Apr 24, 2026
Let X be the value of an underlying at expiration with current price x having put option prices p_i = E[(k_i - X)^+] for increasing strikes k_i, i = 1,\ldots,m. How do we use this to find a risk-neutral distribution of X?
Assume P(X = x_j) = \pi_j, j = 1,\ldots,n where n > m.
Minimize the logical entropy given the constraints. \begin{aligned} \Phi(\pi_j, \lambda, \mu_j) &= \frac{1}{2}(1 - \sum_j \pi_j^2) - \lambda(E[X] - x) - \sum_i \mu_i(E[(k_i - X)^+] - p_i) \\ &= \frac{1}{2}(1 - \sum_j \pi_j^2) - \lambda(\sum_j x_j\pi_j - x) - \sum_i \mu_i([\sum_j (k_i - x_j)^+\pi_j] - p_i) \\ \end{aligned}
0 = \partial_{\pi_k}\Phi(\pi_j, \lambda, \mu_i) = -\pi_k - \lambda x_k - \sum_i \mu_i (k_i - x_k)^+ so \pi_j = - \lambda x_j - \sum_i \mu_i (k_i - x_j)^+ We have \begin{aligned} x &= \sum_j x_j(- \lambda x_j - \sum_i \mu_i (k_i - x_j)^+) \\ &= -\lambda\sum_j x_j^2 - \sum_i\mu_i \sum_j x_j(k_i - x_j)^+ \\ &= -\lambda\sum_j x_j^2 - \sum_i\mu_i \sum_{x_j < k_i} x_j(k_i - x_j) \\ \end{aligned} and \begin{aligned} p_i &= \sum_j (k_i - x_j)^+(- \lambda x_j - \sum_l \mu_l (k_l - x_j)^+) \\ &= -\lambda \sum_j (k_i - x_j)^+ x_j - \sum_l \mu_l \sum_j (k_i - x_j)^+(k_l - x_j)^+ \\ &= -\lambda \sum_{x_j < k_i} (k_i - x_j) x_j - \sum_l \mu_l \sum_{x_j < \min\{k_i,k_l\}} (k_i - x_j)(k_l - x_j) \\ \end{aligned} Let A = \sum_j x_j^2, B_i = \sum_{x_j < k_i} x_j(k_i - x_j) and C_{il} = \sum_{x_j < \min\{k_i,k_l\}} (k_i - x_j)(k_l - x_j). Note C_{il} = C_{li}. \begin{aligned} x &= -\lambda A - \sum_i \mu_i B_i \\ p_i &= -\lambda B_i - \sum_l \mu_l C_{il} \\ \end{aligned} and \begin{bmatrix} x \\ p_1 \\ \vdots \\ p_m \\ \end{bmatrix} = -\begin{bmatrix} A & B_1 & \cdots & B_m \\ B_1 & C_{11} & \cdots & C_{1m} \\ \vdots & \vdots &\ddots & \vdots\\ B_m & C_{m1} & \cdots & C_{mm} \\ \end{bmatrix} \begin{bmatrix} \lambda \\ \mu_1 \\ \vdots \\ \mu_m \\ \end{bmatrix}
If there is only one option with strike k then \pi_j = -\lambda x_j - \mu (k - x_j)^+. We have \begin{aligned} x &= \sum_j x_j(-\lambda x_j - \mu (k - x_j)^+) \\ &= \sum_{x_j < k} x_j(-\lambda x_j - \mu (k - x_j)) +\sum_{x_j \ge k} x_j(-\lambda x_j) \\ &= -\lambda \sum_j x_j^2 - \mu(\sum_{x_j < k} x_j(k - x_j)) \\ &= -\lambda A - \mu B \\ \end{aligned} where A = \sum_j x_j^2 and B = \sum_{x_j < k} x_j(k - x_j). We also have \begin{aligned} p &= \sum_j (k - x_j)^+(-\lambda x_j - \mu (k - x_j)^+) \\ &= \sum_{x_j < k} (k - x_j)(-\lambda x_j - \mu (k - x_j)) \\ &= -\lambda (\sum_{x_j < k} (k - x_j)x_j) - \mu \sum_{x_j < k} (k - x_j)^2 \\ &= -\lambda B - \mu C \\ \end{aligned} where C = \sum_{x_j < k} (k - x_j)^2. Solving for \lambda and \mu \begin{aligned} \lambda &= (-Cx + Bp)/D \\ \mu &= (Bx - Ap)/D \\ \end{aligned} where D = AC - B^2.
We have D\pi_j = (Cx - Bp)x_j + (-Bx + Ap)(k - x_j)^+ so D\pi_j = (Cx - Bp + Bx - Ap)x_j + (-Bx + Ap)k when x_j < k and D\pi_j = (Cx - Bp)x_j if x_j \ge k.
Define \bm{x} = (x_1,\ldots,x_n) and \bm{k} = (k - \bm{x})1(\bm{x} < k) so A = \bm{x}\cdot\bm{x}, B = \bm{x}\cdot\bm{k} and C = \bm{k}\cdot\bm{k}. The Cauchy-Schwartz inequality shows D > 0.
In vector form \begin{aligned} D\bm{\pi} &= (\bm{k}\cdot\bm{k} x - \bm{x}\cdot\bm{k} p) \bm{x} + (-\bm{x}\cdot\bm{k} x + \bm{x}\cdot\bm{x} p) \bm{k} \\ &= (\bm{k}\cdot\bm{k}\,\bm{x} - \bm{x}\cdot\bm{k}\,\bm{k})x + (- \bm{x}\cdot\bm{k}\,\bm{x} + \bm{x}\cdot\bm{x}\,\bm{k})p \\ \end{aligned}