The Geometry of the Payoff Space

Lorenzo Naranjo

Fall 2026

Introduction

I describe the mathematical structure of the payoff space that we will use to characterize the space of traded payoffs and stochastic discount factors.
Even though the results described in this note apply to infinite dimensional Hilbert spaces, we will restrict our attention to the study of finite dimensional Euclidean spaces.

Uncertainty is represented by a finite set \mathcal{S} = \{1, \ldots, S\} of states, defining a finite probability space (\mathcal{S}, q).
The set of all random variables defined in \mathcal{S} is denoted by L and is called the payoff space.
Thus, for any x \in L we have that the vector (x(1), x(2), \ldots, x(S)) \in \mathbb{R}^{S} defines all the possible payoffs in each state, and the probability of getting a payoff in a particular state is given by \Pr(x(s)) = q(s) for all s \in S.

The payoff space is clearly a linear vector space since for any x, y \in L and \alpha, \beta \in \mathbb{R} we have that \alpha x + \beta y \in L.
We endow the payoff space with an inner product \langle{\cdot, \cdot}\rangle: L \times L \rightarrow \mathbb{R} defined such that for any x, y \in L, we have that \langle{x, y}\rangle = \operatorname{E}(xy) = \sum_{s = 1}^{S} q(s) x(s) y(s).
In finite-dimensional spaces, we can use the inner product to define the Euclidean norm \lVert{\cdot}\rVert: L \rightarrow \mathbb{R}^{+} for all x \in L as \lVert{x}\rVert = \sqrt{\langle{x, x}\rangle}.

Given x, y \in L, consider the vectors y_{x} = \alpha x and z = y - y_{x}.
We say that y_{x} is the projection of y on the subspace generated by \{x\} if the norm of z is minimal.
To obtain the projection, we need to compute the \alpha that minimizes \Vert z \Vert = \Vert y - \alpha x \Vert = \operatorname{E}(y - \alpha x)^{2}.
The first-order condition of this problem is: 0 = \operatorname{E}((y - \alpha x) x) = \langle{y - \alpha x, x}\rangle = \langle z, x \rangle, which implies that \alpha = \dfrac{\langle{x, y}\rangle}{\langle{x, x}\rangle} and \langle z, y_{x} \rangle = 0.

We say that two vectors x, y \in L are orthogonal if their inner product is equal to zero.
Thus, we have that y_{x} \mathrel\bot z, implying that the vector y can be decomposed into two orthogonal components.
Indeed, we have that \lVert{y}\rVert^{2} = \lVert{z + y_{x}}\rVert^{2} = \lVert{z}\rVert^{2} + 2 \langle z, y_{x} \rangle + \lVert{y_{x}}\rVert^{2} = \lVert{z}\rVert^{2} + \lVert{y_{x}}\rVert^{2}, which is a generalization of the classical Pythagorean theorem.

The orthogonal decomposition implies that \Vert y \Vert^{2} \geq \Vert y_{x} \Vert^{2}, with equality occuring whenever y is proportional to x.
Therefore, we have that \lVert{y}\rVert^{2} \geq \lVert{y_{x}}\rVert^{2} = \left\lVert \frac{\langle{x, y}\rangle}{\langle{x, x}\rangle} x \right\rVert^{2} = \frac{\langle{x, y}\rangle^{2}}{\lVert{x}\rVert^{2}}.
The previous expression is known as the Cauchy-Schwartz inequality and is fundamental in the study of Euclidean vector spaces.
Given x, y \in L we have that |\langle{x, y}\rangle| \leq \lVert{x}\rVert \lVert{y}\rVert. \tag{1}

Given x, y \in L and \alpha, \beta \in \mathbb{R}, a linear functional f: L \rightarrow \mathbb{R} satisfies f(\alpha x + \beta y) = \alpha f(x) + \beta f(y).
We say that the linear functional f : L \rightarrow \mathbb{R} is bounded if |f(x)| \leq M \lVert{x}\rVert for all x \in L.
A bounded linear functional is also called a continuous linear functional.
The smallest M for which this inequality remains true is called the norm of f, i.e., \lVert{f}\rVert = \inf \{M: |f(x)| \leq M \lVert{x}\rVert, \text{ for all } x \in L\}.

For a given m \in L and any x \in L, the functional f(x) = \langle{m, x}\rangle = \operatorname{E}(m x) = \sum_{s = 1}^{S} q(s) m(s) x(s) is linear.
Furthermore, the Cauchy-Schwartz inequality implies that |f(x)| = |\langle{m, x}\rangle| \leq \lVert{m}\rVert \lVert{x}\rVert, showing that the linear functional f is bounded and hence continuous.
Since the previous inequality is an equality whenever x is proportional to m, we have that \lVert{m}\rVert is the smallest bound of f, showing that \lVert{f}\rVert = \lVert{m}\rVert.

Consider a linear functional f: L \rightarrow \mathbb{R}.
The set K = \{x \in L: f(x) = 0\} describes a hyperplane that can be described by a normal vector z.
Thus, \langle{x, z}\rangle = 0 for all x \in K.
Without loss of generality, assume that z has been appropriately scaled so that f(z) = 1.

Given any x \in L, we have that x - f(x) z \in K since f(x - f(x) z) = f(x) - f(x) f(z) = 0.
Moreover, z \mathrel\bot K, implying that 0 = \langle{x - f(x) z, z}\rangle = \langle{x, z}\rangle - f(x) \langle{z, z}\rangle.
The previous expression implies that f(x) = \frac{\langle{x, z}\rangle}{\langle{z, z}\rangle} = \langle{x, m}\rangle, where m = \dfrac{z}{\lVert{z}\rVert^{2}}.

If f: L \rightarrow \mathbb{R} is a bounded linear functional, there exists a unique vector m \in L such that for all x \in L, f(x) = \langle{m, x}\rangle.
Furthermore, we have \lVert{f}\rVert = \lVert{y}\rVert and every m determines a unique bounded linear functional.