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Assume $\sigma \ne 0$ and the usual Brownian filtration, so the single-risky-asset market is a complete market. With market price of risk $\kappa =(\mu -r)/\sigma$, the normalized state-price density isThe Itô formula shows that ${\zeta }_t{w}_t$ is a local martingale for a self-financing portfolio. For nonnegative admissible wealth it is a supermartingale, yielding the state-price budget constraintEvery integrable nonnegative terminal claim with equality is attainable in the complete market, by the Brownian martingale representation theorem. Its wealth process is $w_t={\zeta }_t^{-1}\mathbb{E}[{\zeta }_{T}X\mid {\mathcal{F}}_t]\ge 0$.

The Inada conditions $u^{\prime }(0+)=\infty$ and $u^{\prime }(\infty )=0$, together with strict concavity, make $I=(u^{\prime }{)}^{-1}$ a decreasing map from $(0,\infty )$ onto $(0,\infty )$. Pointwise optimization of $u(x)-y{\zeta }_{T}x$ gives the optimal terminal wealthwith scalar multiplier $y>0$. The question assumes a multiplier giving the required budget; utility expectations must also be well defined.

For any admissible terminal wealth $X$, the supporting-line inequality for a concave function givesTaking expectations and using the state-price budget constraint proves optimality:Strict concavity makes the optimal terminal wealth unique up to almost-sure equality.

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.

Use positive initial asset prices and the usual augmentation of the natural Brownian filtration. Define the market price of riskContinuity and strict positivity of $\sigma$ imply ${\int }_0^T{\lambda }_t^{2}dt<\infty$ on every finite horizon almost surely, because each path has a positive minimum of $\sigma$ there. Hence the strictly positive stochastic exponentialis defined and has $Y_0=1$ and $d{Y}_t=-Y_t(r_{t}dt+{\lambda }_{t}d{W}_t)$. The Itô product rule givesso it is a local martingale deflator.

For uniqueness, let $\tilde{Y}$ be any normalized strictly positive local martingale deflator. The Brownian martingale representation theorem makes $B\tilde{Y}$ a continuous local martingale with a Brownian motion integral representation. Dividing by $B$ therefore gives $d\tilde{Y}=-r\tilde{Y}dt+\eta dW$. The vanishing drift of $S\tilde{Y}$ requires $\tilde{Y}(\mu -r)+\eta \sigma =0$, so $\eta =-\tilde{Y}\lambda$. This scalar linear stochastic differential equation has exactly the exponential solution above, proving the normalized local martingale deflator is unique.

The assumptions do not make $\lambda$ deterministically bounded: $\sigma$ need not be bounded away from zero uniformly over outcomes. Thus a true martingale density or Novikov condition is not inferred here; the required conclusion is local.

Consider irreversible optimal stopping of employment paying income $\epsilon >0$ and imposing disutility $\lambda >0$. Assume $r,\rho >0$, nonzero market price of risk $\kappa$, a positive Merton consumption constant, and solvency allowing borrowing against perpetual income up to $\epsilon /r$ before retirement. Let $m<0$ solve ${\kappa }^{2}m(m-1)/2+(\rho -r)m-\rho =0$. The retirement threshold is the displayed wealth level.

To derive it, apply the wealth-variable Legendre dual. The working dual equals the retired dual plus $h(z)=\epsilon z/r-\lambda /\rho +D{z}^m$. Excluding the growing homogeneous solution enforces the solvency asymptotic. Value matching and smooth fit at the retirement marginal value $z_{\ast }$ give $h(z_{\ast })=h^{\prime }(z_{\ast })=0$, hence $z_{\ast }=(\lambda r/(\epsilon \rho ))m/(m-1)$ and $D=-\epsilon z_{\ast }^{1-m}/(rm)>0$. The retired inverse marginal value gives $\overline{w}={\gamma }_M^{-1}{z}_{\ast }^{-1/R}$.

The root equation implies $z_{\ast }/(\lambda /\epsilon )=1+{\kappa }^{2}m/(2\rho )<1$. Thus the boundary exceeds the instantaneous gain-zero level ${\gamma }_M^{-1}(\epsilon /\lambda {)}^{1/R}$. A tighter borrowing constraint changes this free-boundary problem. Smooth fit matches the first derivative; matching the second derivative would incorrectly remove the income option.

A positive normalized continuous local martingale deflator requires a locally square-integrable Brownian diffusion coefficient. If the forced market price of risk is $-t^{-1/2}$, the coefficient has magnitude $Z_t/\sqrt{t}$. Since $Z_0>0$, continuity makes its squared integral diverge near zero. The bank account can nevertheless exist because $t^{-1/2}$ itself is integrable; finite integrated interest is weaker than square-integrability of the required risk compensation.

If a positive terminal state-price density has essential infimum zero, let $A_n=\{{\zeta }_T<1/n\}$ and $X_n=w_0{1}_{A_n}/\mathbb{E}[{\zeta }_T{1}_{A_n}]$. Whenever $\mathbb{P}(A_n)>0$, these claims cost $w_0$ and satisfy $\mathbb{E}{X}_n>n{w}_0$. In a Brownian complete market with constant nonzero market price of risk, ${\zeta }_T$ has a lognormal distribution and all these events have positive probability. Thus bounded market coefficients and an increasing concave utility do not alone ensure a finite attained optimum: the linear utility $U(x)=x$ already fails. Usual inverse-marginal-utility formulas require further utility and integrability hypotheses.