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11

Let's relabel this as What (TF) is SABR? Alpha, Beta and Rho are the point of the model. So explaining them is explaining the model. A model of two processes Unlike earlier models in which the volatility was modelled as a constant (Vasicek, Hull-White, etc), SABR assumes that as well as the price of the thing being stochastic, so is its volatility. That ...


6

The SABR process is a strict martingale for all values of beta < 1 (in particular, negative betas are fine). If beta = 1, the process is a strict martingale if and only if rho < 0. Under all other circumstances, i.e. beta > 1, or beta = 1 and rho >= 0, the SABR process is a local martingale but not a martingale (it may explode in finite time).


6

I think you did something wrong in translating the input to numerics. As pointed out by dm63 normal vols are quoted in basis points. Using equation A.67a) from the Hagan paper you linked we see (setting $\beta = 0$) $$\sigma_N(K) = \alpha\frac{\xi}{x(\xi)}\left[1+\frac{2-3\rho^2}{24}\nu^2\tau_{exp}\right]$$ where $\tau_{exp} = 0.25$ in your example and ...


5

If I got this right, you get a SABR model from fitting market implied volatilites (from market price via Black) to SABR volatilites (from SABR parameters via formula above). Then you step back and think the SABR distribution needs improvement because it is not arbitrage free. Instead you use the collocation method to replace it with its projection onto a ...


5

1) The paper Explicit SABR Calibration Through Simple Expansions explains how to calibrate the SABR model in practice. 2) The role of alpha, beta and rho is well explained in the original SABR paper Managing Smile Risk. Beta is most often chosen in advance, to represent a specific dynamic. Although one can find references where people calibrate it to option ...


3

In short , this claim does not hold under all circumstances. There are a few ways to break down such approximation. The options under consideration have very long expiry, i.e. $T$ is very large As expiration date approaches, the volatility smile becomes more pronounced, i.e. $v$ becomes relatively large. Under extreme market condition, the magnitude of $\...


3

This is indeed an important issue if you use SABR in production. If I am correct, you'll need this term $$ \chi(\zeta) = \log \left( \frac{\sqrt{1-2\rho\zeta+\zeta^2}-\rho+\zeta}{1-\rho} \right) $$ and $\zeta=0$ if $F_0=K$. When $\zeta$ is VERY small (e.g., $|\zeta|<10^{-8}$), you can use the Taylor expansion $$\sqrt{1+\varepsilon} \approx 1+\...


3

First and foremost it is important to clarify that the underlying is not necessarily normal/lognormal but for the special cases of $\beta$ the underlying is normal/lognormal Conditioned on a realization of the volatility. As mentioned in the answer by @ilovevolatility. Simple stochastic calculus will show the properties you mentioned. For realized ...


3

In fact, this is a confusion caused by a sloppy notation. The rigorous version of the setup should be $$A(K)\rightarrow \epsilon A(K).$$ Then we let $x:=\frac{f-K}\epsilon$. The rest is the usual singular perturbation operation.


3

The simplest answer is that prices are not derived from the models. Prices are a result of trading in the market and express the sum of the information and opinions of all market participants at the time. Despite what many academics would like you to believe, markets participants are not always perfectly rational and not always 100% motivated by the maximum ...


3

Here you can refer to the previously asked question regarding the advantages and disadvatanges of the volatility surface:- What are the advantages/disadvantages of these approaches to deal with volatility surface? To my knowledge, the main advantage of using SABR in comparison to other models like Heston is that, SABR assumes lognormal distribution whereas ...


3

I have found the answer to my own question during the last month where the question have been unanswered The main question: Look at page 9 from "the ODE can be rearranged to $f'(y)=....=F(y,f)$". For given $\alpha, \beta, rho, \epsilon$ and $\gamma $ we have all the relevant information to solve this ODE, hence finding $f(y(t))$ which is denoted as $f(y)$ ...


3

Here are a few FX structured product examples: All of these can be notes or swaps, notes will pay back the notional at the end and carry no credit risk (and are normally set so that they are worth 100% at inception - i.e. they'll be worth 99% and the seller will take some profit/hedging costs). Swaps will either be set to be worth 0% (same deal as above, ...


