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?

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Casiopea · External communityPost link
External question — Quantitative Finance Stack Exchange Author: Casiopea Original post: https://quant.stackexchange.com/questions/85591 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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?
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D Stanley · External communityPost link
External answer — Quantitative Finance Stack Exchange Author: D Stanley Original post: https://quant.stackexchange.com/a/85592 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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 $\sigma$ .
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Michael Hastings · External communityPost link
External answer — Quantitative Finance Stack Exchange Author: Michael Hastings Original post: https://quant.stackexchange.com/a/85842 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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√T
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Quoted from Forex.com.bd-Editorial External question — Quantitative Finance Stack Exchange Author: Casiopea Source score (net votes, not local likes): 0 Original post: https://quant.stackexchange.com/questions/85591 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. 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?

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