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For constant relative risk aversion utility with parameter $\gamma >0$, terminal weight $\eta >0$ and utility discount rate $\delta$, in a Black-Scholes model put $\lambda =(\mu -r)/\sigma$, $k=[\delta +(\gamma -1)(r+{\lambda }^2/(2\gamma ))]/\gamma$ and $b(\tau )={\int }_0^{\tau }{e}^{-ks}ds+{\eta }^{1/\gamma }{e}^{-k\tau }$. The displayed optimal consumption and dollar stock position follow by inverse marginal utility and the state-price budget constraint. For $\gamma \ne 1$ the value is $b(T-t{)}^{\gamma }{W}_t^{1-\gamma }/(1-\gamma )$ with discount measured from $t$. Finite horizons do not require $k>0$. The concave supporting-tangent inequality and replication of the optimal budget prove global optimality, rather than only a formal differential-equation solution.

Homogeneity of degree $k$ means that multiplying all arguments by a common positive factor $a$ multiplies the output by $a^k$. A homogeneous function is the function possessing this property. This is different from degree-one positive homogeneity; investment values with constant relative risk aversion utility usually have degree $1-R$.

In the infinite-horizon Merton consumption-investment problem with constant relative risk aversion utility, this is the optimal ratio of consumption to wealth. A finite value requires ${\gamma }_M>0$. The state-price budget constraint then fixes consumption's multiplier because $\mathbb{E}[{\xi }_t(y{e}^{\rho t}{\xi }_t{)}^{-1/R}]=y^{-1/R}{e}^{-{\gamma }_{M}t}$, whose time integral is $y^{-1/R}/{\gamma }_M$. When this integral diverges, the value is $+\infty$ for $0<R<1$ and $-\infty$ for $R>1$.

For a constant-coefficient single-asset investment-consumption problem with constant relative risk aversion utility, put $\kappa =(\mu -r)/\sigma$ and $\delta =[\rho -(1-R)(r+{\kappa }^2/(2R))]/R$. When $\delta >0$, the infinite-horizon value is ${\delta }^{-R}{w}^{1-R}/(1-R)$ and the optimal controls are $c^{\ast }=\delta w$ and ${\theta }^{\ast }=(\mu -r)w/(R{\sigma }^2)$. The case $R=1$ uses logarithmic utility.

Use the physical dynamics of the Black-Scholes model, $d{S}_t/S_t=\mu dt+\sigma d{W}_t$, with bank rate $\rho$ and $\sigma >0$. Write $\lambda =(\mu -\rho )/\sigma$ for the market price of risk. If $w_t$ is portfolio wealth, ${\pi }_t$ the dollar amount in stock, and $c_t\ge 0$ the consumption rate, the budget dynamics areAdmissible controls are progressively measurable, locally integrable in the required drift and diffusion norms, and keep $w_t\ge 0$. Such a solvency restriction excludes doubling-type strategies. In a complete Brownian market, the investment-consumption problem is to maximizewhere the utility functions are increasing and concave. A terminal-only problem omits the integral. Strict concavity gives uniqueness of optimal consumption and terminal wealth when an optimum exists.

The state-price density isThe Itô formula givesThus $H_t{w}_t+{\int }_0^t{H}_s{c}_{s}ds$ is a nonnegative local martingale, hence a supermartingale, yielding the state-price budget constraintConversely, a nonnegative consumption plan and terminal payoff $\xi$ satisfying budget equality and the necessary integrability can be financed in the Brownian complete market. DefineThe Brownian martingale representation theorem for the total future budget supplies a stochastic-integral coefficient; comparing it with $H_t(\sigma {\pi }_t-\lambda w_t)$ defines ${\pi }_t$. This reconstructs the wealth equation with $w_t\ge 0$. Localization gives the representation for integrable positive budgets when an $L^2$ assumption is not available.

