Long and Short Volatility: Delta Hedging and Option Gamma

# What does is mean by buy(long) volatility or sell(short) volatility in option trading specifically?


# What does is mean by buy(long) volatility or sell(short) volatility in option trading specifically?












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)?

## Answer by siou0107 (score 5)

https://quant.stackexchange.com/a/49957

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: $$\frac{1}{2}\int_0^T{e^{rt}\Gamma_tS_t^2\left(\sigma_t^2 - \tilde{\sigma}^2 \right)dt}$$