Multiple states and assets

Now we will extend the above analysis when there are multiple states where the world can be tomorrow, and a number of assets available for investment. We need to specify the model in the following lines:

• The initial date is $t=0$ and the terminal date is $t=1$. Trading is possible in both dates, and the agent wishes to maximize the payoffs;

• The state space is finite with $K<\infty$ elements. We denote it by
  $$\Omega =\left\{ \omega _{1},\omega _{2},\ldots ,\omega _{K}\right\}$$   .
  Each element of $\Omega$ represents a different state of the world. This is unknown at time $t=0$ but becomes observed at time $t=1$;

• A probability measure $P$ on $\Omega$, where $P\left( \omega \right) >0$ for all $\omega \in \Omega$;

• A bank account process $B=\left\{ B_{t}:t=0,1\right\}$, where $B_{0}=1$ and $B_{1}$ is a random variable depending on tomorrow's state. In most applications $B_{1}$ is not a random variable since it is usually the same no matter what the state is, but we can consider the general case here.^4.2 The value of $B_{1}$ is strictly positive unlike the value of the other securities, in fact usually we have $B_{1}\geq 1$. Then $r=B_{1}-1$ must be thought of as the interest rate; and

• A price process $\mathbf{S}=\left\{ \mathbf{S}%% _{t}:t=0,1\right\}$, where $\mathbf{S}_{t}$ is a vector of the $N$ security prices, $\mathbf{S}_{t}=\left( S_{1}\left( t\right) ,S_{2}\left( t\right) ,\ldots ,S_{N}\left( t\right) \right)$, $N<\infty$. The price at time $t=0$ is known where the price at time $t=1$ is again a random variable depending on the state of the world.

Now we can define a number of variables that are interesting. A trading strategy describes an investors portfolio which is carried forward from time $t=0$ to $t=1$. It is a $\left( N+1\right) \times 1$ vector $%% \mathbf{H}=\left( H_{0},H_{1},\ldots ,H_{N}\right)$, the first element $%% H_{0}$ describes the amount invested in the bank account while the rest represent the number of each security in the portfolio held between times 0 and $1$. Negative values mean either borrowing or short selling, restrictions can be imposed if these are not allowed.

The value process $V=\left\{ V_{t}:t=0,1\right\}$ will describe the value of the portfolio across time. This is

$$V_{t}=H_{0}B_{t}+\sum_{n=1}^{N}H_{n}S_{n}\left( t\right)$$   , $$t=0,1$$.

The value will depend on the trading strategy, and $V_{1}$ is obviously a random variable. The gains process will describe the total gain or loss realized between time 0 and $1$. For every security the gain will be $%% H_{n}\left[ S_{n}\left( 1\right) -S_{n}\left( 0\right) \right]$, and the total gain will be

$$G=H_{0}r+\sum_{n=1}^{N}H_{n}\Delta S_{n}$$.

It is easily verified that $V_{1}=V_{0}+G$, which means that all changes in the value of the portfolio are due to changes in the investment value, no money has been added or withdrawn.

Our interest is in the relative movements of the security prices, therefore we will normalize them using a numeraire. A natural choice for the numeraire is the bank account, therefore we define the discounted price processes $\mathbf{S}_{t}^{\star }=\left( S_{1}^{\star }\left( t\right) ,S_{2}^{\star }\left( t\right) ,\ldots ,S_{N}^{\star }\left( t\right) \right)$, where $S_{n}^{\star }\left( t\right) =S_{n}\left( t\right) /B_{t}$. The discounted value process and the discounted gains process are defined naturally

$$V_{t}^{\star } = H_{0}+\sum_{n=1}^{N}H_{n}S_{n}^{\star }\left( t\right)$$

$$G^{\star } = \sum_{n=1}^{N}H_{n}\Delta S_{n}^{\star }$$

One can easily verify that $V_{t}^{\star }=V_{t}/B_{t}$ and that $V_{1}^{\star }=V_{0}^{\star }+G^{\star }$.

