How Volatility Models Affect Double Knock-Out Option Prices

# Local Vol vs Stoch Vol Option Pricing


# Local Vol vs Stoch Vol Option Pricing












This is an interview question: Imagine you have a double knock-out barrier option: the current spot is 100, the lower barrier is 80, and upper barrier is 120. The barrier is continuous, meaning that once the spot goes out side the range of 80-120 anytime before time to maturity, you got nothing. If the spot stays within the range 80-120 the whole time before the option matures, you will get paid a fixed amount of 10 as the final payoff. The question is, if you need to price this option, which volatility model will give a higher option price? Local vol or Stoch vol? Why? The interviewer said it's the Stoch vol gives a higher option price, but I didn't get the reason. Does anyone know why?

## Answer by Peter A (score 5)

https://quant.stackexchange.com/a/70126

To simplify things, let's begin by considering a one touch option (that pays cash if the barrier is touched) rather than the double no touch option of the question. Also, let's assume interest rate and dividend yield are zero.

And before starting, we can remember that we are given a volatility surface at the start. Then both local and stochastic volatility models will value vanillas and digitals the same.

As dm63 pointed out, an approximate hedge for the one touch is a digital with strike equal to the barrier level, and double the notional. At the moment the barrier touches, the digital is worth around 50% of its notional, so 100% of the one touch notional. As soon as the barrier touches, the digital hedge must be unwound.

In practice, this is not a perfect hedge for two reasons. Firstly, in a log-normal world, an at-the-money digital is not worth exactly 50%. That is easily dealt with by adding a vanilla option to the hedge that exactly corrects for this.

The second and more interesting reason is that a digital option has a correction to its Black-Scholes value equal to minus its Black-Scholes vega, multiplied by the derivative of the implied volatility (at the money) with respect to strike (hereafter, the "skew").

The upshot is that the value of the one touch at touch time can be replicated from a digital and a vanilla (whose values are independent of the model) plus a constant multiplied by the expected value of the skew at touch time.

In a stochastic volatility model, the smile tends to float when spot moves, meaning the at-the-money skew stays constant. In a local volatility model, the smile tends to stick in strike space when spot moves. As the smile is convex and at-the-money is near the bottom at the outset, this means the skew increases in magnitude. If one is careful with the sign of the factor multiplying the skew term, this gives the result.

It's important to be aware this is not a mathematical proof, and I have heard of claimed counter examples. For all practical cases though, it seems to be true.

Full details of the above calculation, and of the assumptions made are in chapter 9 of the book Smile Pricing Explained.

Returning to the specific question, we have one touch plus no touch equals cash. Then as a one touch is higher in local vol, a no touch is lower in local vol and higher in stochastic vol. Deterministic interest rates and dividend yield can be included back in fairly easily. Finally, we are asked to take on trust that the model behaviour of a double no touch is similar to a no touch. It is, and in fact, it is much more strongly model dependent than a single no touch. Even on a relatively modest FX volatility surface, a double no touch price can vary by a factor of two according to the model.

Finally it is worth mentioning that both volatility swaps and forward volatility agreements (forward starting vanilla options) are valued higher in local volatility than stochastic volatility. For the volatility swap, there is a beautiful proof (due to Dupire) in the very special case of an "instantaneous" volatility swap.

## Answer by dm63 (score 2)

https://quant.stackexchange.com/a/70124

I'm more of a rates person than equities, but i'll give it a shot: first of all one assumes that both models have been calibrated so that vanilla options at all strikes and maturities are aligned, otherwise the question doesn't make sense. Since European digital options are the limit of European call spreads, it must also be true that the European digital options at all maturities are the same in both models. For example, the price of a digi 120 call is the same in both models. Furthermore, there is a well known result due to the reflection theorem, that the price of an American digital call is (under no drift conditions at least (r=q)), equal to twice the value of a European digital call, under any model. Hence, the single knockout 120 call and the single knockout 80 put must be the same under both models. So, if there is a difference in price for the double knockout option in the OP, it must be because there is a difference between the 2 models in the probability of hitting BOTH barriers.

Why should this be? My proposed logic would be to suppose that under some path of the underlying, we have hit one of the barriers (say the 120). Then , I claim that the probability of later hitting the 80 barrier (given that we are at 120 right now) is higher under the local vol model. Why? because under that path, we expect that implied volatilities have gone up (because that is the way that a local vol model explains the smile). In contrast, in a stoch vol model, vol may have gone up or down because it is driven by a separate random variable. So, under local vol we are more likely to hit both barriers, so the double no touch is worth less.

I did assume here that we have a typical positive smile in the underlying. I'm happy to be corrected by a equities or fx practitioner.

EDIT: I stand corrected by the comment of @Peter A. I give below a link with what appears to be an explanation consistent with his comment. http://www.matthiasthul.com/wordpress/2015/06/26/stochastic-vs-local-volatility-barrier-options/#:~:text=Consequently%2C%20they%20also%20agree%20on,hitting%20probabilities%20that%20they%20generate.