Math Summaries

Implementing Vanna Volga

Background.

It is possible to calculate analytically the values of vanillas or barrier options using the Black-Scholes model, however, they are far from quoted prices. This is because the BS-model is based on the assumption that the volatility $\sigma$ of the stock price process remains constant throughout the lifetime of the option.

The vanna-volga method also known as the trader’s rule of thumb is based on adding an analytical correction to the Black-Scholes price of the instrument. In this note, I derive and implement the original paper The vanna-volga method for implied volatilities, by Castagna and Mercurio.

Replicating Portfolio.

Consider a Black-Scholes world with two assets : a locally risk free domestic bank account $B(t)$ and a stock $S(t)$. We assume that the volatility of the stock is stochastic, but strike-independent(flat). We have the asset dynamics:

$$ \begin{align*} dB_t &= r_{DOM}B_t dt \\ dS_t &= \mu S_t dt + \sigma_t S_t dW_t \end{align*} $$

Our aim is to value an arbitrary option contract $O=f(t,S_t,\sigma_t;K)$ with a strike $K$. We price $O$ using a standard hedging argument. We build a hedge aka replicating portfolio such that it zeroes out the greeks of our net position upto the second order.

Consider a self-financing portfolio $\Pi_t$ consisting of:

The pivot options have strikes $K_1 = K_{25P}$, $K_2 = K_{ATM}$ and $K_3 = K_{25C}$ and implied volatility quotes (market prices) $\sigma_1$, $\sigma_2$ and $\sigma_3$ which are known to us.

The value of the portfolio at time $t$ is:

$$\Pi_t = O_t - \Delta_t S_t - \sum_{i=1}^{3} x_i C_t^i \tag{1}$$

By self-financing, I mean, there is no exogenous infusion or withdrawal of cash, once the portfolio has been setup at time zero. Therefore, the changes in the portfolio are solely due to gains/losses on the constituents. The self-financing condition is:

$$d\Pi_t = dO_t - \Delta_t dS_t - \sum_{i=1}^{3} x_i dC_t^i \tag{2}$$

By Ito’s lemma, the differential of the option price $O_t$ can be written as:

$$ \begin{align*} dO_t &= \frac{\partial O}{\partial t} dt + \frac{\partial O}{\partial S_t} dS_t + \frac{\partial O}{\partial \sigma_t} d\sigma_t \\ &+ \frac{1}{2}\frac{\partial^2 O}{\partial t^2} (dt)^2 \\ &+ \frac{1}{2}\frac{\partial^2 O}{\partial S_t^2} (dS_t)^2 \\ &+ \frac{1}{2}\frac{\partial^2 O}{\partial \sigma_t^2} (d\sigma_t)^2 \\ &+ \frac{\partial^2 O}{\partial t \partial S_t} dt \cdot dS_t \\ &+ \frac{\partial^2 O}{\partial S_t \partial \sigma_t} dS_t \cdot d\sigma_t \\ &+ \frac{\partial^2 O}{\partial t \partial \sigma_t} dt \cdot d\sigma_t \tag{3} \end{align*} $$

Since $(dt)^2$, $dt \cdot dS_t$, $dt \cdot d\sigma_t$ are equal to zero, we can write:

$$ \begin{align*} dO_t &= \frac{\partial O}{\partial t} dt + \frac{\partial O}{\partial S_t} dS_t + \frac{\partial O}{\partial \sigma_t} d\sigma_t \\ &+ \frac{1}{2}\frac{\partial^2 O}{\partial S_t^2} (dS_t)^2 \\ &+ \frac{1}{2}\frac{\partial^2 O}{\partial \sigma_t^2} (d\sigma_t)^2 \\ &+ \frac{\partial^2 O}{\partial S_t \partial \sigma_t} dS_t \cdot d\sigma_t \tag{4} \end{align*} $$

