If volatility goes up, what happens to the value of an option? Did you say the value goes up? Oh dear, bottom of the class for you! I didn't ask what happens to the value of a vanilla option, I just said 'an' option, of unspecified terms.^{[1]}

Your boss asks you to price an up-and-out call option. What could be easier? You get out your well-used copy of Espen Haug's Complete Guide to Option Pricing Formulas (Haug, 2007) and code up the relevant option price and greeks. You've got to plug in a number for volatility so you look at some vanilla options, and they all seem to be around 20% implied volatility. So you put 20% into the formula and tell your boss the resulting option value. A small profit margin is added on top, the client is happy, the deal is done and the option sold. All down to Corney and Barrow to celebrate.

At three o'clock in the morning you wake up in a cold sweat, and not due to excessive alcohol intake for once. What if volatility turns out to be something other than 20%? You completely forgot to test the sensitivity of the price to your volatility assumption. What an idiot you've been! You get out of bed and take out your home copy of Espen Haug's book (everyone should own two copies). You code up the formula again and see how the price varies as you change volatility from 17% to 23%. The price seems to be insensitive to volatility, and is anyway within the profit margin (a.k.a. margin for error). You breathe a sigh of relief, Phew!, and go back to bed.

That morning you go into work, perhaps looking a bit more tired than usual after all the champagne and the early-morning panic. Your boss calls you into his office, tells you that a fortune has been lost on your barrier option and you are fired.

Apart from the short time between the pricing and the loss and some risk-management issues this scenario has happened in the past, and looks like it will continue to happen in the future. So what went wrong? How could money have been lost after all that stress testing?

What went wrong was that you assumed volatility to be constant in the option formula/model and then you changed that constant. This is only valid if you know that the parameter is constant but are not sure what that constant is. But that's not a realistic scenario in finance. In fact, I can only think of a couple of scenarios where this makes sense...

The first scenario is when every contract in your portfolio has gamma of the same sign, either all have positive gamma everywhere or all have negative gamma everywhere. We'll see the significance of the sign of gamma in a moment. But, anyway, who only buys options or only sells options? Most people buy some and sell some, even Nassim Taleb.

The other scenario is ...

The telephone rings, you answer. At the other end of the line a deep, manly voice says 'This is God here. I've got a hot tip for you. The volatility of IBM will be constant for the next year with value ... And the line goes dead. Damn, a hot tip from the top man and my battery dies! Never mind, all is not lost. We may not know what volatility is going to be, but at least we know it is going to be constant, and that is useful information.

Ok, so that s not a realistic scenario, unless you are an ex President of the US or an ex Prime Minister of the UK.

By varying a constant parameter you are effectively measuring

This is what you are doing when you measure the 'greek vega:

But this greek is misleading. Those greeks which measure sensitivity to a 'variable' are fine, those which supposedly measure sensitivity to a 'parameter are not. Plugging different constants for volatility over the range 17% to 23% is not the same as examining the sensitivity to volatility when it is allowed to roam freely between 17 and 23% without the constraint of being constant. I call such greeks 'bastard greeks because they are illegitimate.

Figure 5.1: The value of some up-and-out call option using volatilities 17% and 23%.

The following example demonstrates this.

Example In Figure 5.1 is shown the value of some up-and-out call option using the two volatilities 17% and 23%. Notice that at an asset value of around 80 the two curves cross. This is because the higher volatility increases the option's value for lower asset prices but decreases the option's value above. If you are sitting around the 80 asset value you would conclude that the option value is insensitive to the volatility. Vega here is zero.

The problem arises because this option has a gamma that changes sign. For lower asset values it is positive and for higher it is negative. Generally speaking, if you increase volatility where the gamma is positive then the price will rise. If you increase it where gamma is negative the price will fall.^{[2]}

The relationship between sensitivity to volatility and gamma is because they always go together. In the Black-Scholes equation we have a term of the form

The bigger this combined term, the more the option is worth. But if gamma is negative then large volatility makes this big in absolute value, but negative, so it decreases the option s value.

So what happens if there is a negative skew in our barrier-option problem? Increase volatility where gamma is positive and the price will rise. Decrease volatility where the gamma is negative and the price will ... rise. The result is that with a negative skew the option value rises everywhere. You should have sold the option for significantly more than you did, hence your loss of money and job.

