## FX Conventions

We list and derive the various FX conventions.

### FX Volatility Smile - Vanna Volga method

This section is a bit longer than the other sections, and as such we split the discussion into sub-sections to simplify the presentation:

• In the first sub-section, we explain and derive the formulae relating to Vega-Vanna-Volga hedging, often known as Vanna-Volga hedging.
• In the next sub-section, we explain how the cost of the Vanna-Volga hedging, in addition to the cost of Delta hedging, can be used to compute the price of an option.
• In the final sub-section, we show how the Vanna-Volga pricing formula can be used to construct the volatility smile (volatility as a function of strike for a given maturity), which is the main objective of this section.

### Vanna Volga - Hedging

We know from the Black Scholes section that the Vega of a European option is:

$$Vega = S e^{- r_{f} \tau} \sqrt{\tau} n{\left (d_{1} \right)}$$

Differentiating Vega with respect to S and $$\sigma$$ give us Vanna and Volga, respectively:

$$Vanna =\frac{\partial Vega}{\partial S} =- e^{- r_{f} \tau} \frac{n{\left (d_{1} \right)} d_{2}}{\sigma} = - Vega \frac{ d_{2}}{S \sigma \sqrt{\tau}}$$ $$Volga =\frac{\partial Vega}{\partial \sigma} = S e^{- r_{f} \tau} \sqrt{\tau} \frac{n{\left (d_{1} \right)} d_{1} d_{2}}{\sigma} = Vega \frac{d_{1} d_{2}}{\sigma}$$

Let us assume we want to hedge the Vega, Vanna, and Volga of a European option with the usual parameters. As we need to match three characteritics, we have three requirements, and from algebra we know that we will need to come up with three equations. Assume we have three options available with strikes $$K_{1}, K_{2}$$, and $$K_{3}$$. Let $$w_{1}, w_{2}$$, and $$w_{3}$$ represent the weight of the three options in the portfolio that we build to hedge our option's Vega, Volga, and Vanna. Then, we require:

$$Vega= w_{1} Vega_{1}+w_{2} Vega_{2}+w_{3} Vega_{3}$$ $$Vanna= w_{1} Vanna_{1}+w_{2} Vanna_{2}+w_{3} Vanna_{3}$$ $$Volga= w_{1} Volga_{1}+w_{2} Volga_{2}+w_{3} Volga_{3}$$

Where the subscripts identify the options and note K is the only factor that differentiate the options. Now expressing the Vanna and Volga in terms of Vega, we get:

$$Vega= w_{1} Vega_{1}+w_{2} Vega_{2}+w_{3} Vega_{3}$$ $$- Vega \frac{ d2}{S \sigma \sqrt{\tau}}= - w_{1} Vega_{1} \frac{ d2_{1}}{S \sigma \sqrt{\tau}} - w_{2} Vega_{2} \frac{ d2_{2}}{S \sigma \sqrt{\tau}}- w_{3} Vega_{3} \frac{ d2_{3}}{S \sigma \sqrt{\tau}}$$ $$Vega \frac{d1 d2}{\sigma}= w_{1} Vega_{1} \frac{d1_{1} d2_{1}}{\sigma} +w_{2} Vega_{2} \frac{d1_{2} d2_{2}}{\sigma}+w_{3} Vega_{3} \frac{d1_{3} d2_{3}}{\sigma}$$

Canceling terms, shortening Vega to V, and expressing in matrix form, we get:

$$\begin{bmatrix} V_1 & V_2 & V_3 \\V_1 d2_1 & V_2 d2_2 & V_3 d2_3 \\V_1 d1_1 d2_1 & V_2 d1_2 d2_2 & V_3 d1_3 d2_3 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} = \begin{bmatrix} V \\ V d2 \\ V d1 d2 \end{bmatrix}$$

The above is a linear system of the form $$A w=b$$. By Cramer's rule, the solution is $$w_i=\frac{det (A_i)} {det(A)}$$, where $$A_i$$ is formed by repalcing the i-th column of A with b.