Example 14 (Discounted process I)   Suppose that there are $K=2$ possible states of the world and $N=1$ risky security. The interest rate is $r=1/9$.

$$\begin{tabular}{cc\vert cc} $n$ & $S_{n}\left( 0\right) $ & \mu... ...\omega _{2}$ \hline $1$ & $5$ & $20/3$ & $40/9$ \hline \end{tabular}$$

The value $B_{1}=r+1=10/9$ and the discounted process is

$$\begin{tabular}{cc\vert cc} $n$ & $S_{n}^{\star }\left( 0\right)... ...$ & $\omega _{2}$ \hline $1$ & $5$ & $6$ & $4$ \hline \end{tabular}$$

For a trading strategy $\mathbf{H}$ we have that

$$V_{0}=V_{0}^{\star }=H_{0}+5H_{1}$$, and

$$V_{1} = \frac{10}{9}H_{0}+H_{1}S_{1}\left( 1\right)$$

$$V_{1}^{\star } = H_{0}+H_{1}S_{1}^{\star }\left( 1\right)$$

$$G = \frac{1}{9}H_{0}+H_{1}\left[ S_{1}\left( 1\right) -5\right]$$

$$G^{\star } = H_{1}\left[ S_{1}^{\star }\left( 1\right) -5\right]$$

Therefore in state $\omega _{1}$

$$V_{1}\left( \omega _{1}\right) = \frac{10}{9}H_{0}+\frac{20}{3}H_{1}$$

$$V_{1}^{\star }\left( \omega _{1}\right) = H_{0}+6H_{1}$$

$$G\left( \omega _{1}\right) = \frac{1}{9}H_{0}+\frac{5}{3}H_{1}$$

$$G^{\star }\left( \omega _{1}\right) = H_{1}$$

and in state $\omega _{2}$

$$V_{1}\left( \omega _{2}\right) = \frac{10}{9}H_{0}+\frac{40}{9}H_{1}$$

$$V_{1}^{\star }\left( \omega _{2}\right) = H_{0}+4H_{1}$$

$$G\left( \omega _{2}\right) = \frac{1}{9}H_{0}-\frac{5}{9}H_{1}$$

$$G^{\star }\left( \omega _{2}\right) = -H_{1}$$

Example 15 (Discounted process II)   Just take $K=3$ and set in state $\omega _{3}$ $S_{1}\left( 1\right) =30/9$ so that $S_{1}^{\star }\left( 1\right) =3$. Examine how things change now.

Example 16 (Discounted process III)   Suppose that there are $K=3$ possible states of the world and $N=2$ risky securities. The interest rate is $r=1/9$.

$$\begin{tabular}{cc\vert ccc} $n$ & $S_{n}\left( 0\right) $ & \m... ... $40/9$ \\ $2$ & $10$ & $40/3$ & $80/9$ & $80/9$ \hline \end{tabular}$$

The value $B_{1}=r+1=10/9$ and the discounted process is

$$\begin{tabular}{cc\vert ccc} $n$ & $S_{n}^{\star }\left( 0\right... ...$ & $6$ & $4$ \\ $2$ & $10$ & $12$ & $8$ & $8$ \hline \end{tabular}$$

Example 17 (Discounted process IV)   Suppose that $K=4$ and the prices in state $\omega _{4}$ are $S_{1}\left( 1\right) =20/9$ and $S_{2}\left( 1\right) =20/9$. The discounted process is now

$$\begin{tabular}{cc\vert cccc} $n$ & $S_{n}^{\star }\left( 0\righ... ...$ & $2$ \\ $2$ & $10$ & $12$ & $8$ & $8$ & $2$ \hline \end{tabular}$$

In order for the model to be appropriate from the economic point of view, it has to satisfy some criteria. The most important of them is that it must be impossible for one to start with nothing and not run the risk of loosing. This would happen if there existed a dominant strategy.

