# Option: link between Vega and Gamma

I am reading Dynamic Hedging by N. Taleb and I do not understand this statement in Chapter 9:

The vega is the integral of the gamma profits over the duration of the option at one volatility minus the same integral at a different volatility. The vega P/L that results from the volatility going higher for a long option holder should be equal to the expected sum of the gamma profits over the period should the market goess his way.

What are the "gamma profits"? How do you set up this integral?

I would like to have both an intuitive and a mathematical, answer if possible.

I found this question linked to this but it is not clear in the answers to me why this statement hold: Link between Vega and Gamma

Thanks for your help.

• Briefly "gamma p&L" refers to the P&L on a dynamic replication strategy, a strategy of trading the underlying to maintain the correct delta over time. Sometimes referred to as gamma scalping (q.v.). Commented May 30 at 14:53

## 4 Answers

When you hedge an option's delta at implied volatility, the resulting PnL over timestep, dt, is: $$PnL = (\Delta_i * dS + \frac{\Gamma_i dS^2}{2} + \Theta * dt) - \Delta_i * dS$$ Leaving: $$\frac{\Gamma_i dS^2}{2} + \Theta * dt$$ In the BSM framework, where $$\sigma_i == \sigma_r$$, Theta = Gamma, but when there is a mismark between $$\sigma_i$$ & $$\sigma_r$$, the resulting PnL over timestep, dt, is $$\frac{\Gamma_i * S^2}{2} * (\sigma_r^2 - \sigma_i^2) * dt$$

Which translates to the difference between realised and implied variance over dt, scaled by the cash gamma priced at implied volatility. Where the gamma is negative for a short option postion and vice versa for a long. So the cumulative PnL becomes: $$\sum_{t=0}^{T} \frac{\Gamma_{i,t} * S_t^2}{2} * (\sigma_{i, t}^2 - \sigma_i^2) * dt$$

As you increase the hedging frequency, reducing dt, in the limit at $$dt \rightarrow 0$$, the cumulative PnL can be written: $$\int_{t}^{T} \frac{\Gamma_{i,t} * S_t^2}{2} * (\sigma_{i, t}^2 - \sigma_i^2) * dt$$

Which leads us to Taleb's statement that Vega PnL is the integral of the gamma PnL over T-t at two different volatilties.

One can also plot out the expected PnLs; Vega * ($$\sigma_i - \sigma_r$$), Cash Gamma * ($$\sigma_i^2 - \sigma_r^2$$) to find that vega PnL locally approximates the expected gamma PnL.

Why? Take the BSM gamma formula: $$\Gamma = \frac{n(d1)}{S * \sigma * \sqrt{T}}$$

Then let's approximate for an ATMF option, where d1 = 0: $$\Gamma = \frac{1}{S * \sigma_i * \sqrt{2*\pi * T}}$$

Integrating w.r.t T: $$\int \frac{1}{S * \sigma_i * \sqrt{2*\pi * T}} dT = \frac{\sqrt{2T}}{S * \sigma_i * \sqrt{\pi}}$$

So, the approximate Gamma PnL (0.5 * Cash Gamma * (rv^2 - iv^2)) simplifies to: $$\frac{S * \sqrt{T}}{\sigma_i * \sqrt{2 \pi}} * (\sigma_r^2 - \sigma_i^2)$$ So where $$\sigma_i \approx \sigma_r$$, this approximates to: $$\frac{S * \sqrt{T}}{\sqrt{2 \pi}} * (\sigma_r - \sigma_i)$$

BSM vega is $$Vega = S * n(d1) * \sqrt{T}$$, thus Vega PnL is $$S * n(d1) * \sqrt{T} * (\sigma_r - \sigma_i)$$, and we again approximate the vega to be the vega at the ATMF strike (approximating n(d1) to 1/sqrt(2*pi)), we are left with:

Vega PnL = $$\frac{S * \sqrt{T}}{\sqrt{2\pi}} * (\sigma_r - \sigma_i)$$

Comparing with the gamma PnL approximation:

Gamma PnL = $$\frac{S * \sqrt{T}}{\sqrt{2\pi}} * (\sigma_r - \sigma_i)$$

So locally for ATMF options, Vega PnL = Gamma PnL. Where Vega PnL is the change in option value marked at different IVs, and Gamma PnL is the integral (realistically a cumulative sum) of spread between realised and implied variance, scaled by cash gamma multiplied by dt.

