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.


n <- 5

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 <-diag(n)


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)


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.

  • 1
    $\begingroup$ In MATLAB backslash is a linear equation solver, possibly using least squares, mathworks.com/help/matlab/ref/… Also called "mldivide". $\endgroup$
    – nbbo2
    May 2, 2017 at 12:05
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    $\begingroup$ I guess it is the vector composed of ones only, of size equal to the number of rows in $R$. $\endgroup$ May 2, 2017 at 12:06
  • 1
    $\begingroup$ Left argument of '\' is a possibly non-square matrix, right argument is a column vector (in this case all ones). $\endgroup$
    – nbbo2
    May 2, 2017 at 12:14
  • $\begingroup$ @JejeBelfort, thanks you for comment. If '1' is the vector what is '+'? Union operation? $\endgroup$
    – Nick
    May 2, 2017 at 22:54
  • $\begingroup$ @Nick Do you mean the $\intercal$ operator? This should represent the matrix transpose operator applied to $(1+R)$. $\endgroup$ May 3, 2017 at 8:39

1 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$.
  • $\begingroup$ Can you explain what you mean when you say that 1 is "the price of a return"? I didn't get that part. Thanks. $\endgroup$
    – nbbo2
    May 5, 2017 at 2:51
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    $\begingroup$ @noob2 Let $X$ be some future payoff. Let $p(X)$ be today's price for that future payoff. Any payoff $X$ with a price of 1 is a return. I can pay 1 dollar and get the risk free rate tomorrow. I can pay 1 dollar and get the return on Apple tomorrow etc... For any arbitrary payoff $X$, you can construct the associated return as $R = \frac{X}{p(X)}$. Then $p\left(R \right) = p\left( \frac{X}{p(X)} \right) = \frac{p(X)}{p(X)} = 1$ where the 2nd to last step comes from the linearity of the pricing function (an assumption). $\endgroup$ May 5, 2017 at 3:27

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