We say that the trading strategy $\mathbf{\hat{H}}$ is a dominant strategy if there exists another strategy $\mathbf{\check{H}}$ such that $%% \hat{V}_{0}=\check{V}_{0}$ and $\hat{V}_{1}\left( \omega \right) >\check{V}%% _{1}\left( \omega \right)$ for all $\omega \in \Omega$. Both strategies start with the same amount of money, but one will certainly end up with more. It is easy to see that by employing the strategy $\mathbf{\bar{H}=\hat{%% H}-\check{H}}$ one will start will nothing and end up with something positive, no matter what the state of the world is tomorrow. This is obviously unreasonable. In fact the following are true

• There exists a dominant strategy if and only if there exists a trading strategy satisfying $V_{0}=0$ and $V_{1}\left( \omega \right) >0$ for all $\omega \in \Omega$; and

• There exists a dominant strategy if and only if there exists a trading strategy satisfying $V_{0}<0$ and $V_{1}\left( \omega \right) \geq 0$ for all $\omega \in \Omega$.

It is easy to note now the relationships between the value processes and the state prices introduced in the previous sections. The quantity $V_{1}\left( \omega \right)$ is the payoff of the strategy $\mathbf{H}$ if state $\omega$ prevails whereas $V_{0}$ is the price of this claim. The pricing of claims is logically consistent if there exists a linear pricing measure $\pi :\Omega \rightarrow \left[ 0,1\right]$ regardless of the trading strategy fostered. Then for the trading strategy $\mathbf{H}$ one would have

$$V_{0}^{\star }=\sum_{\omega }\pi \left( \omega \right) V_{1}^{\st... ...right) \frac{V_{1}\left( \omega \right) }{B_{1}\left( \omega \right) }\text{.}$$

In addition, one has

$$H_{0}+\sum_{n=1}^{N}H_{n}S_{n}^{\star }\left( 0\right) =\sum_{\om... ...) \left[ H_{0}+\sum_{n=1}^{N}H_{n}S_{n}^{\star }\left( 1,\omega \right) \right]$$   ,

for all trading strategies. By choosing the strategy $H_{1}=\cdots =H_{N}=0$, one can see that

$$\pi \left( \omega _{1}\right) +\cdots +\pi \left( \omega _{K}\right) =1$$.

Therefore the linear pricing measure can be thought of as a probability measure on the sample space $\Omega$. Taking now for an arbitrary $i\in \left\{ 1,\ldots ,N\right\}$ a trading strategy with $H_{n}=0$ for all $%% n\neq i$, the above imply that

$$S_{n}^{\star }\left( 0\right) =\sum_{\omega }\pi \left( \omega \right) S_{n}^{\star }\left( 1,\omega \right)$$ (LPM)

One therefore proves that

• The vector $\pi$ is a linear pricing measure if and only if it is a probability measure on $\Omega$ satisfying the equation (LPM) above.

One can see that since the pricing measure can be thought of as a probability measure, the initial price of each security is equal to the expectation under this measure of the final discounted price. This is denoted as

$$S_{n}\left( 0\right) =\mathbf{E}_{\pi }S_{n}^{\star }\left( 1\right)$$   .

It turns out that there is a very close relationship between the concepts of dominant strategies and linear pricing measures.^4.3 In particular

• There exists a linear pricing measure if and only if there are no dominant strategies.

The above result is very important, because it states that if one can find a linear pricing measure, she does not have to check whether or not dominant strategies exist. Therefore in that sense a pricing measure is sufficient to ensure that the model does not permit unreasonable results. In that stage we can see that there exist cases when the securities market model is even more unreasonable than the one that permits dominant strategies. This happens if the law of one price fails to hold.

It is said that the law of one price holds for a securities market model if there do not exist two trading strategies $\mathbf{\hat{H}}$ and $%% \mathbf{\check{H}}$ such that $\hat{V}_{1}\left( \omega \right) =\check{V}%% _{1}\left( \omega \right)$ for all $\omega \in \Omega$, but $\hat{V}_{0}>%% \check{V}_{0}$. Therefore there is no ambiguity about the price at time $t=0$ of a claim. One has to note though that if there are no trading strategies that yield the same payoffs at time $t=1$ then the law of one price will hold automatically. The above can be translated in the statement that

• If there are no dominant strategies then the law of one price holds. The converse is not necessarily true.

