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For $g(0)=0$ and strict concavity, per-capita recruitment $g(N)/N$ strictly decreases because $g(N)={\int }_0^N{g}^{\prime }(s)ds>N{g}^{\prime }(N)$. Thus a positive fixed point is unique if it exists. A slope $g^{\prime }(0)>1$ makes extinction unstable but does not guarantee a positive intersection: $g(N)=2N+1-e^{-N}$ stays above the identity. A large-population restriction supplies existence; the local fixed-point stability condition remains $|g^{\prime }(N_{\ast })|<1$.

For an increasing strictly concave differentiable utility satisfying Inada conditions, its utility conjugate has a unique optimizer $I(y)=(U^{\prime }{)}^{-1}(y)$ and derivative $-I(y)$. The dual is strictly decreasing and strictly convex. Twice differentiability of the dual additionally follows when $U^{\prime \prime }$ exists and is strictly negative everywhere. Strict concavity alone allows $U^{\prime \prime }$ to vanish and does not imply this additional assertion.

The log-likelihood is the natural logarithm of the likelihood function:For positive finite likelihood, the natural logarithm is a strictly increasing function, so the maximizing parameter values are unchanged. Set $\ell =-\infty$ where $L=0$; the same ordering then includes zero-likelihood boundaries whenever a positive finite maximum exists. For independent and identically distributed random variables, a product becomes a sum,This makes derivatives simpler and avoids forming products of tiny numbers in floating-point arithmetic. These equal-maximizer and additive formulas underlie the worked calculations in MIT 18.05, Reading 10b, section 3.1.

For the ordered sample of eight successes and two failures in likelihood function, on $0<p<1$,The score equation ${\ell }^{\prime }(p)=0$ gives $8(1-p)=2p$, hence $\hat{p}=8/10=0.8$. The negative second derivative proves strict concavity, so this is the unique interior maximum. Both counts are positive, so $L(0)=L(1)=0$, while $L(p)>0$ inside: the same point is the unique global maximum on $[0,1]$. Counting only successes adds $\log 45$ to the log-likelihood, leaving both derivatives and the MLE unchanged. For all-success or all-failure samples, the boundary analysis in maximum-likelihood estimator applies instead of an interior score equation.

The two panels below compare $k=2,5,8$ successes out of ten Bernoulli trials. Their peaks are at $\hat{p}=k/10$. The left panel uses relative height; the right panel uses the log-likelihood difference from each peak. These are the same comparison on two scales:A difference of $-2$ means relative height $e^{-2}$, and a difference of $-10$ means $e^{-10}$; the log-likelihood scale makes small relative heights easier to see. Neither vertical axis is a parameter probability density function. The zero likelihood endpoints have log-likelihood $-\infty$ and are outside the finite vertical range in the right panel.

Multiplying a likelihood function by a positive parameter-independent constant adds a constant to its log-likelihood. Subtracting its peak removes that constant. This is why the ordered-data and count-data curves coincide in this figure after rescaling. When differentiable, the gradient of the log-likelihood is the score function; the negative Hessian matrix is the observed information. A solution of a score equation still needs checking against other stationary points and boundaries: maximizing the log-likelihood does not automatically make it concave.

For a ferromagnetic Ising model with coordination number of a lattice $z$, its mean-field self-consistency equation is $M=\tanh (zJM/(k_{B}T))$. Put $A=T_c/T$. For $A\le 1$, the only solution is zero. For $A>1$, strict concavity of $\tanh (AM)-M$ on $M>0$ gives one positive solution and its spin-reversed partner. Expanding $\tanh (AM)=AM-(AM{)}^3/3+O(M^5)$ gives $M^2=3(A-1)/A^3+O((A-1{)}^2)$, proving a continuous onset with order-parameter critical exponent $\beta =1/2$. The Bragg-Williams free energy of the Ising model has a positive quartic coefficient at $T_c$, so these are the stable ordered minima rather than a first-order jump.

Exchanging the first and third inputs, together with the corresponding outputs, leaves the channel unchanged. Concavity therefore shows that averaging an input distribution with its exchanged distribution cannot reduce mutual information. We may take input probabilities $(a,1-2a,a)$.

