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For unrestricted final portfolio wealth in an $n$-period binomial market, write $R$ for the bank growth factor, $a>0$ for absolute risk aversion, and $Z=dQ/dP$ for the binomial-market probability density. The displayed optimizer has the state-price budget constraint ${\mathbb{E}}_Q{H}^{\ast }=R^n{w}_0$. Its exponential utility derivative satisfies $e^{-a{H}^{\ast }}=\lambda Z$ for a constant $\lambda >0$. The concave supporting-tangent inequality bounds any competing expected utility by this value, because the expected linear error is $\lambda {\mathbb{E}}_Q(H-H^{\ast })=0$. Strict concavity proves uniqueness, and market completeness supplies replication. The formula is altered by a constraint requiring nonnegative final wealth. With physical up probability $p$ and risk-neutral probability $q$, ${\mathbb{E}}_Q\log Z=n[q\log (q/p)+(1-q)\log ((1-q)/(1-p))]$.

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 $qu+(1-q)d=R$, givingFor 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 areThe cost of this replicating portfolio in a binomial market isApply 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 valueAny 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, isThis last expression makes $Z$ a function of the terminal stock price. Summing over paths gives ${\mathbb{E}}_{P}Z=1$, and strict positivity proves equivalence of the two probability measures.

For unrestricted terminal portfolio wealth $H$, its initial replication cost is $R^{-n}{\mathbb{E}}_{Q}H$. Hence feasible wealth satisfies the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. For the stated exponential utility, $v^{\prime }(x)=e^{-ax}$ and $v^{\prime \prime }(x)=-a{e}^{-ax}<0$. The binomial exponential-utility terminal wealth iswhereIt satisfies the budget, and its marginal utility is $v^{\prime }(H^{\ast })=\lambda Z$ with $\lambda =\exp (-a{R}^n{w}_0-{\mathbb{E}}_Q\log Z)>0$. The concave supporting-tangent inequality givesTaking expectations under $P$ makes the last term zero by the budget. Strict concavity makes equality possible only when $H=H^{\ast }$ 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.

Let the one-period riskless gross return be $R>0$, with $0<d<R<u$, and let the physical up probability be $\pi \in (0,1)$. The binomial market has risk-neutral probabilityUnder its equivalent martingale measure $Q$, successive multipliers $J_k$ are independent, taking $u,d$ with probabilities $q,1-q$. The risk-neutral pricing function has the extensionFor every fixed positive multiplier $A$, the function $s\mapsto f(As)$ is convex when $f$ is convex. Taking a positive weighted sum preserves convexity. More explicitly, for $0\le \lambda \le 1$,Thus the pricing extension is convex, and so are its values on the attainable stock-price grid.

The replicating portfolio in a binomial market holdsshares over the next interval. If $r+1<n$, put $A={\Delta }_{r+1}(us)$ and $D={\Delta }_{r+1}(ds)$. These are the secant slopes of $g_{r+2}$ on $[uds,u^{2}s]$ and $[d^{2}s,uds]$, respectively. For a convex function, the slope of a secant on a later adjacent interval is no smaller than that on the earlier interval: apply the defining convexity inequality to the middle point $uds$, then rearrange. Hence $A\ge D$. Using backward option pricing twice givesThe two coefficients are positive and sum to one. Therefore convex-payoff delta monotonicity in a binomial market yieldsIn particular an up move cannot reduce the stock holding. “Increases” here means nondecreases: an affine payoff gives equality, so strict increase cannot be claimed for every convex payoff.

For the expected utility maximization problem, let $P$ be the physical measure and $Z=dQ/dP$ on terminal paths. If a path has $j$ up moves, its binomial-market probability density isEvery terminal payoff is replicable in this complete market. Buying a positive payoff $H$ with initial wealth $w_0$ therefore imposes the state-price budget constraint ${\mathbb{E}}_{Q}H=R^n{w}_0$. Since the utility function is increasing, the full budget is used. Its derivative and second derivative arePut $a=\gamma /(\gamma -1)$. The first-order condition $v^{\prime }(H_{\ast })=\lambda Z$ for a positive multiplier $\lambda$ gives $H_{\ast }=C{Z}^{-a}$. The budget determines $C$ uniquely. Independence of the successive moves givesThus the optimal claim, as a function of the number of up moves, isBecause $S_n=S_0{u}^j{d}^{n-j}$ uniquely determines $j$, this is also a terminal-stock-price payoff, attainable by backward option pricing and claim replication. It rewards states relatively more probable under the investor's physical measure than under the pricing measure. If $\pi =q$, the optimum is the riskless payoff $R^n{w}_0$.

To prove global optimality rather than merely assert a stationary point, use the tangent inequality for the strictly concave function $v$:Taking physical expectations makes the last term zero for any fully financed competing claim, and nonpositive for any claim costing less. Therefore ${\mathbb{E}}_{P}v(H)\le {\mathbb{E}}_{P}v(H_{\ast })$, with equality only when $H=H_{\ast }$ almost surely. Finiteness causes no difficulty on the finite binomial tree, and $H_{\ast }>0$ in every terminal state.

Suppose under the physical measure $P$ the up indicators are independent with probability $p\in (0,1)$. If $N_k$ is the number of up moves through date $k$, a particular history has probability $p^{N_k}(1-p{)}^{k-N_k}$ under $P$ and $q^{N_k}(1-q{)}^{k-N_k}$ under $Q$. Thus the binomial-market probability density, the Radon-Nikodym derivative on ${\mathcal{F}}_k$, isIt is strictly positive. Its next multiplier has conditional expectation $p(q/p)+(1-p)((1-q)/(1-p))=1$, so $L_k$ is a mean-one martingale and $L_k={\mathbb{E}}_P[L_n\mid {\mathcal{F}}_k]$. These facts verify normalization and equivalence of the two probability measures.

The state-price density for terminal cash is ${\zeta }_n=L_n/B_n=b^{-n}{L}_n$. In particular $V_0={\mathbb{E}}_P[{\zeta }_{n}H]$. At an intermediate date the conditional change-of-measure identity givesIf physical up probabilities depend on the past, replace the constant-probability expression by the product over periods of $q/p_k$ on an up move and $(1-q)/(1-p_k)$ on a down move. Each multiplier still has conditional mean one, provided $0<p_k<1$, and the pricing measure retains conditional up probability $q$.