SF Option Payoff

Let's consider vanila call option. The corresponding payoff looks as following:

\(\text{Payoff}_{vanila} = max(S_T-K,0)\)

However Stochastic.Finance options have a bit of non-trivial payoff as a fraction of the supplied collateral:

\(\text{Payoff}_{SF} = \frac{\max(S_T - K, 0)}{\max(S_T - K, 0) + K}\)

for every ETH supplied collateral, so that in case the collateral is 10 ETH the total payoff would be \(P_{SF}\) * 10 ETH

This payoff function can be interpreted as:

\(\text{Payoff}_{SF} = \begin{cases}\frac{S_T - K}{S_T} & \text{if } S_T > K \\0 & \text{if } S_T \leq K\end{cases}\)

The price of the option \(V\) is given by the expected value of the discounted payoff under the risk-neutral measure \(\mathbb{Q}\):

\(V = e^{-rT} \mathbb{E}^{\mathbb{Q}}\left[\frac{\max(S_T - K, 0)}{\max(S_T - K, 0) + K}\right]\)

Substituting the explicit form of the payoff:

\(V = e^{-rT} \mathbb{E}^{\mathbb{Q}}\left[\frac{S_T - K}{S_T} \cdot \mathbb{I}(S_T > K)\right]\)

where \(\mathbb{I}(S_T > K)\) is an indicator function that equals 1 when \(S_T > K\) and 0 otherwise.

Let’s express the expected value in terms of the distribution of \(S_T\), the stock price at time \(T\) under the risk-neutral measure:

\(V = e^{-rT} \int_{K}^{\infty} \frac{S_T - K}{S_T} f_{S_T}(S_T) dS_T\)

Log-Normal Distribution of \(S_T\):

Under the Black-Scholes model, the stock price \(S_T\) is log-normally distributed:

\(S_T = S_0 \exp\left(\left(r - \frac{\sigma^2}{2}\right)T + \sigma W_T\right)\)

where \(W_T\) is a standard Brownian motion under the risk-neutral measure.

The PDF of \(S_T\) is:

\(f_{S_T}(S_T) = \frac{1}{S_T \sigma \sqrt{2\pi T}} \exp\left(-\frac{\left(\ln\left(\frac{S_T}{S_0}\right) - \left(r - \frac{\sigma^2}{2}\right)T\right)^2}{2\sigma^2 T}\right)\)

Rewriting the Integral:

We can now rewrite the option price as:

\(V = e^{-rT} \int_{K}^{\infty} \left(1 - \frac{K}{S_T}\right) f_{S_T}(S_T) dS_T\)

This can be split into two integrals:

\(V = e^{-rT} \left[ \int_{K}^{\infty} f_{S_T}(S_T) dS_T - K \int_{K}^{\infty} \frac{1}{S_T} f_{S_T}(S_T) dS_T \right]\)

The first integral

\(\int_{K}^{\infty} f_{S_T}(S_T) dS_T\)

is simply the probability \(\mathbb{P}^{\mathbb{Q}}(S_T > K)\), which is given by \(\Phi(d_2)\) where: \(d_2 = \frac{\ln\left(\frac{S_0}{K}\right) + \left(r - \frac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}}\)

The second integral can be expressed as:

\(\int_{K}^{\infty} \frac{1}{S_T} \cdot \frac{1}{S_T \sigma \sqrt{2\pi T}} \exp\left(-\frac{\left(\ln\left(\frac{S_T}{S_0}\right) - \left(r - \frac{\sigma^2}{2}\right)T\right)^2}{2\sigma^2 T}\right) dS_T\)

\(= \frac{1}{\sigma \sqrt{2\pi T}} \int_{K}^{\infty} \frac{1}{S_T^2} \exp\left(-\frac{\left(\ln\left(\frac{S_T}{S_0}\right) - \left(r - \frac{\sigma^2}{2}\right)T\right)^2}{2\sigma^2 T}\right) dS_T\)

Fortunatly such integral has a closed-form solution. The final formula for option price looks as following:

\(\text{V}_{SF} = \exp(-rT)\left(\Phi(d_2) - \frac{1}{2}\frac{K}{S_0}\text{erfc}(\zeta)\exp\left(-(r -\sigma^2)T\right)\right)\)

where:

\(\zeta = \frac{3 \sqrt{2} T \sigma^2 - 2^{3/2} T r + 2^{3/2} \ln\left(\frac{K}{S_0}\right)}{4 \sqrt{T} \sigma}\)