For $p<1$, put $b=1-p$ and $t=2ab$, so $0\le t\le b$. The output distribution is $(t/2,1-t,t/2)$ and the mean conditional output Shannon entropy is $2ah(p)=th(p)/b$. HenceIts strict concavity gives the unique stationary value $t_{\ast }=[1+2^{h(p)/b-1}{]}^{-1}$. This lies in the allowed interval: for $0<p<1$, the derivative at $t=b$ is $({\log }_{2}p+b)/b<0$, while it tends to positive infinity at zero. At $p=0$ the stationary value is $2/3<1$ directly. Consequently an achieving input distribution isThe resulting capacity is $C={\log }_2(1+2^{1-h(p)/(1-p)})$. In particular at $p=1/2$, $a=1/3$ andAt $p=1$ all outputs are the middle symbol, so $C=0$ and every input distribution achieves it; the divided formula is not used at that endpoint.

The conditional output distributions are permutations of the same triple, so both conditional entropies equalWith input probability $p$, the three output probabilities are $q_1=(2-p)/6$, $q_2=(1+2p)/6$, and $q_3=(3-p)/6$. ThereforeDifferentiating, the constant terms cancel because $\sum q_j^{\prime }=0$:Thus there is a unique maximum. Its stationary equation reduces to $3p^2+9p-5=0$, whose root in $[0,1]$ isThe other root is negative; strict concavity proves this is the global maximum.

Risk aversion means that a sure mean is preferred to the corresponding risky portfolio wealth. An increasing concave utility function, with $u^{\prime }>0$ and $u^{\prime \prime }\le 0$, has this property by the Jensen inequality; strictly concave utility gives strict preference for nondegenerate risks. Differentiating the expected utility from part (a) givesAn interior optimum therefore satisfiesThis balances the expected loss of marginal utility in the good state against its expected gain in the bad state. The borrowing constraint also requires checking the endpoints: $x_{\ast }=0$ if $U^{\prime }(0)\le 0$, while $x_{\ast }=T$ would require $U^{\prime }(T)\ge 0$. At $x=T$ both state wealths equal $T(1+r)$, soHence an optimum never puts all wealth into deposits. If $U^{\prime }(0)>0$, continuity and the displayed endpoint sign give an interior root, and concavity makes every such root globally optimal. Under strict concavity it is unique.

In a two-person Nash bargaining problem, an agreement specifies a jointly feasible pair of payoffs $(u,v)$, measured in each player's own utility function. For a finite game, binding agreements can randomize jointly among action pairs, so the feasible payoff region $F$ is the convex hull of the pure-action payoff. This permits correlated joint actions; it is generally larger than the payoff region generated by independent mixed strategies. A disagreement point $d=(d_1,d_2)$ specifies what each player receives if negotiations fail. Assume a compact convex set of feasible agreements and at least one agreement strictly improving both disagreement payoffs.

A vector $(u^{\prime },v^{\prime })$ jointly dominates $(u,v)$ if $u^{\prime }\ge u$, $v^{\prime }\ge v$, and at least one inequality is strict. This is joint dominance of payoff vectors; strict improvement for both is unnecessary. A Pareto optimal agreement is one that has no feasible joint dominator. A jointly dominated agreement wastes a possible gain, but Pareto efficiency alone does not settle how gains should be divided. The negotiation set in two-person bargaining isIts individual-rationality constraints keep either player from accepting less than disagreement; its Pareto efficiency constraint rules out wasted opportunities. This negotiation set is distinct from objection-and-counterobjection bargaining sets for coalitional games.

The Nash arbitration procedure selects the Nash bargaining solution by maximizing gains relative to disagreement:The objective function is the Nash product. Essentiality makes its maximum positive; compactness gives attainment. On positive gains, its logarithm is $\log (u-d_1)+\log (v-d_2)$, whose Hessian is diagonal and negative definite. Thus strict concavity on the feasible convex set makes the maximizing payoff vector unique. Positive gain in either coordinate increases the product, so the solution is Pareto optimal.

The procedure respects feasibility, bargaining individual rationality, Pareto efficiency, bargaining symmetry, positive affine invariance in bargaining, and bargaining independence of irrelevant alternatives. In particular, replacing each player's utility by a separate positive affine rescaling multiplies the product by a positive constant; it does not compare one player's raw utility scale to the other's. Deleting other feasible agreements cannot change a unique maximizer that remains feasible. Symmetry under interchange of the players gives equal selected coordinates when the bargaining problem itself is symmetric.

