Problem 24.4 Easy Free 2 variations

Stock Against an Even-Money Bet

A stock currently worth $1 will either double in value or fall by half by tomorrow. You may also bet with a friend on the stock’s direction, at even (1:1) odds. Let $A$ be the smallest whole-dollar amount you invest in (long) the stock, and $B$ the smallest whole-dollar amount you bet with your friend that the stock falls, chosen so that your total profit is the same regardless of which way the stock moves. Find this guaranteed profit, or report $0$ if no such guaranteed profit is possible.

Work it out, then compare with the answer.

SolutionShow the worked solution

Hold the stock and bet that it falls, which is the opposite side to the stock. The bet on a fall is the right hedge, because a long stock position loses only half its value in the down state, while an even-money bet loses all of its stake.

Invest $A$ in the stock and bet $B$ that it falls. The profits are:

\text{up: } A - B, \qquad \text{down: } B - \tfrac12 A .

For the profit to be the same in both states we need

A - B = B - \tfrac12 A \iff \tfrac32 A = 2B \iff A = \tfrac43 B .

The smallest whole-dollar solution is $B = 3$ , $A = 4$ . Check the two states:

up: $4 - 3 = 1$ ;
down: $3 - 2 = 1$ .

So the profit is $1$ in each state, and in general it is $A - B = \tfrac13 B > 0$ . The profit is positive, so a genuine arbitrage exists.

\boxed{\$1}

Key idea. The stock prices the up state with risk-neutral probability $\tfrac13$ , since $\tfrac13\cdot 2 + \tfrac23\cdot\tfrac12 = 1$ . The bet prices it at $\tfrac12$ . Two inconsistent prices for the same event always allow a Dutch book.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 24.4a

Suppose instead all positions must be taken in whole units of $100. Investing one unit in the stock gains you $100 if it rises and loses $50 if it falls. Separately, you may bet one unit at a casino on the stock’s direction, winning $100 if correct and losing $100 if wrong. Find the guaranteed profit of a strategy that is certain to profit regardless of the stock’s movement and uses the smallest possible total number of units.

Work it out, then compare with the answer.

Solution

Take $s$ stock units and $b$ casino units on a fall. The profits are:

up: $100(s-b)$ ;
down: $100b - 50s$ .

Both are positive exactly when $b < s < 2b$ .

Other combinations cannot help:

Betting on a rise only adds to the up-state exposure.
Shorting the stock together with betting on a rise requires $b > s$ and $s > 2b$ , which is impossible.

Now minimise $s+b$ . For $b=1$ no integer satisfies $1 < s < 2$ . For $b=2$ we get $s=3$ , so the minimum is $5$ units, and this is the only $5$ -unit solution.

The strategy earns $100$ if the stock rises and $50$ if it falls, so the guaranteed profit is $50$ . The equal-payoff hedge ( $s=4$ , $b=3$ ) needs $7$ units to lock in $100$ .

\boxed{\$50}

Variation 24.4b

You and a friend each hold a $100 bill, each of which may be genuine or counterfeit with some unknown probability. Your friend is willing either to buy your bill from you for $20, or to sell you theirs for $20. You may also bet any amount with your friend, at even (1:1) odds, on whether the bills are genuine or counterfeit. Find the largest profit you can guarantee regardless of the true status of the bills.</statement>

Work it out, then compare with the answer.

Solution

Interpretation. Take the standard reading that both bills share the same status, so the bet is on “genuine” versus “counterfeit”. Profit means the change in your wealth, where a genuine bill is worth $100 and a counterfeit one $0. The answer is the same if the bills are independent and the bet is on your friend’s bill.

Let $x\in\{-1,0,1\}$ be the trade: sell your bill, do nothing, or buy your friend’s bill. Let $b$ be your net stake on “counterfeit”; a negative $b$ is a bet on “genuine”. Relative to your starting position, your profit is:

\text{genuine: } 100x - 20x - b = 80x - b, \qquad \text{counterfeit: } -20x + b .

As $b$ changes, the two profits move in opposite directions. So the smaller of the two is largest when they are equal, which gives $b = 50x$ . Both profits are then $30x$ .

The best trade is therefore $x=1$ : buy your friend’s bill for $20 and bet $50 that the bills are counterfeit.

Genuine: $100 - 20 - 50 = 30$ .
Counterfeit: $-20 + 50 = 30$ .

Key idea. The price of $20 implies a probability of $\tfrac15$ that the bill is genuine, while the even-odds bet implies $\tfrac12$ . Buying the bill and hedging with the bet exploits this gap. You can trade only one bill, which caps the profit at $30. Selling your bill instead, hedged with a $50 bet on “genuine”, locks in a loss of $30.

\boxed{\$30}

Spotted an error? Email us