Long Call, Delta Neutral: Gains Both Ways, Yet No Arbitrage
You buy a European call on GM (assume GM pays no dividends) and make the position delta neutral by shorting $\Delta = N(d_1)$ shares of GM stock against it.
(a) GM's stock price immediately jumps up. What happens to the value of your portfolio? What if it immediately jumps down instead?
(b) If the portfolio gains value whichever way the stock moves, isn't this an arbitrage? Explain why or why not, using the relationship between theta, gamma and delta implied by the Black-Scholes-Merton PDE.
Hints
- Expand the call in a Taylor series in $S$: the delta term is cancelled by the short stock, leaving $\tfrac12\Gamma(\Delta S)^2$, and a long call has $\Gamma > 0$.
- Now include the passage of time. The Black-Scholes-Merton PDE, $\Theta + rS\Delta + \tfrac12\sigma^2S^2\Gamma = rV$, ties theta to gamma.
- For the delta-neutral portfolio $\Pi = c - \Delta S$, the PDE gives $\Theta_\Pi + \tfrac12\sigma^2S^2\Gamma = r\Pi$: positive gamma must be paid for with negative theta, and the expected gamma gain over $dt$ equals the theta loss when the stock moves at its implied volatility.
Worked Solution
How to Think About It: Delta neutrality kills the first-order exposure to $S$, so the next term in the Taylor expansion, gamma, decides the sign of an instantaneous move's P&L. For a long call gamma is positive, so both directions look profitable. The catch is the other variable, time: the Black-Scholes-Merton PDE forces a long-gamma position to have negative theta of exactly the size that makes the position fair when the stock moves at its implied volatility.
Quick Estimate: Take $S = K = 100$, $r = 5\%$, $\sigma = 20\%$, $\tau = 0.5$: $c = 6.89$, $\Delta = 0.598$, $\Gamma = 0.0274$, $\Theta = -8.12$ per year. An instantaneous move of $\pm 5$ changes the hedged portfolio by about $\tfrac12\Gamma(5)^2 = 0.34$ (exact: $+0.32$ up, $+0.35$ down). But a day of theta costs $8.12/365 = 0.022$, and a typical daily move is $100 \times 0.2/\sqrt{252} = 1.26$, whose gamma gain is $\tfrac12 \times 0.0274 \times 1.26^2 = 0.022$. The two match: on an average day the hedged position breaks even.
Formal Solution:
Part (a): Instantaneous moves
*Step 1 -- Taylor expansion of the portfolio.* Let $\Pi = c - \Delta S$ with $\Delta = N(d_1)$ fixed at the moment of the move. For an instantaneous change $\Delta S$ (no time passes), $$\Delta\Pi = \left[\Delta\cdot\Delta S + \tfrac12\Gamma(\Delta S)^2 + O((\Delta S)^3)\right] - \Delta\cdot\Delta S = \tfrac12\Gamma(\Delta S)^2 + O((\Delta S)^3).$$
*Step 2 -- Sign.* For a long call $\Gamma = N'(d_1)/(S\sigma\sqrt{\tau}) > 0$, so $\Delta\Pi > 0$ whether the stock jumps up or down. (In the example, $+10$ gives $+1.21$ and $-10$ gives $+1.44$; the asymmetry is the third-order term.)
Part (b): Why this is not an arbitrage
*Step 3 -- Bring in time.* The gain in Step 1 assumes no time passes. Over an interval $dt$ the call also loses time value. The Black-Scholes-Merton PDE for any derivative $V$ on a non-dividend stock is $$\Theta + rS\frac{\partial V}{\partial S} + \tfrac12\sigma^2S^2\Gamma = rV.$$ Apply it to the delta-neutral portfolio $\Pi = c - \Delta S$ (whose delta is zero and whose gamma equals the call's): $$\Theta_\Pi + \tfrac12\sigma^2S^2\Gamma = r\Pi.$$ Numerically: $-8.12 + \tfrac12(0.04)(10^4)(0.0274) = -8.12 + 5.47 = -2.64 = 0.05 \times (6.89 - 59.77)$, as the identity requires.
*Step 4 -- Interpretation.* Over $dt$ the P&L of the hedged position is approximately $$d\Pi \approx \Theta_\Pi\,dt + \tfrac12\Gamma(dS)^2 = \tfrac12\Gamma\left[(dS)^2 - \sigma^2S^2\,dt\right] + r\Pi\,dt.$$ The term $r\Pi\,dt$ is just interest on the (negative) cash value of the hedged package; the rest is $\tfrac12\Gamma S^2$ times realized minus implied variance over the interval. Under the risk-neutral measure $E[(dS)^2] = \sigma^2S^2dt$, so the expected excess P&L is zero: the positive gamma is exactly paid for by negative theta. You profit only if the stock actually moves more than the implied volatility predicts, and you lose if it moves less. That is a volatility bet, not an arbitrage.
*Step 5 -- Practical caveat.* Even with continuous rehedging, the position is exposed to the difference between realized and implied volatility and to jumps; with discrete rehedging there is additional hedging error. Long gamma, short theta positions are common but they are risk positions.
Answer:
(a) The delta-neutral portfolio gains approximately $\tfrac12\Gamma(\Delta S)^2 > 0$ for an immediate jump in either direction, because a long call has positive gamma.
(b) No. Time decay offsets the gamma gain: the PDE gives $\Theta_\Pi + \tfrac12\sigma^2S^2\Gamma = r\Pi$ for the delta-neutral portfolio, so the expected gamma profit $\tfrac12\Gamma\sigma^2S^2\,dt$ is exactly cancelled by theta when the stock moves at its implied volatility. The position is a bet that realized variance exceeds implied variance, not a riskless profit.
Intuition
A delta-hedged long call is a bet on movement, not direction: to second order its P&L is $\tfrac12\Gamma(\Delta S)^2$, positive for any move. There is no free lunch because the same option bleeds theta every instant, and the PDE says the bleed is exactly the gamma gain you would expect if the stock moved at its implied volatility: $\tfrac12\sigma^2S^2\Gamma$ per unit time, with the loan on the hedge accounting for the $r\Pi$ term. You make money only if realized variance beats implied variance. This gamma-theta trade-off is the core economics of every options desk and of volatility trading in general.