These properties also explain why the product rule is distinguished. Normalize a positive-gain maximizer to $(1,1)$ and disagreement to $(0,0)$ by positive affine maps. The first-order inequality along a segment from $(1,1)$ to any feasible $(a,b)$ gives $a+b\le 2$. A sufficiently large symmetric triangle, bounded below in each coordinate and above by $a+b=2$, contains the normalized feasible region. Symmetry and Pareto efficiency select $(1,1)$ in that triangle. Independence of irrelevant alternatives then selects the same point in the original normalized region, and affine invariance transforms it back. This is the supporting triangle for Nash bargaining argument; the assumptions of convexity and a strict joint improvement are important. Without strict improvement, a zero product need not select a unique agreement.

For a finite game, a natural disagreement convention is each player's security level payoff:where $A,B$ are the two payoff matrices and $p,q$ are mixed strategies. Each player can independently guarantee at least its own security level. The maximin bargaining solution applies the Nash product rule using this $d$. In this convention, “maximin” determines the disagreement payoffs; it does not replace the arbitration objective function by $\min (u,v)$, which would depend improperly on interpersonal utility scales. The resulting agreement can strictly benefit both players relative to their guarantees, yet need not benefit both relative to a particular noncooperative Nash equilibrium.

The complete market makes every state-contingent consumption vector attainable. Its budget is ${\sum }_s{p}_s{c}_s=100$, and expected logarithmic utility under the equally likely states is $\frac{1}{4}{\sum }_s\log c_s$, with $c_s>0$. Increasing utility makes the budget bind. For a Lagrange multiplier $\eta$, the first-order conditions areSubstitution in the budget gives $1/\eta =100$, so $\eta =0.01$ andEvery state receives an expenditure $p_s{c}_s=25$, rather than equal consumption. Strict concavity of logarithmic utility makes this feasible stationary point the unique global maximizer. The physical probabilities enter the utility objective; the supporting prices enter the budget, and they must not be interchanged.

For active-route per-flow rates $x_r>0$, feasibility means ${\sum }_{r:j\in r}{n}_r{x}_r\le C_j$ for every resource. The defining condition for a proportionally fair allocation is that every feasible alternative $\tilde{x}$ satisfiesThis is equivalent to maximizing weighted logarithmic utility ${\sum }_{r:n_r>0}{n}_r\log x_r$ over the feasible set. Its gradient is $(n_r/x_r)$, so the displayed inequality is precisely the first-order condition for this concave function. Strict concavity gives unique active-flow rates. Routes with $n_r=0$ have no utility term or actual per-flow rate to determine; set their aggregate service to zero. This is the inactive routes in proportional fairness convention.

Abbreviate the route populations as $n_1,n_2,n_3,n_4$ in cyclic order and let $z_r=n_r{x}_r$ denote aggregate route service. The resource constraints areEquivalently, $\max (z_1,z_3)+\max (z_2,z_4)\le 1$. Fix these two maxima as $u,v$. Because utility is increasing in each active allocation, every active odd route attains $u$ and every active even route attains $v$. Dropping terms constant in the populations, the objective becomesIf both groups are nonempty, the maximum has $u+v=1$ and derivative $P/u-Q/(1-u)=0$, giving $u=P/M,v=Q/M$. If one group is empty, every route in the other group gets aggregate service one. For $M>0$, all cases are summarized, on active routes, byAt $M=0$ there are no flows and all aggregate services are zero. In particular $n_1{x}_1=(n_1+n_3)/M$ whenever $n_1>0$, as printed in the PDF. The per-flow rate is not generally $1/M$: opposite routes can use the same resource capacities in parallel.

An exponential document size with parameter ${\mu }_r$ has completion hazard ${\mu }_r{x}_r$ when a flow is served at rate $x_r$. Summing over its $n_r$ flows gives the transition ratesThus an active odd route departs at rate ${\mu }_{r}P/M$, and an active even route at rate ${\mu }_{r}Q/M$. An empty route has departure rate zero even if its opposite route is populated.

