Get expected joint-payoff price of digital options from individual payoffs

Question

I am trying to model a joint distribution $f(X_1,X_2)$

(where $X_1$ and $X_2$ are market prices of the options) and then find from it the value of joint payoff price:

$F(X_1, X_2; B_1, B_2) = E[ max(X_1-B_1,0) * max(B_2 -X_2,0)] $

where $B_1$ and $B_2$ are corresponding strike prices.

I have limited samples of individual payoffs $Q_1$ and $Q_2$ for different values of $B_1$ and $B_2$

$Q_1(X_1,X_2;B2) = E[X_1 * max(B_2-X_2,0)]$

$Q_2(X_1,X_2;B1) = E[X_2 * max(X_1-B_1,0)]$

$C(X;k) = E[max(0,X_T - k)]$

Is there a way to solve this with or without assuming that $X_1$ and $X_2$ are independent.

I am new to modelling option-prices.

Answer 1

Suppose you would like to compute \begin{align} Q_1(x_1,x_2;B) &= \Bbb{E}[X_1\max(B-X_2,0)]\\ Q_2(x_1,x_2;B) &= \Bbb{E}[X_2\max(X_1-B,0)] \end{align}

where you know the marginal probability density functions $p_{X_1}(u)$ and $p_{X_2}(v)$.

Let's start by focusing on $Q_1$. By definition, the expectation equivalently writes: $$ Q_1(x_1,x_2;B) = \int_0^\infty \int_0^\infty u \max(B-v,0)\, p_{X_1X_2}(u,v) du dv $$ where $p_{X_1X_2}$ figures the joint probability density function of random variables $X_1$ and $X_2$: $$ p_{X_1X_2}(u,v) = p_{X_1 \vert X_2 = v}(u) p_{X_2}(v) $$

[Case 1: $X_1$ and $X_2$ are independent]

Then by definition $$ p_{X_1 \vert X_2 = v}(u) = p_{X_1}(u) $$ and we have (Fubini) \begin{align} Q_1(x_1,x_2;B) &= \int_0^\infty \int_0^\infty u \max(B-v,0)\, p_{X_1}(u) p_{X_2}(v) du dv \\ &= \int_0^\infty u p_{X_1}(u) du \int_0^\infty \max(B-v,0) p_{X_2}(v) dv \\ &= \Bbb{E}[X_1] \Bbb{E}[\max(B-X_2,0)] \end{align} which you know how to solve since you know the marginals.

[Case 2: $X_1$ and $X_2$ are not independent]

The marginals do not suffice: you need an assumption concerning the dependence structure of $X_1$ and $X_2$. Your question amounts then to saying, "can I infer a unique dependence structure from the knowledge of $Q_1$ and $Q_2$", the answer is no. The intuition behind that is given in this SE question.