# 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.

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

$$d_1^2=C$$

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.