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A Bellman equation is the recursive optimality relation for the value function of a dynamic programming problem.

Fix the first tape of a group multiplier automaton and use dynamic programming over automaton states and output-padding status, retaining predecessor pointers. The pumping of a group multiplier automaton bounds the search by $|u|+N$ layers, each of fixed size. This constructs a representative of $\overline{u}x$ in time $O(|u|+1)$ without assuming uniqueness of representatives.

Wealth and admissibility. The dollar holding $\theta$ earns the risky return, while $w-\theta$ earns the continuous-time bank account return. Removing consumption therefore givesHere $\theta$ is a dollar amount, not a number of shares; the number of shares is ${\theta }_t/S_t$. Both controls must use available information: take $\theta$ predictable and $c\ge 0$ progressively measurable, withalmost surely on every finite interval. Require a well-defined objective and an admissible trading strategy satisfying $w_t\ge 0$. At zero portfolio wealth this excludes continued risky gambling or positive consumption. The state-price budget constraint rules out doubling strategies. Throughout the diffusion calculations take $\sigma \ne 0$, positive discount $\rho$, positive decay $\lambda$, and finite value; degeneracies are discussed where they affect the conclusions.

Satisfaction and dynamic programming. Write the consumption satisfaction stock asThe product rule gives, almost everywhere in time,This state is a finite-variation process, so it has no quadratic covariation with portfolio wealth. Applying the Itô formula to the discounted value function over a short interval, then using dynamic programming, gives the interior HJB equationAssuming $V_{ww}<0$, completing the square givesThe last supremum is zero if $V_{\xi }-V_w\le 0$ and infinite otherwise. Thus the gradient constraint for unbounded consumption is $V_{\xi }\le V_w$, and $c=0$ wherever the inequality is strict. Since there is no direct penalty for a very large consumption rate, an active boundary can involve singular consumption control. In that relaxed interpretation the HJB equation isWith ordinary rate controls, this describes the supremum and its limiting transfer policy; it does not promise that an instantaneous transfer is attained by a finite rate.

Power reduction. An additive constant in the utility function only adds a control-independent constant divided by $\rho$ to the value, so normalize $U(\xi )={\xi }^{1-R}/(1-R)$. Put $p=1-R$ and $x=w/\xi$ for $\xi >0$. Scaling portfolio wealth, consumption satisfaction and the controls by the same positive number gives the wealth-to-satisfaction reductionConsequently the reduced HJB equation isIn the strict waiting region the first expression vanishes and consumption is zero. On a transfer region the second vanishes; integrating it gives $v(x)=K(1+x{)}^p$. This reflects preservation of $w+\xi$ during an instantaneous wealth-to-satisfaction transfer.

Why a waiting threshold is expected, and its qualification. The gradient constraint for unbounded consumption compares the benefit of increasing consumption satisfaction with the opportunity cost of spending financial portfolio wealth. When consumption satisfaction is already large relative to cash, waiting lets consumption satisfaction decay while financial portfolio wealth earns returns; consuming immediately can be wasteful. Homogeneity makes the comparison depend only on $x$. For fixed $s=w+\xi$, joint concavity of the value function makes $q_s(a)=V(s-a,a)$ concave in financial portfolio wealth $a$. It is nondecreasing because an immediate transfer can reproduce any smaller financial allocation. Consequently its derivative $V_w-V_{\xi }$ is nonnegative and nonincreasing in $a$: a strict waiting region, if present, starts at zero financial portfolio wealth and ends at a single transfer boundary. Homogeneity makes the corresponding boundary a ratio $x_{\ast }$. In the usual finite-boundary regime this gives $c=0$ for $x<x_{\ast }$, and transfers push a larger ratio down towards $x_{\ast }$. The condition is a comparison of marginal values, not the ordinary formula $c=(V_w{)}^{-1/R}$: current utility function here depends on consumption satisfaction, not on current consumption.

