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How Theta and Gamma Determine Delta-Hedging P&L

Article Quant Q&A · Author: gill

Summary

The note addresses an apparent funding gap when rebalancing a short call hedge as time passes and the underlying price stays unchanged. Its answer uses the Black–Scholes pricing equation with zero interest and dividends, where theta offsets the gamma contribution implied by the assumed volatility. For a small time interval, delta-hedged option P&L is approximated by a gamma term from the underlying’s squared move plus a theta term from elapsed time.

Combining those expressions shows that realized movement relative to the volatility used for pricing determines the hedge’s gain or loss. A long-gamma position loses when the underlying does not move, and gains when its move exceeds the model’s volatility-scaled threshold; the short-gamma position has the opposite exposure. This explains the role of realized movement rather than identifying a separate cash source for theta. The approximation assumes small time steps and the stated Black–Scholes setup, so it is not a general account of discrete hedging costs or model error.

Key ideas

  • Under the stated Black–Scholes assumptions, theta and gamma are linked by the pricing equation.
  • Delta-hedged P&L over a short interval combines a gamma contribution and a theta contribution.
  • Long gamma loses when the underlying remains still over the interval.
  • Movement greater than the model’s volatility-scaled threshold benefits long gamma in the approximation.
  • The result abstracts from discrete hedging costs and model error.

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Full text
# How do you finance theta decay when replicating an option?


# How do you finance theta decay when replicating an option?












When constructing a replicating portfolio for a short position in a call option under Black Scholes, I am not able to pinpoint the source of gains from theta decay. When theta decay materializes, I don't understand how the replicating portfolio generates the gains necessary to finance the new, higher value (less negative), position.

To be concrete, I will try illustrate the question with a numerical example. Suppose we have the following parameters: $K=10$, $S(0)=10$, $\sigma=0.1$, $r=0$, $T=10, q=0$.

The call price formula from Wikipedia is

$$C(S,t)=S\Phi(d_1)-K\Phi(d_2)$$

where

$$d_1 = \frac{\log(S/K)+ (\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \\ d_2 = d_1 - \sigma \sqrt{T-t} $$

At $t=0$, we have $C(10,0)=1.256$. To replicate a short position in the option, we use that the initial delta is $C_S(10,0)=\Phi(d_1)=0.5628$.

We begin by taking a short position of delta in the underlying and invest the remaining proceeds in the money market. We therefore have a short position worth $-10*C_S(10,0)=-5.628$ in the risky asset and a money market position of $-C_S(10,0)+10*C(10,0)=4.372$. We can check that $4.372-5.628=-1.256=-C(10,0)$.

Now suppose we are at $t=1$ and the price of the underlying doesn't change, i.e. $S(1)=10$. The value of the call is now $C(10,1)=1.192$, which means that a short position in the call should net 0.064 in profit. However, before adjusting the hedge, the value of our replicating portfolio seems to be the same: We have a short position of -5.628 and a money market position of 4.372, giving $4.372-5.628=-1.256 \leq -C(10,1)=-1.192$.

To continue the replication, we would need a short position in the underlying equal to $-10*C_S(10,1)=-5.596$ and a money market position of $C_S(10,1)+10*C(10,1)=4.404$. But we need an additional 0.064 which we don't seem to have: where do we get the 0.0032 units of the risky asset to decrease the short position and the 0.032 units of numeraire to increase the money market position? It would seem like we are missing 0.064 and can't continue self-financing replication.

Perhaps this has something to do with using a discrete time increment in the example?

## Answer by Antoine Conze (score 2)

https://quant.stackexchange.com/a/61678

I haven't checked your numbers, but the delta hedging principle is that if the realized stock volatility is equal to the pricing/hedging volatility, then theta and gamma compensate each other.

Since you have assumed $r = q = 0$ the pricing PDE is $$ \frac{\partial C}{\partial t} + \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 C}{\partial S^2} = 0 $$ which you can rewrite using the option greeks notation as $$ \Theta + \frac{1}{2}\sigma^2 S^2 \Gamma=0 $$ Now assume you are delta hedging your option on a small time step $\delta t$. Your PnL is $$ \delta \text{pnl}=C(S+\delta S, t + \delta t)-C(S,t)-\Delta \delta S \approx \frac{1}{2}\Gamma \delta S^2+\Theta \delta t $$ Combining with the pricing PDE you obtain $$ \delta \text{pnl}\approx \frac{1}{2}\sigma^2 S^2 \Gamma\left(\left(\frac{\delta S}{\sigma S}\right)^2 - \delta t\right) $$

So if the market does not move ($\delta S/S=0$) and you are long gamma, you lose money. Conversely if the market move (up or down) is $|\delta S/S| > \sigma \sqrt{\delta t}$ you make money on a long gamma position.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.