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
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Author: Casiopea
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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
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Author: D Stanley
Original post: https://quant.stackexchange.com/a/85592
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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
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Author: Michael Hastings
Original post: https://quant.stackexchange.com/a/85842
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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 answer — Quantitative Finance Stack Exchange Author: D Stanley Source score (net votes, not local likes): 0 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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