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edited May 29, 2018 at 20:10

To get uniqueness of the SDF, you need complete markets. In a world with $n$ possible outcomes, complete markets would require $n$ linearly independent securities. Without complete markets, the SDF is not unique.

Simple Example: 2 possible outcomes, 1 security

Imagine there are two states of the world and I represent probability measures and random variables as two-dimensional vectors. Let:

$$ \mathbf{p} = \begin{bmatrix}\frac{1}{2} \\ \frac{1}{2} \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix}3 \\ 1 \end{bmatrix} $$

$$ \mathbf{m}^{(a)} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \quad \quad \mathbf{m}^{(b)} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2}\end{bmatrix}$$

Let's imagine the price of security $x$ is 1. Both $\mathbf{m}^{(a)}$ and $\mathbf{m}^{(b)}$ correctly price the security since $\sum_i p_i m^{(a)}_i x_i = \sum_i p_i m^{(b)}_i x_i = 1$.

Simple Example

Imagine there are two states of the world and I represent probability measures and random variables as two-dimensional vectors. Let:

$$ \mathbf{p} = \begin{bmatrix}\frac{1}{2} \\ \frac{1}{2} \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix}3 \\ 1 \end{bmatrix} $$

$$ \mathbf{m}^{(a)} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \quad \quad \mathbf{m}^{(b)} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2}\end{bmatrix}$$

Let's imagine the price of security $x$ is 1. Both $\mathbf{m}^{(a)}$ and $\mathbf{m}^{(b)}$ correctly price the security since $\sum_i p_i m^{(a)}_i x_i = \sum_i p_i m^{(b)}_i x_i = 1$.

Simple Example: 2 possible outcomes, 1 security

Imagine there are two states of the world and I represent probability measures and random variables as two-dimensional vectors. Let:

$$ \mathbf{p} = \begin{bmatrix}\frac{1}{2} \\ \frac{1}{2} \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix}3 \\ 1 \end{bmatrix} $$

$$ \mathbf{m}^{(a)} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \quad \quad \mathbf{m}^{(b)} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2}\end{bmatrix}$$

Let's imagine the price of security $x$ is 1. Both $\mathbf{m}^{(a)}$ and $\mathbf{m}^{(b)}$ correctly price the security since $\sum_i p_i m^{(a)}_i x_i = \sum_i p_i m^{(b)}_i x_i = 1$.

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created May 29, 2018 at 20:01

Matthew Gunn

Simple Example

Imagine there are two states of the world and I represent probability measures and random variables as two-dimensional vectors. Let:

$$ \mathbf{p} = \begin{bmatrix}\frac{1}{2} \\ \frac{1}{2} \end{bmatrix} \quad \mathbf{x} = \begin{bmatrix}3 \\ 1 \end{bmatrix} $$

$$ \mathbf{m}^{(a)} = \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} \quad \quad \mathbf{m}^{(b)} = \begin{bmatrix} \frac{1}{2} \\ \frac{1}{2}\end{bmatrix}$$

Let's imagine the price of security $x$ is 1. Both $\mathbf{m}^{(a)}$ and $\mathbf{m}^{(b)}$ correctly price the security since $\sum_i p_i m^{(a)}_i x_i = \sum_i p_i m^{(b)}_i x_i = 1$.