Skip to content
All library documents

Put–Call Symmetry and Logarithmic Strike Distance in Black–Scholes

Article Quant Q&A · Author: Ang Yiwei
Read the original

Summary

The document explains why an out-of-the-money call and put placed the same absolute distance above and below spot need not have equal prices. Under Black–Merton–Scholes, a call price can be related to a put price at a reciprocal strike, with a scaling factor involving the strikes and spot. This relationship is known as put–call symmetry and also holds for a limited class of stochastic volatility models when volatility is uncorrelated with the asset.

The practical point is that equal absolute strike offsets do not represent equal distances in the model’s logarithmic price space. A comparison based on symmetric log differences is more appropriate for examining the stated symmetry. The document offers a theoretical explanation rather than a numerical demonstration or empirical test, and the relationship depends on model assumptions; it should not be treated as a universal equality across option models or market conditions.

Key ideas

  • Put–call symmetry relates a call to a put at a reciprocal strike, with a price scaling factor.
  • Equal absolute strike offsets around spot are not symmetric in logarithmic price space.
  • The cited symmetry applies under Black–Merton–Scholes and a limited class of uncorrelated stochastic volatility models.
  • Model assumptions determine whether this relationship can explain observed option prices.

Tags

OptionsDerivatives pricingVolatility
Full text
# Will OTM Vanilla Put equal to OTM Vanilla Call with different relative strike?


# Will OTM Vanilla Put equal to OTM Vanilla Call with different relative strike?












I wanted to test if my strike moves 0.01 away from my current spot for OTM Call and Put. Say I have the following parameters:

For put: $Spot = 1, \sigma = 1, K_p = 0.99 , r = 0, q = 0, $

For call: $Spot = 1, \sigma = 1, K_c = 1.01 , r = 0, q = 0$

However, using fOption package in R yields the following result:

OTM Call option will have the higher price than OTM put.

I am wondering what is reason behind. Could it be the Geometric Brownian Motion assumption with lognormal distribution of spots?

Appreciate if anyone could provide some ideas on this. Thanks.

## Answer by user34971 (score 2, accepted)

https://quant.stackexchange.com/a/49946

Under the Black-Merton-Scholes model, and also for a limited class of stochastic volatility models (when volatility is not correlated to the asset), the following relationship holds:

$$ C(S,K) = \frac{K}{S} P(S, S^2/K) $$

This relationship is called put-call symmetry. Here is a short introduction:

PCS

Also, note that under geometric Brownian motion, $1+x$ is not equidistant from 1 as $1 - x$ where in your example $x=0.01$. You need to look at log-differences.

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.