It is quite simple to measure the true sensitivity of an option value to a range of volatilities as above and that is to solve the Black-Scholes equation with volatility being 17% whenever gamma is positive and 23% whenever gamma is negative. This will give the lowest possible option value. And then price again using 23% when gamma is positive and 17% when gamma is negative. This will give the highest possible option value. This is easily done, but not by Monte Carlo, you'll have

Figure 5.2: Uncertain volatility model, best and worst cases.

to solve by finite-difference methods.^{[3]} This model is called the Uncertain Volatility Model (see Avellaneda, Levy & Paras, 1995).

In Figure 5.2 are shown the best and worst cases for this up-and-out call option. Note that at the point where vega is zero there is actually a factor of 3 difference between the worst and best cases. That is an enormous range considering volatility is only allowed between 17 and 23%. Yet it is far more realistic than what you get by 'varying a constant.'

Figure 5.3: Value versus constant volatility.

As well as looking out for gamma changing sign you can spot potential hazards of the above form by plotting the value of an option versus a constant parameter. This is shown in Figure 5.3 for the up-and-out call option. If you ever see non-monotonicity then it should set alarm bells ringing. Non-monotonicity is telling you that sometimes the higher the parameter value the better, sometimes the lower the better. Beware.^{[4]} And, by the way, you get the same thing happening in some CDO tranches.

Traditionally minded quants have fudges to try to address this problem.^{[5]} These fudges involve measuring delta and seeing how delta changes as volatility changes. This is a very poor substitute for doing the job properly.

Example Another obvious example is the cliquet option. With some of these cliquet contracts you find that they are insensitive to volatility in classical models. Suppose that you price a cliquet using a volatility of 20%. You find that the price is $17.1. You then use a volatility of 15% and the price becomes $17.05. Continuing with the test for robustness you use a volatility of 25% and find a price of $17.07. Seems pretty insensitive to volatility. You now use a volatility of 10%, and then 30%, both times the theoretical price hardly changes. You then use the model that everyone else thinks of as 'cutting edge,' the Heston model (Heston, 1993), again $17 give or take a few cents. Finally, the ultimate test, you call up another bank, disguising your voice, and ask their price. Yet again, $17.

By now you are thinking that the cliquet is insensitive to volatility, that its price is close to $17. So you add on a tiny profit margin (after all, these contracts are so popular, lots of demand and therefore not much room for a profit margin), sell it to your client and relax after a job well done. Risk management will be proud of all the stress testing you've done.

Christ! A few weeks later you're fired, again, after huge losses on cliquet options. What went wrong?

There's a big clue that you found during your stress testing. It seems that the price was insensitive to simple volatility models. From this what can you conclude?

The incorrect and naive, although common, conclusion is that indeed your volatility assumption does not matter. You could not be more wrong. There is a lot of volatility risk in this contract, it just happens to be cleverly hidden.

What contract is insensitive to volatility? Simply stock. Is the cliquet option the same as stock? No way, it is far more complicated than that!

Therefore the apparent insensitivity to volatility is masking the 'change of sign of gamma that we ve seen above. See Wilmott (2002) for details about these contracts, how sensitive they really are and why the traditional 'fudges will not help you.

[1] If Bill Clinton can ask what the meaning of 'is' is then I can ask how important is an 'an.

[2] I say 'Generally speaking because this is not exactly true. We are dealing with diffusion equations here and for them any change to a parameter in one place will affect the solution everywhere.

[3] If you have a finite-difference code working for a constant volatility model then rewriting the code to price in this model should take less than a minute. To modify your Monte Carlo code to do the same things will takes weeks!

[4] People also use plots like this to back out implied volatilities. This is meaningless when there is non-monotonicity. You can get multiple implied volatilities, or no implied volatility at all. The naive would say that the latter means arbitrage. It doesn't, unless you live in a world where the parameter is definitely constant. That's not my world.

[5] It is common practice to fudge your way out of a hole in this business. Although the subject could be far more 'scientific' this tendency to apply fixes without addressing fundamental issues restricts the subject to being a branch of 'carpentry.'

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