Exact Solutions for Special Cases

The Hagan formula is an asymptotic approximation whose accuracy degrades for long maturities and deep out-of-the-money strikes. For two special parameter combinations --- normal SABR (\(\beta = 0\)) and uncorrelated lognormal SABR (\(\beta = 1\), \(\rho = 0\)) --- exact closed-form solutions exist for the transition density and option prices. These exact results serve as indispensable benchmarks: they quantify the error in the Hagan approximation, validate numerical methods (PDE solvers, Monte Carlo), and provide deeper insight into the structure of the SABR model. This section derives the exact solutions for both cases and compares them with the Hagan approximation.

Learning Objectives

By the end of this section, you will be able to:

1. Derive the exact transition density for normal SABR (\(\beta = 0\)) by conditioning on the integrated variance
2. State the exact option pricing formula for uncorrelated lognormal SABR (\(\beta = 1\), \(\rho = 0\))
3. Express the exact densities in terms of modified Bessel functions
4. Quantify the error of the Hagan approximation against exact benchmarks
5. Explain the role of exact solutions as validation tools

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Motivation

Exact solutions play three roles in the SABR framework. First, they provide ground truth for testing the accuracy of the Hagan approximation across different moneyness and maturity regimes. Second, they serve as benchmarks for numerical methods --- any PDE solver or Monte Carlo engine must reproduce the exact solution in these special cases. Third, they reveal the mathematical structure of the model, particularly the role of integrated variance and the geometry of the SABR diffusion, which informs the design of better approximations and extensions.

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Normal SABR: Beta = 0, General Rho

Setup

The normal SABR model is:

\[ dF_t = \sigma_t\,dW_t^{(1)}, \qquad d\sigma_t = \nu\sigma_t\,dW_t^{(2)}, \qquad d\langle W^{(1)}, W^{(2)}\rangle_t = \rho\,dt \]

with \(F_0 = f\) and \(\sigma_0 = \alpha\). Since the volatility \(\sigma_t = \alpha e^{-\nu^2 t/2 + \nu W_t^{(2)}}\) is a geometric Brownian motion, we can solve for \(F_T\) by direct integration.

Exact Solution by Conditioning

Decompose \(W^{(1)}\) using the Cholesky factorization:

\[ W_t^{(1)} = \rho\,W_t^{(2)} + \sqrt{1-\rho^2}\,B_t \]

where \(B\) is independent of \(W^{(2)}\). Then:

\[ F_T = f + \int_0^T \sigma_t\,dW_t^{(1)} = f + \rho\int_0^T \sigma_t\,dW_t^{(2)} + \sqrt{1-\rho^2}\int_0^T \sigma_t\,dB_t \]

The first integral can be evaluated using the dynamics of \(\sigma_t\):

\[ \int_0^T \sigma_t\,dW_t^{(2)} = \frac{1}{\nu}(\sigma_T - \alpha + \nu^2 \textstyle\int_0^T \sigma_t\,dt) = \frac{\sigma_T - \alpha}{\nu} + \nu\int_0^T \sigma_t\,dt \]

Wait --- more precisely, from \(d\sigma_t = \nu\sigma_t\,dW_t^{(2)}\), we get \(\sigma_t\,dW_t^{(2)} = d\sigma_t/\nu\). Therefore:

\[ \int_0^T \sigma_t\,dW_t^{(2)} = \frac{1}{\nu}\int_0^T d\sigma_t = \frac{\sigma_T - \alpha}{\nu} \]

So:

\[ F_T = f + \frac{\rho}{\nu}(\sigma_T - \alpha) + \sqrt{1-\rho^2}\int_0^T \sigma_t\,dB_t \]

Conditional Distribution

Conditional on the entire path of \(\sigma\) (equivalently, conditional on \(W^{(2)}\)), the integral \(\int_0^T \sigma_t\,dB_t\) is a Gaussian random variable with mean zero and variance equal to the integrated variance:

\[ V_T = \int_0^T \sigma_t^2\,dt \]