Now we turn back to the models where no dominant strategies exist. We will show that this family of models is still not perfectly reasonable: they allow models where investors have the possibility of making a profit on a transaction without being exposed to the risk of incurring a loss. This would happen if there exists a trading strategy that starts with nothing, cannot loose any money and can end up with a positive amount in at least one of the states. This strategy is called an arbitrage opportunity.

An arbitrage opportunity is a trading strategy $\mathbf{H}$ such that $V_{0}=0$, $V_{1}\left( \omega \right) \geq 0$ for all $\omega \in \Omega$, and $\mathbf{E}V_{1}>0$. In fact it is a riskless way of making money and therefore something that we want to rule out. It follows that

• If there exists a dominant strategy, then there exists an arbitrage opportunity. The converse is not necessarily true.

• The trading strategy $\mathbf{H}$ is an arbitrage opportunity if and only if $V_{0}^{\star }=0$, $V_{1}^{\star }\left( \omega \right) \geq 0$ for all $\omega \in \Omega$, and $\mathbf{E}V_{1}^{\star }>0$.

• The trading strategy $\mathbf{H}$ is an arbitrage opportunity if and only if $G^{\star }\geq 0$ and $\mathbf{E}G^{\star }>0$.

Therefore the securities markets have been classified as described in figure 4.5

Figure 4.5: Classification of security markets

The connection of linear pricing measures and the ruling out of arbitrage opportunities is not sufficient. The existence of a linear pricing measure might be enough to rule out dominant strategies but it is not enough to rule out arbitrage opportunities. What we need is the risk neutral probability measure, a special member of the linear pricing measure family. The difference is tiny: a risk neutral probability measure just has to attach a strictly positive probability to every event $\omega \in \Omega$.

A probability measure $Q:\Omega \rightarrow \left( 0,1\right)$ on $\Omega$ is called risk neutral if $Q\left( \omega \right) >0$ for all $\omega \in \Omega$, and $\mathbf{E}_{Q}\Delta S_{n}^{\star }=0$ for all $n=1,\ldots ,N$.^4.4 The latter is equivalent to $S_{n}\left( 0\right) =S_{n}^{\star }\left( 0\right) =\mathbf{E}_{Q}S_{n}^{\star }\left( 1\right)$. This is effectively the same measure we introduced in the beginning of the chapter, but from a different starting point. We can therefore now present the very important result

• There are no arbitrage opportunities if and only if there exists a risk neutral probability measure $Q$.

Example 18 (Risk neutral measure II)   Say that we have the situation described in the Discounted process II example. In order for a risk neutral measure to exist there must be

$$\left\{ \begin{tabular}{l} $5=6Q\left( \omega _{1}\right) +4Q\lef... ...Q\left( \omega _{2}\right) +Q\left( \omega _{3}\right) $%% \end{tabular}\right.$$   .

The result is of the form $Q=\left( \lambda ,2-3\lambda ,-1+2\lambda \right)$ for $1/2<\lambda <2/3$. Since we have two equations with three unknowns, the problem will admit an infinite number of probability measures as solutions. Each one will be sufficient to rule out arbitrage opportunities.

Example 19 (Risk neutral measure III)   In this case one has to solve the system

$$\left\{ \begin{tabular}{rl} $5=$ & $6Q\left( \omega _{1}\right) ... ...Q\left( \omega _{2}\right) +Q\left( \omega _{3}\right) $%% \end{tabular}\right.$$   .

The unique solution is $Q\left( \omega _{1}\right) =Q\left( \omega _{3}\right) =1/2$, and $Q\left( \omega _{2}\right) =0$. Although this is a linear pricing measure, it is not a risk neutral measure since it is not strictly positive. Therefore there is an arbitrage opportunity.

Remember now the definition of a contigent claim $X$: it a random variable representing a payoff at time $t=1$. A contigent claim is called attainable or marketable if there exists some trading strategy $%% \mathbf{H}$ called the replicating portfolio, such that $V_{1}=X$. Denote with $p$ the price of this contigent claim at time $t=0$. If $%% p<>V_{0}$, then there are obvious arbitrage opportunities. If $%% p=V_{0}$ then a riskless profit cannot be created. Nevertheless this is not sufficient to say that $p$ is the arbitrage free price of the contigent claim, unless the law of one price holds. We have seen before that the existence of a risk neutral probability measure is sufficient for the law of one price to hold, therefore in this case we have the valuation concept

• If the law of one price holds then the time $t=0$ value of an attainable contigent claim $X$ is
  $$V_{0}=H_{0}B_{0}+\sum_{n=1}^{N}H_{n}S_{n}\left( 0\right)$$   ,
  where $\mathbf{H}$ is the trading strategy that generates $X$.

