When we derive the P&L of a delta hedged option we obtain:

$$ \text{P&L}=\dfrac{1}{2}\Gamma(\delta S)^{2}-\theta\delta t $$

and setting equal to zero and rearranging we obtain:

$$ \dfrac{1}{2}\Gamma(\delta S)^{2}-\theta\delta t =0 $$ $$ \implies\dfrac{1}{2}\Gamma(\delta S)^{2} =\theta\delta t\ $$ $$ \implies(\delta S)^{2} =2\dfrac{\theta\delta t}{\Gamma} $$ $$ \implies\delta S =\sqrt{\dfrac{2\theta\delta t}{\Gamma}} $$

Suppose $$ \delta t = 1\text{ day}$$

so that we obtain the break-even daily move in the underlying to be:

$$ \delta S_{Break-Even} =\sqrt{\dfrac{2\theta}{\Gamma}}. $$ My question is this. Suppose for concreteness $\Gamma$ and $\theta$ are such that our break-even daily move, $\delta S_{Break-Even}$, is 10. Does this mean that we break even when the end of day price on the underlying has changed by 10 or more OR does this mean that we merely need to realize a total change of 10 across that day (but not necessarily end the day with a price that is different by 10). For example: if the underlying starts the day at 20 and follows the path : 20 -> 22 -> 20 -> 26, then we have realized a total of 2 + 2 + 6 = 10 if we sum up each individual incriment (vs 26-20=6 as the total change).


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