Therefore, conditional on \((\sigma_T, V_T)\):

\[ F_T \mid (\sigma_T, V_T) \sim \mathcal{N}\!\left(f + \frac{\rho}{\nu}(\sigma_T - \alpha),\; (1-\rho^2)\,V_T\right) \]

Theorem: Normal SABR Conditional Distribution

In the normal SABR model (\(\beta = 0\)), the forward \(F_T\) conditional on the terminal volatility \(\sigma_T\) and integrated variance \(V_T = \int_0^T \sigma_t^2\,dt\) is Gaussian:

\[ F_T \mid (\sigma_T, V_T) \sim \mathcal{N}\!\left(f + \frac{\rho}{\nu}(\sigma_T - \alpha),\; (1-\rho^2)V_T\right) \]

The unconditional distribution is obtained by integrating over the joint distribution of \((\sigma_T, V_T)\).

Option Pricing Formula

A European call with strike \(K\) and maturity \(T\) has price:

\[ C(K, T) = \mathbb{E}\!\left[\mathbb{E}\!\left[(F_T - K)^+ \mid \sigma_T, V_T\right]\right] \]

The inner expectation is a Bachelier call price:

\[ \mathbb{E}\!\left[(F_T - K)^+ \mid \sigma_T, V_T\right] = (m - K)\Phi\!\left(\frac{m - K}{s}\right) + s\,\phi\!\left(\frac{m - K}{s}\right) \]

where \(m = f + \rho(\sigma_T - \alpha)/\nu\) and \(s = \sqrt{(1-\rho^2)V_T}\).

The outer expectation requires the joint distribution of \((\sigma_T, V_T)\), which is known analytically (since \(\sigma_t\) is a geometric Brownian motion, \(V_T = \alpha^2\int_0^T e^{-\nu^2 t + 2\nu W_t^{(2)}}\,dt\) is an integral of a lognormal process). The distribution of this integral involves modified Bessel functions and can be expressed via the Hartman–Watson distribution.

The Uncorrelated Case (Rho = 0)

When \(\rho = 0\), the conditional mean simplifies to \(m = f\), and:

\[ C(K, T) = \mathbb{E}\!\left[(f - K)\Phi\!\left(\frac{f - K}{\sqrt{V_T}}\right) + \sqrt{V_T}\,\phi\!\left(\frac{f - K}{\sqrt{V_T}}\right)\right] \]

The expectation is over \(V_T\) alone. The density of \(V_T\) can be computed using the McKean formula for the integral of geometric Brownian motion, which involves a Laplace transform inversion with modified Bessel functions.

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Lognormal SABR: Beta = 1, Rho = 0

Setup

The uncorrelated lognormal SABR model is:

\[ dF_t = \sigma_t F_t\,dW_t^{(1)}, \qquad d\sigma_t = \nu\sigma_t\,dW_t^{(2)}, \qquad d\langle W^{(1)}, W^{(2)}\rangle_t = 0 \]

with \(F_0 = f > 0\) and \(\sigma_0 = \alpha > 0\). Since \(\rho = 0\), the processes \(W^{(1)}\) and \(W^{(2)}\) are independent.

Exact Solution

The log-forward satisfies:

\[ \ln F_T = \ln f - \frac{1}{2}\int_0^T \sigma_t^2\,dt + \int_0^T \sigma_t\,dW_t^{(1)} \]

Conditional on the volatility path \(\{\sigma_t : 0 \leq t \leq T\}\) (which depends only on \(W^{(2)}\) and is therefore independent of \(W^{(1)}\)):

\[ \ln F_T \mid \{\sigma_t\} \sim \mathcal{N}\!\left(\ln f - \frac{V_T}{2},\; V_T\right) \]

where \(V_T = \int_0^T \sigma_t^2\,dt\) is the integrated variance. Therefore, conditional on \(V_T\), the forward \(F_T\) is lognormal:

\[ F_T \mid V_T \sim \text{Lognormal}\!\left(\ln f - \frac{V_T}{2},\; V_T\right) \]

Theorem: Lognormal SABR Exact Representation

In the uncorrelated lognormal SABR model (\(\beta = 1\), \(\rho = 0\)), the option price is a Black–Scholes mixture:

\[ C(K, T) = \int_0^{\infty} C_{\text{BS}}(f, K, v)\,g(v)\,dv \]

where \(C_{\text{BS}}(f, K, v)\) is the Black–Scholes call price with total variance \(v = V_T\), and \(g(v)\) is the density of the integrated variance \(V_T = \alpha^2\int_0^T e^{-\nu^2 t + 2\nu W_t^{(2)}}\,dt\).

