Vanna Volga Price of an Up and In Put

Question

In the Vanna-Volga approach to pricing first generation exotics, such as single barriers, as I understand it the pricing is as follows:

Let $K,S_t < B$. I'll choose the ATM IV $I_{ATM}$ as the reference volatility for the VV price. Then, $$ UIP(S_t,K,B) = UIP^{BS} (S_t,K,B, I_{ATM}) + p \times \text{Hedge cost} $$ where $UIP^{BS} (S_t,K,B, I_{ATM})$ is the UIP price in a Black-Scholes world with a flat volatility equal to $I_{ATM}$, and $p$ is the touch probability. If I were considering an Up and Out (UOP) then $p$ would be the no-touch probability, right?

So here is my question: the UIP price in a Black-Scholes world with flat vol $I_{ATM}$ is $$ UIP^{BS} (S_t,K,B, I_{ATM}) = \frac{K}{B} C^{BS}(S_t, B^2/K, I_{ATM}) $$

The hedge cost would be $$ \sum_{i=1}^3 x_i \left( C(S_t,K_i) - C^{BS}(S_t,K_i,I_{ATM}) \right) $$ where the $C(S_t,K_i)$ are the market prices of options (i.e. using the actual IVs of the strikes $K_i$), and $x_i$ are determined such that \begin{align} \frac{\partial}{\partial\sigma} UIP^{BS} (S_t,K,B, I_{ATM}) &= \sum_{i=1}^3 x_i \frac{\partial}{\partial\sigma}C^{BS}(S_t,K_i,I_{ATM}) \\ \frac{\partial^2}{\partial\sigma\partial\sigma} UIP^{BS} (S_t,K,B, I_{ATM}) &= \sum_{i=1}^3 x_i \frac{\partial^2}{\partial\sigma\partial\sigma}C^{BS}(S_t,K_i,I_{ATM}) \\ \frac{\partial^2}{\partial S_t\partial\sigma} UIP^{BS} (S_t,K,B, I_{ATM}) &= \sum_{i=1}^3 x_i \frac{\partial^2}{\partial S_t\partial\sigma}C^{BS}(S_t,K_i,I_{ATM}) \end{align}

Is my understanding, in particular regarding the touch probability, correct?

Answer 1

I do not think you should (can) use the opposite probability (going from p touch to p no touch) because there exists a so called In-out parity: $$European \ vanilla\ option = European\ KI + European\ KO$$

The justification is simple:

• assume you hold both a KI and KO option
• if the barrier is untouched, the KO pays a vanilla payoff at expiry
• if the barrier is touched, the KI pays a vanilla payoff at expiry
• since the payoff is identical to a vanilla option, its price must also be equal due to no arbitrage

This is also what is mentioned in the Bloomberg article that is linked in the answer I linked in the comment. I'll just quote the relevant section below:

The compromise adjustment described in Equation 11 is justified for a knock-out option, with the vanna portion of the adjustment forced to zero in the limit of the option certainly being knocked out. For knock-in options, one could use a similar formula, replacing psym in Equation 11 with 1− psym, which is the probability of hitting the barrier. Unfortunately, this will not satisfy the no-arbitrage condition that a knock-out option plus a knock-in option equals a vanilla option. With this in mind, we will instead price knock-in options as the difference between the vanna-volga price of a vanilla option and the vannavolga price of the knock-out option.

Equation 11 prices a KO option and looks as follows:

and provides the adjustment needed to the Black-Scholes price. I have replicated this in computer code (more or less since I used Bloomberg's OVML to do the bulk work of getting the various values needed to compute the options values). If needed, I can for sure find that somewhere or redo again (not in the near future I am afraid). The weights are chosen to match market prices (at the time this model was built) and are a compromise from opposing views where vega and volga should either be unweighted or weighted by a function that goes to zero as the barrier is approached.

Now, what the paper suggest is to decompose a KI option into: $$ KI\ option = BS\ - KO\ option$$

The Bloomberg article mentions several such adjustments and the same can be found in Wystup: Vanna-volga pricing.

For options with strike K, barrier B and type φ = 1 for a call and φ = −1 for a put, we use the following pricing rules which are based on no-arbitrage conditions. KI is priced via KI = vanilla − KO.

To summarize, I think you are right and you should not use the touch probability. Instead, you should decompose it and compute it via the KO option logic.