A positive threshold is not guaranteed by the printed hypotheses alone. A useful sufficient local test illustrates the intended argument. Let $a=\rho +\lambda p>0$. At zero portfolio wealth,Starting with a small extra portfolio wealth $\epsilon$, holding it in the continuous-time bank account until a fixed time $t$, then transferring it into consumption satisfaction, has right derivative in $\epsilon$ at zero equal toTherefore, if $\rho <r+\lambda R$, the portfolio wealth marginal value is strictly larger than the consumption satisfaction marginal value at zero. With the usual continuity of marginal values, there is a positive interval on which the gradient constraint for unbounded consumption is strict. The fixed-total-resource concavity argument then gives the threshold structure.

For a concrete counterexample to an unconditional positive threshold, take $R=1/2$, $\lambda =1$, $r=\mu =1/10$, $\sigma =1$ and $\rho =2$. There is zero market price of risk. The relaxed value isIndeed $F_{\xi }=F_w$, $F_{ww}<0$, and the optimized waiting residual, after division by ${\xi }^p$, isbecause $(r+\lambda )/a<1$ and the integrand is decreasing. The Itô formula gives an upper bound by $F$, while transferring all portfolio wealth into consumption satisfaction over intervals tending to zero attains that bound in the limit. Hence the transfer boundary is $x_{\ast }=0$ in this example. The positive-threshold explanation needs a parameter regime supporting a genuine waiting region. For $R>1$, even the zero-wealth value is finite only if $\rho >\lambda (R-1)$; otherwise decaying consumption satisfaction gives value $-\infty$.

The joint generator. With $\theta$ again measured in dollars, portfolio wealth obeysIts noise and the factor noise are driven by the same Brownian motion. Their quadratic covariation is $\sigma (x)\theta \alpha (x)dt$, so the diffusion generator has a cross derivative. Dynamic programming and the Itô formula giveAll coefficient functions in this formula are evaluated at $x$. The cross derivative is essential: it gives intertemporal hedging demand. Normalizing CRRA utility as $U(c)=c^p/p$, where $p=1-R$, the two optimizations give, at nonzero volatility,and henceHere $p(-1/R)=1-1/R$.

Homogeneity and the reduced equation. Scaling initial portfolio wealth and both controls preserves the portfolio wealth constraint and multiplies the objective by the positive number $b^p$. ThusThis expression is valid for both signs of $p$: $V$ is negative when $R>1$, but its portfolio wealth derivative is positive. Its derivatives areSubstitution yieldsThe resulting feedback isThe first portfolio term is myopic, and the second is intertemporal hedging demand.

Constant market price of risk. If $\mu -r=\sigma \kappa$ and volatility is nonzero, the portfolio term becomes $p(\kappa f+\alpha f^{\prime }{)}^2/(2Rf)$, so the magnitude of stock volatility disappears. Applying the power transformation of a complete-market investment equation $f=g^R$ cancels the squared-gradient terms and gives the further reductionFor ${\gamma }_M>0$ its economic solution is $g=1/{\gamma }_M$. ThusTo see why this solves the investment-consumption problem, optimize directly over Brownian portfolio exposures, writing $y=\sigma (x)\theta$. The portfolio wealth equation becomes $dw=ydW+(rw+\kappa y-c)dt$, which no longer contains $X$. With nonzero volatility the same admissible exposure processes are available for every factor state, so the attainable wealth-consumption pairs, and therefore the value, are exactly those of the Merton consumption-investment problem. This also excludes extraneous solutions of the linear equation without imposing artificial factor boundary data.

The printed boundedness assumptions do not ensure nonzero volatility or a finite value. The unsimplified HJB equation remains the correct control equation at a zero of $\sigma$. There the hedge term vanishes; if $\mu \ne r$, the riskless excess return gives arbitrage with unrestricted holdings. Under $\mu -r=\sigma \kappa$, a zero-volatility state offers only the bank exposure at that instant. For example $\sigma \equiv 0$, $\mu \equiv r$ satisfies this relation for any chosen $\kappa$, but its value uses ${\gamma }_0=[\rho -pr]/R$, not a fictitious nonzero risk premium. The constant-value formula using $\kappa$ presupposes access to the Brownian exposure, with sufficient integrability for the corresponding holdings. Additive utility constants again only shift $V$ by a constant divided by $\rho$.

