Confused About Ex-Ante vs. Ex-Post Pricing Representation

Question

This is going to be a really simple question, but I am confused by it. The basic pricing formula is $p_t=E^p_t(m_{t+1}X_{t+1})$, where $p$ is the physical measure. We can also say that $R_{t+1}=\frac{X_{t+1}}{p_{t}}$ ex-post. This is just the definition of returns. This means that $E^l_t(R_{t+1})=\frac{E^l_t(X_{t+1})}{p_t}$, and thus $p_t=\frac{E^l_t(X_{t+1})}{E^l_t(R_{t+1})}$. This is with respect to any measure $l$, including the physical measure(it is just a weighted average). If we choose $l=p$(the arbitrary measure equal to the physical measure) then, by uniqueness of the SDF we can say that $m_{t+1}=\frac{1}{E_t^p(R_{t+1})}$. Obviously there must be something wrong with this reasoning, since it concludes that $m_{t+1}$ is not time-t random. I think the problem must be an ex-post vs. ex-ante kind of thing, basically taking the arbitrary measure expectation on the third line is not as simple as it sounds, but I was wondering if anyone had a better explanation.\

I found this especially confusing because my understanding of the basic discounted cash flow model in corporate finance comes from the $p_t=\frac{E^l_t(X_{t+1})}{E^l_t(R_{t+1})}$ expression. But there must be some reason why this expression only makes economic sense if the arbitrary measure is the risk neutral measure.

Answer 1

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 $\sum_i p_i m^{(a)}_i x_i = \sum_i p_i m^{(b)}_i x_i = 1$.