Why an Out-of-the-Money Binary Call Can Fall as Volatility Rises
Summary
The note explains why an out-of-the-money binary call price need not rise monotonically with implied volatility. It suggests plotting the option price across increasing volatility values and gives an example with spot just below the strike, zero interest rates, and one year to expiry. In that setup, the price first rises and then falls, reaching a peak at a low volatility level.
The explanation points to differentiating the Black–Scholes d₂ term with respect to volatility: the binary call’s value depends on the probability of finishing above the strike, and that probability can decline as volatility increases when spot is below strike. The example illustrates the behavior but does not provide the derivative or a full proof. Its conclusion concerns an out-of-the-money binary call under the stated setup; other moneyness, rates, and maturities may change the behavior.
Key ideas
- An out-of-the-money binary call price can be non-monotonic in implied volatility.
- Plotting price against volatility can reveal a rise followed by a decline.
- The behavior can be analyzed by differentiating d₂ with respect to volatility.
- The cited numerical illustration uses spot below strike, zero rates, and one year to expiry.
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Full text
# Is the price of a binary call not monotonous with vol for OTM # Is the price of a binary call not monotonous with vol for OTM Is this true and how would you prove it ## Answer by Mark Joshi (score 2) https://quant.stackexchange.com/a/15393 The OTM binary is not increasing with volatility. Simply plot the price as a function of implied volatility. eg take S = 0.99 K = 1 r=0 T =1 and $\sigma$ vary upwards in steps of 0.01 starting at 0. It peaks around 0.04. Differentiating $d_2$ with respect to $\sigma$ makes this behaviour obvious.
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