Put ${\alpha }_r={\nu }_r/{\mu }_r$ andFor an active odd route and an active even route respectively,Hence aggregate service is the neighboring-state ratio of this balance function of a flow-level network. Consider weights $w(n)=\Phi (n){\prod }_r{\alpha }_r^{n_r}$. For every birth edge,These are detailed balance equations. The total departure rate is bounded by ${\sum }_r{\mu }_r$ and total arrival rate by ${\sum }_r{\nu }_r$, so the process is nonexplosive. Provided the normalizing sum is finite, its stationary distribution is thereforeThe large parentheses in the original PDF represent a binomial coefficient, not a ratio. The verified adjacent-state ratios explain why that distinction matters.

The required stability condition can be established directly from the normalizing sum. Set $a={\alpha }_1,c={\alpha }_3,b={\alpha }_2,d={\alpha }_4$, and $h_p(a,c)={\sum }_{i=0}^p{a}^i{c}^{p-i}$. Grouping opposite-route counts givesIf $A_{\ast }=\max (a,c)$ and $B_{\ast }=\max (b,d)$, then $A_{\ast }^p\le h_p\le (p+1)A_{\ast }^p$, with the same bounds for the other group. Restriction to the routes attaining the maxima bounds $B$ below by ${\sum }_m(A_{\ast }+B_{\ast }{)}^m$. If $A_{\ast }+B_{\ast }<1$, the upper bound is at most ${\sum }_m(m+1{)}^2(A_{\ast }+B_{\ast }{)}^m$, which converges. ThusEquivalently every resource's offered load is strictly below its unit capacity. With positive arrival rates the irreducible chain then has the unique stationary probability distribution above. Without this condition the weights remain an invariant measure but are not normalizable; the question's probability-distribution claim implicitly needs this hypothesis.

For completeness, the normalizing constant has the four-cycle partition function expressionTo derive it when $a\ne c$ and $b\ne d$, substitute $h_p=(a^{p+1}-c^{p+1})/(a-c)$ into the double sum and use ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. The resulting expression iswhich simplifies to the displayed formula. Continuity within the stability region covers equal opposite loads and zero loads.

Take a riskless gross return $R>0$ per period and a stock whose independent multiplicative moves are $u$ and $d$, with physical up probability $p\in (0,1)$. Assume $0<d<R<u$. With $K_k$ up moves by date $k$,A self-financing portfolio chooses each stock holding from information available before the next move, and invests the remainder in the bank account. At any node the two possible next payoffs can be matched by two holdings because $u\ne d$. Backward induction therefore makes the binomial market complete.

The unique risk-neutral probability of an up move is determined by the discounted-stock martingale equation:The risk-neutral measure assigns product probabilities to the successive moves. A path with $K=K_n$ up moves has physical probability $p^K(1-p{)}^{n-K}$ and pricing probability $q^K(1-q{)}^{n-K}$. Its Radon-Nikodym derivative is thereforeSince $S_n=S_0{d}^n(u/d{)}^K$, eliminate $K$ to obtain the requested terminal-price expression:The pricing density depends only on the terminal stock price even though different paths to that price have different intermediate histories.

For $w_0>0$, a nonnegative terminal wealth $X$ with initial cost $w_0$ satisfiesThe utility function has derivative $v^{\prime }(x)=x^{-(\gamma -1)/\gamma }$. Put $\eta =\gamma /(\gamma -1)>1$. Its first-order condition makes $v^{\prime }(X)$ proportional to $Z$, so $X=C{Z}^{-\eta }$. Choosing $C$ to satisfy the pricing budget constraint gives the binomial power-utility terminal wealthAll states have positive probability and the state space is finite, so this payoff is positive, integrable and attainable. The normalizing constant is explicit: independence givesCombining these two formulas with the terminal-price expression for $Z$ gives $X_{\ast }$ entirely as a function of $S_n$.

To prove global optimality, strict concavity gives $v(X)-v(X_{\ast })\le v^{\prime }(X_{\ast })(X-X_{\ast })$. Because $v^{\prime }(X_{\ast })=C^{-1/\eta }Z$, the expectation of the right side is zero for any other payoff satisfying the same pricing budget constraint. Equality requires $X=X_{\ast }$ almost surely. Backward replication supplies its self-financing strategy, so this static optimum is attainable by trading. If $w_0=0$ and terminal wealth must be nonnegative, the only feasible terminal wealth is zero. For $p=q$, the density is one and the optimizer is the riskless wealth $R^n{w}_0$.