Let's compute the determinants then. The determinant of the left matrix, which we call A, is:

$$Det \left( A \right) = V_1 \left( V_2 V_3 d2_2 d1_3 d2_3 - V_2 V_3 d2_3 d1_2 d2_2 \right) - V_2 \left( V_1 V_3 d2_1 d1_3 d2_3 - V_1 V_3 d1_1 d2_1 d2_3 \right) + V_3 \left( V_1 V_2 d2_1 d1_2 d2_2 -V_1 V_2 d2_2 d1_1 d2_1 \right)$$ $$\quad \quad \quad = V_1 V_2 V_3 \left( d2_2 d1_3 d2_3 - d2_3 d1_2 d2_2 - d2_1 d1_3 d2_3 + d1_1 d2_1 d2_3 + d2_1 d1_2 d2_2 - d2_2 d1_1 d2_1 \right)$$

Substituting the expressions for $$d_1$$ and $$d_2$$, which we simplify:

$$d_1 =\frac{1}{\sigma \sqrt{\tau}} \left(\ln{\left (\frac{S}{K} \right ) + \left(r_{d} - r_{f} + \frac{\sigma^{2}}{2}\right) \tau}\right)$$ $$\quad = \frac{ ln(S)+ \left( r_{d} - r_{f} \right) \tau}{\sigma \sqrt{\tau}}-\frac{ln(K)}{\sigma \sqrt{\tau}} + \frac{\sigma \sqrt{\tau}}{2}$$ $$\quad = f- k + s$$ $$d_2 = f- k - s$$

Where we substituted: $$f=\frac{ ln(S)+ \left( r_{d} - r_{f} \right) \tau}{\sigma \sqrt{\tau}}$$, $$k=\frac{ln(K)}{\sigma \sqrt{\tau}}$$, and $$s=\frac{\sigma \sqrt{\tau}}{2}$$

Hence:

$$d2_2 d1_3 d2_3=\left(f - k_{2} - s\right) \left(f - k_{3} - s\right) \left(f - k_{3} + s\right)$$ $$\quad \quad \quad \quad= f^{3} - f^{2} k_{2} - 2 f^{2} k_{3} - f^{2} s + 2 f k_{2} k_{3} + f k_{3}^{2} + 2 f k_{3} s - f s^{2} - k_{2} k_{3}^{2} + k_{2} s^{2} - k_{3}^{2} s + s^{3}$$ $$d2_3 d1_2 d2_2=\left(f - k_{2} - s\right) \left(f - k_{2} + s\right) \left(f - k_{3} - s\right)$$ $$\quad \quad \quad \quad = f^{3} - 2 f^{2} k_{2} - f^{2} k_{3} - f^{2} s + f k_{2}^{2} + 2 f k_{2} k_{3} + 2 f k_{2} s - f s^{2} - k_{2}^{2} k_{3} - k_{2}^{2} s + k_{3} s^{2} + s^{3}$$ $$d2_1 d1_3 d2_3=\left(f - k_{1} - s\right) \left(f - k_{3} - s\right) \left(f - k_{3} + s\right)$$ $$\quad \quad \quad \quad = f^{3} - f^{2} k_{1} - 2 f^{2} k_{3} - f^{2} s + 2 f k_{1} k_{3} + f k_{3}^{2} + 2 f k_{3} s - f s^{2} - k_{1} k_{3}^{2} + k_{1} s^{2} - k_{3}^{2} s + s^{3}$$ $$d1_1 d2_1 d2_3=\left(f - k_{1} - s\right) \left(f - k_{1} + s\right) \left(f - k_{3} - s\right)$$ $$\quad \quad \quad \quad = f^{3} - 2 f^{2} k_{1} - f^{2} k_{3} - f^{2} s + f k_{1}^{2} + 2 f k_{1} k_{3} + 2 f k_{1} s - f s^{2} - k_{1}^{2} k_{3} - k_{1}^{2} s + k_{3} s^{2} + s^{3}$$ $$d2_1 d1_2 d2_2=\left(f - k_{1} - s\right) \left(f - k_{2} - s\right) \left(f - k_{2} + s\right)$$ $$\quad \quad \quad \quad = f^{3} - f^{2} k_{1} - 2 f^{2} k_{2} - f^{2} s + 2 f k_{1} k_{2} + f k_{2}^{2} + 2 f k_{2} s - f s^{2} - k_{1} k_{2}^{2} + k_{1} s^{2} - k_{2}^{2} s + s^{3}$$ $$d2_2 d1_1 d2_1=\left(f - k_{1} - s\right) \left(f - k_{1} + s\right) \left(f - k_{2} - s\right)$$ $$\quad \quad \quad \quad = f^{3} - 2 f^{2} k_{1} - f^{2} k_{2} - f^{2} s + f k_{1}^{2} + 2 f k_{1} k_{2} + 2 f k_{1} s - f s^{2} - k_{1}^{2} k_{2} - k_{1}^{2} s + k_{2} s^{2} + s^{3}$$