Similarly, we can apply Ito’s lemma to the European vanilla pivot options to find the differential $dC^i_t$. Putting it together we have:

$$ \begin{align*} d\Pi_t &= \left(\frac{\partial O(t;K)}{\partial t} - \sum_{i=1}^{3}x_i\frac{\partial C^i(t;K_i)}{\partial t} \right) dt \\ &+ \left(\frac{\partial O(t;K)}{\partial S_t} - \Delta_t - \sum_{i=1}^{3}x_i\frac{\partial C^i(t;K_i)}{\partial S_t}\right) dS_t \\ &+ \left(\frac{\partial O(t;K)}{\partial \sigma_t} - \sum_{i=1}^{3}x_i\frac{\partial C^i(t;K_i)}{\partial \sigma_t} \right)d\sigma_t\\ &+ \frac{1}{2}\left(\frac{\partial^2 O(t;K)}{\partial S_t^2} - \sum_{i=1}^{3}x_i\frac{\partial^2 C^i(t;K_i)}{\partial S_t^2}\right)(dS_t)^2 \\ &+ \frac{1}{2}\left(\frac{\partial^2 O(t;K)}{\partial \sigma_t^2} - \sum_{i=1}^{3}x_i\frac{\partial^2 C^i(t;K_i)}{\partial \sigma_t^2} \right)(d\sigma_t)^2 \\ &+ \left(\frac{\partial^2 O(t;K)}{\partial S_t \partial \sigma_t} - \sum_{i=1}^{3}x_i\frac{\partial^2 C^i(t;K_i)}{\partial S_t \partial \sigma_t}\right) dS_t \cdot d\sigma_t \tag{5} \end{align*} $$

We claim that we can choose the weights $\Delta_t$ and $\mathbf{x}=(x_1,x_2,x_3)$ of the replicating portfolio, such that the coefficient of the terms $dS_t$, $d\sigma_t$, $(d\sigma_t)^2$ and $dS_t \cdot d\sigma_t$ are zeroed out.

We are therefore left with:

$$ \begin{align*} d\Pi_t &= \left(\frac{\partial O}{\partial t} - \sum_{i=1}^{3}x_i\frac{\partial C^i_t}{\partial t} \right) dt \\ &+ \frac{1}{2}\left(\frac{\partial^2 O}{\partial S_t^2} - \sum_{i=1}^{3}x_i\frac{\partial^2 C^i_t}{\partial S_t^2}\right)\sigma_t^2 S_t^2 dt \tag{6} \end{align*} $$

The portfolio value process has no driving Brownian motion $dW_t$ term, and hence the source of randomness has been eliminated. Therefore, $\Pi_t$ must be a locally risk-free portfolio. That is, it satisfies:

$$d\Pi_t = r_{DOM}\Pi_t dt \tag{7}$$

Calculating the VV weights

We assume hereafter, that the constant BS volatility is the at-the-money one; $\sigma = \sigma_2 = \sigma_{ATM}$. We assume $t=0$, so we can drop the argument $t$ in the call prices $C_i(t;K)$ in equation (5). The weights $\mathbf{x}=(x_1,x_2,x_3)$ are determined by solving the system of equations $A\mathbf{x}=\mathbf{b}$ where:

$$ A = \begin{bmatrix} \frac{\partial C_1(K_1)}{\partial \sigma_t} & \frac{\partial C_1(K_2)}{\partial \sigma_t} & \frac{\partial C_3(K_3)}{\partial \sigma_t} \\ \frac{\partial^2 C_1(K_1)}{\partial S_t \partial \sigma_t} & \frac{\partial^2 C_2(K_2)}{\partial S_t \partial \sigma_t} & \frac{\partial^2 C_3(K_3)}{\partial S_t \partial \sigma_t}\\ \frac{\partial^2 C_1(K_1)}{\partial \sigma_t^2} & \frac{\partial^2 C_2(K_2)}{\partial \sigma_t^2} & \frac{\partial^2 C_3(K_3)}{\partial \sigma_t^2} \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} \frac{\partial O(K)}{\partial \sigma_t} \\ \frac{\partial^2 O(K)}{\partial S_t \partial \sigma_t} \\ \frac{\partial^2 O(K)}{\partial \sigma_t^2} \end{bmatrix} $$

The entries in the first, second and third rows of $A$ and $\mathbf{b}$ are the option vega, the option vanna and the option volga.