For smooth strictly concave utilities with the usual marginal-utility range, let $I_u=(u^{\prime }{)}^{-1}$ and $I_U=(U^{\prime }{)}^{-1}$ be the inverse marginal utility functions. The Lagrange-multiplier solution iswhere $y>0$ makes the budget an equality. This is a global argument, not merely formal differentiation: the concave supporting-tangent inequality giveswith the analogous terminal inequality. Taking expectations, integrating, and applying the budget bound makes the difference in objectives nonpositive. Replication of the candidate budget proves attainability and optimality whenever these expectations are well-defined and finite. If the marginal-utility ranges or integrability conditions fail, existence must be established separately rather than inferred from the first-order equations.

A detailed example is constant relative risk aversion utility $u(c)=c^{1-\gamma }/(1-\gamma )$ and $U(w)=\eta w^{1-\gamma }/(1-\gamma )$, with $\gamma >0$, $\gamma \ne 1$, $\eta >0$, and $\delta \ge 0$. PutThe optimal plans from the marginal-utility equations are $c_t^{\ast }=(y{e}^{\delta t}{H}_t{)}^{-1/\gamma }$ and ${\xi }^{\ast }={\eta }^{1/\gamma }(y{e}^{\delta T}{H}_T{)}^{-1/\gamma }$. The normal distribution of $W_t$ gives, for $a=1-1/\gamma$,The budget therefore becomes $y^{-1/\gamma }b(T)=w_0$, determining the multiplier explicitly. Independent Brownian increments in the conditional budget formula give $w_t^{\ast }=c_t^{\ast }b(T-t)$. The diffusion coefficient of $c_t^{\ast }$ is $(\lambda /\gamma )c_t^{\ast }$, so matching the wealth diffusion determines the dollar stock holding. Consequently the finite-horizon power-utility investment and consumption optimum isAll candidate moments are finite on a finite horizon, since powers of a lognormal state-price density have finite moments. The budget and tangent argument above therefore verify optimality in this example, including admissibility and positive wealth.

For another verification of the formulas, the Hamilton-Jacobi-Bellman equation for a value function with discount measured from the current time isSubstitute $V(w,t)=b(T-t{)}^{\gamma }{w}^{1-\gamma }/(1-\gamma )$. The two concave maximizations give the displayed controls, and the remaining equation is $b^{\prime }(\tau )=1-kb(\tau )$ with $b(0)={\eta }^{1/\gamma }$, exactly solved by the formula above. ThusFor $k=0$, interpret the integral as $\tau$, so $b(\tau )=\tau +{\eta }^{1/\gamma }$. Unlike the infinite-horizon Merton consumption-investment problem, finite horizons do not require $k>0$. Consumption vanishes in the terminal-only formulation; then the integral term in $b$ is absent and the optimal stock fraction remains $(\mu -\rho )/(\gamma {\sigma }^2)$. Logarithmic utility is handled directly with inverse marginal utility $1/z$ and the same wealth-to-consumption formula at $\gamma =1$, with $k=\delta$.

For constant relative risk aversion utility, $I(z)=z^{-1/R}$. Define the Merton consumption constantThe Gaussian exponential moment givesWhen ${\gamma }_M>0$, the budget is $w_0=y^{-1/R}/{\gamma }_M$. Conditional pricing then gives $c_t^{\ast }={\gamma }_M{w}_t^{\ast }$, and the Merton consumption-investment problem hasThe resulting wealth is a geometric Brownian motion, so positive initial wealth remains positive. Substitution in the Hamilton-Jacobi-Bellman equation verifies the first-order conditions and the value; the budget argument in part (a) supplies global optimality.

The problem has a finite value precisely when ${\gamma }_M>0$. For $0<R<1$ and ${\gamma }_M\le 0$, the supremum is $+\infty$. Indeed, use the displayed risky fraction and consumption $c_t=k{w}_t$, $k>0$. Its expected utility isFor ${\gamma }_M<0$, sufficiently small $k$ already gives infinite utility. For ${\gamma }_M=0$, the finite values tend to infinity as $k\downarrow 0$.