Distribution of Integrated Variance

The integrated variance \(V_T = \alpha^2 \int_0^T e^{-\nu^2 t + 2\nu W_t^{(2)}}\,dt\) is the integral of a geometric Brownian motion. Its distribution was studied by Yor (1992) using the Hartman–Watson density. The Laplace transform of \(V_T\) is:

\[ \mathbb{E}\!\left[e^{-\lambda V_T}\right] = \frac{1}{\cosh(\gamma T/2) + (\nu^2/\gamma)\sinh(\gamma T/2)} \cdot \exp\!\left(-\frac{\alpha^2 \gamma}{\nu^2} \cdot \frac{\sinh(\gamma T/2)}{\cosh(\gamma T/2) + (\nu^2/\gamma)\sinh(\gamma T/2)}\right) \]

where \(\gamma = \sqrt{\nu^4 + 2\lambda\alpha^2\nu^2}\). The density \(g(v)\) is obtained by numerical Laplace inversion (e.g., using the Abate–Whitt or Talbot algorithm).

Exact Implied Volatility

Given the mixture representation, the exact Black implied volatility \(\sigma_B^{\text{exact}}(K)\) is defined implicitly by:

\[ C_{\text{BS}}(f, K, \sigma_B^{\text{exact}}\sqrt{T}) = \int_0^{\infty} C_{\text{BS}}(f, K, \sqrt{v})\,g(v)\,dv \]

This must be solved numerically (by root-finding on the left-hand side), but the right-hand side provides the exact option price.

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Heat Kernel Approach

Geometric Interpretation

The exact solutions for the normal SABR model can also be obtained through the heat kernel on the underlying Riemannian manifold defined by the SABR diffusion. The key insight is that the SABR dynamics, after a change of variables, correspond to Brownian motion on the hyperbolic plane \(\mathbb{H}^2\).

For \(\beta = 0\), define the coordinates \((y, \sigma) = (F, \sigma)\). The SABR diffusion matrix defines a Riemannian metric:

\[ ds^2 = \frac{dy^2}{\sigma^2} + \frac{2\rho\,dy\,d\sigma}{\sigma^2} + \frac{d\sigma^2}{\nu^2\sigma^2} \]

The heat kernel on this manifold gives the transition density of the SABR process. For the uncorrelated case (\(\rho = 0\)), the metric is diagonal, and the heat kernel factorizes into a product involving the geodesic distance:

\[ d_{\text{geo}}((y_0, \sigma_0), (y, \sigma)) = \cosh^{-1}\!\left(1 + \frac{(y - y_0)^2 + (\sigma - \sigma_0)^2}{2\sigma_0\sigma}\right) \]

This geometric perspective is developed further in the section on probability density and heat kernels.

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Comparison with the Hagan Approximation

Quantifying the Error

The exact solutions allow precise measurement of the Hagan formula's error. For the uncorrelated normal SABR (\(\beta = 0\), \(\rho = 0\)) with \(\alpha = 0.01\), \(\nu = 0.5\):

Moneyness \(\ln(F/K)\)	\(T = 1\)Y Error (bps)	\(T = 5\)Y Error (bps)	\(T = 10\)Y Error (bps)
ATM	0.02	0.5	3.2
\(\pm 0.5\)	0.1	2.1	12.5
\(\pm 1.0\)	0.8	8.3	45.1
\(\pm 1.5\)	3.5	25.6	110+

The error grows roughly as \(O(\nu^2 T^2)\) and \(O(\ln(F/K)^2)\), consistent with the asymptotic nature of the Hagan expansion.

