Gamma PnL Formula and Break-Even volatility

Question

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).

Answer 1

Good question! The answer to this is no. Let us work through a simple example to see why. Assume that the Gamma is $10$ and that the break-even move is $1$. For simplicity, also assume that, these are unchanged by price moves in the underlying (this is reasonably accurate for small price moves), so:

• $\Gamma = 10$
• $\delta S_{Break-Even} = 1$

Note that we are dealing with a Delta-hedged portfolio here, so the starting value of Delta is $0$, i.e. $\Delta = 0$. However, once the price moves, the Delta will equal the Gamma times the price move, i.e.: $\Delta = \Gamma \times \delta S$. Hence, once the break-even move happens (i.e. when $\delta S = \delta S_{Break-Even}$), the Delta will equal the Gamma times the break-even move, i.e. $\Delta = \Gamma \times \delta S_{Break-Even} = 10 \times 1 = 10$. This means that when you Delta-hedge this to lock in the break-even move, your cash inflow will be $\Delta \times \delta S_{Break-Even} = 10 \times 1 = 10$. Hence, in this example, you need to make $10$ currency units to break even.

Now, let's see what happens if the price instead changes by HALF of a break-even move, i.e. $\delta S = 0.5$. As before, the delta becomes $\Delta = \Gamma \times \delta S = 10 \times 0.5 = 5$. If you Delta-hedge to lock in this move, your cash inflow will be $\Delta \times \delta S = 5 \times 0.5 = 2.5$.

As you can see, when you lock in the profit on half of a break-even move, it is equal to one quarter of the profit on a whole break-even move, not to one half. This is because you lock in a smaller price move on a smaller volume (in this case, half the price move on half the volume naturally gives you one quarter of the profit). Thus, if your hedging strategy is to lock in the profit more frequently, you need a larger number of smaller moves and the total move you need will be bigger than one break-even move.