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 Incoming links: Two-stock binomial market without a supplied bond

Past exam of the martingales and financial mathematics course of the University of Oxford / 2009 / B / 6 / a / Solution 2026-10-31

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d_{1} > d_{2}

, buy

1/100

of stock 1 and sell

1/100

of stock 2. Initial cost is zero; the one-period payoff is

u_{1} - u_{2} > 0

in the up state and

d_{1} - d_{2} > 0

in the down state. This is an arbitrage. If terminal rather than time-one arbitrage is required, reinvest the positive receipts in either positive-price stock.

For

d_{1} < d_{2}

, define the one-period Arrow state prices

π_{u} = \frac{d _{2} - d _{1}}{u _{1} d _{2} - u _{2} d _{1}}, π_{d} = \frac{u _{1} - u _{2}}{u _{1} d _{2} - u _{2} d _{1}} .

(1)

The denominator is positive because

u_{1} / u_{2} > 1 > d_{1} / d_{2}

. Both state prices are strictly positive and satisfy

π_{u} u_{i} + π_{d} d_{i} = 1

for

i = 1, 2

. Thus every portfolio's present value equals the positive state-price-weighted value of its two future payoffs.

Use the same strictly positive pricing rule at every node of the two-period tree. A nonnegative terminal payoff that is positive in at least one state has strictly positive price by backward induction, so a zero-cost self-financing strategy cannot generate an arbitrage. Therefore the market is arbitrage free when $d_{1} < d_{2}$ , despite having no initially supplied bond. This is a two-stock binomial market without a supplied bond.

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