For $R>1$ and ${\gamma }_M\le 0$, every admissible stream has utility $-\infty$. To see this, put $q=(R-1)/R$ and apply Holder inequality on $[0,T]\times \Omega$:The left side diverges as $T\to \infty$, forcing the positive utility-cost integral to diverge. Thus there is no finite-value optimization problem in either ill-posed case.

Put $b=1-R>0$ and $m_b=\mathbb{E}{X}_0^b$. Homogeneity of the constant relative risk aversion utility gives $U(z_N)=U(z_n){\prod }_{j=n}^{N-1}{X}_j^b$. ThusAs before this is an extended-valued conditional expectation if the required moment is infinite and days remain.

Normalize the constant relative risk aversion utility as $U(c)=c^{1-R}/(1-R)$. The printed specification of $U^{\prime }$ alone also permits an additive constant $K$; that would add $K/\rho$ to the value, so the stated homogeneity holds for the normalized value. SetMultiplying initial wealth, the historical maximum, dollar investments, and consumption by $a>0$ multiplies every term of the wealth equation by $a$, including the tax term. It maps admissible policies bijectively and multiplies the normalized reward by $a^{1-R}$. Therefore the value scales asThe historical maximum means ${\bar{w}}_t=\max ({\bar{w}}_0,{\sup }_{0\le s\le t}{w}_s)$ when initially $w<\bar{w}$.

Inside the state region $w<\bar{w}$, the maximum is locally constant. The Hamilton-Jacobi-Bellman equation isAt $w=\bar{w}$, the finite-variation contribution in the Itô formula is $(V_{\bar{w}}-\tau V_w)d\bar{w}$. The high-water mark tax boundary condition is consequentlyFor an increasing, strictly concave function of wealth, optimizing the two controls givesThe consumption convex conjugate isSubstitution of the homogeneity derivatives gives the reduced equation and boundary conditionThe controls in scaled variables are $c^{\ast }/\bar{w}=(v^{\prime }{)}^{-1/R}$ and ${\theta }^{\ast }/\bar{w}=-(\mu -r)v^{\prime }/({\sigma }^2{v}^{\prime \prime })$.

Use the wealth-variable Legendre dualSince $v^{\prime \prime }<0$, the inverse map satisfies $J^{\prime }=-x$ and $J^{\prime \prime }=-1/v^{\prime \prime }>0$. Inserting $v=J-z{J}^{\prime }$ into the reduced Hamilton-Jacobi-Bellman equation gives a linear Euler equationFor the Euler differential equation operator on the left, its action on $z^t$ is $Q(t)z^t$. Write $b=1-1/R$. Under ${\gamma }_M=-Q(b)>0$, the particular solution is $-\tilde{U}(z)/Q(b)$. For the nondegenerate case $\kappa \ne 0$, the two characteristic roots are $-\alpha <0<t_{+}$, with $t_{+}>1$ because $Q(0)=-\rho$ and $Q(1)=-r$. Thus the general solution also contains $A(z/z_{\ast }{)}^{-\alpha }+B{z}^{t_{+}}$.

Here $z_{\ast }=v^{\prime }(1)$ and the relevant dual domain is $z\ge z_{\ast }$. The inverse wealth ratio must have $x=-J^{\prime }(z)\to 0$ as $z\to \infty$. The particular solution and the negative-root term have derivatives tending to zero, whereas $B{z}^{t_{+}}$ has an unbounded derivative unless $B=0$. Hence the admissible dual solution has the claimed formThe quadratic-root formulation presupposes a nonzero market price of risk. If $\kappa =0$, the dual equation becomes first order; it is treated directly or by an appropriate nondegenerate limit rather than by assuming two quadratic roots.