Substituting into the $$det(A)$$ and simplifying, we get:

$$Det \left( A \right) = V_1 V_2 V_3 \left( d2_2 d1_3 d2_3 - d2_3 d1_2 d2_2 - d2_1 d1_3 d2_3 + d1_1 d2_1 d2_3 + d2_1 d1_2 d2_2 - d2_2 d1_1 d2_1 \right)$$ $$\quad = V_1 V_2 V_3 \left( k_{1}^{2} k_{2} - k_{1}^{2} k_{3} - k_{1} k_{2}^{2} + k_{1} k_{3}^{2} + k_{2}^{2} k_{3} - k_{2} k_{3}^{2} \right)$$ $$\quad = V_1 V_2 V_3 \left(k_{1} - k_{2}\right) \left(k_{1} - k_{3}\right) \left(k_{2} - k_{3}\right)$$ $$\quad = V_1 V_2 V_3 \left(\frac{ln K_1}{\sigma \sqrt{\tau}} - \frac{ln K_2}{\sigma \sqrt{\tau}}\right) \left(\frac{ln K_1}{\sigma \sqrt{\tau}} - \frac{ln K_3}{\sigma \sqrt{\tau}}\right) \left(\frac{ln K_2}{\sigma \sqrt{\tau}} -\frac{ln K_3}{\sigma \sqrt{\tau}} \right)$$ $$\quad = \frac{V_1 V_2 V_3}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K_1}{K_2} ln \frac{K_1}{K_3} ln \frac{K_2}{K_3}$$

The determinants of the $$A_1, A_2$$, and $$A_3$$ can be inferred from the above as these matrices differ from A only in one column:

$$Det \left( A_1 \right)= \frac{V V_2 V_3}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K}{K_2} ln \frac{K}{K_3} ln \frac{K_2}{K_3}$$ $$Det \left( A_2 \right)=\frac{V_1 V V_3}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K_1}{K} ln \frac{K_1}{K_3} ln \frac{K}{K_3}$$ $$Det \left( A_3 \right)=\frac{V_1 V_2 V}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K_1}{K_2} ln \frac{K_1}{K} ln \frac{K_2}{K}$$

Hence the solution is:

$$w_1 = \frac{Det \left( A_1 \right)}{Det \left( A \right)}= \frac{\frac{V V_2 V_3}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K}{K_2} ln \frac{K}{K_3} ln \frac{K_2}{K_3}}{\frac{V_1 V_2 V_3}{{\sigma}^3 \tau \sqrt{\tau}} ln \frac{K_1}{K_2} ln \frac{K_1}{K_3} ln \frac{K_2}{K_3}} = \frac{ V ln \frac{K}{K_2} ln \frac{K}{K_3}}{V_1 ln \frac{K_1}{K_2} ln \frac{K_1}{K_3}}$$ $$w_2 = \frac{Det(A_2)}{Det(A)}= \frac{ V ln \frac{K_1}{K} ln \frac{K}{K_3}}{V_2 ln \frac{K_1}{K_2} ln \frac{K_2}{K_3}}$$ $$w_3 = \frac{Det(A_3)}{Det(A)}= \frac{ V ln \frac{K_1}{K} ln \frac{K_2}{K}}{V_3 ln \frac{K_1}{K_3} ln \frac{K_2}{K_3}}$$

We write the above as:

$$w_i = \frac{V}{V_i} y_i \mbox{ where } i=1,2,3$$

### Vanna Volga - Pricing

Let us assume, in line with the assumptions in the previous sub-section, that we have three options available with strikes $$K_1, K_2$$, and $$K_3$$. Let the market prices of the three options be $$C^{Mkt}(K_1), C^{Mkt}(K_2)$$, and $$C^{Mkt}(K_3)$$. And we have been asked to quote a price for an option with some strike K.

One thing to note before we proceed is that if we were to use the standard Black Scholes model to price these options, it is very likely that the model will give very different prices. But if we use different implied volatilities for the different options (which is inconsistent with the BS assumptions but we are only after approximation here), then we can make our Black Scholes formula reproduce the market prices(in most cases of interest anyway):

$$C^{Mkt}(K_i) = C^{BS}(K_i,\sigma_i) \quad \mbox{ for } i=1,2,3$$ $$C^{Mkt}(K) = C^{BS}(K,v)$$

Now we can use the Vanna Volga method to hedge the Vega, Vanna, and Volga (and as a by-product Gamma given the relationship between Volga and Gamma) of the option we have been asked to price using the three available options. If we ignore higher order Greeks, then the only other Greek that we need to hedge is Delta, which we can do by taking a position in the underlying. Our hedged portfolio is then:

$$\Pi= C(K) - \Delta S - \sum_{i=1}^{3}{w_i C^{Mkt}(K_i)}$$

The above equation simply says that we hedge a long position in the option with strike K, which we have been asked to price, with a short position of size $$\Delta$$ in the underlying, and short positions of $$w_1, w_2, \mbox{ and } w_3$$ in the three options whose market prices are available to us.

