State-Price Vectors and No-Arbitrage Tests in Return Matrices
Summary
The document explains the notation in a paper’s test for whether a simulated return sample is arbitrage-free. In the linear system, the symbol 1 denotes a vector of ones, and adding it to a return matrix converts simple returns into gross returns. Solving the transposed system for state prices checks whether those prices can reproduce the unit price of each asset’s return across states.
A nonnegative solution is linked to the absence of arbitrage, while Farkas’s lemma gives a complementary portfolio condition: if no such state-price vector exists, a portfolio may have nonnegative payoffs and negative initial cost. The discussion distinguishes satisfying the Law of One Price from establishing no arbitrage. It gives the conceptual conditions and a MATLAB-style interpretation, but does not work through a full numerical solution; the exact dimensions depend on how states and assets are arranged in the return matrix.
Key ideas
- The symbol 1 represents a vector of ones with one entry for each asset equation.
- Adding that vector to a return matrix forms gross returns by increasing each simple return by one.
- A state-price vector solves a linear system that prices each asset’s return at one.
- A nonnegative state-price vector implies no arbitrage under the stated setup.
- Farkas’s lemma relates failure of the state-price condition to a portfolio with nonnegative payoffs and negative cost.
Tags
Full text
# What is a notation '1' in risk neutral probabilities paper?
# What is a notation '1' in risk neutral probabilities paper?
I'm reading the paper by Zhao et al (2008) and have a problem with used definitions in the text on the page 1535.
> First, we generate a sample, $R$, of a given size from the distribution (21). Let $\hat{\mu}$ and $\hat{\sigma}^2$ be the sample mean vector and the sample variance covariance matrix. Then, we modify the generated sample to $$\hat{R}=\mu + (R-\hat{\mu})\hat{\sigma}^{-1}\sigma.$$ Then, the modified sample $\hat{R}$ has the same first and second moments as the original distribution. To further test, whether the modified sample is arbitrage free, we use the Matlab backslash function to examine whether the solution to $(1 + R)^{\top} \backslash 1$ is componentwise positive.
Question. What means $1$ in the last line? Is it an identity matrix or a column vector of ones? And what is dimensions of this $1$?
I have tried to examine the solution from Table 1.
```
library(pracma)
n <- 5
R<-matrix(
c(
0.0025, 0.0377, 0.0110, 0.0769, 0.0047,
0.0025, 0.0431, 0.0001, 0.0045, 0.0562,
0.0025, 0.0469, 0.0643, 0.0400, 0.0370,
0.0025, 0.0504, 0.0422, 0.0169, 0.0333,
0.0025, 0.0596, 0.0038, 0.1896, 0.0663), ncol=n)
#I<-ones(n)
I <-diag(n)
mldivide(t(I+R),I)
```
Add after JejeBelfort's comment. If the $1$ is a column vector of ones then what is $+$? Union operation or Kronecker product operator?
```
I <- rep(1, n)
cbind(I, R)
```
Reference.
Yonggan Zhao, William T. Ziemba (2008) Calculating risk neutral probabilities and optimal portfolio policies in a dynamic investment model with downside risk control. European Journal of Operational Research 185 (2008) 1525–1540.
## Answer by Matthew Gunn (score 3, accepted)
https://quant.stackexchange.com/a/34051
### The 1 you are referring to is a vector of ones
The expression $(1 + R)^T \backslash 1$ appears to be shorthand for a MATLAB equation such as:
`s = (ones(n,k) + R)' \ ones(n, 1)`
where `R` is an `n` by `k` matrix (i.e. specifying returns for $n$ periods of $k$ assets).
`1 + R` adds 1 to each element of the matrix R (eg. makes it a return like 1.02 instead of .02). The MATLAB command `\` would solve the below system for $\mathbf{s}$ in the least squares sense:
$$ ( 1 + R)^T \mathbf{s} = \begin{bmatrix} 1 \\ 1 \\ \ldots \\ 1 \end{bmatrix} $$
### A non-negative solution $\mathbf{s}$ guarantees the absence of arbitrage
Any solution $\mathbf{s}$ to this linear system will be a vector of state prices for the $n$ states that satisfies the Law of One Price (LOOP). For any asset $i$, the inner product of the return series and the state price vector gives the value 1 which is the price of a return.
$$ (1 + \mathbf{r}_i)^T \mathbf{s} = 1$$
If $\mathbf{s}$ correctly prices each asset, the Law of One Price is satisfied. No arbitrage is a somewhat different condition though.
The existence of a non-negative state price vector $\mathbf{s}$ implies the absence of arbitrage. A positive payoff with a strictly negative cost is called an arbitrage. If the state price density (stochastic discount factor) is positive, then one cannot construct an arbitrage.
(Note: I'm using bold letters for vectors.)
#### Farkas's Lemma and No Arbitrage
By Farkas's Lemma, exactly one of the following conditions holds:
- There exists a state price vector $\mathbf{s} \in \mathbb{R}^n$ such that $(1+R)^T \mathbf{s} = \mathbf{1}$ and $\mathbf{s} \geq 0$
- There exists a $\mathbf{w} \in \mathbb{R}^k$ (giving investments in the $k$ assets) such that $(1+R)\mathbf{w} \geq 0$ and $\mathbf{1}^T \mathbf{w} < 0$
Condition (1) is the existence of a non-negative state price vector $\mathbf{s}$.Condition (2) implies it's possible to construct an arbitrage:
- Investing $w_i$ in asset $i$ gives the non-negative payoff $(1 + R) \mathbf{w} \geq 0$
- The portfolio has a negative cost since $ \mathbf{w}^T \mathbf{1} < 0$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.