2

You would simply calculate the prices of various strike options using your parameters, then calculate the black scholes implied vol of each option. Did I miss the point of your question ?


2

It comes from Heat Kernel expansion and differential geometry. See Theorem 6 and Section 8 of http://papers.ssrn.com/sol3/papers.cfm?abstract_id=1717676&download=yes


2

I would say that SABR is overkill for inflation options, because due to the scarcity of prices for inflation options, there isn't enough information to fit all the SABR parameters. It is probably better to adopt a simple Normalized model of inflation rates, which you can calibrate by looking at historical normalized volatility. This will also take care of ...


2

Try to redesign your object function, your difvol, so it's a function of a two dimensional vector dilvol<-function(X){ rho<-X[1] nu<-X[2] ##type in your function here and use rho and nu normally.. } nlm(difvol,c(0.1,0.1) This might work


2

The relationship between the two models is described in details in Implied Volatility Formulas for Heston Models by Hagan et al. In particular an expansion of the implied volatility under the Heston model that matches the one of a SABR model is described. It gives an explicit correspondence between the parameters of each model.


2

If you want to use the normal SABR ($\beta=0$), my paper, Hyperbolic normal stochastic volatility model (arXiv | SSRN | DOI) might give you a solution. It reports an exact closed-form MC simulation scheme for the normal SABR model. Better than that, it shows that Johnson's $S_U$ distribution ($\sinh$ transformation of normal variate) is a close cousin of the ...


2

You don't want to use the SABR (or an extension) to price equity options or FX options. The lag of mean-reversion in the model's volatility dynamics leads to explosive behavior and to a implied distribution that is absolutely not in line with empirics -- especially on longer time horizons. To my knowledge people use it mostly for interest rate derivatives. ...


2

The SABR model itself is arbitrage-free even for high vol of vol. The question is whether the Hagan et al formula for implied volatility under the SABR model is arbitrage free - it isn't actually. For very low strikes arbitrage can occur using the Hagan et al formula for implied volatility, and perhaps also for very high vol of vol. Question: how do you ...


2

I can confirm there is no error in @Sanjay graph. I obtain the same plot with Obloj correction for the SABR formula. In fact, the popular SABR approximation formulas (Hagan or the further corrections) use as hypothesis a small vol of vol. In your case, the vol of vol $\nu$ is very large ($\nu=7$) and it is not too surprising that the approximations break ...


2

No, the simulation is not exact in general, precisely for the reason you mentioned. By "exact", it is meant that there is no discretization error in time. Of course, there will always be a Monte-Carlo sampling error. For the Black-Scholes model, the simulation is exact if you simulate the log asset, as it is a standard arithmetic Brownian motion, and then ...


2

I am going to try to answer your more general question "Do all SV models generate a smile?" which you put in one of the comments. (Maybe edit also the title of your question if you want, if my answer is satisfactory.) I will take zero correlation between asset and the volatility process to start with. The generalisation to non-zero correlation is ...


1

my two cent's worth. It depends ultimately what you are trying to achieve, and calibrate to. The key tests for implied vols on option prices, relate to their pdf (probability density functions). The Hagan expansion allows us to have a analytical form, by which one can compute the implied vols, and subsequently feed that into a Black76 equation. If the ...


1

Given (conditional on) a realisation of the volatility, it is normal for $\beta = 0$ and lognormal for $\beta = 1$


1

If you have a few options quotes, preferably a few around ATM, and a few in the call and put wings, then you might want to try using SVI for the smoothing of the IV smile (and hence get a value for the ATM IV). SVI calibration is also relatively quick, so you don't need to use stale parameters, you could update the parameters intra-day to the available ...


1

Apparently Hagan et al original formula has a problem, and strictly speaking is incorrect. This was later corrected by Obloj 2008 using the BBF paper giving the implied volatilities directly from the Dupire PDE: https://arxiv.org/pdf/0708.0998.pdf I did not check your claims against the original paper but it is a possibility that you have precisely bumped ...


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