 Solution

A zero-cost stock holding $\theta$ has terminal payoff $\theta (S_1-1)$. If $a\le 1$, shorting the share and holding a bond gives $1-S_1\ge 0$ and a strictly positive payoff almost surely. If $a>1$, both signs of $S_1-1$ occur with positive probability, so no nonzero holding is an arbitrage. Hence arbitrage freedom is equivalent to $a>1$.

The equivalent martingale measures are exactly the densities $Z(s)>0$ almost everywhere on $(0,a)$ satisfyingThey exist for $a>1$: exponential tilts of the uniform law have means ranging continuously from zero to $a$, so one has mean one.

Under the usual utility assumptions $U^{\prime }>0$, $U^{\prime \prime }<0$, and existence of an admissible optimum, $J(\theta )=E[U(w+\theta (S_1-1))]$ is strictly concave and $J^{\prime }(0)=U^{\prime }(w)(a/2-1)$. Its decreasing derivative implies that the optimizer is positive exactly when $a>2$, negative when $a<2$, and zero when $a=2$, provided the feasible interval contains zero in its interior. The hypothesis $C^2$ alone is insufficient: a linear increasing utility can give an unbounded objective rather than a finite optimizer. Concavity and existence are implicit in the standard utility formulation.

For $a=2$, $\theta =0$ makes marginal utility constant, so the pricing measure is the original uniform law andFor any equivalent pricing measure, the call expectation is positive and strictly less than $E_Q{S}_1/2=1/2$, since $(s-1{)}^{+}<s/2$ for $0<s<2$. Both endpoints can be approached. Symmetric densities concentrated in $[1-\delta ,1+\delta ]$ give prices tending to zero; symmetric densities concentrated near zero and two give prices tending to one half. Mix either density with an arbitrarily small amount of the uniform law to keep it everywhere positive; symmetry keeps its mean one. The set of positive mean-one densities is convex and the call price is linear, so every intermediate value is attained. Thus the exact price range is $(0,1/2)$, with neither endpoint attained.