Only active routes, those with $n_r>0$, enter the proportional fairness inequality and the weighted logarithmic utility. Their rates must be positive; inactive routes have no per-flow rate to determine and may be assigned zero. This convention removes otherwise meaningless terms $0\log 0$ and $0/0$.

For positive active rates, concavity of the logarithm gives, for any positive feasible competitor $y$,Thus the proportional fairness inequality proves that $x$ maximizes the log-utility objective. A competitor with an active zero rate has utility $-\infty$ and cannot improve it. Conversely, if $x$ maximizes utility, the right directional derivative along $x+\epsilon (y-x)$ at $\epsilon =0$ is nonpositive, giving exactly that inequality. Positive resource capacities and finitely many active routes allow a feasible allocation with every active rate positive, so an optimizer cannot have an active zero rate. Strict concavity gives uniqueness of the active rates. Hence proportional fairness is equivalent to the stated log-utility maximization.

Index the four adjacent-pair routes in cyclic order and put $z_i=n_i{x}_i$, the aggregate service on route $i$. The resource constraints areThey involve every odd-even route pair, so they are equivalent toFor given maxima, increasing each active odd service to $p$ and each active even service to $q$ remains feasible and improves utility. Apart from the constant $-{\sum }_i{n}_i\log n_i$, the objective therefore reduces toWhen both groups are present the constraint is tight, and differentiation gives $P/p=Q/(1-p)$. If one group is empty, the other receives unit aggregate service on each of its active routes. Thus the opposite-route reduction for a proportionally fair four-cycle yields, with $T=P+Q>0$,Inactive rates can be set to zero, including every rate in the empty state. In particular $n_{\{1,2\}}{x}_{\{1,2\}}=(n_{\{1,2\}}+n_{\{3,4\}})/{\sum }_r{n}_r$ whenever that route is active. This proves the required formula even when other routes are empty.

For the flow-level network model, an active document's residual size retains an exponential distribution with parameter ${\mu }_i$. Receiving service at rate $x_i(n)$ gives completion intensity ${\mu }_i{x}_i(n)$, so $n_i$ independent active documents give total completion intensity ${\mu }_i{n}_i{x}_i(n)$. Consequently the population continuous-time Markov chain hasAll death intensities are zero at $T=0$. Their total is bounded by ${\sum }_i{\mu }_i$, and the total arrival rate is constant, proving nonexplosion.

Put ${\rho }_i={\nu }_i/{\mu }_i$. The required stationary weight uses a binomial coefficient, as printed in the PDF; the converted TeX incorrectly rendered it as an ordinary fraction. DefineFor an odd route, the neighboring binomial coefficient ratio iswhile its death intensity at the neighboring state is ${\mu }_i(P+1)/(T+1)$. Their product is ${\nu }_i$, so $w(n)q(n,n+e_i)=w(n+e_i)q(n+e_i,n)$. For an even route, replace $P$ by $Q$ in the same calculation. This proves detailed balance and the stationary law of a proportionally fair four-cycle:A finite $B$ is essential for this expression to be a probability distribution. We determine its exact convergence condition rather than asserting stationarity for arbitrary arrival rates. LetGrouping the states by odd and even population givesWith $A=\max (a,c)$ and $E=\max (b,d)$, the bounds $A^p\le h_p(a,c)\le (p+1)A^p$ and $E^q\le h_q(b,d)\le (q+1)E^q$ show that convergence holds exactly when $A+E<1$. For necessity, the lower bound sums to ${\sum }_{k\ge 0}(A+E{)}^k$, which diverges at or above one. For sufficiency, choose slightly larger $A^{\prime },E^{\prime }$ with $A^{\prime }+E^{\prime }<1$; the polynomial factors are absorbed into constants times $(A^{\prime }{)}^p(E^{\prime }{)}^q$, giving a convergent geometric generating series. Thus the stability region for the reversible four-cycle flow network isEquivalently, the offered load on each resource, ${\rho }_1+{\rho }_4$, ${\rho }_1+{\rho }_2$, ${\rho }_2+{\rho }_3$, or ${\rho }_3+{\rho }_4$, is strictly below its unit capacity. For positive arrival rates this finite normalizing constant gives the unique stationary distribution of the irreducible process. At or beyond the boundary, the displayed invariant weights are not summable and do not define a stationary probability law.