The inverse map and the tax boundary provide two equations at $z_{\ast }$:Because ${\tilde{U}}^{\prime }(z)=-z^{-1/R}$, these becomeLet $D=\alpha R+R-1$. Eliminating $z_{\ast }^{1-1/R}/{\gamma }_M$ gives $-DA=\tau z_{\ast }$. Also $D=R(\alpha +b)>0$, since $Q(b)<0$ puts $b$ strictly between the two roots. The derivative boundary then gives $z_{\ast }^{-1/R}/{\gamma }_M=(D+\alpha \tau )/D$. Therefore the constants areThere is an important admissibility qualification in the printed conclusion. With $x(z)=-J^{\prime }(z)$,The second term is negative, but decays faster than the first because $\alpha +1>1/R$. Direct differentiation at the boundary givesA wealth-variable Legendre dual of a concave value must be a convex function. Thus the printed smooth tax-paying solution requires $\alpha \tau <1$, with the limiting case $\alpha \tau =1$ allowed as a degenerate boundary. The assumptions ${\gamma }_M>0$ and $0<\tau <1$ do not imply this: for example $R=2$, $r=0.03$, $\rho =0.1$, $\kappa =0.2$, and $\tau =0.5$ give $\alpha =(5+\sqrt{105})/4>2$ and ${\gamma }_M=0.07$, hence negative boundary dual curvature.

For higher tax the investor can avoid raising the historical maximum. The wealth-cap investment boundary replaces the tax-paying equality by $J^{\prime \prime }(z_{\ast })=0$, alongside $J^{\prime }(z_{\ast })=-1$. Solving these equations givesThese are the same constants with $\tau$ replaced by ${\tau }_{\mathrm{eff}}=1/\alpha$. More generally the admissible two-regime expression usesin the constants, while the claimed formulas themselves describe the tax-paying regime. In the cap regime the portfolio volatility vanishes at $x=1$ and the consumption rate there is $r+{\kappa }^2(\alpha +1)/(2R)>r$, so the wealth drift points inward and no new maximum is required. The remaining boundary inequality is $(V_{\bar{w}}-\tau V_w)/V_w=1/\alpha -\tau \le 0$, consistent with avoiding costly maximum increases. For $z>z_{\ast }$, dual curvature is positive because the negative term in $J^{\prime \prime }$ decays faster than the positive term. Thus this qualification repairs an actual missing parameter restriction rather than a TeX transcription error.

Use dollar holdings ${\theta }_0,\dots ,{\theta }_n$, including the traded index, and let $m=\log M$. The Itô formula givesDefine the state-dependent correlation $q(m)=\rho (e^m)$, $d(m)=\sqrt{1-q(m{)}^2}>0$, and constants $p_i=({\mu }_i-r)/{\sigma }_i$. The self-financing portfolio with consumption has wealth equationSince all these assets are traded and $d(m)>0$, the volatility map is invertible. It is useful to optimize over Brownian portfolio exposureswhose market price of risk vector isIndeed ${\sum }_i({\mu }_i-r){\theta }_i=\ell \cdot y$. Put $L^2(m)=p_0^2+{\sum }_{i=1}^n{\ell }_i(m{)}^2$.

For $R\ne 1$, normalize constant relative risk aversion utility to $U(c)=c^{1-R}/(1-R)$. Scaling wealth, holdings, and consumption by $a>0$ leaves the index state unchanged and multiplies reward by $a^{1-R}$. ThusFor a finite smooth value with $V_{ww}<0$, the Hamilton-Jacobi-Bellman equation, including the shared-noise cross derivative, isThe last term is essential: index changes and the $W^0$ component of wealth are correlated. The first-order conditions yieldBecause $V_w=w^{-R}f$, $V_{ww}=-R{w}^{-R-1}f$, and $V_{wm}=w^{-R}{f}^{\prime }$, the optimized portfolio contribution isSubstitution gives the requested second-order nonlinear equationRecovering dollar holdings from the exposures gives the optimal controlsOnly the index holding carries the extra intertemporal hedging demand, because the index is the source of state variation. The stock positions hedge their common exposure through the index.

