Can we calculate Implied ATM volatility with Vega?

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

  • $\begingroup$ The ATMF point is where the vega reaches its maximum. $\endgroup$ – byouness Dec 19 '19 at 12:30
  • $\begingroup$ @Olorin, it's the price that's bijective, not the vega. The vega is bell shaped. $\endgroup$ – byouness Dec 19 '19 at 12:30

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


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.


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