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Why do you get long vega when you buy an option and short vega when you sell an option?

I would have thought that for both buying and selling options the vega would change according to whether the option was ITM or OTM. This would be because as an option participant an increase/decrease in volatility would change the chance of expiring at a financially beneficial position. However this not appear to explain the first statement (which I believe is correct).

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The risk exposures/sensitivities of long and short positions always have different signs. This has to hold since derivatives are zero sum games.

Vega is always positive for a long position in a European plain vanilla option (or any convex payoff in general). This is true even when the option is already in-the-money. As volatility increases, the probability of very positive and very negative returns increases. As the holder of the option, you are protected against moves in one direction but participate in the other.

You can construct a very simple binomial example to illustrate this. Consider a one period setting. You are long a call with strike 90. The current stock price is 100 and there are no rates or dividends.

  1. First consider a "low volatility" scenario, where the stock either goes up to 105 or down to 95. The payoff is either 15 or 5 and the initial price is 10 with a time value of 0.
  2. Now consider a "high volatility" scenario, where the stock either goes up to 120 or down to 80. They payoff is either 30 or 0 and the initial price is 15 with a time value of 5.

You see that due to the convexity of the payoff, a higher volatility is advantageous even when the option is already in-the-money since losses are limited. European vanilla options derive their time value (ignoring rates, dividends, ...) from the possibility of crossing the strike, no matter whether they are already in-the-money or not.

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  • $\begingroup$ I still dont understand how if you own an option which is deep ITM surely the price of that option goes down and volatility increases as the probability of the option expiring OTM increases, surely this means that vega is negative despite the fact you are long the option $\endgroup$ – Permian Sep 18 '16 at 11:08
  • $\begingroup$ dm63 provides a nice explanation. I added the binomial example to illustrate in a simple setting why higher volatility is desirable for the holder even when the option is already in-the-money. $\endgroup$ – LocalVolatility Sep 18 '16 at 13:30
  • $\begingroup$ it is a nice answer but am trying to understand this heuristicly without considering put call parity $\endgroup$ – Permian Sep 22 '16 at 22:22
  • $\begingroup$ Here is a different way of looking at my explanation: If you have a linear contract such as a forward, then the vega is zero. Why? Because there is no optionality. I.e. you can statically hedge a short position in the forward by buying the underlying. A deep in-the-money call under a low volatility regime is like a forward. The chance of ending in the money is very high, there is almost no time value attached to the optionality. When volatility increases the value of the optionality increases and thus vega is positive. $\endgroup$ – LocalVolatility Sep 22 '16 at 22:47
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No, you are incorrect. A deep in the money option is long vega. It's not just about the probability of being in the money, it's about how far in the money it is. Your reasoning is correct if we are talking about digital options which pay a fixed amount if the option expires in the money, but incorrect for regular options.

One way to prove this explicitly: a deep in the money call is equal (by put call parity) to a long position in the underlying + a deep out of the money put. Since the position in the underlying has no vega, the deep ITM call has the same vega as the deep OTM put.

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  • $\begingroup$ do you have a more heuristic explanation ignoring put call parity that explains why a deep ITM option is long vega. As far as I can see any lower movement decreases the chance of finshes ITM $\endgroup$ – Permian Sep 22 '16 at 22:21

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