Volatility Surfaces and Time Scaling in Black–Scholes d₁ and d₂
Summary
The document asks whether using strike- and maturity-dependent volatility removes the usual square-root-of-time scaling in Black–Scholes d₁ and d₂. One response explains that the paper may define its volatility input as total volatility, combining annualized implied volatility with the square root of time to expiry. Under that convention, the time factor is already included in the input, and the notation differs from annualized volatility commonly used elsewhere.
A second response stresses that a volatility surface does not by itself remove time scaling: the standard formula still uses volatility multiplied by the square root of time. The excerpts do not resolve which convention the cited paper actually uses; checking the paper’s definitions is necessary. The discussion also notes that alternative notation for total volatility can reduce confusion. It gives no numerical example or broader treatment of volatility-surface option pricing.
Key ideas
- A volatility surface describes volatility as a function of strike and maturity.
- The standard Black–Scholes d₁ and d₂ expressions include volatility scaled by the square root of time.
- A paper may instead define its volatility input as total volatility, with the time scaling already incorporated.
- The notation alone does not establish which volatility convention a paper uses.
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Full text
# Does using volatility surfaces instead of constant vol dismisses you from scaling your vol terms in d1/2?
# Does using volatility surfaces instead of constant vol dismisses you from scaling your vol terms in d1/2?
I am working on the paper Valuation and Hedging of cryptocurrency inverse options by A. Sepp and V. Lucic (https://www.researchgate.net/publication/382233254_Valuation_and_hedging_of_cryptocurrency_inverse_options) and it seems that in page 5, they use σ(K,T) for the volatility terms of d1 and d2, but do not scale it by the square root of time. Does it mean that the $\sqrt{T-t}$ is included in $\sigma$(K,T) or that by using a vol surface, we actually don't need to adjust volatility since it's already not considered constant?
## Answer by D Stanley (score 0)
https://quant.stackexchange.com/a/85592
It appears that they are using a convention by defining $\sigma(K,T) = \sigma\sqrt{T-t}$ where $\sigma$ would be the annualized implied volatility. This just simplifies the black-scholes formula which defines $\sigma$ as a constant instead of as a function of time (and strike).
According to Claude (take it for what it's worth), $v(K,T) = \sigma\sqrt{T-t}$ is more commonly used for this simplification instead of $\sigma(K,T)$ to avoid (understandable) confusion from re-defining $\sigma$.
## Answer by Michael Hastings (score 0)
https://quant.stackexchange.com/a/85842
No. Volatility surfaces don't dismiss you from the scaling.
In Black–Scholes, d₁ and d₂ are:
d₁ = [ln(S/K) + (r + σ²/2)T] / (σ√T) d₂ = d₁ − σ√TShown 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.