commit 4af538

Commit `4af538`

2024-08-24 22:16:53 Stanislav Kubrak: SF Option Payoff - analytical solution finally

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`sf option payoff.md` ..
@@ 3,10 3,74 @@
	Let's consider vanila call option.
	The corresponding payoff looks as following:

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

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

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

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

Commit `4af538`