How do we choose $$\Delta$$? We determine it by setting the total Delta of our position to zero (Note we use constant volatility in the calculation of Delta to be consistent with the calculation of Vega, Vanna, and Volga in the previous sub-section):

$$\frac{\partial \Pi}{\partial S}= 0$$ $$\frac{\partial C(K)}{\partial S} - \Delta \frac{\partial S}{\partial S} - \sum_{i=1}^{3}{w_i \frac{\partial C^{BS}(K_i,\sigma)}{\partial S}} = 0$$ $$\Delta = \frac{\partial C(K)}{\partial S}- \sum_{i=1}^{3}{w_i \frac{\partial C^{BS}(K_i,\sigma)}{\partial S}}$$

Our portfolio is locally hedged up to second order. We can set the price of the option to the cost of the above hedging strategy, sum of Vanna-Volga hedge and Delta Hedge. We know the cost of Vanna-Volga hedge is $$\sum_{i=1}^{3}{w_i C^{Mkt}(K_i)}$$. But what is the cost of delta hedge? Recall that the Black Scholes price is essentially the cost of dynamic Delta hedging, and hence we can write the cost of our Delta hedging as:

$$\Delta= \frac{\partial C(K)}{\partial S}- \sum_{i=1}^{3}{w_i \frac{\partial C^{BS}(K_i,\sigma)}{\partial S}}$$ $$\mbox{Cost of} \; \Delta \; \mbox{hedge}= C^{BS}(K,\sigma) - \sum_{i=1}^{3}{w_i C^{BS}(K_i,\sigma)}$$

Note the inconsistency in the treatment of volatility: in the calculation of the Delta hedge, we assume volatility is constant, whereas in Vanna-Volga, we assume volatility is stochastic. But it is ok as we did warn you that Vanna Volga is only an approximation.

The required price is then:

$$C(K)= \mbox{Cost of} \; \Delta \; \mbox{hedge} + \mbox{Cost of Vega Vanna Volga hedge}$$ $$C(K) = C^{BS}(K,\sigma) - \sum_{i=1}^{3}{w_i C^{BS}(K_i,\sigma)} + \sum_{i=1}^{3}{w_i C^{Mkt}(K_i)}$$ $$\quad \quad = C^{BS}(K,\sigma) + \sum_{i=1}^{3}{w_i \left( C^{Mkt}(K_i) -C^{BS}(K_i,\sigma) \right)}$$

### Vanna Volga - FX Volatility Smile

We saw in the previous sub-section that we can price an option with an arbitrary strike K using the Vanna-Volga method if we know the market prices of three options, and the constant volatility to plug into the Black Scholes model (usually the ATM volatility). We can then invert the strike K option price, which we computed, to get volatility for strike K, assuming we know all other parameters such as discount rate etc. As K is arbitrary, we can repeat the procedure for all Ks within the range of interest to construct the FX smile curve for the given maturity. This is all fine, but will involve a lot of computations and repeated use of numerical root finding algorithms as we do not have analytical formula for computing volatility from option price (recall the implied volatility calculation).

To simplify and accelerate the calculation of FX smile, we use Taylor series approximation of price, which enables us to compute the desired volatility directly, without having to resort to numerical root finding methods. We give the derivation for first order and second order approximation, where the second order should be more accurate than the first order.

The first order Taylor series expansion of the Vanna Volga price relationship from the previous sub-section is: (recall Taylor series: $$f(x+h)=f(x) + f^{\prime}(x)h +...$$ )