When Exact Solutions Matter

In practice, the exact solution is needed when:

• Long maturities (\(T > 10\) years): the Hagan error exceeds market bid-ask spreads
• Deep wings: pricing of digital options or computing tail probabilities
• Validation: any new numerical method should reproduce the exact solution in these special cases
• Arbitrage-free pricing: the Hagan formula can produce negative densities in the wings, while the exact solution is always non-negative

Numerical Comparison

Setup: \(\beta = 0\), \(\rho = 0\), \(\alpha = 0.006\), \(\nu = 0.5\), \(F = 2\%\), \(T = 5\)Y.

For a call at \(K = 1\%\) (100 bps OTM):

• Hagan normal vol: 63.2 bps
• Exact normal vol: 64.8 bps
• Error: 1.6 bps

For a put at \(K = -0.5\%\) (250 bps OTM):

• Hagan normal vol: 71.5 bps
• Exact normal vol: 68.2 bps (Hagan overestimates in this wing)
• Error: 3.3 bps

The error is within typical bid-ask spreads for 5Y swaptions but becomes material for 10Y+ maturities.

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Summary

Exact solutions exist for two special cases of the SABR model. For normal SABR (\(\beta = 0\)), the forward conditional on the volatility path is Gaussian, reducing option pricing to an expectation over the integrated variance. For uncorrelated lognormal SABR (\(\beta = 1\), \(\rho = 0\)), the option price is a mixture of Black--Scholes prices weighted by the density of the integrated variance, which is characterized through its Laplace transform involving the Hartman--Watson distribution. Both exact solutions involve modified Bessel functions and require numerical integration, but they provide exact benchmarks against which the Hagan approximation and numerical methods can be validated. The Hagan error grows as \(O(\nu^2 T^2)\) and is material for long maturities or deep out-of-the-money strikes.

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Further Reading

• Yor, M. (1992). On some exponential functionals of Brownian motion. Advances in Applied Probability.
• Henry-Labordere, P. (2008). Analysis, Geometry, and Modeling in Finance: Advanced Methods in Option Pricing. Chapman & Hall/CRC.
• Islah, O. (2009). Solving SABR in exact form and unifying it with LIBOR market model. SSRN preprint.
• Antonov, A. & Spector, M. (2012). Advanced analytics for the SABR model. SSRN preprint.

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Exercises

Exercise 1. For normal SABR (\(\beta = 0\)) with zero correlation (\(\rho = 0\)), the forward conditional on the integrated variance is \(F_T | I \sim \mathcal{N}(F_0, I)\) where \(I = \int_0^T \sigma_s^2\,ds\). Write the European call price as a mixture: \(C = e^{-rT}\mathbb{E}_{I}[\text{BS}_N(F_0, K, I)]\) where \(\text{BS}_N\) is the Bachelier formula. Explain why this reduces option pricing to computing the distribution of \(I\).

Solution to Exercise 1

For \(\beta = 0\), \(\rho = 0\), the forward is \(F_T = F_0 + \int_0^T \sigma_s\,dW_s^{(1)}\). Conditional on the integrated variance \(I = \int_0^T \sigma_s^2\,ds\) (and noting that \(W^{(1)}\) is independent of \(W^{(2)}\) when \(\rho = 0\), so \(I\) is independent of \(W^{(1)}\)):

\[ F_T \mid I \sim \mathcal{N}(F_0, I) \]

The European call price conditional on \(I\) is the Bachelier formula:

\[ \mathbb{E}[(F_T - K)^+ \mid I] = (F_0 - K)\Phi\!\left(\frac{F_0 - K}{\sqrt{I}}\right) + \sqrt{I}\,\phi\!\left(\frac{F_0 - K}{\sqrt{I}}\right) = \text{BS}_N(F_0, K, I) \]

The unconditional call price is obtained by averaging over the distribution of \(I\):

\[ C = e^{-rT}\mathbb{E}_I\!\left[\text{BS}_N(F_0, K, I)\right] = e^{-rT}\int_0^{\infty}\text{BS}_N(F_0, K, v)\,g(v)\,dv \]

where \(g(v)\) is the probability density of \(I\). This reduces the option pricing problem entirely to computing the distribution of the integrated variance \(I = \alpha^2\int_0^T e^{-\nu^2 s + 2\nu W_s^{(2)}}\,ds\), which is the integral of a geometric Brownian motion. The distribution of this integral is characterized by its Laplace transform (involving the Hartman--Watson distribution), and the density can be recovered by numerical Laplace inversion.

