Delta hedging in practice: how often, and what it costs

Black–Scholes assumes continuous costless hedging. Neither holds, and the gap is the whole of an options desk’s day-to-day P&L: you hedge discretely, which introduces error, and each hedge costs a spread, which means hedging more is not always better.

Equation 3.1

What a hedged position earns

Over a day, a delta-hedged long option earns the difference between the variance the stock delivered and the variance you paid for — weighted by dollar gamma.

\text{P\&L} \approx \tfrac{1}{2}\Gamma S^2\left(\sigma_{\text{realised}}^2 - \sigma_{\text{implied}}^2\right)\Delta t

$\tfrac12\Gamma S^2$: Dollar gamma, the sensitivity to squared *percentage* moves.
$\sigma^2_{\text{realised}} - \sigma^2_{\text{implied}}$: The whole bet, in variance terms.

Proposition 3.2

The break-even move

Set the gamma term equal to the theta and solve: the position breaks even on a daily move of roughly $\sigma_{\text{implied}}\sqrt{\Delta t}$ — that is, on exactly the move the implied volatility was pricing. Move more and gamma wins; move less and theta does. This is the cleanest statement of what a long option position needs.

Holds when

A $16\%$ implied volatility needs about a $1\%$ daily move to break even.
The relation is about the *size* of the move, not its direction.
It holds per rebalancing interval, so hedging twice a day changes the relevant $\Delta t$ .

Figure 3.3 · More rebalancing trades error against cost. Illustrative scaling, with cost normalised to 0.1 per hedge. Error falls slowly as 1/√n while trading cost rises with n. The crossing is not a universal optimum; the desk chooses a band using its own spreads and risk tolerance.

Example 3.4

You are long a delta-hedged option with dollar gamma such that a $1\%$ move earns $\$500$ of gamma P&L. Implied volatility is $16\%$ and the stock moves $1.5\%$ today. What do you make?

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Worked solution

Formula $\text{gamma P\&L} \propto (\text{move})^2, \qquad \text{theta} = \text{gamma P\&L at the break-even move}$
Substitute $\text{break-even} = 16\%/16 = 1\% \Rightarrow \theta = -\$500$
Solve $\text{gamma at } 1.5\% = 500 \times \left(\tfrac{1.5}{1.0}\right)^2 = \$1{,}125$
$\text{net} = 1125 - 500$
Answer $+\$625$

Sanity check. A move half as large again produces gamma P&L a bit over twice as big, because the payoff is quadratic. That convexity is exactly what you bought, and it is why a few large days dominate a long-gamma month.

Proposition 3.5

Hedging error from discreteness

Hedging $n$ times over the life of the option leaves a random hedging error whose standard deviation falls as $1/\sqrt{n}$ . It is mean-zero for a fairly priced option, so discrete hedging does not bias your P&L — it adds variance. The consequence is that a "hedged" position still has a distribution of outcomes, and a desk sizes for that rather than treating the hedge as exact.

Holds when

Quadrupling the hedge frequency halves the error, and multiplies the transaction costs by four.
Hedging on a fixed *move* rather than a fixed *time* is usually better, because it concentrates effort where gamma is being realised.

Delta hedging in practice: how often, and what it costs

After this lesson you should be able to

What a hedged position earns

The break-even move

Hedging error from discreteness

The rest of this lesson is in Premium