I derived the expressions for option vega, vanna and volga here. They are:

$$ \begin{align*} \text{Vega} &= S_0 e^{-r_{FOR}T} \phi(d_{+}) \sqrt{T} \\ \text{Vanna} &= -e^{-r_{FOR}T} \phi(d_{+})\frac{d_{-}}{\sigma}\\ \text{Volga} &= S_0 e^{-r_{FOR}T}\sqrt{T}\phi(d_{+}) \frac{d_{+}d_{-}}{\sigma} \end{align*} $$

We can re-phrase the other greeks in terms of vega $\mathcal{V}$. Recall, that $d_{+}$ varies with the option strike $K$, so all other things equal, we can write $\mathcal{V} = \mathcal{V}(K)$.

$$ \begin{align*} \text{Vanna} &= -\frac{d_{-}}{\sigma S_0 \sqrt{T}} \mathcal{V}(K)\\ \text{Volga} &= \frac{d_{+}d_{-}}{\sigma} \mathcal{V}(K) \end{align*} $$

The augmented matrix $[A | b]$, therefore is:

$$ \begin{bmatrix} \mathcal{V}(K_1) & \mathcal{V}(K_2) & \mathcal{V}(K_3) & | & \mathcal{V}(K)\\ -\frac{d_{-}(K_1)}{\sigma S_0 \sqrt{T}}\mathcal{V}(K_1) & -\frac{d_{-}(K_2)}{\sigma S_0 \sqrt{T}}\mathcal{V}(K_2) & -\frac{d_{-}(K_3)}{\sigma S_0 \sqrt{T}}\mathcal{V}(K_3) & | &\frac{d_{-}(K)}{\sigma S_0 \sqrt{T}}\mathcal{V}(K)\\ \frac{d_{+}(K_1) d_{-}(K_1)}{\sigma}\mathcal{V}(K_1) & \frac{d_{+}(K_2) d_{-}(K_2)}{\sigma}\mathcal{V}(K_2) & \frac{d_{+}(K_3) d_{-}(K_3)}{\sigma}\mathcal{V}(K_3) & | & \frac{d_{+}(K) d_{-}(K)}{\sigma}\mathcal{V}(K) \end{bmatrix} $$

Cancelling out the constant terms, we get:

$$ \begin{bmatrix} \mathcal{V}(K_1) & \mathcal{V}(K_2) & \mathcal{V}(K_3) & | & \mathcal{V}(K)\\ d_{-}(K_1)\mathcal{V}(K_1) & d_{-}(K_2)\mathcal{V}(K_2) & d_{-}(K_3)\mathcal{V}(K_3) & | & d_{-}(K)\mathcal{V}(K)\\ d_{+}(K_1) d_{-}(K_1)\mathcal{V}(K_1) & d_{+}(K_2) d_{-}(K_2)\mathcal{V}(K_2) & d_{+}(K_3) d_{-}(K_3)\mathcal{V}(K_3) & | & d_{+}(K) d_{-}(K)\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_3 \leftarrow R_3 - d_{+}(K_1) R_2$, we get:

$$ \begin{bmatrix} \mathcal{V}(K_1) & \mathcal{V}(K_2) & \mathcal{V}(K_3) & | & \mathcal{V}(K)\\ d_{-}(K_1)\mathcal{V}(K_1) & d_{-}(K_2)\mathcal{V}(K_2) & d_{-}(K_3)\mathcal{V}(K_3) & | & d_{-}(K)\mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1} d_{-}(K_2)\mathcal{V}(K_2) & \log \frac{K_3}{K_1} d_{-}(K_3)\mathcal{V}(K_3) & | & \log \frac{K}{K_1} d_{-}(K)\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_2 \leftarrow R_2 - d_{-}(K_1)R_1$, we get:

