How Delta-Hedged Gamma P&L Relates to Option Vega

# Option: link between Vega and Gamma


# 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.

## Answer by Newquant (score 2, accepted)

https://quant.stackexchange.com/a/79603

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.

## Answer by Arshdeep (score 2)

https://quant.stackexchange.com/a/79553

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

## Answer by JohnGalt (score 2)

https://quant.stackexchange.com/a/79554

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.

## Answer by Arshdeep (score 1)

https://quant.stackexchange.com/a/79582

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.