What is a notation '1' in risk neutral probabilities paper?

Question

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 1

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:

1. There exists a state price vector $\mathbf{s} \in \mathbb{R}^n$ such that $(1+R)^T \mathbf{s} = \mathbf{1}$ and $\mathbf{s} \geq 0$
2. 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$.