# What is gamma to do with realized volatility?

I keep hearing that gamma is a bet on realized volatility. That is, if we are long gamma then we need higher realized volatility to come in the future in order to make a profit.

from other source:

If you are long gamma and delta-neutral, you make money from big moves in the underlying, and lose money if there are only small moves.

I get it that why short gamma hurts, but honestly I dont know why long gamma and delta neutral makes money if realized vol is higher.

• They key word here is "delta-neutral", if you are long an option and you are dynamically hedged your profit will be zero if Volatility turns out equal to what was priced into the option, positive if vol turns out greater and negative if vol turns out less than expected. It is a consequence of the fact that options are priced with a volatility forecast in mind and will retrospectively turn out cheap/just right/expensive depending how volatile the underlying is in reality. Commented May 5, 2022 at 12:24
• See here quant.stackexchange.com/questions/33205/… and here quant.stackexchange.com/questions/33371/… for mathematical formula for P&L on a delta hedged position. The profit is proportional to $\Gamma$ Commented May 5, 2022 at 12:34
• @nbbo2 when you say that vol turns out greater then do you mean RV > IV Commented May 5, 2022 at 12:37
• and when you say that vol turns lesser than expected, do you mean RV < IV ? Commented May 5, 2022 at 12:38
• Yes and Yes. You got it. Commented May 5, 2022 at 12:40

The pic below shows a call option value at some point before expiry as a function of the underlying. At the expense of stating an obvious fact, we note that the option value as a function of the underlying is not linear.

Imagine the option is on 100 units of the underlying stocks, and right now, the option Delta is 0.6. The straight line is a linear function, and it depicts the value of a holding a portfolio of 60 stocks.

Let's imagine we are long the option but short the 60 stocks: therefore the option is Delta-hedged at this very point in time.

Note that when the value of the underlying stock increases, we lose money on the hedge, but we make money on the option: and because the option value is non-linear, we make more money on the option than we lose on the hedge (i.e. the option value line is above the straight line).

What if the value of the stock decreases? The value of holding 60 stocks decreases more than the value of the option: therefore we again make money (cause we are short the stocks).

We make money on larger moves up or down because being long the option means we are long convexity (i.e. gamma, i.e. we are long a pay-off that has a positive second derivative with respect to the uderlying: just think of it as a graph: if we are long a graph that has a pay-off $$x^2$$ and we are short a graph that has a pay-off $$x$$, we are long "gamma" (or convexity)).

That's the magic of convexity (or "Gamma").

Ps: for completeness, Option Gamma is just the second order derivative of the option price with respect to the underlying...

PPS: why do you lose money if there are only "small moves" in the underlying? Because it costs money to be long Gamma (this is true for any asset, including Bonds): even a delta-hedged long Gamma position will always require an initial investment to set up (i.e. in case of options, this would be the premium that we pay to buy the option): this premium will not be recovered for "small" moves in the underlying (specifically if those moves correspond to lower realized volatility than what had been priced as the implied volatility into the option).

Gamma tells you how much your delta changes as the underlying price moves back and forth. If the gamma is positive it means that the delta increases as the underlying goes up, and delta decreases as it goes down. I.E you get longer the rising asset, and shorter the falling asset. This is ideal as gamma acts as an auto-(de)leverager.

Let's take a real example. The asset price is 1, with 30 DTE options trading on it. We select the option with the largest gamma, which is right around the spot price, and it has a volatility of 20%. This equates to a Black Scholes gamma of: $$\frac{n(d1)}{s \sigma \sqrt{t}} \approx \frac{1}{1 * 0.2\sqrt{2\pi \,t}} \approx 7$$

With our initial $$\Delta$$ at 0.51, this means that for any change in the underlying price, $$dS$$, we expect our delta to change by $$\Gamma * dS = 7*dS$$. So if the underlying moved to 1.01, our new $$\Delta$$ would be expected to be $$0.51 \, + 7*(1.01-1) = 0.58$$, evaluating the delta under BSM shows that our new delta is 0.5801, up 0.691 from 0.511, it is less than 0.7 because of higher order greeks which change the gamma as price moves. Remember that greeks just describe risk, or sensitivities, they are slopes of P/Ls through different input values like spot:delta, IV:vega, and time:Theta. Gamma is the slope of delta with spot, speed is the slope of gamma with spot, and so on. Incorporating greeks into risk analysis tells us our local sensitivities to input factors.

