Why Driftless GBM Can Price Binary Puts Above Calls

Summary

This note asks why a long-dated cash-or-nothing put can be worth more than a comparable call when the underlying follows a geometric Brownian motion with zero drift. It points to Black–Scholes binary option pricing, where call and put values depend on opposite tail probabilities under the risk-neutral distribution.

The explanation is the asymmetry of the lognormal distribution. Zero drift in the stated setup keeps the expected underlying value at its initial level, but it does not make the probabilities above and below that level equal. The distribution can place more probability below its mean, while the less frequent outcomes above it have larger values; those larger outcomes offset the lower probability in the expectation. The note is a qualitative explanation rather than a worked derivation, and it does not specify maturity, volatility, strike, or discounting assumptions needed to compare particular prices.

Key ideas

• Zero drift does not make a lognormal distribution symmetric around its mean.
• A binary option pays according to whether the underlying finishes on one side of its strike.
• More probability can lie below the mean even when rarer upper outcomes are larger.
• The expectation balances probability and outcome size, while a binary payoff focuses on probability.

Tags

# Why is long term binary put option more expensive than call assuming driftless GBM?


# Why is long term binary put option more expensive than call assuming driftless GBM?












Says X follows a driftless geometric brownian motion(GBM) given a volatility ($\mu = 0$). It gives the expected value of its initial spot. (Source: https://en.wikipedia.org/wiki/Geometric_Brownian_motion)

$E(X) = X_0$

Since Black Scholes Pricing Model assumes spots following GBM,

$Binary\ Cash \ or \ Nothing \ Call = e^{rt}N(d_2)$

and

$Binary\ Cash \ or \ Nothing \ Put = e^{rt}N(-d_2)$

My question is by referring to Black Scholes formula, why would cash or nothing put is supposed to be more expensive than call, provided that both were on driftless GBM? Would Black Scholes assumed downside probability has higher than upside probability?

## Answer by siou0107 (score 7, accepted)

https://quant.stackexchange.com/a/50339

This is due to the asymetry in the lognormal distribution.

You have a higher probability of being below the mean, but since values of $X$ are lower that compensates, in the expectation computation, for the lower probability of being above the mean with higher values of $X$.

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