Deriving the Gamma–Theta Relationship from Expected Hedged PnL

Summary

The document derives the familiar link between theta and gamma by examining the expected profit and loss of a delta-hedged, vega-hedged position. After the delta and volatility terms are removed, the second-order price move contributes a gamma term, while elapsed time contributes theta. Under the no-arbitrage argument used in the answer, expected PnL is set to zero. For an underlying with instantaneous volatility σ, the expected squared price change over a short interval is σ squared times the interval, yielding the relationship between theta and gamma in the stated notation.

The answer also flags a convention issue: the gamma in the formula is dollar gamma, whereas many texts define gamma per unit of the underlying; converting between them introduces a factor of the underlying price squared. This is a local, continuous-time approximation that depends on the volatility and hedging assumptions. The brief exchange does not address dividends, discrete hedging, transaction costs, or changes in volatility over the interval.

Key ideas

• Delta and vega hedging remove the corresponding first-order PnL exposures in the stated setup.
• The second-order price contribution is governed by gamma and the expected squared price move.
• For instantaneous volatility σ, the expected squared price change over a short interval is proportional to σ squared and time.
• Setting expected hedged PnL to zero gives the gamma–theta relation.
• Dollar gamma differs from per-unit gamma by a factor involving the underlying price squared.

Tags

# How to derive the relationship between gamma and theta?


# How to derive the relationship between gamma and theta?












I am trying to derive this formula Θ = –0.5 × Γ × S^2 × σ^2 to see where it comes from.

My thinking is that PnL = delta dS + Vdσ + 0.5Γ(dS)^2 + Θdt. Assume we delta hedged and vega hedged, first and second term drops off, so we have PnL = 0.5Γ(dS)^2 + Θdt.

Now assuming no free lunch, PnL = 0. Hence, 0 = 0.5Γ(dS)^2 + Θdt.

Here I am stuck

## Answer by dm63 (score 1, accepted)

https://quant.stackexchange.com/a/67829

Continuing in your notation, it is the expected PnL=0, so

E[0.5$\Gamma $dS^2 +$ \theta $dT]=0

Now, E[dS^2]= $\sigma$^2 dT

So we are done, except to note that your $\Gamma$ is the dollar gamma whereas most textbooks use the gamma per unit of stock, and they are related by a factor of S^2.

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.