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Inverting Black–Scholes Vega to Recover At-the-Money Implied Volatility

Article Quant Q&A · Author: Turtle203
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Summary

The document derives a way to recover implied volatility from Black–Scholes vega when the option is at the money. Starting from the vega expression, it isolates the squared Black–Scholes d-one value using the option’s vega, spot price, and time to expiry. It then substitutes the standard expression for d-one in terms of moneyness and total volatility, producing an equation that can be solved for total volatility.

The derivation expresses the general case as a quadratic in squared total volatility, then notes that at-the-money log-moneyness is zero, which simplifies the equation. The response does not discuss numerical stability, input constraints, or handling cases where the algebra yields multiple or non-real solutions. Its use is therefore an analytic inversion under the stated Black–Scholes assumptions, with the at-the-money case receiving the clearest simplification.

Key ideas

  • Black–Scholes vega can be rearranged to express squared d-one in terms of observed vega and option inputs.
  • Substituting the d-one formula yields an equation in total volatility.
  • The general inversion reduces to a quadratic in squared total volatility.
  • At-the-money log-moneyness is zero, simplifying the algebra.
  • The derivation does not address solution selection or numerical edge cases.

Tags

OptionsVolatilityDerivatives pricing
Full text
# Calculating Implied ATM Volatility with Vega


# Calculating Implied ATM Volatility with Vega












Can we calculate Implied ATM volatility with Vega?

Normally, Vega is derived from Volatility, but I wonder the availability of the opposite way.

## Answer by Magic is in the chain (score 3, accepted)

https://quant.stackexchange.com/a/50378

Assuming you mean inverting the Black Scholes Vega, it does seem possible:

Take the Vega formula:

$V = S \sqrt{\tau} n{\left (d_{1} \right)}= S \sqrt{\tau} \frac{1}{\sqrt{2 \pi}}e^{-0.5 d_1^2}$

Rearrange to isolate $d_1$:

$d_1^2=-2\ln \left(V \frac{\sqrt{2 \pi}}{S \sqrt{\tau}} \right)$

To simplify, let's call the right hand side $ C=-2\ln \left(V \frac{\sqrt{2 \pi}}{S \sqrt{\tau}} \right)$, so

$d_1^2=C$

Now let's recall the famous expression:

$d_1= \frac{\ln S_0 -\ln Ke^{-r \tau} }{\sigma \sqrt{\tau}}+\frac{1}{2}\sigma \sqrt{\tau}$

Which we can abbreviate (M is the money-ness and v is the total volatility):

$d_1= \frac{\ln M }{v}+\frac{1}{2}v$

Plugging into the previous expression,

$d_1^2=\left( \frac{\ln M }{v}+\frac{1}{2}v\right)^2=C$

Now we need to solve for v (which is the implied vol times square root of time to maturity, $\sigma \sqrt{\tau}$), so let's expand the square, and simplify:

$ \frac{\left(\ln M\right)^2 }{v^2}+\frac{1}{4}v^2+2 \frac{\ln M }{v}\frac{1}{2}v=C$

$v^4+4 \left(\ln M -C\right)v^2+4 \left(\ln M\right)^2=0$

So all set for the quadratic formula:

$v^2=\frac{-4 \left(\ln M -C\right)\pm \sqrt{16\left(\ln M -C\right)^2-16 \left(\ln M\right)^2}}{2}$

Which we can simplify:

$v^2=-2 \left(\ln M -C\right)\pm 2\sqrt{\left(\ln M -C\right)^2- \left(\ln M\right)^2}$

For ATM, ln M will be zero, so the formula simplifies considerably: 4C.

Shown 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.