The power transformation of a complete-market investment equation gives a useful further simplification. Let $g=f^{1/R}$, and defineExpanding the squared term and substituting $f=g^R$ cancels the two $(g^{\prime }{)}^2$ terms. The power-transformed investment equation is linear:Then $c^{\ast }=w/g$ and $f^{\prime }/f=R{g}^{\prime }/g$. A practical finite difference method solves this linear differential equation on an expanding truncated interval in $m$, enforcing the economically relevant positive solution and checking domain and mesh convergence. If correlation approaches limits strictly inside $(-1,1)$ and the corresponding ${\delta }_{\pm }>0$, the constant-coefficient Merton consumption-investment problem gives endpoint approximations $g\to 1/{\delta }_{\pm }$. Without such asymptotics one must determine appropriate growth/transversality conditions; arbitrary fixed endpoint values are not justified. The statement's smooth decreasing correlation alone does not ensure globally bounded market prices of risk or a finite value. Any numerical candidate must also satisfy admissibility and the investment value transversality condition.

The printed $R>0$ includes $R=1$. In that case choose $U(c)=\log c$, whose scaling is additive:There is no shared-noise cross derivative because $V_{wm}=0$. The logarithmic case isThus the dollar holdings are the preceding formulas with $R=1$ and the $f^{\prime }/f$ hedge term omitted. This linear differential equation can be solved by the same truncation and convergence strategy.

For a complete-market investment-consumption problem with constant relative risk aversion utility, the nonlinear wealth-homogeneity coefficient equation may contain $f^{\prime 2}/f$. Writing $f=g^R$ cancels this gradient square against the one from $f^{\prime \prime }$. In the index-driven correlation model the result is ${\sigma }_0^2{g}^{\prime \prime }/2+B{g}^{\prime }-\delta (m)g+1=0$, a linear differential equation; the positive economic solution gives consumption $w/g$.

For an observed finite-state continuous-time Markov chain with rate matrix $Q$, constant bank rate $r$, and state-dependent stock drifts ${\mu }_i$ and positive volatilities ${\sigma }_i$, constant relative risk aversion utility gives $V_i(w)=A_i{w}^{1-R}/(1-R)$. Set ${\kappa }_i=({\mu }_i-r)/{\sigma }_i$ and $d_i=\rho -(1-R)(r+{\kappa }_i^2/(2R))$. The Hamilton-Jacobi-Bellman equation reduces to the displayed nonlinear algebraic system with $A_i>0$. Its feedback controls are $c_i=A_i^{-1/R}w$ and ${\theta }_i=({\mu }_i-r)w/(R{\sigma }_i^2)$.

A damped Newton method in the logarithmic variables $A_i=e^{x_i}$ preserves positivity. Solving these equations produces a candidate value; admissibility, stochastic-integral bounds and the investment value transversality condition still need verification.

A fee proportional to terminal portfolio wealth but reduced by accumulated portfolio quadratic variation changes the optimal risk exposure of a manager with constant relative risk aversion utility. Write $A=\sigma {\sigma }^T$, $b=\mu -r1$ and assume $A$ is invertible. The fee process has drift $r+b^T\pi -\epsilon {\pi }^{T}A\pi /2$ and volatility ${\sigma }^T\pi$. Its Hamilton-Jacobi-Bellman equation therefore maximizes $b^T\pi -(R+\epsilon ){\pi }^{T}A\pi /2$, giving ${\pi }^{\ast }=A^{-1}b/(R+\epsilon )$. The fee increases effective relative risk aversion from $R$ to $R+\epsilon$, independently of its positive scale $a$.

There is also a direct global bound. Put $p=1-R$, $v^{\ast }={\sigma }^T{\pi }^{\ast }$, and tilt probability by $\exp (p{v}^{\ast }\cdot W_T-p^2|v^{\ast }{|}^{2}T/2)$. Under this probability let $Z$ be the stochastic exponential of $(1+\epsilon )\int ({\sigma }^T\pi -v^{\ast })\cdot d{W}^{\ast }$. The terminal fee ratio satisfies $(y_T/y_T^{\ast }{)}^p=Z_T^{p/(1+\epsilon )}$. A nonnegative local martingale has expectation at most one. The Jensen inequality, applied to a concave positive power when $p>0$ and a convex negative power when $p<0$, proves that the constant exposure maximizes expected CRRA utility. Effective risk aversion one uses logarithmic utility.