We can take this one step further when we have the stronger condition that the market is free of arbitrage. The risk neutral valuation principle states that

• If the simple period model is free of arbitrage opportunities, then the time $t=0$ value of an attainable contigent claim $X$ is
  $$p=\mathbf{E}_{Q}\left[ \frac{X}{B_{1}}\right]$$   ,
  where $Q$ is any risk neutral probability measure.

Example 20 (State prices)   The state prices that we introduced in the previous sections can be seen in this context as the contigent claim

$$X_{\hat{\omega}}\left( \omega \right) =\left\{ \begin{tabular}{ll... ...hat{\omega}$ \\ $0$ & , if $\omega \neq \hat{\omega}$%% \end{tabular}\right.$$   ,

for some $\hat{\omega}\in \Omega$. If $X_{\hat{\omega}}$ is attainable this leads to the state price

$$q_{\hat{\omega}}=\mathbf{E}_{Q}\left[ \frac{X_{\hat{\omega}}}{B_{... ...rac{Q\left( \hat{\omega}\right) }{%% B_{1}\left( \hat{\omega}\right) }\text{.}$$

Example 21 (Option price)   Suppose that there is only one security $N=1$ and $X$ has the form

$$X=\max \left\{ 0,S\left( 1\right) -e\right\}$$   ,

where $e$ denotes the exercise price. If $X$ is attainable, then the time $%% t=0$ price of the option is

$$\mathbf{E}_{Q}\left[ \frac{X}{B_{1}}\right] =\sum_{\omega \in \Om... ...right) \frac{S\left( 1,\omega \right) -e}{B_{1}\left( \omega \right) }\text{,}$$

where the set $\Omega ^{\star }=\left\{ \omega \in \Omega :S\left( 1,\omega \right) \geq e\right\}$.

We will denote now with $\mathcal{Q}$ the set of all risk neutral probability measures, and assume that $\mathcal{Q}\neq \emptyset$. This fact does not ensure that all contigent claims can be priced, since they might not be marketable. If they are not marketable then of course there are no arbitrage opportunities. Now we have to create a method of checking whether or not a contigent claim is indeed marketable. We say that a market is complete if all contigent claims can be generated by some trading strategy, otherwise the model is incomplete. We can show that

• Suppose that there are no arbitrage opportunities. Then the model is complete if and only if the number of states in $\Omega$ is equal to the number of the independent vectors in $\left\{ B_{1},S_{1}\left( 1\right) ,\ldots ,S_{N}\left( 1\right) \right\}$, the matrix
  $$\mathbf{A}=\left[ \begin{tabular}{cccc} $B_{1}\left( \omega _{1}\... ...$ & $%% \cdots $ & $S_{N}\left( 1,\omega _{K}\right) $%% \end{tabular}\right]$$   .
  Completeness of course means that the system
  $$\left[ \begin{tabular}{cccc} $B_{1}\left( \omega _{1}\right) $ &... ...ht) $ \\ $\vdots $ \\ $X\left( \omega _{K}\right) $%% \end{tabular}\right]$$
  has a solution for every vector
  $$\left[ \begin{tabular}{llll} $X\left( \omega _{1}\right) $ & $X\... ... & $\cdots $ & $ X\left( \omega _{K}\right) $\end{tabular}\right] ^{\prime }$$

To complete this discussion, there are a couple of very strong results, namely

• The contigent claim $X$ is attainable if and only if $\mathbf{E}_{Q} \left[ \frac{X}{B_{1}}\right]$ takes the same value for every $Q\in \mathcal{Q}$; and

• The model is complete if and only if $\mathcal{Q}$ consists of exactly one probability measure.