Prediction Markets

Prediction markets trade in contracts that pay off according to whether some future event occurs. If the event occurs the contract pays off 1 unit, otherwise the contract becomes worthless. As a result, the price of a contract can be interpreted as the market's estimate of the probability that the future event will occur. Assume frictionless markets: one can buy (risking $r$ units to win $1-r$ units) and sell (risking $1-r$ units to win $r$ units) contracts at no cost.

QUESTIONS

1.) Individuals $A$ and $B$ think the probability that a future event will occur is $p$ and $q$ respectively. $A$ and $B$ have risk capital $X$ and $Y$. $A$ and $B$ invest so as to maximize the expected log of their capital. What price $r$ for the contract will assure that the sum of the demand for the contract from $A$ and $B$ is 0?

2.) A future event has $n$ possible outcomes each with associated contract $c_{i}$ which will pay off 1 unit if the $i$th possibility occurs while the other $n-1$ contracts will become worthless. Each contract has a market price $m_{i}$ and the market is efficient: $\sum_{i}^{n} m_{i}=1$.

You believe the true probability of the $i$th possibility occurring is $q_{i}$ for $i=1,...,n$ (where the $q_{i}$ also sum to 1). You have $X$ units to invest and you wish to do this in such a way that the expected value of the logarithm of your capital is maximized. How many of each of the $n$ contracts should you buy?

ANSWERS

1.) Let the demand for the contract from $A$ and $B$ be $a$ and $b$ respectively, where negative values of $a$ or $b$ correspond to a desire to sell the contract short. Then the expected value for $A$ is

$E_{A}=p*\log(X+a*(1-r))+(1-p)*\log(X-a*r)$


Maximizing this (by taking the derivative with respect to $a$ and setting equal to zero) we see that

$a=\frac{(p-r)*X}{(r*(1-r))}$

Similarly

$b=\frac{(q-r)*Y}{(r*(1-r))}$

Setting $a+b=0$ and solving for $r$ we obtain

$r=\frac{p*X+q*Y}{X+Y}$

So the market clearing price is just the weighted average of the opinions of $A$ and $B$ as to the correct price.

2.) We may assume all of $X$ is invested: if we buy an equal number of each of the $n$ contracts we will break even since the market is efficient. Let a_{i} be the amount invested in the $i$th contract. Since the price of the $i$th contract is $m_{i}$ we are buying $a_{i}/m_{i}$ contracts. If event $i$ occurs these contracts will each be worth 1 unit otherwise they will be worthless. Hence the payout if the $i$th event occurs is $a_{i}/m_{i}$. The expected log value is 

$\sum_{i}^{n} q_{i}*\log(a_{i}/m_{i})$

We wish to maximize this over the constraint that the sum of the a(i) is X. So using the method of Lagrange multipliers we look at the derivatives of (Sum q(i)*log(a(i)/p(i))) - y * (Sum a(i)) with respect to a(i). The ith derivative is q(i)/a(i) - y. So setting the derivatives to 0 we have a(i)=q(i)/y. So Sum a(i) = Sum q(i)/y = 1/y. So to satisfy the constraint we should set y=1/x. So a(i)=q(i)*X. So in other words we should spread our capital X over the n contracts in proportion to our estimate of the probability that each of the n associated mutually exclusive events will occur.

The problem as stated asks for the number of each contract we should buy. Let the number for the ith contract be n(i). Then n(i)=a(i)/p(i)=X*q(i)/p(i).