= Solution
Take bank-account values $B_k=R^k$ with $R>0$, and let the <stock> multiply independently by $u$ or $d$ at each step, where $0<d<R<u$. Under the physical <probability measure> $P$, the up probability is $p\in(0,1)$; the <filtration> records the successive moves. These assumptions specify the standard <binomial market>. If $R$ lies outside the strict interval $(d,u)$, a long or short stock position financed through the <bank account> gives an <arbitrage>.
The <risk-neutral probability in a binomial market> is fixed by $qu+(1-q)d=R$, giving
$$
\boxed{q=\frac{R-d}{u-d}\in(0,1).}
$$
For successor claim values $V_u,V_d$ at a node with current stock price $S$, solve the two replication equations. The current stock holding and current cash value are
$$
\Delta=\frac{V_u-V_d}{S(u-d)},\qquad
C=\frac{uV_d-dV_u}{R(u-d)}.
$$
The cost of this <replicating portfolio in a binomial market> is
$$
\Delta S+C=\frac{qV_u+(1-q)V_d}{R}.
$$
Apply this <backward option pricing> recursion from any terminal <contingent claim> $H$, including claims depending on the whole sequence of moves. It constructs a <self-financing portfolio> and proves <market completeness> by induction on the remaining number of periods. Taking successive <conditional expectations> under the measure $Q$ with independent up probability $q$ gives the unique <risk-neutral pricing> value
$$
V_k=R^{-(n-k)}\mathbb E_Q[H\mid\mathcal F_k].
$$
Any other price would give an <arbitrage> by buying the cheaper of the claim and its replica and selling the dearer.
If $K$ is the number of up moves, a particular path has probabilities $p^K(1-p)^{n-K}$ under $P$ and $q^K(1-q)^{n-K}$ under $Q$. Therefore the <binomial-market probability density>, namely the <Radon-Nikodym derivative>, is
$$
\boxed{Z=\frac{dQ}{dP}=\left(\frac qp\right)^K\left(\frac{1-q}{1-p}\right)^{n-K},\qquad
K=\frac{\log(S_n/(S_0d^n))}{\log(u/d)}.}
$$
This last expression makes $Z$ a function of the terminal <stock> price. Summing over paths gives $\mathbb E_PZ=1$, and strict positivity proves equivalence of the two <probability measures>.
For unrestricted terminal <portfolio wealth> $H$, its initial replication cost is $R^{-n}\mathbb E_QH$. Hence feasible wealth satisfies the <state-price budget constraint> $\mathbb E_QH=R^nw_0$. For the stated <exponential utility>, $v'(x)=e^{-ax}$ and $v''(x)=-ae^{-ax}<0$. The <binomial exponential-utility terminal wealth> is
$$
\boxed{H^*=R^nw_0+\frac{\mathbb E_Q\log Z-\log Z}{a},}
$$
where
$$
\mathbb E_Q\log Z
=n\left[q\log\frac qp+(1-q)\log\frac{1-q}{1-p}\right].
$$
It satisfies the budget, and its <marginal utility> is $v'(H^*)=\lambda Z$ with $\lambda=\exp(-aR^nw_0-\mathbb E_Q\log Z)>0$. The <concave supporting-tangent inequality> gives
$$
v(H)\leq v(H^*)+\lambda Z(H-H^*).
$$
Taking <expectations> under $P$ makes the last term zero by the budget. Strict <concavity> makes equality possible only when $H=H^*$ in every positive-probability state. The already constructed <complete market> supplies its <replicating strategy>, so this proves global optimality and uniqueness. The formula allows negative final wealth, as permitted by the unrestricted trading problem.
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