I often hear this term quite lot from traders, what does it really mean?

And some additional question: In option trading, is "buying vol" equivalent to "buying option" (no matter it's call, put or even straddle)? on the other hand, is "selling vol" equivalent to "selling option" (no matter it's call, put or even straddle)?

  • $\begingroup$ All else being equal, the price of an option is higher if volatility is higher, so you're pretty much correct that long(short) options positions are long(short) vol. But they might be long and short other things as well, so getting pure vol exposure without exposure to anything else takes hedging, for example delta hedging eliminates exposure to movement in the underlying in theory. $\endgroup$
    – ontic
    Nov 27, 2019 at 12:36
  • $\begingroup$ "Buying vol" means having a position with positive Vega, and yes, both a long call and a long put have positive Vega (speaking here of vanilla calls and puts) as you can verify. $\endgroup$
    – Alex C
    Nov 27, 2019 at 14:09
  • 2
    $\begingroup$ Alex C - Careful though, some say "long volatility" for long gamma positions, and don't care about vega. This is because some trades are "long volatility" as "long option prices", hoping for implied volatilities to increase, and some are "long volatility" on the underlying moves. $\endgroup$
    – siou0107
    Nov 27, 2019 at 15:20
  • $\begingroup$ Interesting point, thanks. $\endgroup$
    – Alex C
    Nov 27, 2019 at 21:08
  • $\begingroup$ @siou0107 thanks for the answer! I have the exact doubt of what u are saying. "Long Vol" can mean (1) long gamma - profit by hoping the the underlying to move. or (2) long vega - profit by hoping the IV to move. However, "long vol" essentially does both at the same time, so how would these 2 strategy interfere each other? any comments? $\endgroup$
    – Jeremy
    Nov 28, 2019 at 3:25

1 Answer 1


Not necessarily. It is true for vanilla options : all else being equal, their price increases with implied volatility, and if you delta-hedge them the residual P&L is a function of the difference between realised and implied volatility (variance). This is expressed in the so-called Black-Scholes robustness formula : for a long, delta-hedged option position, we have

$$\text{P&L} = \frac{1}{2}\int_0^T{e^{rt}\Gamma_tS_t^2\left(\sigma_t^2 - \tilde{\sigma}^2 \right)dt}$$ where $\sigma_t$ is the realised volatility (squared return) and $\tilde{\sigma}$ is your implied volatility (used to compute option price and gamma).

However, when you talk about more exotic instruments (e.g., digital or barrier options), this becomes more subtle. Take an out-of-the money knock-in option : you are totally long volatility, since you need 1) to knock in and 2) the spot to move the furthest away from the strike so that you get the biggest payoff

Now take an out-of the money up-and-out call : you want high volatility near the strike, to get into the money, but once in the money you want low volatility not to trigger the knockout barrier. Thus, you are not uniformly long or short volatility. In fact, such options have a gamma that changes sign.

Proof Suppose you just bought a European option with terminal payoff $g(S_T)$ considering a BS model with constant volatility $\tilde{\sigma}$, and delta-hedged it on the basis of that model. The PDE satisfied by your model's price $p^M(t,S_t)$ is: $$\frac{\partial p^M}{\partial t}\left(t, S_t\right) + rS_t\frac{\partial p^M}{\partial x}\left(t, S_t\right) + \frac{1}{2}\tilde{\sigma}^2S_t^2 \frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right) - rp^M(t,S_t) = 0$$ $$\Leftrightarrow \frac{\partial p^M}{\partial t}\left(t, S_t\right) + rS_t\frac{\partial p^M}{\partial x}\left(t, S_t\right) = rp^M(t,S_t) - \frac{1}{2} \tilde{\sigma}^2S_t^2 \frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right)$$ $$p^M(T, x) = g(x)$$ (If it is not the case, your model is arbitrageable.) The value of your hedging portfolio (stock and cash) has the following dynamics: $$dV_t = \frac{\partial p^M}{\partial x}\left(t, S_t\right)dS_t + r\left[ V_t - S_t \frac{\partial p^M}{\partial x}\left(t, S_t\right)\right]dt$$

If you apply Itô's lemma to $p^M$, you get: $$dp^M(t,S_t) = \frac{\partial p^M}{\partial t}\left(t, S_t\right) dt + \frac{\partial p^M}{\partial x}\left(t, S_t\right) dS_t + \frac{1}{2} \frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right) d\langle S \rangle_t = \left[\frac{\partial p^M}{\partial t}\left(t, S_t\right) + \frac{1}{2} \sigma_t^2S_t^2\frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right)\right]dt + \frac{\partial p^M}{\partial x}\left(t, S_t\right) dS_t$$

If you denote by $Z_t := p^M\left(t, S_t\right) - V_t$ the value of the hedging P&L, you have $$dZ_t = \left[\frac{\partial p^M}{\partial t}\left(t, S_t\right) + \frac{1}{2} \sigma_t^2S_t^2\frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right)\right]dt - r\left[ V_t - S_t \frac{\partial p^M}{\partial x}\left(t, S_t\right)\right]dt$$ $$Z_0 = 0$$ The initial condition corresponds to the assumption that you invest the whole premium $p^M\left(0, S_0\right)$ into the hedging portfolio $V_0$. You replace the terms that you have in your model PDE and get: $$dZ_t = \left[rp^M(t, S_t) + \frac{1}{2} \left(\sigma_t^2 - \tilde{\sigma}^2\right)S_t^2\frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right)\right]dt - rV_tdt = \left[rZ_t + \frac{1}{2} \left(\sigma_t^2 - \tilde{\sigma}^2\right)S_t^2\frac{\partial^2 p^M}{\partial x^2}\left(t, S_t\right) \right]dt$$ Since $p^M\left(T, S_T\right) = g\left(S_T\right)$, the final P&L is: $$\frac{1}{2}\int_0^T{e^{rt}\Gamma_tS_t^2\left(\sigma_t^2 - \tilde{\sigma}^2 \right)dt}$$

  • $\begingroup$ +1 Good answer. I would just say sigma tilda is the static option hedge volatility. $\endgroup$
    – ir7
    Nov 28, 2019 at 23:02
  • $\begingroup$ Well, not necessarily. If you use a different implied volatility to compute your Greeks (so delta and gamma) at each time step, that result will still hold. But that is not crystal clear in my formula, I will add the time index to the implied volatility. Thanks for the feedback! $\endgroup$
    – siou0107
    Nov 30, 2019 at 1:16
  • $\begingroup$ I’m not sure. I mean the technical formula itself in non-Black-Scholes (non-constant volatility) world, not practical applications. Time dependent sigma tilda wouldn’t be an implied volatility at time t (what expiry? what strike?), but maybe a Dupire local volatility surface recalibrated periodically to implied vol surface. $\endgroup$
    – ir7
    Nov 30, 2019 at 3:31
  • $\begingroup$ It would be the implied volatility of your option, recalibrated at each time period :) I’ll make a quick proof soon $\endgroup$
    – siou0107
    Dec 1, 2019 at 11:17
  • $\begingroup$ Do it! 😄 I’ll validate it myself. Carr-Madan, flat hedging implied vol, is clear. Google pointed to resource below for extensions. It should be a good start. pdfs.semanticscholar.org/c1ab/… $\endgroup$
    – ir7
    Dec 1, 2019 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.