# Return to Answer

7 deleted 5 characters in body

## 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 exist stochastic discount factorexists 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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 exist stochastic discount factor $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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 $$\mathbf{w}^T \mathbf{1} < 0$$.
6 added 197 characters in body

## 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 exist state pricesstochastic discount factor $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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.

(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 exist state prices $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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 exist stochastic discount factor $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.
5 added 2 characters in body

## 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.

(Note: I'm using bold letters for vectors.)

### Farkas'Farkas's Lemma and No Arbitrage

By Farkas'Farkas's Lemma, exactly one of the following conditions holds:

1. There exist state prices $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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.

(Note: I'm using bold letters for vectors.)

### Farkas' Lemma and No Arbitrage

By Farkas' Lemma, exactly one of the following conditions holds:

1. There exist state prices $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.

## 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.

(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 exist state prices $$\mathbf{s} \in \mathbb{R}^n$$ 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 $$\mathbf{w}^T \mathbf{1} < 0$$.
4 Add latex backslash
3 added 94 characters in body
2 added 138 characters in body
1