If an explicit normalizing constant is desired, $h_p(a,c)=(a^{p+1}-c^{p+1})/(a-c)$ for $a\ne c$, and ${\sum }_{p,q}\left(\genfrac{}{}{0ex}{}{p+q}{p}\right)u^p{v}^q=(1-u-v{)}^{-1}$. Hence for $a\ne c$, $b\ne d$ in the stable region,Equal-parameter cases are obtained by continuity or directly from the positive series. In particular the empty-traffic limit is $B=1$.

Independence and the Poisson probabilities give the log-likelihoodIts score and curvature areIf some $x_i\ne 0$, this is strictly concave. A finite maximum is therefore the unique zero of the score when such a zero exists. Newton iteration givesStart from a finite value, compute the score and curvature, and iterate until both the score and step are small. A line search that shortens a step until the likelihood increases improves robustness. Strict concavity identifies any converged stationary point as the global maximum. A finite root need not exist, for example when all observations are zero and all $x_i>0$, in which case the likelihood supremum is at $\beta \to -\infty$. If all $x_i=0$, the parameter is unidentifiable.

Let $I=(U^{\prime }{)}^{-1}$ be the inverse marginal utility. The derivative $U^{\prime }$ is continuous and strictly decreasing: concavity makes it nonincreasing, and equality at two distinct points would make $U$ affine between them, contradicting strict concavity. The Inada conditions therefore make $U^{\prime }$ a bijection from $(0,\infty )$ onto $(0,\infty )$. For each $y>0$, the derivative of $U(x)-xy$ is positive below $I(y)$ and negative above it. Consequently its unique global maximum is finite andThere is no need to differentiate $I$ to find the first derivative of this utility conjugate. Comparing the maxima at $y$ and $y+h$ gives, for $h>0$,Continuity of $I$, together with the corresponding inequalities for $h<0$, givesThus $V$ is strictly decreasing and strictly convex, since its derivative is strictly increasing. This proves all the asserted properties except the second differentiability, which does not follow from the printed hypotheses.

Here is an explicit Inada utility with vanishing curvature. PutThe positive integral defining $p$ makes $U$ strictly increasing. Also $p^{\prime }=-w$ is negative except at one point, so $p$ is strictly decreasing and $U$ is strictly concave. Near zero $w(s)$ is asymptotic to $s^{-2}$, and at infinity it is asymptotic to $s^{-2}$; thus $p(0+)=\infty$ and $p(\infty )=0$. This smooth utility function satisfies every printed positive-domain assumption, but $U^{\prime \prime }(1)=0$. At $y_0=p(1)=3-4\log 2$, differentiability of $I$ would imply, by differentiating $p(I(y))=y$,which is impossible. Hence $V^{\prime }=-I$ has no finite derivative there. More explicitly, $p(x)-p(1)=-(x-1{)}^3/12+o((x-1{)}^3)$, so its inverse function has a cube-root singularity. The twice-differentiable conclusion is false as printed. With the additional assumption $U^{\prime \prime }(x)<0$ everywhere, the inverse function theorem proves the intended dual differentiability with nonvanishing utility curvature:

For the financial assertions, use the positive-wealth admissible trading strategy class appropriate to the positive-domain utility function, and expectations for which the expressions are defined. A normalized terminal state-price density satisfies $Z_N>0$, $\mathbb{E}{Z}_N=1$, and defines an equivalent martingale measure $Q$ by $dQ=Z_{N}dP$. A self-financing portfolio obeyswhere the holdings are measurable before the next price increment. Under $Q$ the gains are a martingale when integrable; for the usual lower-bounded admissible class, localization gives a supermartingale. In particular the state-price budget constraint is $\mathbb{E}[Z_N{X}_N]\le x$. The positive-domain qualification matters: assumptions only on $U$ at positive arguments do not control its values at nonpositive wealth.

The definition of the utility conjugate gives the pointwise inequalityTaking expectations, using the state-price budget constraint, and then taking the supremum over strategies gives $u(x)\le \mathbb{E}V(y{Z}_N)+yx$ for every $y>0$ and every state-price density. Taking the two infima proves the utility duality with martingale deflators upper bound:If an expectation is infinite, use the same inequality in the extended-real sense whenever the negative parts make it well defined; no subtraction of two infinities is needed.