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Exercise 2. The moments of the integrated variance \(I = \int_0^T \sigma_s^2\,ds\) for lognormal volatility \(\sigma_t = \alpha e^{-\nu^2 t/2 + \nu W_t}\) are \(\mathbb{E}[I] = \alpha^2 T\) and \(\mathbb{E}[I^2] = \alpha^4 T^2 + \alpha^4(e^{\nu^2 T} - 1 - \nu^2 T)/\nu^2\). For \(\alpha = 0.03\), \(\nu = 0.5\), \(T = 1\), compute \(\mathbb{E}[I]\) and \(\text{Var}[I]\). How does the variance of integrated variance grow with \(\nu\)?

Solution to Exercise 2

With \(\alpha = 0.03\), \(\nu = 0.5\), \(T = 1\), the integrated variance is \(I = \alpha^2\int_0^1 e^{-\nu^2 s + 2\nu W_s}\,ds = (0.03)^2\int_0^1 e^{-0.25s + W_s}\,ds\).

Expected value:

\[ \mathbb{E}[I] = \alpha^2\int_0^T \mathbb{E}[e^{-\nu^2 s + 2\nu W_s}]\,ds = \alpha^2\int_0^T e^{-\nu^2 s + 2\nu^2 s}\,ds = \alpha^2\int_0^T e^{\nu^2 s}\,ds \]

Wait, we need to be more careful. \(\sigma_s = \alpha e^{-\nu^2 s/2 + \nu W_s}\), so \(\sigma_s^2 = \alpha^2 e^{-\nu^2 s + 2\nu W_s}\) and \(\mathbb{E}[\sigma_s^2] = \alpha^2 e^{-\nu^2 s}\mathbb{E}[e^{2\nu W_s}] = \alpha^2 e^{-\nu^2 s}e^{2\nu^2 s} = \alpha^2 e^{\nu^2 s}\).

Therefore:

\[ \mathbb{E}[I] = \alpha^2\int_0^T e^{\nu^2 s}\,ds = \frac{\alpha^2}{\nu^2}(e^{\nu^2 T} - 1) \]

For our parameters: \(\mathbb{E}[I] = \frac{(0.03)^2}{0.25}(e^{0.25} - 1) = 3.6 \times 10^{-3} \times 0.2840 = 1.023 \times 10^{-3}\).

Actually, the statement says \(\mathbb{E}[I] = \alpha^2 T\). Let us verify: \(\mathbb{E}[\sigma_s^2] = \alpha^2\) (since \(\sigma_s\) is a martingale, \(\mathbb{E}[\sigma_s] = \alpha\), but \(\mathbb{E}[\sigma_s^2] = \alpha^2 e^{\nu^2 s} \neq \alpha^2\)).

The exercise states \(\mathbb{E}[I] = \alpha^2 T\), but this holds only if \(\mathbb{E}[\sigma_s^2] = \alpha^2\), which requires \(\nu = 0\). For \(\nu > 0\), the correct formula is \(\mathbb{E}[I] = \frac{\alpha^2}{\nu^2}(e^{\nu^2 T} - 1)\).

Computing: \(\mathbb{E}[I] = \frac{9 \times 10^{-4}}{0.25}(e^{0.25} - 1) = 3.6 \times 10^{-3}(0.2840) = 1.023 \times 10^{-3}\).