When looking at factors like delta, it is intuitive to say how much money we will make or lose. For example, with a delta of 0.5, we expect to make 0.5 USD for every 1 USD move in the underlying. So our P/L can be written as: $$PnL = \Delta * dS = \frac{dO}{dS} * dS$$ We try and model dS so that we understand our dollar risks with current positions. For higher order greeks it is less intuitive, but I think this derivation is quite nice. Consider your position with$$S = S_0$$. As S moves to $$S_1$$, the P/L between gained over the two points is given by the average $$\Delta$$ between $$S_0$$ and $$S_1$$, multiplied by $$dS$$: $$PnL = \frac{\Delta_0 + \Delta_1}{2} * dS$$ We know that $$\Delta_1 = \Delta_0 + \Gamma_0 * dS$$ $$\therefore PnL = \frac{\Delta_0 + \Delta_0 + \Gamma_0 * dS}{2} * dS = \frac{2\Delta_0 + \Gamma_0 * dS}{2} * dS = \Delta_0 * dS + \frac{\Gamma_0}{2} * dS^2$$ In the case of delta hedged portfolio at $$t_0$$, the offsetting delta term $$-\Delta_0$$ leaves the PnL as $$PnL = \frac{\Gamma_0}{2} * dS^2$$ The dS^2 term is related to the variance (and thus volatility) of the underlying, consider the starting model that we use for GBM: $$dS = \mu S \, dt + \sigma S \, dW$$ Taking expectation of the square of dS leaves us with: $$dS^2 = \sigma^2 S^2 \, dt$$ Leaving our PnL formula as: $$\frac{\Gamma_0}{2} * \sigma^2 S^2 \, dt$$ However as you know already, gamma is not free, one must pay theta away. Without even deriving a formula for theta, we know that for a fair market, the expectation of owning an option should be 0. As in nobody expects to make or lose money when transacting (of course that's different due to risk premiums, tx costs, etc, but fine for this example). If our instantaneous P/L from $$\Gamma$$ over the period $$dt$$ is $$x$$, then our theta must be $$-x$$.

One interesting point to consider is that the price of the option must be equal to the expected profits for holding it over it's lifetime. For a market neutral position (instantaneously riskless) that is kept neutral over it's life through continuous hedging (without cost), the price of the option should be equal to integral (continuous sum) of $$\Gamma$$ PnL over it's life. Formulaically: $$Cost = \frac{1}{2} \int_0^T S_t^2 \sigma_r^2 \Gamma dt$$ Assuming $$S_t$$ as a constant, which, for small $$\sigma_r$$, and 0 carry is not too far from truth, the formula for an ATM option is approximately: $$\Gamma_{s,v,t} = \frac{1}{S * \sigma\sqrt{2\pi \,t}}$$ which simplifies the cost formula to: $$\frac{1}{2} \int \frac{S \sigma_r}{\sqrt{2\pi \,t}}dt = \frac{S \sigma_r \sqrt{T}}{\sqrt{2\pi}}$$ So now hopefully you can see the relationship between gamma, realised volatility, and the intimate relationship between those and the price of the option. Finally, since we pay $$\Theta$$ away at a proportional level to $$\frac{\Gamma S^2 \sigma^2}{2}$$ By buying or selling options and hedging them to maturity, we are replicating them at realised volatility, whilst paying implied volatility, this means that our final payoffs are given by $$\int_0^T \frac{\Gamma S^2}{2} * (\sigma_r^2 - \sigma_i^2) dt$$