A complete market is one in which every terminal contingent claim in the specified claim class is attainable by a self-financing portfolio. Bounded terminal claims suffice for the uniqueness argument. In the standard finite-horizon discrete-time market, absence of arbitrage gives existence of an equivalent martingale measure by the fundamental theorem of asset pricing. To prove uniqueness, let $Z_N$ and ${\tilde{Z}}_N$ be two normalized state-price densities. Every bounded claim $H$ has a replicating price $p(H)$, and its replication has the same price under either equivalent martingale measure:Take $H=1_A$ for every terminal event $A$. The integrable difference $Z_N-{\tilde{Z}}_N$ has zero integral over every event, hence vanishes almost surely: its positive and negative sets would otherwise give a nonzero integral. The normalized terminal state-price density is unique up to null sets. Its density process is then uniquely $Z_n=\mathbb{E}[Z_N\mid {\mathcal{F}}_n]$. This uses the usual pricing replication class, where bounded claims have bounded replicating value processes; arbitrary doubling strategies are not an admissible definition of completeness.

For logarithmic utility, direct maximization gives $V(y)=-\log y-1$. In the complete market, write $Z=Z_N$. The unique-density utility conjugate bound is minimized at $y=1/x$, givingThe candidate terminal portfolio wealth is $H^{\ast }=x/Z$, whose cost is exactly $\mathbb{E}[Z{H}^{\ast }]=x$. If completeness includes nonnegative $Q$-integrable claims, it is replicable and attains the bound. There is also an approximation proof using only bounded-claim completeness. DefineThen $0<b_n\le 1$, $b_n\uparrow 1$, $H_n$ is positive and bounded, and $\mathbb{E}[Z{H}_n]=x$. Replicating $H_n$ therefore gives an admissible terminal claim of initial cost $x$. Since $\mathbb{E}(\log Z{)}^{+}\le \mathbb{E}Z=1$, the negative parts of $-\log (Z\vee n^{-1})$ have an integrable common bound. Monotone convergence after adding this lower bound givesConsequently the logarithmic terminal wealth in a complete market isThe value may be $+\infty$ if $\mathbb{E}(\log Z_N{)}^{-}=\infty$; the displayed formula remains meaningful, while attainment by $H^{\ast }$ depends on the precise admissible claim class.

For a discrete memoryless channel, its channel capacity in bits per use is $C={\max }_{P_X}I(X;Y)$. The Shannon second coding theorem identifies this with the supremum of rates achievable by arbitrarily long block codes with error probability tending to zero: every rate below $C$ is achievable, and no rate above $C$ can have vanishing error. The channel is used independently at each transmission, and coding rate is measured using base-two logarithms.

Let $q$ be the probability of using the second input symbol. The second output symbol occurs with probability $q/2$, so $H(Y)=h_2(q/2)$. The first input has deterministic output and the second has one bit of conditional output Shannon entropy; hence $H(Y\mid X)=q$ andFor $0<q<1$, its derivative is $\frac{1}{2}{\log }_2[(2-q)/q]-1$, which vanishes at $q=2/5$. Strict concavity of binary entropy makes this the unique global maximum. Substituting givesThus Shannon's theorem supplies operational coding rates arbitrarily close to this value from below, and rules out reliable transmission above it.

Choose a point of $K$ with both coordinates positive, so its product is positive. The set $K\cap [0,\infty {)}^2$ is nonempty and compact; the continuous function $xy$ therefore attains its maximum there. A maximizer cannot lie on either axis, where the product is zero. It consequently belongs to $K\cap P^{+}$ and also maximizes the product on that set.

For uniqueness, the function $\Phi (x,y)=\log x+\log y$ is strictly concave on the positive quadrant: its Hessian is $\mathrm{diag}(-1/x^2,-1/y^2)$, negative definite. If two distinct positive points attained the same maximal product, their midpoint would belong to $K\cap P^{+}$ by convexity and strict concavity would give a larger value of $\Phi$, hence a larger product. This contradiction proves existence and uniqueness of the positive product maximizer. This argument is positive-product maximization on a convex set; compactness should be applied to the closed quadrant intersection, not directly to the potentially noncompact set $K\cap P^{+}$.