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Why Option Gamma and Theta Often Have Opposite Signs

Article Quant Q&A · Author: tcquant
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Summary

The document examines the relationship between gamma and theta in the Black–Scholes equation. Rewriting the pricing equation in terms of the Greeks shows that theta and the volatility-weighted gamma term are linked, but their sum also includes interest-rate carry through the difference between option value and delta times the underlying price. The question is whether one must assume a zero interest rate to infer opposite signs.

The responses say that the carry term is often small in practice, so gamma and theta commonly move in opposite directions, while stopping short of claiming an exact universal sign rule. A second explanation offers intuition: time decay moves the option’s P&L profile toward its expiration profile, whereas volatility keeps it farther away. This is a qualitative picture, and the discussion does not lay out the conditions for every option type, position, or market assumption. The Black–Scholes relationship and the small-carry observation provide the main analytical basis.

Key ideas

  • The Black–Scholes equation links theta, gamma, and an interest-rate carry term.
  • Opposite gamma and theta signs are an approximation when the carry contribution is small, not an unconditional identity.
  • Time decay moves an option's value toward its expiration payoff, while volatility supports value away from expiration.
  • The intuitive explanation does not specify conditions covering every option position or market setup.

Tags

OptionsDerivatives pricingVolatilityStatistics
Full text
# Why gamma and theta have opposite signs?


# Why gamma and theta have opposite signs?












I saw some textbooks use B-S equation to explain why gamma and theta have opposite signs in most of the cases. For example, John Hull's classic book.

The explanation is, first write B-S equation in terms of greeks:

$\frac{\partial V}{\partial t}+rS\frac{\partial V}{\partial S}+\frac{1}{2}\sigma^2S^2\frac{\partial^2 V}{\partial S^2}=rV$

$\Theta+rS\Delta+\frac{1}{2}\sigma^2S^2\Gamma=rV$

$\Theta+\frac{1}{2}\sigma^2S^2\Gamma=r(V-S\Delta)$

Do we need to assume r=0, in order to draw the conclusion that gamma and theta have opposite signs?

## Answer by jaamor (score 4)

https://quant.stackexchange.com/a/16533

I think you are answering your own question.

Hull states: "When $\Theta$ is large and positive, $\Gamma$ tends to be large and negative and vice versa."

In practice, you can expect $r(V-S \Delta)$ to be quite small.

## Answer by x4444 (score 0)

https://quant.stackexchange.com/a/57157

As time passes P/L line is getting closer and closer to expiration P/L line.

Volatility resists P/L line to get closer to expiration P/L line. You can think about Volatility as an elastic element btw P/L line and expiration P/L line. The higher is Volatility the bigger is the gap btw P/L line and the expiration P/L line.

Time and Volatility are opposite forces on P/L diagram.

Time pushes P/L line toward expiration P/L line.

Volatility pushes P/L line away from expiration P/L line.

Volatility affects daily P/L slices thickness. Higher volatility - thicker the daily slice - https://www.dropbox.com/s/s7pscwm2yp31sha/Screenshot%202020-08-05%2018.51.18.png?dl=0

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.