$$ \begin{bmatrix} \mathcal{V}(K_1) & \mathcal{V}(K_2) & \mathcal{V}(K_3) & | & \mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1}\mathcal{V}(K_2) & \log \frac{K_3}{K_1}\mathcal{V}(K_3) & | & \log \frac{K}{K_1}\mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1} d_{-}(K_2)\mathcal{V}(K_2) & \log \frac{K_3}{K_1} d_{-}(K_3)\mathcal{V}(K_3) & | & \log \frac{K}{K_1} d_{-}(K)\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_3 \leftarrow R_3 - d_{-}(K_2) R_2$, we get:

$$ \begin{bmatrix} \mathcal{V}(K_1) & \mathcal{V}(K_2) & \mathcal{V}(K_3) & | & \mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1}\mathcal{V}(K_2) & \log \frac{K_3}{K_1}\mathcal{V}(K_3) & | & \log \frac{K}{K_1}\mathcal{V}(K)\\ 0 & 0 & \log \frac{K_3}{K_1} \frac{K_3}{K_2}\mathcal{V}(K_3) & | & \log \frac{K}{K_1} \log \frac{K}{K_2}\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_1 \leftarrow \log(K_2/K_1) R_1 - R_2$, we get:

$$ \begin{bmatrix} \log \frac{K_2}{K_1}\mathcal{V}(K_1) & 0 & - \log \frac{K_3}{K_2}\mathcal{V}(K_3) & | & -\log \frac{K}{K_2}\mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1}\mathcal{V}(K_2) & \log \frac{K_3}{K_1}\mathcal{V}(K_3) & | & \log \frac{K}{K_1}\mathcal{V}(K)\\ 0 & 0 & \log \frac{K_3}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_3) & | & \log \frac{K}{K_1} \log \frac{K}{K_2}\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_1 \leftarrow \log(K_3/K_1)R_1 + R_3$, we get:

$$ \begin{bmatrix} \log \frac{K_3}{K_1}\log \frac{K_2}{K_1}\mathcal{V}(K_1) & 0 & 0 & | & \log \frac{K}{K_3}\log \frac{K}{K_2}\mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1}\mathcal{V}(K_2) & \log \frac{K_3}{K_1}\mathcal{V}(K_3) & | & \log \frac{K}{K_1}\mathcal{V}(K)\\ 0 & 0 & \log \frac{K_3}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_3) & | & \log \frac{K}{K_1} \log \frac{K}{K_2}\mathcal{V}(K) \end{bmatrix} $$

Performing the elementary row operation $R_2 \leftarrow \log(K_3/K_2) R_2 - R_3$, we get:

$$ \begin{bmatrix} \log \frac{K_3}{K_1}\log \frac{K_2}{K_1}\mathcal{V}(K_1) & 0 & 0 & | & \log \frac{K}{K_3}\log \frac{K}{K_2}\mathcal{V}(K)\\ 0 & \log\frac{K_2}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_2) & 0 & | & \log \frac{K}{K_1} \log \frac{K_3}{K}\mathcal{V}(K)\\ 0 & 0 & \log \frac{K_3}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_3) & | & \log \frac{K}{K_1} \log \frac{K}{K_2}\mathcal{V}(K) \end{bmatrix} $$

Thus, the solution vector $\mathbf{x}$ is:

$$ \begin{align*} x_1(K) &= \frac{\log \frac{K_3}{K}\log \frac{K_2}{K}\mathcal{V}(K)}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}\mathcal{V}(K_1)}\\ x_2(K) &= \frac{\log \frac{K}{K_1} \log \frac{K_3}{K}\mathcal{V}(K)}{\log\frac{K_2}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_2)}\\ x_3(K) &= \frac{\log \frac{K}{K_1} \log \frac{K}{K_2}\mathcal{V}(K)}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}\mathcal{V}(K_3)} \tag{9} \end{align*} $$

The VV Option price.

