3
$\begingroup$

Definition. An arbitrage is a portfolio $H$ ∈ $R^n$ such that

• $H⋅P_0≤0≤H⋅P_1$ almost surely, and

• $P(H⋅P_0=0=H⋅P_1)<1$.

where $P_0$ and $P_1$ ∈ $R^n$ represent the prices at time $t=0,1$ respectively.

Now, my question is why do we need the first condition. Suppose there is only one asset $A$ which at time $t=0$ costs $3$. Then, at $t=1$ we have $P(A=2)=\frac{1}{2}$ and $P(A=1)=\frac{1}{2}$. The portfolio $H=−1$ should be an arbitrage because it yields certain profit with no risk attached but it isn't because $H⋅P_1<0$.

$\endgroup$
0

2 Answers 2

6
$\begingroup$
  • Conceptually, an arbitrage gives you something for nothing.

  • This is a different idea than making or losing money almost surely. A risk free bond allows you to make money almost surely, but it isn't an arbitrage.

What's going wrong, the source of confusion in your example?

You've implicitly assumed the existence of a cash security, with interest rate 0 (but haven't made it explicit)

In your example, you've assumed that cash can be moved between $t=0$ and $t=1$ with an interest rate of 0. You need a security to do this.

If there exists a risk free security returning 0, then making money almost surely violates the law of one price (because you effectively have 2 different risk free rates). With unrestricted buying & selling, you can then construct an arbitrage by going long the high rate and short the low rate.

See my comment, below @noob2's quality answer.

The Law of One Price (linearity) and No Arbitrage

Two different concepts that often get conflated (even in texts) are:

  • The law of one price, that is, that the pricing function is linear. If $X$ and $Y$ are random variables representing payoffs, $\alpha$, and $\beta$ are scalars, and $f$ is the pricing function, linearity of the pricing function $f$ implies: $$ f(\alpha X + \beta Y) = \alpha f(X) + \beta f(Y) $$ The idea here is that the price of a portfolio should be linear in the price of its components.

  • The absence of arbitrage. Loosely speaking, no arbitrage requires that any security with strictly positive payoffs should have a positive price, that you can't get something for nothing.

The law of one price allows you to write the pricing function as an inner product with state prices. The absence of arbitrage implies those state prices are positive.

How the two different concepts get conflated is that violations of linearity can allow you to construct an arbitrage. People often say no arbitrage when they really mean linearity and no arbitrage.

References

Cochrane, John, Asset Pricing, 2005

$\endgroup$
0
3
$\begingroup$

Your example is not an arbitrage, it is just an economy where interest rates are negative, where a fierce 50% a year deflation is taking place.

In these models arbitrage is when you have negative net worth in one period (you owe) and positive in the next (you no longer owe, other people may even pay you money). The first condition expresses this.

For example in one period asset 1 is worth 7 and asset 2 is worth 5. In the second period both assets are worth 5 in every case. Then the arbitrage is to sell a1 and buy a2. You owe 2. Lo and behold your debt disappears in the next period, so you made an arb profit. (Note that it involves two or more securities, not just one. Arbitrage is always one thing relative to another (or others)).

[We could easily reformulate your example to fit this framework. Suppose we have 2 assets: cash (which is always worth 1, in every period and in every state) and the asset A which you described. Now we can make an arbitrage by selling Asset A and going long 3 units of cash in period zero. We owe nothing in period zero and we are worth something in pperiod 1. Arbitrage. Note that it is the existence of an asset which keeps its value while Asset A depreciates that makes the arbitrage possible].

$\endgroup$
3
  • $\begingroup$ Thanks for your reply. What about the following case: There are assets $A$ and $B$ worth $(3,4)$ at $t=0$. Then at $t=1$ the prices are $(4,7)$. The portfolio $(-4,3)$ is an arbitrage, but the portfolio $(-4,2)$ isn't according to the conditions. However, the second portfolio means we short 4 assets $A$ so we have $12$ cash and we are long in 2 assets $B$ and we have spent $8$ cash. At $t=1$ we buy 4 assets $A$ for $16$ and sell 2 assets $B$ for $14$ so overall we have $2$ cash (we have $12-8=4$ at end of $t=0$ and we lose $2$ at $t=1$) so we have made money. $\endgroup$
    – tergarg
    Aug 16, 2018 at 14:17
  • 1
    $\begingroup$ @tergarg Your example in the comment above assumes there is a third asset $C$ called cash worth $1$ at $t=0$ and $1$ at $t=1$! In your example, your second portfolio isn't $H = \begin{bmatrix} -4 \\ 2 \end{bmatrix}$, it is $H = \begin{bmatrix} -4 \\ 2 \\ 4 \end{bmatrix}$. Your time 0 vector or prices is $P_0 = \begin{bmatrix}3 \\ 4 \\ 1 \end{bmatrix}$. Your time 1 vector or prices is $P_1 = \begin{bmatrix}4 \\ 7 \\ 1 \end{bmatrix}$. Then $H \cdot P_0 = -4\cdot 3 + 2 \cdot 4 + 4 \cdot 1 = 0$ but $H \cdot P_1 = -4 \cdot 4 + 2 \cdot 7 + 4 \cdot 1 = 2$ so it is an arbitrage. $\endgroup$ Aug 16, 2018 at 16:55
  • $\begingroup$ To clarify further, your example has security prices which violate the assumption of a linear pricing function, and this allows you to construct an arbitrage. In your example in the comments $A$, $B$, and $C$ are all risk free assets giving you risk free rates of 33%, 75%, and 0% respectively. $\endgroup$ Aug 16, 2018 at 16:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.