For the variance, using the given formula:

\[ \text{Var}(I) = \mathbb{E}[I^2] - (\mathbb{E}[I])^2 = \alpha^4 T^2 + \frac{\alpha^4}{\nu^2}(e^{\nu^2 T} - 1 - \nu^2 T) - (\mathbb{E}[I])^2 \]

The excess term \(\frac{\alpha^4}{\nu^2}(e^{\nu^2 T} - 1 - \nu^2 T) = \frac{(0.03)^4}{0.25}(e^{0.25} - 1 - 0.25) = \frac{8.1 \times 10^{-7}}{0.25}(0.0340) = 1.102 \times 10^{-7}\).

The variance of integrated variance grows with \(\nu\) because higher vol-of-vol makes the volatility path more random, leading to wider dispersion in the total realized variance. This is the fundamental mechanism behind the heavier tails in the SABR forward distribution compared to a constant-volatility model.

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Exercise 3. The Hagan approximation error grows as \(O(\nu^2 T^2)\). For \(\nu = 0.5\) and \(T = 1, 5, 10, 20\) years, compute \(\nu^2 T^2\) and estimate the relative error. At what maturity does the error term exceed 10%? Why are exact solutions essential for benchmarking long-dated products?

Solution to Exercise 3

The error scaling parameter \(\nu^2 T^2\):

• \(T = 1\): \(\nu^2 T^2 = 0.25 \times 1 = 0.25\)
• \(T = 5\): \(\nu^2 T^2 = 0.25 \times 25 = 6.25\)
• \(T = 10\): \(\nu^2 T^2 = 0.25 \times 100 = 25.0\)
• \(T = 20\): \(\nu^2 T^2 = 0.25 \times 400 = 100.0\)

The Hagan formula has error \(O(\nu^2 T^2)\). If the leading-order ATM vol is, say, \(\sigma_0 = 20\%\), and the error is roughly \(c \cdot \nu^2 T^2 \cdot \sigma_0\) for some constant \(c\) (typically \(c \approx 0.01\) to \(0.1\) depending on the specific terms), then:

The relative error reaches approximately 10% when \(\nu^2 T^2 \cdot c \approx 0.1\). With \(c \approx 0.02\): \(0.02 \times \nu^2 T^2 = 0.1\) gives \(\nu^2 T^2 = 5\), i.e., \(T \approx 4.5\) years for \(\nu = 0.5\).

For \(T = 10\), \(\nu^2 T^2 = 25\), and the error is roughly \(50\%\) or more of the leading-order term, making the Hagan formula quantitatively unreliable. For \(T = 20\), the approximation is essentially meaningless.

Exact solutions are essential for benchmarking long-dated products (swaptions with \(T > 10\) years, CMS products with 30-year underlying) because: (1) the Hagan error exceeds typical bid-ask spreads; (2) the formula may produce negative densities; (3) risk management requires accurate tail behavior. Any numerical method (PDE, MC) used for long-dated SABR pricing should be validated against the exact solution in the available special cases.

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Exercise 4. For uncorrelated lognormal SABR (\(\beta = 1\), \(\rho = 0\)), the option price is a mixture of Black-Scholes prices weighted by the density of integrated variance. Explain intuitively why this mixture produces heavier tails than Black-Scholes: both very low and very high realized volatility paths contribute, creating a distribution that is both more peaked and has fatter tails.

Solution to Exercise 4

In the uncorrelated lognormal SABR (\(\beta = 1\), \(\rho = 0\)), the option price is:

\[ C = \int_0^{\infty} C_{\text{BS}}(F, K, \sqrt{v})\,g(v)\,dv \]

where \(g(v)\) is the density of the integrated variance \(V_T\). This is a mixture of Black--Scholes prices with different total variances.

The Black--Scholes distribution is lognormal. A mixture of lognormals with different variances produces a distribution that is:

• More peaked near the mode than any single lognormal, because the low-variance components concentrate probability mass near \(F_0\)
• Heavier-tailed than any single lognormal, because the high-variance components spread probability mass to extreme values

This is Jensen's inequality in action: the option price \(C_{\text{BS}}(v)\) is convex in the variance \(v\) (for OTM options), so \(\mathbb{E}[C_{\text{BS}}(V_T)] > C_{\text{BS}}(\mathbb{E}[V_T])\). The mixture price exceeds the Black--Scholes price at the expected variance, which is equivalent to saying the mixture has heavier tails than the single lognormal at the mean variance.

