0)$ . These describe the formula, which is the approximation, not the exact smile of the model.

In plain words At the forward the quotient $z/x(z)$ is a zero over zero whose limit is $1$ , so the at-the-money value is the leading volatility times a small time correction. Near the forward, a negative correlation tilts the formula’s smile down toward higher strikes, and, when $\lvert \rho \rvert$ is below $\sqrt{2/3} \approx 0.8165$ , the volatility of volatility bends it upward.

Example In the constructed example $H_0 = 0.0055333$ , so $H_0 T = 0.0110667$ and

\begin{aligned} \sigma_{\mathrm{ATM}} &\approx 0.2 \cdot 1.0110667 \\[0.4em] &\approx 0.2022133. \end{aligned}

The slope in $\ln K$ is $-0.3 \cdot 0.4 \cdot 1.0110667/2 \approx -0.060664$ and the second derivative is about $0.233219$ . At $K = 0.044$ , where $\ln(K/F) \approx 0.095310$ , the quadratic estimate is $0.197491$ , close to the value $0.197562$ of the full formula. These are outputs of the asymptotic formula for constructed inputs, not exact SABR prices or market quotes.

Check yourself

For $\beta = 1$ , $a_0 = 0.3$ , $\nu = 0.6$ , $\rho = -0.5$ and $T = 1$ , find the at-the-money value of the Hagan formula. Is it above or below $a_0$ , and what does the formula return at the forward when $\nu = 0$ ?

Show the answer

The correlation term is $\rho\nu a_0/4 = -0.5 \cdot 0.6 \cdot 0.3/4 = -0.0225$ , and the volatility of volatility term is $(2 - 0.75) \cdot 0.36/24 = 0.01875$ . So $H_0 = -0.00375$ and

\begin{aligned} \sigma_{\mathrm{ATM}} &\approx 0.3\,(1 - 0.00375) \\[0.4em] &= 0.298875. \end{aligned}

The value lies below $a_0$ , because the negative correlation term outweighs the positive volatility of volatility term. With $\nu = 0$ both terms vanish and the formula gives exactly $0.3$ , which is also the exact implied volatility, since the model is then Black’s model with volatility $0.3$ . For $\nu = 0.6$ the number $0.298875$ is an approximation of the model’s implied volatility.

Two fits of the same at-the-money quote use $\beta = 1$ with $a_0 = 0.2$ and $\beta = 0.5$ with $a_0 = 0.04$ , both with $\nu = 0.4$ , $\rho = -0.3$ , $T = 2$ and $F = 0.04$ . Compare their at-the-money values. Does the larger $a_0$ mean the first fit is more volatile, and what happens to each at-the-money value if the forward falls to $0.03$ with the parameters held fixed?

Show the answer

For $\beta = 1$ the value is $0.2022133$ , as in the constructed example. For $\beta = 0.5$ the leading factor is $0.04/\sqrt{0.04} = 0.2$ , and

\begin{aligned} H_0 &= 0.0004167 - 0.003 \\[0.4em] &\quad + 0.0115333 \\[0.4em] &= 0.00895, \end{aligned}

so $\sigma_{\mathrm{ATM}} \approx 0.2 \cdot 1.0179 = 0.20358$ . The two fits nearly agree. The values of $a_0$ differ by a factor of five only because their units differ, rate units to the power $0.5$ per square root of a year against a pure number per square root of a year, and both give the instantaneous lognormal volatility $a_0 F^{\beta - 1} = 0.2$ . If the forward falls to $0.03$ , the $\beta = 1$ value stays $0.2022133$ , since it does not depend on $F$ . For $\beta = 0.5$ the leading factor rises to $0.04/\sqrt{0.03} \approx 0.230940$ and $H_0 \approx 0.0086248$ , so the value becomes about $0.234924$ . A single smile often pins down $\beta$ only weakly, because $\beta$ and $\rho$ both tilt it, but the two fits imply different behavior when the forward moves, and so different hedges.

For $\beta = 1$ , $a_0 = 0.3$ , $\nu = 1$ and $\rho = -0.9$ , evaluate the at-the-money Hagan value at $T = 10$ and at $T = 15$ . A colleague concludes from the second value that the SABR model has no sensible price for the at-the-money caplet. What is wrong with this conclusion?

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Here

\begin{aligned} H_0 &= -\frac{0.9 \cdot 0.3}{4} + \frac{2 - 2.43}{24} \\[0.4em] &\approx -0.0675 - 0.0179167 \\[0.4em] &= -0.0854167. \end{aligned}

At $T = 10$ the factor $1 + H_0 T$ is about $0.145833$ , which gives $0.04375$ . At $T = 15$ it is $-0.28125$ , which gives $-0.084375$ , a negative number that is not a volatility. The sign changes at $T = 1/0.0854167 \approx 11.71$ years. The failure belongs to the approximation, not to the model. The time correction is the first term of an expansion in quantities such as $\nu^2 T$ , which here equals $15$ . In the model with $\beta = 1$ the forward stays positive, it is a nonnegative local martingale and so has $\mathbb{E}^{\mathbb{Q}^{T_p}}[F_T] \le F_0$ , and given the path of the scale its logarithm is Gaussian with variance $(1 - \rho^2)\int_0^T a_t^2\, dt > 0$ . So the at-the-money call has an undiscounted price strictly between $0$ and $F_0$ , and that price has a positive Black implied volatility. It has to be computed another way, for example by simulation or by solving the pricing equation with the generator. The positive value at $T = 10$ is not evidence of accuracy either, because the formula carries no error bound.

SABR and smile asymptotics

Key terms

SABR model

Elasticity parameter

Volatility of volatility

Implied volatility smile

Hagan formula

At-the-money limit

Check yourself

How this lesson connects

Builds on