Implied volatility model-free

I know that $$\operatorname{IV-model \space free}=2 \int_{0}^{+\infty}\frac{c_0(T,Ke^{r(T-t)})-c_0(t,Ke^{r(T-t)})}{K^2}\operatorname{d}K$$ is calculated using an iterative procedure, i.e. setting a volatility value and comparing theoretical price obtained by software (with that volatility value) and the market price: if theoretical price is less than market price we have to increase the volatility (and vice-versa). This would be the reason:

"You can not invert the BS-equation because implied volatility is repeated both in 1) the PDF of Normal standard like standard deviation of log-returns distribution and 2) the argument of Normal standard in $$d_1$$ and $$d_2$$ terms."

Deriving BS-close form I have $$\varphi(S_t)=e^{-r(T-t)}\int_{z_0}^{+\infty}((S_te^{(r-\frac{\sigma^2}{2})(T-t)+\sigma\sqrt{T-t}z})-K)^+\Phi(z)\operatorname{d}z$$ with $$z_0=\frac{\operatorname{ln}(\frac{S_t}{K})+(r+\frac{\sigma^2}{2})(T-t)}{\sigma\sqrt{T-t}}$$ and $$\Phi(z)\doteq \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$$. So, knowing that $$d_1:=-z_0+\sigma\sqrt{T-t}$$ and $$d_2:=d_1-\sigma\sqrt{T-t}$$ (so the point 2) is trivially true), what is the $$\sigma=\operatorname{IV}$$ to which it refers the point 1)?

Thanks in advance for any help!

The "IV model-free" formula you wrote is the implied volatility of the log contract, which is the fair strike of a variance swap. That is the meaning of your IV model-free.

The IV in the Black-Scholes vanilla options formula is the volatility parameter you need to input into the formula to match the market price of vanilla options.

So these are two different things.

• Or, to add a bit: You may want to approximate your model free IV integral from observed option prices, no need for the BS formula here. Jan 5, 2021 at 14:43
• @ilovevolatility My question is (and I apologize to you for my English): how can I justify the statement in quotation marks? Point 2) is because we find $\sigma$ in $d_1$ and $d_2$, by definition. But for point 1), my PDF is $\frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$ and there's no sign of $\sigma$ here. To which $\sigma$ the statement it refers the point 1)? Jan 6, 2021 at 9:18
• @ilovevolatility In any case, referring to your answer, I'd have said that $\operatorname{IV-model \space free}$ is obtained by the equation $c_t(T,K)=e^{-r(T-t)}\int_{0}^{+\infty}(S_T-K)^+\mathbb{Q}(S_T \equiv S)\operatorname{d}s$ and deriving two times the Call value for $K$. So $\frac{\partial^2c}{\partial K^2}=-e^{-r(T-t)}\frac{\partial}{\partial K}\int_{K}^{+\infty}\mathbb{Q}(S_T \equiv S)\operatorname{d}s = -e^{-r(T-t)}\mathbb{Q}(S_T \equiv K)\operatorname{d}s$: this probability describes $\operatorname{IV}$. It reflects the expectations that market has to exercise the option at maturity Jan 6, 2021 at 9:35