How big a daily move breaks even on your theta?

Question

You hold a delta-hedged long option, gamma $\Gamma = 0.05$ per share and theta $\Theta = -\$ 0.04 $per share per day, on a$ $100 $stock priced at$ 20%$ implied vol.

What size daily move exactly breaks even on the day, and why does that number equal the move the option was priced for?

Accepted Answer

Over one day, a delta-hedged option earns gamma P&L 	frac12\Gamma(\Delta S)^2 and pays theta \Theta\,dt. Break even where they cancel: 	frac12\,\Gamma\,(\Delta S^)^2 = \Theta \;\Longrightarrow\; \Delta S^ = \sqrt{\frac{2\,\Theta}{\Gamma}} = \sqrt{\frac{2 	imes 0.04}{0.05}} = \sqrt{1.6} \approx \1.26. If the stock moves more than \approx\1.26 today, you profit; less, you lose. Why this equals the implied daily move. Black–Scholes ties theta to gamma through implied vol: \Theta \approx 	frac12\,\sigma{	ext{impl}}^2\,S^2\,\Gamma\,dt. Substitute into the break-even formula: \Delta S^ = \sqrt…

How big a daily move breaks even on your theta?

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