Gamma, Volatility, and the Slope of Delta

Companion to the delta note: raise volatility enough and gamma goes to zero — because a flat delta curve has no slope.

The intuition (the right one)

The argument in one breath:

vol rises and rises ⇒ every delta $\to 0.5$ ⇒ OTM = ATM = ITM = 0.5 everywhere ⇒ the deltas are all equal, so delta stops changing ⇒ $\Gamma = 0$.

Keep raising the volatility input. In the delta note we saw that every delta gets pulled toward the middle: out-of-the-money, at-the-money and in-the-money all compress toward $\sim$0.5. So the delta profile flattens — it becomes almost the same number wherever spot sits relative to the strike.

Now recall what gamma is: the rate at which delta changes as spot moves,

$$\Gamma \;=\; \frac{\partial \Delta}{\partial S}.$$

If delta is the same at every moneyness, then sliding the spot changes nothing — delta does not move — so its slope is zero. Flat delta ⇒ no gamma. That is the whole argument, and it is correct.

The formula agrees

The Black–Scholes gamma is

$$\Gamma \;=\; \frac{\varphi(d_1)}{S\,\sigma\sqrt{\tau}}, \qquad \varphi(x)=\frac{1}{\sqrt{2\pi}}\,e^{-x^{2}/2},$$

and as $\sigma$ grows both pieces push it toward $0$: the factor $1/(\sigma\sqrt{\tau})$ shrinks, and $\varphi(d_1)$ collapses because $d_1\to+\infty$. For an at-the-money option ($S=100$, one year, $r=q=0$):

Gamma falls by more than two orders of magnitude and heads to zero — exactly what the flat-delta argument predicts.

The trader's picture

At low vol, the delta curve is a steep S-ramp: it climbs from 0 to 1 across a narrow band of spot right around the strike. That steepness is gamma — tall and concentrated at the strike (the source of pin risk and the hedging pain near expiry). Raise vol and the ramp stretches out: the same rise is now spread over a huge range of spot, so the slope everywhere becomes gentle. Push vol to the extreme and the curve is flat — zero slope, zero gamma. High vol smears the gamma spike into a low, broad bump and finally into nothing.

The honest caveat

Strictly, at infinite vol the deltas flatten toward 1, not 0.5 (the drift term of the delta note). But it does not matter for the argument: flat at 0.5 or flat at 1, a flat delta curve has no slope, so $\Gamma\to 0$ either way. And at any finite vol gamma is small, not exactly zero — the deltas compress toward the middle but keep a faint tilt, and that residual tilt is the residual gamma.

One consequence worth carrying: as vol (or maturity) rises, gamma dies but vega takes over. Short-dated, low-vol options are "all gamma"; long-dated, high-vol options are "all vega." Same underlying, opposite risk profile.

Takeaway. Gamma is the slope of the delta curve. Raising volatility flattens that curve — every delta drifts toward the middle — so the slope, and with it gamma, collapses toward zero. Low vol: a tall gamma spike at the strike. High vol: flat delta, no gamma.

Assumes $r=q=0$ and a flat volatility surface (the exam convention); $S=100$ in the table. Non-zero carry shifts the terms but not the mechanism.