# What does it mean to be “long or short in volatility”?

I've heard a question regarding pricing of european calls. The question is:

Is the call long or short in volatility when it is (deep) OTM? What is the profile of the implied volatility?

I know that in that case the answer is "long". Conversely the call would be short in vol if it was ITM.

I see a relation between long and the wish that volatility is high in order to go ITM if you hold the call. Also if you are ITM, I can see that your interest is to continue ATM. Therefore you want the volatility to be low.

I don't understand what exactly this term means neither where it comes from. I guess it is related to volatility trading/arbitrage.

Could someone please help me out and give me a precise definition for the term "long/short in volatility"?

I'll expand on Mark's and SRKX's answers which are both correct but brief. To be clear the words long and short have been generalized in finance. They used to mean that you owned a stock or had sold a stock short. Now they are often used to say you make money when a value goes up (long) or make money when some value goes down (short).

In this case whenever you own a call or a put you are "long" volatility. Meaning that as volatility increases the value of your position increases (holding everything else the same). How much added value that you get for a certain increase in volatility (called vega) depends on how in/out of the money the option is at currently among other things, but if you own the call/put it is always positive as more volatility means more possible upside.

When you sell calls or puts, then volatility decreases are good for your position so you are called "short" vol.

• I do not agree. If the stockprice is close to zero, the put has its maximum value. Hence any additional volatility can only reduce value i.e. negative vega. – emcor Jan 22 '15 at 23:47
• @emcor negative vega? I don't think that's possible, is it? I mean, the whole idea is that you can replicate an option payoff by holding a portion of the stock and that portion of the stock is determined among other things by volatility I understand. The more volatility the more of the Stock you need to hold and the more the option is worth then right? – SRKX Jan 23 '15 at 7:13
• put-call parity guarantees that a put and call with the same strike have the same vega. a very OTM call definitely has positive vega – Mark Joshi Jan 23 '15 at 20:38
• @emcor interesting example, though I think you would have trouble trading on it. However, theoretically (infinitesimally small moves) you are not correct as though it is near its maximum value it is not quite there. The vega would be nearly zero but still positive. – rhaskett Jan 26 '15 at 20:05
• The concept of vega breaks down at this point. Remember vega is the change in the value for a small change in the volatility of an option holding all else (including the price of the underlying) fixed. As the price is "fixed" it being close to zero doesn't really matter. In your extreme case, none of the greeks really apply and even measuring volatility would be funny as well. At this point, people often use state models looking at the probability of various futures like bankruptcy and the option payoff in those cases. – rhaskett Jan 26 '15 at 20:28

the vega of a call is always positive. The holder of a call option is therefore long volatility whatever the spot price.

Mark Joshi's answer is absolutely right, but just to elaborate a bit:

The Vega of an option is the sensitivity of its value with respect to volatility $\nu = \frac{\partial V}{\partial \sigma}$.

For calls, it makes sense that the Vega is always positive, not matter the level of the underlying.

If you take the Black-Sholes model, you can find the theoretical value of Vega here.

• Thank you for your answer.I am OK with that. Could you develop a bit and answer the questions please? Why the terms "long/short" and when do you employ each of them? Also by your's and Joshi's answers a call is always long in vol no matters its moneyness or I misunderstood it? – Cooper Jan 22 '15 at 12:31
• "it makes sense that the Vega is always positive" how does that work?^^ – emcor Jan 22 '15 at 23:44

There are a number of ways to look at this:

• BSM model vega: $$Vega_{call}=Se^{-q\tau}\phi(d_1)\sqrt{\tau}$$ Now if you buy the arguments and assumptions of the model, that means it must be positive mathematically. No matter the value of $$d_1 = \frac{\log\left(\frac{S}{K}\right) + (r-q+\frac{1}{2}\sigma^2)(T-t)}{\sigma\sqrt{T-t}},$$ the value of $$\phi(d_1)\ge 0$$. The same applies to the exponential function. Negative Spot values are ruled out because $$\frac{S}{K}$$ would be undefined. Since your main claim was that ITM calls are short vol, it is safe to ignore negative spot prices anyways. Therefore, if you are long (which means you bought the option), you are always long volatility as more vol will never decrease the value of your option.
The image below plots the call option price for given inputs and varying volatility on the horizontal axis.
Being deep ITM just shifts the value of the option up, but it still increases with increasing vol.

• Put-call parity: $$Vega_{put}=Ke^{-r\tau}\phi(d_2)\sqrt{\tau}$$ which is the same for as for the call (not immediately obvious but easy to test if you plug in the values (or just trust existing derivations).

• Using intuition: you have a hockey stick payoff. More time and more vol will always increase your option price (or at least leave never decrease it). Your downside is limited to the premium, your upside is uncapped. Using an insurance analogy, if you get car insurance for 1 or 5 years, and on average crash it more often, what will be more expensive? The one for 1 year or the one for 5 years. The one where it is likely that there will be a crash or the insurance for a safe driver with a car model that generally has low probability of being involved in an accident per miles driven? I guess the answer is obvious. The same applies for options. If you have more time and more vol, it benefits the owner of the option (long position) and puts the seller at greater risk. The premium is adjusted upwards to account for this.

The images below have varying spot on the horizontal axes. The first one is for short time and low vol. You can see how vega is quickly becoming zero for deep ITM and OTM calls. Increasing time and vol has largely the same effect, it increases the area where vega is greater than zero and adds time value on top of the intrinsic value.

Or similarly, in a 3D graph, you can also see how more time, or more vol increases the value of the call (vega is never negative).

• The time value argument can be taken a step further: It can be shown that early exercise of a non dividend paying American call option is never optimal. Check here for dividend paying stocks. Here more vol means that spot could be even higher or also substantially lower. If it falls below strike (however unlikely that may be for deep ITM), it would have been unwise to exercise and pay more than the actual value at expiry. If a sharp decline is expected, an investor will still always be better off to sell the option rather than to exercise it (which would mean the time value is lost and only the intrinsic value is gained), or even keep the option and short the stock. In any case, the argument is the same as with the insurance logic provided before. Once you exercise, you have no insurance anymore.

• Really deep ITM (or OTM) options will be unaffected by vol changes. The first point shows this mathematically, but think of it intuitively. If you have the right to buy an iphone for 3 million pounds, would you use that option or rather buy it online for a fraction? Volatility in the price of the iphone will not change that, as it is reasonable to assume it will be well below 3 million in the foreseeable future. Hence, your option is worthless and does not change no matter the vol (ignoring Weimar republic inflation scenarios).