We can now proceed to the definition of an option price that is consistent with the market prices of the basic options. The above replication argument shows that a portfolio comprising of $x_i(K)$ units of the option with strike $K_i$ and $\Delta_0$ units of the underlying gives a local perfect hedge in a Black-Scholes world. The hedging strategy has to be implemented at prevailing market prices, which generates an extra cost with respect to the Black-Scholes value of the options portfolio. Such a cost has to be added to the Black-Scholes price $O^{BS}(K)$, with $t=0$, to produce an arbitrage free price which is consistent with the quoted option prices $C_1^{MKT}(K_1)$, $C_2^{MKT}(K_2)$ and $C_3^{MKT}(K_3)$.

In fact, in the case of a short-maturity $T$, the equation (7) can be written as:

$$ \begin{align*} &((S_T - K)^{+} - O^{BS}(K)) - \Delta_0(S_T - S_0)\\ -& \sum_{i=1}^{3} x_i(K) (C_i^{MKT}(K_i) - C_i^{BS}(K_i))\\ &= r_{DOM} (O^{BS}(K) - \Delta_0 S_0 - \sum_{i=1}^{3} x_i(K_i)C_i^{BS}(K_i))T \tag{10} \end{align*}$$

Therefore, setting

$$O_{VV}^{MKT}(K) = O^{BS}(K) + \sum_{i=1}^{3}x_i(K)(C_i^{MKT}(K_i) - C_i^{BS}(K_i))\tag{11}$$

We get:

$$ \begin{align*} (S_T - K)^{+} &= O^{MKT}{VV}(K) + \Delta_0(S_T - S_0) \\ &+ r{DOM} (O^{BS}(K) - \Delta_0 S_0 - \sum_{i=1}^{3} x_i(K)C_i^{BS}(K_i))T \tag{12} \end{align*}$$

Thus, when actual market prices are considered, the option payout $(S_T-K)^{+}$ can still be replicated by starting with an initial capital of $O_{VV}^{MKT}(K)$, buying $\Delta_0$ units of the stock and $x_i(K)$ units of the pivot options with strike $K_i$, and investing the rest at the cash rate $r_{DOM}$.

Hence, implicitly ignoring the replication error over longer maturities, the price of the option must the initial capital required to setup the hedge portfolio $O_{VV}^{MKT}(K)$.

The option premium $O_{VV}^{MKT}(K)$ equals the Black-Scholes price of the option $O^{BS}(K)$ plus a vanna-volga correction term, or overhedge $O_{VV}$, which is the difference between the market quoted prices of the pivot options and the Black-Scholes prices of the pivot options under the constant BS volatility $\sigma = \sigma_{ATM}$.

Deriving the implied volatility smile.

We can now derive an easy approximation for the vanna-volga implied volatility smile curve $\xi(K)$.

In formula (11), we Taylor expand the market quotes $C_1^{MKT}(K_1)$, $C_2^{MKT}(K_2)$, $C_3^{MKT}(K_3)$ and $O^{MKT}{VV}(K)$, about $\sigma = \sigma{2}$. We have:

$$ \begin{align*} C_i^{MKT}(\sigma_i,K_i) &= C_i^{BS}(\sigma,K_i)+ \frac{\partial C_i^{BS}(\sigma,K_i)}{\partial \sigma}(\sigma_i - \sigma)\\ O_{VV}^{MKT}(\xi(K),K) &= O^{BS}(\sigma,K)+ \frac{\partial O^{BS}(\sigma,K)}{\partial \sigma}(\xi(K) - \sigma) \tag{13} \end{align*} $$

Equivalently,

$$ \begin{align*} C_i^{MKT}(\sigma_i,K_i) - C_i^{BS}(\sigma,K_i) &= \mathcal{V}(K_i)(\sigma_i - \sigma)\\ O_{VV}^{MKT}(\xi(K),K) - O^{BS}(\sigma,K) &= \mathcal{V}(K)(\xi(K) - \sigma) \tag{14} \end{align*} $$

Substituting (13) and (14) in formula (11), we get:

$$ \begin{align*} \mathcal{V}(K)(\xi(K) - \sigma) &= \sum_{i=1}^{3} x_i(K) \mathcal{V}(K_i)(\sigma_i - \sigma) \end{align*} $$