• Thanks, very clear, especially the approximations, I will remember them for sure. I see that you start directly from the P/L that show the link between realized volatility and implied used to "start" the position. Could you explain how the realized volatility is linked with the quadratic variation process and why we divide it by the square of the price? Here an answer in which they recover the P/L I mentioned: quant.stackexchange.com/questions/33371/…. Thank for your time. Commented Jun 3 at 11:03
• That's just the notation for the 2nd derivative of the option position w.r.t the underlying (d2v/ds2). Realised volatility is the square root of the variance of the returns process (or the change in log price process) Commented Jun 3 at 12:08
• Thanks. My fault, here is the correct link: quant.stackexchange.com/questions/39010/…. Here there is this realized volatility as the quadratic variation divided by the square fo the price. Commented Jun 3 at 12:12
• Maybe I'm wrong but I'm certain that that part of the equation is identical to saying (dS/S)^2. I.e the squared return. Commented Jun 3 at 15:22
• I do not know. I looked but I can't find anything about that term involving the quadratic variation and its interpretation when divided by the square of the underlying asset. Commented Jun 6 at 14:00

$$Vega*(Vol1-Vol2)=C(t,S(t),vol1)-C(t,S(t),vol2)$$ (1=2)

where vol1 and vol2 are close enough for 1 and 2 to be the same.

Now we delta hedge both calls at implied volatility. We ignore that change in delta and theta while shifting vol slightly as they are second order.

Look at $$dC(t,S(t),vol1)-dC(t,S(t),vol2)+delta*dS-delta*dS$$ (3)

This is itself a long comment so I am linking the answer here.PnL of a delta hedge at implied vol.

Integrate (3), and realize at expiry it is 0. So initial value equals the integral of (3), which the link shows to be gamma profits.

• Is it right to say that, if the realized volatility is equal to the implied one, this latter used for the delta hedging at the beginnig of position, the gamma P/L are zero? Thank you. Commented May 31 at 11:52
• Yes, though it is very theoretical. Commented May 31 at 12:28
• Well, I meant that the error in the P/L, made when the realized volatility is different from the one used to hedge, is zero. I suppose there is always some gamma P/L or not because there would be arbitrage opportunity? I'm having difficulties in computing the expectation of the gamma P/L. Commented May 31 at 17:17
• Gamma PnL is usually what we call the second term in the ito expansion. It is never 0 unless the spot never moved or gamma is 0. If you mean hedging error when realized vol is different, it is the formula. I don't think you can compute it, you can use MC. For intuition take a look at the 2nd answer. Commented Jun 2 at 21:09
• Thank you for your time. The only things I am missing now is the "realized" volatility (see my comment in JohnGalt answer). Do you know, even intuitively, why we are dividing by the square of the price at time $t$ the quadratic variation at time t? I saw this in the delta hedging error evaluation. Commented Jun 3 at 9:51

This is simple. If you are far away from maturity, your option price will more sensitive to volatility on your underlying, effective change on your underlying price won't have any significative impact.

On the opposite, if the maturity is closer your option price will be more sensitive to effective change on the underlying price. for the volatility, since it has annual range, we are days before maturity, you can easily see it won't have much impact.

I like to see Gamma as "realized volatility" and the Vega as "gamma reserve".

Hope this informal explanation helps.

• Could you articulate a bit more the concept about "Gamma is realized volatility and Vega is gamma reserve"? Thank you. Commented May 31 at 8:16
• I let you check this topic: quant.stackexchange.com/questions/70751/… Commented May 31 at 19:52
• If it doesn’t help, we will derive the relation so you can see the link. If you need intuitive / none mathematic point of view just let me know :). Commented May 31 at 19:53
• Forget for a moment Vega and Gamma, look at the option price, at the beginning the underlying volatility will be somewhat more important than your underlying price, because the volatility set a range of possible price. Then, when the time pass, volatility will change but no more than few percents in a normal regime, the time pass, you arrive at 6 months before the expiration. you have 2 choice, does the volatility on the underlying will allow you to go over the strike or not? Commented Jun 5 at 12:16
• If the current volatility won't allow you to get back above the strike, you will have a big vega and a small gamma, because you hope for more volatility and since you are far away from strike a small effective change won't matter so much. on the opposite, if your volatility allows you to easily hit above the strike, you will care for effective change. Does it make sense? (In this example we are long call) Commented Jun 5 at 12:22

I have another answer. We know by the link, the call valued at the incorrect vol (beta) loses gamma PnL. We know the call valued rightly loses no PnL (all strategies are fair in the risk neutral world). So a difference of call prices is the (expected) gamma PnL. The difference is vega times change in vol. This links vega to expected gamma PnL.

I am also adding intuition, which is much simpler if you see prices as expectations

Prices are expectations against density (payoff*mass). If you have a convex payoff (high gamma), then increasing volatility creates a larger price separation because it is taking advantage of the convexity to increase the overall value of the expectation.

Example: $$Payoff: [1,1,3,10,20], Density: [0,0.33,0.33,0.33,0]$$ So expectation is $$14/3$$.

Now density: $$[0.1,0.1,0.1,0.1,0.1]$$ So expectation is $$31/3.$$

If the payoff was not convex (0 gamma), it would look like $$[-1,1,3,5,7]$$

and you can check that increasing vol makes no difference. So 'vega' is a way of taking advantage of gamma.

• Very clear example about volatility and payoff convexity. Commented Jun 3 at 9:51