= Solution
A one-period <self-financing strategy> is a deterministic holdings vector $h=(h^0,\ldots,h^d)$, with initial cost $h\cdot S_0$ and terminal wealth $h\cdot S_1$. In the zero-initial-cost convention, an <arbitrage> satisfies
$$
h\cdot S_0=0,\qquad h\cdot S_1\ge0\text{ a.s.},\qquad\mathbb P(h\cdot S_1>0)>0.
$$
The convention allowing a nonpositive initial cost is equivalent here: invest any initial proceeds in the strictly positive <numéraire> to obtain a zero-cost <portfolio> with nonnegative terminal wealth.
Suppose a strictly positive <state-price density> $\rho$ satisfies the asset-pricing equations, with the indicated products integrable. By linearity,
$$
\mathbb E[\rho(h\cdot S_1)]=h\cdot S_0.
$$
For an <arbitrage>, the right-hand side is zero while the left-hand side is strictly positive, since the integrand is nonnegative and is positive with positive probability. This contradiction proves \b[absence of <arbitrage>].
An <attainable European contingent claim> $X$ is a terminal random payoff for which some holdings vector $h$ satisfies $h\cdot S_1=X$ almost surely; its replication cost is $h\cdot S_0$. To prove uniqueness in a <complete market>, suppose $\rho'$ is another positive <state-price density>. For every event $A$, attainability gives a <portfolio> replicating $S_1^{(0)}\mathbf1_A$. Its initial cost has both representations, so
$$
\mathbb E[(\rho-\rho')S_1^{(0)}\mathbf1_A]=0\qquad\text{for every }A.
$$
The signed random variable here is integrable, because $\mathbb E[\rho S_1^{(0)}]=\mathbb E[\rho'S_1^{(0)}]=S_0^{(0)}<\infty$. Take $A$ to be the event where it is positive and then where it is negative. Both its positive and negative parts have expectation zero, and hence vanish almost surely. Since $S_1^{(0)}>0$,
$$
\boxed{\rho=\rho'\quad\text{a.s.}}
$$
This proves that <complete markets have unique state-price densities> without requiring an extra lower bound on the random terminal <numéraire>.
For the finite-state example, order the states from top to bottom in the diagram and write the density values as $\rho_1,\rho_2,\rho_3$. The two pricing equations are
$$
3\rho_1+5\rho_2+6\rho_3=12,\qquad6\rho_1+4\rho_2+3\rho_3=15.
$$
Adding gives $\rho_1+\rho_2+\rho_3=3$. Put $t=\rho_3$; substitution then gives
$$
\boxed{(\rho_1,\rho_2,\rho_3)=\left(\frac{3+t}{2},\frac{3(1-t)}2,t\right),\qquad0<t<1.}
$$
These are all the strictly positive solutions, and each satisfies both equations. The parameter is a state-price-density value, not a probability.
For a concrete unattainable <contingent claim>, take the payoff vector $(0,1,0)$. A replicating <portfolio> $(a,b)$ would have to satisfy $3a+6b=0$ and $6a+3b=0$ in the first and third states. These imply $a=b=0$, contradicting the required payoff one in the middle state. Thus \b[the market is an <incomplete market>], despite having strictly positive <state-price densities>.
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