Calibrating SABR Parameters to the Implied Volatility Smile

# SABR Implied Volatility and Option Prices


# SABR Implied Volatility and Option Prices












I am trying to understand the SABR model. Specifically, I am having difficulty to understand how to calibrate the model parameters, that is,

- initial variance,

- volatility of variance,

- exponent for the forward rate, and

- correlation between the Brownian motions.

Small example on the above would be useful. Thanks in advance

## Answer by user16651 (score 8)

https://quant.stackexchange.com/a/26268

The SABR model of Hagan is described by the following Stochastic differential equations: $$\begin{align} & d{{f}_{t}}={{\alpha }_{t}}f_{t}^{\beta }d{{W}_{t}}^{1} \\ & d{{\alpha }_{t}}=v\,{{\alpha }_{t}}d{{W}_{t}}^{2} \\ & {{E}^{Q}}[d{{W}_{t}}^{1},d{{W}_{t}}^{2}]=\rho dt \\ \end{align}$$ In these equations, $f_t$ is the forward rate, $\alpha$ is the initial variance, $\beta$ is the exponent for the forward rate and $v$ is the volatility of variance.

It is well-known the prices of European call options in the SABR model are given by Black's model. For a current forward rate $f$, strike $K$, and implied volatility $\sigma_{B}$ the price of a European call option with maturity $T$ is $$C(f,K,{{\sigma }_{\beta }},T)={{e}^{-rT}}(f\,N({{d}_{1}})-K\,N({{d}_{2}}))$$ where \begin{align} & {{d}_{1}}=\frac{\ln \left( \frac{f}{K} \right)+\frac{1}{2}\sigma _{B }^{2}T}{{{\sigma }_{B }}\sqrt{T}} \\ & {{d}_{2}}=\frac{\ln \left( \frac{f}{K} \right)-\frac{1}{2}\sigma _{B}^{2}T}{{{\sigma }_{B }}\sqrt{T}} \\ \end{align} and

Estimating $\alpha$, $\rho$ and v: This can be accomplished by minimizing the errors between the model and market volatilities {$\sigma_{i}^{market}$}(from interest rate derivatives, for example) with identical maturity T. Hence, for example, we can use SSE, which produces $$(\widehat{\alpha },\widehat{\rho },\widehat{v})=\underset{\alpha ,\rho ,v}{\mathop{\arg \min }}\,{{\sum\limits_{i}{\left( \sigma _{i}^{market}-{{\sigma }_{B }}({{f}_{i}},{{K}_{i}};\alpha ,\rho ,v) \right)}}^{2}}$$

Estimating $\beta$:

The at-the-money volatility $\sigma_{ATM}$ is obtained by setting $f = K$ in equation $\sigma (K,\beta)$, which produces $${{\sigma }_{ATM}}={{\sigma }_{\beta }}(f,f)=\frac{\alpha \left( 1+\left[ \frac{{{(1-\beta )}^{2}}}{24}\times \frac{{{\alpha }^{2}}}{{{f}^{2-2\beta }}}+\frac{1}{4}\frac{\rho \beta v\alpha }{{{f}^{1-\beta }}}+\frac{2-3{{\rho }^{2}}}{24}{{v}^{2}} \right]T \right)}{{{f}^{1-\beta }}}$$ Taking logs produces $$\ln {{\sigma }_{ATM}}\approx \ln \alpha -(1-\beta )\ln f$$ Edit for Gordon

In practice, the choice of $\beta$ has little effect on the resulting shape of the volatility curve produced by the SABR model, so the choice of is not crucial. The choice of $\beta$, however, can affect the Greeks. Barlett provides more accurate Greeks and shows that they are less sensitive to the choice of $\beta$.Indeed The case $\beta=0$ produces the stochastic normal model, $\beta=1$ produces the stochastic log-normal model, $\beta=\frac{1}{2}$ produces the stochastic CIR model.

## Answer by Amiro (score 1)

https://quant.stackexchange.com/a/21487

one of my friend recently wrote about SABR model and calibration. I highly recommend you to read it to get your answers http://janroman.dhis.org/stud/EXJOBB/SABR.pdf