SABR with rho less than -1

Soorma · January 13th, 2010, 9:25 pm

Hi Can i have my SABR fit rho less than -1 ?What would be the effect on the dynamics and the distribution f the underlying?Can i have beta negative?Again how does the dynamics change ?Thanks

Alan · January 14th, 2010, 2:07 pm

Here is the SABR process:dS(t) = sigma(t) S(t)^beta dB1(t), (t < T0)d sigma(t) = a sigma(t) [rho dB1(t) + sqrt(1 - rho^2) dB2(t) ]or equivalently,dS(t) = sigma(t) S(t)^beta [rho dB1(t) + sqrt(1 - rho^2) dB2(t) ] , (t < T0)d sigma(t) = a sigma(t) dB1(t)where T0 is the first hitting time of S=0, and (B1,B2) are independent BM's.So, to answer your questions:1. Unless you can make sense of a complex-valued volatility or asset prices (or whatever your S is),|rho| > 1 is nonsense.2. On the other hand, beta < 0 is fine and does not change the process dynamics (much).The S(t) process will hit S=0 with positive probability just like it will for any beta < 1. As time passes, the sigma(t) process will accumulate near, but slightly above, sigma=0, just like it always does.The Feller boundary classification (for the CEV model) is that S=0 is *regular* for (-infinity < beta < 1/2).This means the model admits both reflecting and absorbing behavior there. Certainly for stocks, S=0 must always be taken as absorbing ifit is hit, in the absence of cash flows. Anyway, the model dynamics, for t < T0, are qualitativelythe same for all beta < 1.

Traden4Alpha · January 14th, 2010, 2:48 pm

A beta<0 would mean that the price becomes LESS volatile as price level increases and much more volatile as price level declines.

Alan · January 14th, 2010, 2:53 pm

That's true for any beta < 1

Traden4Alpha · January 14th, 2010, 3:07 pm

QuoteOriginally posted by: AlanThat's true for any beta < 1Are your sure? Beta=+0.1: S^0.1 < (2*S)^0.1Beta=0.0: S^0 = (2*S)^0Beta=-0.1: S^-0.1 > (2*S)^-0.1

Alan · January 14th, 2010, 3:15 pm

Yes, write the S-process asdS = [sigma S^(beta-1)] S dB(t)and think of the 'effective lognormal' volatility as the term in brackets.You can see that the effective sigma will increase as S falls (and vice-versa) for any beta < 1 That's why the CEV model has a negatively-sloped implied volatility vs. strike (skew graph) for any beta < 1. The effective volatlity diverges just as S=0 is being hit, for any beta < 1. The effective volatility falls to 0, for large S, for any beta < 1.

Soorma · January 15th, 2010, 9:36 pm

Hi Alan .Thank you for the explanations. I understand the rho less than -1 gives complex prices , but the Eurodollar market is so steep that keeping a beta of .2 and rho of -.999 still cannot fit the calls( price domain not the yield domain). The sep'10 Eurodollar future is trading at 99.21 , The SABR calibration is good till 99.5 strike but after that the market implied vol drops so steeply that we get a very bad representation of the Skew.I read that SABR is not able to fit at low strikes because of the approximations used for finding a closed form solution but i am in a price domain and not a yield domain .1) How can i workaround to get a better snapshot of the market skew with SABR calibration2) Because i am in the price domain( 99.5(100-.5) , instead of .5) so how can i model the underlying process and the vol process so that i dont get positive probabilities for strikes above 100 .3) I read a couple of papers that claim a better approximation of SABR . Do you have any recommendations on that?

Alan · January 16th, 2010, 12:52 am

Not really my market. To even begin to make a suggestion, I wouldneed to see what you are trying to fit. Suggest you post your smile chartfor that expiration, ideally with the implied vols shown as a bid-ask range so that theacceptable variation is clear. Even better would be two charts,one using rates and one using prices.

Soorma · January 19th, 2010, 5:48 pm

Hi AlanI have added the graphs for both yield as well as price model.red * - underlying priceblack diamonds - ask volsgreen diamonds - bid volsblue diamonds - sabr vols

Soorma · January 19th, 2010, 5:49 pm

Soorma · January 19th, 2010, 5:50 pm

ignore the multiple attachments , both are the same . thanks

Alan · January 19th, 2010, 9:31 pm

Thanks, I took a look. The appeal of the SABR approximation is that it isa closed-form that is easy to fit to a single expiration and has some intuitiveappeal. Looking at your charts, I asked myself what might be almost aseasy to fit and have some good intuition, too.The implied vol vs rates suggested to me a simple Merton's jump-diffusionmodel where the jump distribution is highly skewed to the upside.Playing around a little with the optioncity.net calculator, I couldget a similar looking chart with the "Constant Volatility + Jumps" tab,which implements that model. If you want to see what I saw, try it withyour rates x 10, so take underlying S0 = 7, strike range K in [2,22]Interest rate = Div yield = 0.Current volatility sig0 = 27, lambda = 1, muJ = 0.7, sigJ =0.1, T = 0.64 years.Options = euro-style with the closed form solver. Look at the smile chart.Anyway, it looks to me that you could calibrate to Merton's jump-diffusionmodel using rates as the underlying and likely fit your smile chart better. Look fora best fit where the mean jump size is very positive. Intuitively, this seems quite plausible at this stage in the interest rate cycle.

Soorma · January 19th, 2010, 9:59 pm

Hi AlanActually the Rates vs Vol chart is acceptable in terms of fit for the traders as we dont really trade anything bigger than 99.75(.25).The problem is with the price vs Vol fit, the problem arises because using price instead of yield , doesnot add a boundary at 100, and hence the Vol's for those strikes are way higher than the Market Vols.Is there a way in which under the SABR framework in the price domain i can add a boundary at underlying 100?Traders are more comfortable in price domain , than yield domains and hence the confusionAlso, in the yield domain , although the skew fit is good but the low strike put deltas are way lower than market deltas (Sabr delta -.06,market delta -.15) , how can i fix that ?Thanks

Alan · January 19th, 2010, 10:21 pm

Well, let's call your fitted curve V(K) in the implied vol. vs rates chart.You said that one was acceptable.If so, why don't you just keep the same function V(K) which you like, and plot forthe trader's V(100 - K), relabelling the horizonal axis "the price"?. It will be exactly the same fit, just plotted in the reverse direction.

Soorma · February 9th, 2010, 8:24 pm

Alan, i feel the issue can be in the distribution of the underlying being assumed by the market .Short term Interest rates seem to be more normally distributed than log normal distribution.One question , what would beta greater than one mean ? does that mean the atm vol increases with increase in underlying ?