= Solution
{c}
If $d_1>d_2$, 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>
$$
\pi_u=\frac{d_2-d_1}{u_1d_2-u_2d_1},\qquad
\pi_d=\frac{u_1-u_2}{u_1d_2-u_2d_1}.
$$
The denominator is positive because $u_1/u_2>1>d_1/d_2$. Both state prices are strictly positive and satisfy $\pi_u u_i+\pi_d d_i=1$ 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 \b[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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