Since $\sigma_2 = \sigma$, the second term in the summation vanishes. Simplifying, we have:

$$ \begin{align*} \mathcal{V}(K)(\xi(K) - \sigma_2) &= x_1(K)\mathcal{V}(K_1)(\sigma_1 - \sigma_2) + x_3(K)\mathcal{V}(K_3)(\sigma_3 - \sigma_2)\\ \xi(K) - \sigma_2 &= x_1(K)\frac{\mathcal{V}(K_1)}{\mathcal{V}(K)}(\sigma_1 - \sigma_2) + x_3(K)\frac{\mathcal{V}(K_3)}{\mathcal{V}(K)}(\sigma_3 - \sigma_2)\\ \xi(K) &= \frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}} \sigma_1 + \frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}\sigma_3 \\ &+ \left(1 - x_1(K)\frac{\mathcal{V}(K_1)}{\mathcal{V}(K)} - x_3(K)\frac{\mathcal{V}(K_3)}{\mathcal{V}(K)} \right)\sigma_2 \tag{15} \end{align*} $$

But, we know from the matrix system $A\mathbf{x}=b$, that the weights $\mathbf{x}=(x_1,x_2,x_3)$ satisfy:

$$x_1(K) \mathcal{V}(K_1) + x_2(K) \mathcal{V}(K_2) + x_3(K) \mathcal{V}(K_3) = \mathcal{V}(K) \tag{16}$$

So,

$$ \begin{align*} \left(1 - x_1(K)\frac{\mathcal{V}(K_1)}{\mathcal{V}(K)} - x_3(K)\frac{\mathcal{V}(K_3)}{\mathcal{V}(K)} \right) &= x_2(K) \frac{\mathcal{V}(K_2)}{\mathcal{V}(K)}\\ &=\frac{\log \frac{K}{K_1} \log \frac{K_3}{K}}{\log\frac{K_2}{K_1} \log \frac{K_3}{K_2}} \tag{17} \end{align*} $$

Substituting (17) in the expression (15), we have the result:

$$ \xi^{1}(K) := \xi(K) = \frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}} \sigma_1 + \frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}\sigma_3 + \frac{\log \frac{K}{K_1} \log \frac{K_2}{K_1}}{\log \frac{K_2}{K_1} \log \frac{K_3}{K_2}} \tag{18} $$

The VV-implied volatility smile $\xi(K)$ is thus approximated by a linear combination of the implied vol quotes $\sigma_1$, $\sigma_2$ and $\sigma_3$ of the vanilla pivot options with coefficients that add up to $1$. This approximation is extremely accurate in the interval $[K_1,K_3]$. The wings, however, tend to be overvalued.

A second order approximation for VV-smile.

Let’s Taylor expand the market quotes $C_i^{MKT}(\sigma_i,K_i)$ and $O_{VV}^{MKT}(\xi(K),K)$ upto the second order.

$$ \begin{align*} C_i^{MKT}(\sigma_i,K_i) - C_i^{BS}(\sigma,K_i) &= \frac{\partial C_i^{BS}(\sigma,K_i)}{\partial \sigma}(\sigma_i - \sigma) + \frac{1}{2}\frac{\partial^2 C_i^{BS}(\sigma,K_i)}{\partial \sigma^2}(\sigma_i - \sigma)^2 \\ O^{MKT}_{VV}(\xi(K),K) - O^{BS}(\sigma,K) &= \frac{\partial O^{BS}(\sigma,K)}{\partial \sigma}(\xi(K) - \sigma) + \frac{1}{2}\frac{\partial^2 O^{BS}(\sigma,K)}{\partial \sigma^2}(\xi(K) - \sigma)^2 \tag{19} \end{align*} $$

Equivalently,

$$ \begin{align*} C_i^{MKT}(\sigma_i,K_i) - C_i^{BS}(\sigma,K_i) &= \mathcal{V}(K_i)(\sigma_i - \sigma) + \frac{1}{2}\frac{d_{+}(K_i)d_{-}(K_i)}{\sigma}\mathcal{V}(K_i)(\sigma_i - \sigma)^2 \\ O^{MKT}_{VV}(\xi(K),K) - O^{BS}(\sigma,K) &= \mathcal{V}(K)(\xi(K) - \sigma) + \frac{1}{2}\frac{d_{+}(K) d_{-}(K)}{\sigma}\mathcal{V}(K)(\xi(K) - \sigma)^2 \tag{20} \end{align*} $$

