Optimal Hedging

February 16, 2025

Warmup

First exercise in understanding the market (X,C), trading (\tau,\Gamma), and account (V, A) notation.

Assume interest rate is zero, stock price is standard Brownian motion starting at 1, and no dividends.

Let X_t = (R_t, S_t) = (1, 1 + B_t), C_t = (0, 0).

Find the best least square initial hedge for contractual at-the-money call payoff {\hat{A}_1 = \max\{S_1, 1\} = 1 + B_1^+}.

Initial hedge, \Gamma_0 = (m, n) with unwind \Gamma_1 = -\Gamma_0.

Amount from unwinding the initial hedge is A_1 = -\Gamma_1\cdot X_1 = mR_1 + nS_1 = m + n (1 + B_1).

Find \min_{m,n} E[(A_1 - \hat{A}_1)^2] = \min_{m,n} E[(m + n (1 + B_1) - (1 + B_1^+))^2].

First order conditions taking derivatives with respect to m and n are \begin{aligned} 0 &= E[2(m + n (1 + B_1) - (1 + B_1^+))] \\ 0 &= E[2(m + n (1 + B_1) - (1 + B_1^+))(1 + B_1)] \\ \end{aligned} so \begin{aligned} E[1 + B_1^+] &= m + n E[1 + B_1] \\ E[(1 + B_1^+)(1 + B_1)] &= m E[B_1^+] + n E[(1 + B_1)^2] \\ \end{aligned}

Since E[B_1^+] = 1/\sqrt{2\pi}, E[B_1] = 0, and E[(B_1^+)^2] = E[B_1B_1^+] = 1/2 we have \begin{bmatrix} 1 + 1/\sqrt{2\pi} \\ 1 + 2/\sqrt{2\pi} + 1/2 + \\ \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1/\sqrt{2\pi} & 1/2 \\ \end{bmatrix} \begin{bmatrix} m \\ n \\ \end{bmatrix} hence \begin{bmatrix} m \\ n \\ \end{bmatrix} = 2 \begin{bmatrix} 1/2 & 0 \\ -1/\sqrt{2\pi} & 1 \\ \end{bmatrix} \begin{bmatrix} 1/\sqrt{2\pi} \\ 1/2 \\ \end{bmatrix} so m = 1/\sqrt{2\pi} \approx 0.4 and n = -1/\pi + 1 \approx 0.68.

Note V_0 = \Gamma_0\cdot X_0 = 1/\sqrt{2\pi} = E[B_1^+].

In matrix form, minimizing E[(\Gamma\cdot X - \hat{A})^2] has first order condition {0 = E[2(\Gamma\cdot X - A)X]} so \Gamma = E[XX^*]^{-1}E[AX].