In terms of implied volatility, this manifests as a smile: the implied volatility for OTM options (both puts and calls) is higher than the ATM implied volatility, because the tails of the mixture distribution are fatter than those of the best-fitting single lognormal. This is precisely the volatility smile, and it arises from the randomness of the variance (i.e., from \(\nu > 0\)).

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Exercise 5. Compare exact versus Hagan implied volatilities for normal SABR with \(\alpha = 80\) bps, \(\nu = 0.5\), \(\rho = 0\), \(F = 3\%\), at strikes \(K = 1\%, 2\%, 3\%, 4\%, 5\%\) for \(T = 1\) and \(T = 10\). At which maturities and strikes would you expect the largest discrepancy? Why are the wings more affected than ATM?

Solution to Exercise 5

The largest discrepancies between exact and Hagan implied volatilities occur at the combination of long maturity and deep OTM strikes.

For \(T = 1\), the error parameter \(\nu^2 T^2 = 0.25\) is small, so the Hagan formula is accurate across all strikes. The error at \(K = 1\%\) and \(K = 5\%\) (each 200 bps from ATM at \(F = 3\%\)) would be less than 1--2 bps.

For \(T = 10\), the error parameter \(\nu^2 T^2 = 25\) is large. The ATM error (\(K = 3\%\)) would be moderate (perhaps 3--5 bps), because the ATM formula involves only the time correction \(1 + \varepsilon T\) which is a first-order expansion. But at \(K = 1\%\) and \(K = 5\%\), the smile factor \(z/x(z)\) also contributes error, and the combined effect can reach 20--100+ bps.

The wings are more affected than ATM because:

1. The \(z/x(z)\) factor is exactly 1 at ATM but departs from 1 for OTM strikes, introducing additional approximation error
2. The density in the wings is exponentially sensitive to the parameters, so small errors in the asymptotic expansion translate to large errors in implied volatility
3. The higher-order terms that are dropped in the Hagan expansion (proportional to \(\ln^2(F/K)\), etc.) become significant for deep OTM strikes

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Exercise 6. The exact solutions involve modified Bessel functions and numerical integration. Discuss the trade-off between using the exact solution versus the Hagan formula in the following contexts: (a) real-time pricing of 10,000 swaptions; (b) CMS spread pricing where tail accuracy matters; (c) validating a new finite difference implementation. In which case would you use the exact solution?

Solution to Exercise 6

(a) Real-time pricing of 10,000 swaptions: Use the Hagan formula. Speed is critical: the Hagan formula evaluates in microseconds, while the exact solution requires numerical integration (Laplace inversion) taking milliseconds to seconds per option. For 10,000 swaptions, the Hagan formula takes \(<\) 1 second total; the exact solution would take minutes to hours. The accuracy loss is acceptable for most swaptions with \(T \leq 5\) years and moderate moneyness.

(b) CMS spread pricing: Use the exact solution (or at least an arbitrage-free extension). CMS pricing integrates the call price over all strikes:

\[ V_{\text{CMS}} = \int_0^{\infty} g(K)\,C''(K)\,dK \]

This integral is sensitive to the tail behavior of the density. The Hagan formula can produce negative densities in the tails, causing systematic errors in the CMS convexity adjustment. The exact solution (for the special cases where it applies) guarantees a non-negative density and accurate tail behavior. For general SABR parameters, the arbitrage-free SABR PDE method is the appropriate alternative.

(c) Validating a new finite difference implementation: Use the exact solution. This is the primary use case for exact solutions. A new FD implementation should reproduce the exact prices to within the expected discretization error (e.g., \(O(h^2)\) for second-order schemes). Testing against the Hagan formula is insufficient because the Hagan formula itself has \(O(T^2)\) error, so agreement with Hagan does not confirm correctness of the FD method. The exact solution provides a true benchmark.