Substituting (20) in formula (11):

$$ \mathcal{V}(K)(\xi(K) - \sigma) + \frac{1}{2}\frac{d_{+}(K)d_{-}(K)}{\sigma}\mathcal{V}(K)(\xi(K) - \sigma)^2 \\ = x_1\mathcal{V}(K_1)(\sigma_1 - \sigma) + \frac{1}{2}x_1\frac{d_{+}(K_1)d_{-}(K_1)}{\sigma}\mathcal{V}(K_1)(\sigma_1 - \sigma)^2\\ +(\sigma_3 - \sigma) + \frac{1}{2}x_3\frac{d_{+}(K_3)d_{-}(K_3)}{\sigma}\mathcal{V}(K_3)(\sigma_3 - \sigma)^2 \tag{21} $$

Simplifying we have:

$$ \begin{align*} &(\xi(K) - \sigma_2) + \frac{1}{2}\frac{d_{+}(K)d_{-}(K)}{\sigma}(\xi(K) - \sigma_2)^2 \\ =& \frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}}(\sigma_1 - \sigma_2) + \frac{1}{2}\frac{d_{+}(K_1)d_{-}(K_1)}{\sigma}\frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}}(\sigma_1 - \sigma_2)^2\\ +& \frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}(\sigma_3 - \sigma_2) + \frac{1}{2}\frac{d_{+}(K_3)d_{-}(K_3)}{\sigma}\frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}(\sigma_3 - \sigma_2)^2 \end{align*}\tag{22} $$

Let

$$ \begin{align*} D_1(K) &:= \frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}}(\sigma_1 - \sigma_2) + \frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}(\sigma_3 - \sigma_2)\\ &= \xi^1(K) - \sigma_2 \tag{23} \end{align*} $$

and

$$ \begin{align*} D_2(K) &:= d_{+}(K_1)d_{-}(K_1)\frac{\log \frac{K_3}{K}\log \frac{K_2}{K}}{\log \frac{K_3}{K_1}\log \frac{K_2}{K_1}}(\sigma_1 - \sigma_2)^2 \\ &+ d_{+}(K_3)d_{-}(K_3)\frac{\log \frac{K}{K_1} \log \frac{K}{K_2}}{\log \frac{K_3}{K_1} \log \frac{K_3}{K_2}}(\sigma_3 - \sigma_2)^2 \tag{24} \end{align*} $$

Substituting (23) and (24) in equation (22), we get:

$$(\xi(K) - \sigma_2) + \frac{d_{+}(K)d_{-}(K)}{2\sigma_2}(\xi(K) - \sigma_2)^2 = D_1(K) + \frac{D_2(K)}{2\sigma_2}\tag{25}$$

Multiplying throughout by $2\sigma_2$, we get:

$$2\sigma_2(\xi(K) - \sigma_2) + d_{+}(K)d_{-}(K)(\xi(K) - \sigma_2)^2 = 2\sigma_2 D_1(K) + D_2(K)\tag{26}$$

Solving for $\xi(K) - \sigma_2$, we have:

$$ \begin{align*} \xi(K) - \sigma_2 &= \frac{-2\sigma_2 \pm \sqrt{4\sigma_2^2-4d_{+}(K)d_{-}(K)(2\sigma_2 D_1(K) + D_2(K))}}{2d_{+}(K)d_{-}(K)}\\ \xi^2(K) := \xi(K) &=\sigma_2 + \frac{-\sigma_2 \pm \sqrt{\sigma_2^2-d_{+}(K)d_{-}(K)(2\sigma_2 D_1(K) + D_2(K))}}{d_{+}(K)d_{-}(K)} \tag{27} \end{align*} $$