$$C^{BS}(K,v)-C^{BS}(K, \sigma) = \sum_{i=1}^{3}{w_i \left[ C^{BS}(K_i, \sigma_i) - C^{BS}(K_i, \sigma)\right]}$$ $$\frac{\partial C^{BS}(K, \sigma)}{\partial \sigma}(v-\sigma) = \sum_{i=1}^{3}{w_i \left[ \frac{\partial C^{BS}(K_i, \sigma)}{\partial \sigma}(\sigma_i-\sigma) \right]}$$ $$\frac{\partial C^{BS}(K, \sigma)}{\partial \sigma}(v-\sigma) = \sum_{i=1}^{3}{w_i \left[ \frac{\partial C^{BS}(K_i, \sigma)}{\partial \sigma}(\sigma_i-\sigma) \right]}$$ $$V(v-\sigma) = \sum_{i=1}^{3}{w_i V_i(\sigma_i-\sigma) }$$ $$V(v-\sigma) = \sum_{i=1}^{3}{w_i V_i\sigma_i} - \sigma \sum_{i=1}^{3}{w_i V_i }$$ $$V(v-\sigma) = \sum_{i=1}^{3}{\frac{V}{V_i} y_i V_i \sigma_i} - \sigma V$$ $$v = \sum_{i=1}^{3}{y_i \sigma_i }$$

And the second order Taylor series expansion of the Vanna Volga price formula is:

$$C^{BS}(K,v)-C^{BS}(K, \sigma) = \sum_{i=1}^{3}{w_i \left[ C^{BS}(K_i, \sigma_i) - C^{BS}(K_i, \sigma)\right]}$$ $$\frac{\partial C^{BS}(K, \sigma)}{\partial \sigma}(v-\sigma) + \frac{1}{2} \frac{\partial^2 C^{BS}(K, \sigma)}{\partial^2 \sigma}(v-\sigma)^2 = \sum_{i=1}^{3}{w_i \left[ \frac{\partial C^{BS}(K_i, \sigma)}{\partial \sigma}(\sigma_i-\sigma) + \frac{1}{2} \frac{\partial^2 C^{BS}(K_i, \sigma)}{\partial^2 \sigma}(\sigma_i-\sigma)^2 \right] }$$ $$V (v-\sigma) + \frac{1}{2} Volga \; (v-\sigma)^2 = \sum_{i=1}^{3}{w_i \left[ V_i(\sigma_i-\sigma) + \frac{1}{2} Volga_i \;(\sigma_i-\sigma)^2 \right] }$$ $$V (v-\sigma) + \frac{1}{2} V \frac{d1 d2}{\sigma} \; (v-\sigma)^2 = \sum_{i=1}^{3}{w_i \left[ V_i(\sigma_i-\sigma) + \frac{1}{2} V_i \frac{d1_{i} d2_{i}}{\sigma} \;(\sigma_i-\sigma)^2 \right] }$$ $$V (v-\sigma) + \frac{1}{2} V \frac{d1 d2}{\sigma} \; (v-\sigma)^2 = \sum_{i=1}^{3}{w_i V_i\sigma_i} - \sigma \sum_{i=1}^{3}{w_i V_i } + \frac{1}{2} \sum_{i=1}^{3}{w_i V_i \frac{d1_{i} d2_{i}}{\sigma} \;(\sigma_i-\sigma)^2 }$$ $$V (v-\sigma) + \frac{1}{2} V \frac{d1 d2}{\sigma} \; (v-\sigma)^2 = \sum_{i=1}^{3}{\frac{V}{V_i} y_i V_i\sigma_i} - \sigma V + \frac{1}{2} \sum_{i=1}^{3}{\frac{V}{V_i} y_i V_i \frac{d1_{i} d2_{i}}{\sigma} \;(\sigma_i-\sigma)^2 }$$ $${\sigma}(v-\sigma) + \frac{1}{2} d1 d2 (v-\sigma)^2 = {\sigma} \sum_{i=1}^{3}{ y_i \sigma_i} - \sigma^2 + \frac{1}{2} \sum_{i=1}^{3}{ y_i d1_{i} d2_{i} \;(\sigma_i-\sigma)^2 }$$ $$\frac{1}{2} d1 d2 (v-\sigma)^2 + {\sigma}(v-\sigma) - {\sigma} \left(\sum_{i=1}^{3}{ y_i \sigma_i} -\sigma \right) - \frac{1}{2}\sum_{i=1}^{3}{ y_i d1_{i} d2_{i} \;(\sigma_i-\sigma)^2 } =0$$

This is just quadratic equation in $$v-\sigma=x$$, and its positive root is (recall quadratic formula:$$\frac{-b + \sqrt{b^2-4ac}}{2a}$$):

$$v-\sigma = \frac{-\sigma + \sqrt{\sigma^2 +d1 d2 \left[ {2\sigma} \left(\sum_{i=1}^{3}{ y_i \sigma_i} -\sigma \right) +\sum_{i=1}^{3}{ y_i d1_{i} d2_{i} \;(\sigma_i-\sigma)^2 } \right]}}{d1 d2}$$