Available portfolio wealth and the ruin boundary. Put $h=\bar{r}D$, the constant interest payment on the fixed loan. The loan principal is already included in available portfolio wealth; it is not a growing portfolio holding. ThereforeIn particular the interest outflow is $\bar{r}D$, not $(\bar{r}-r)D$. Writing $w=x+D$ would instead give net portfolio wealth drift $rx+(\mu -r)\theta -(\bar{r}-r)D-c$, which explains the distinction.

Let ${\tau }_0$ denote the ruin time, to avoid confusing it with a fixed terminal horizon. The objective stops at ${\tau }_0$; consequently the absorbing boundary is $V(0)=0$, without an obligation to keep financing the loan after ruin. Dynamic programming gives, for $w>0$,For increasing strictly concave value, put $z=V^{\prime }(w)>0$ and use inverse marginal utility $I$. The optimal controls and the optimized HJB equation arewhere $\tilde{U}(z)={\sup }_{c\ge 0}[U(c)-zc]$. For CRRA utility with $0<R<1$, write $p=1-R$ and $q=1-1/R<0$. Then

Dualization and the printed constant. Use the convex wealth-variable Legendre dualAt an interior maximizing portfolio wealth, $J^{\prime }=-w$, $J^{\prime \prime }=-1/V^{\prime \prime }>0$ and $V=J-z{J}^{\prime }$. The dual HJB equation is the linear Euler differential equationA trial power $z^t$ givesDirect substitution givesThe PDF prints an additional factor $1/2$ before ${\kappa }^2/(2R)$ in its definition of ${\gamma }_M$. That printed definition is inconsistent with its own identity for $Q(q)$. The expression above is the one used here; assume this corrected ${\gamma }_M>0$.

Solution when $\kappa \ne 0$. Assume positive discount $\rho$, nonzero $\sigma$, and $h>0$. Let $m>0$ and $n<0$ be the two roots of $Q$:Since $Q(q)<0$, $n<q<0<m$. Put $C=R/(p{\gamma }_M)$. For $r\ne 0$, the general interior solution isThe appropriate large-wealth condition is the Merton consumption-investment problem boundIndeed any original control consumes $c+h$ in the debt-free comparison model until ruin, and $U(c+h)\ge U(c)$. Dualizing this bound gives $0\le J(z)\le C{z}^q$. Because $n<q$, convexity and this upper bound force $B=0$ as $z\downarrow 0$.

At the other endpoint portfolio wealth reaches zero. If $z_{\ast }=V^{\prime }(0+)$, the dual ruin boundary with debt service requiresSolving these two equations givesFor $r>0$, $m>1$, and for $r<0$, $0<m<1$, so the quantity defining $z_{\ast }$ is positive in either case. These formulas determine the entire value. For each $w>0$, choose the unique $z\in (0,z_{\ast })$ satisfyingThenBoth terms in the bracket are positive, even when $r<0$ and $A<0$. Thus $J^{\prime \prime }>0$, $-J^{\prime }$ decreases from infinity to zero as $z$ increases, and the portfolio wealth inversion really is unique. The extended dual is continuously differentiable and convex.

There is no additional condition $J^{\prime \prime }(z_{\ast }^{-})=0$. In factAvailable portfolio wealth is killed at zero; the portfolio can have a nonzero limiting volatility immediately before ruin. Imposing a reflecting-boundary or zero-curvature condition would solve a different problem.

For $r=0$, the linear forcing resonates with the root $m=1$. Put $b=\rho +{\kappa }^2/2=Q^{\prime }(1)$. The dual and boundary constants instead areExtend by zero for $z\ge z_{\ast }$. Hereand the controls remain $c=z^{-1/R}$ and $\theta =\kappa z{J}^{\prime \prime }/\sigma$, with $J^{\prime \prime }=h/(bz)+Cq(q-1)z^{q-2}>0$.

Verification and transversality. The candidate is nonnegative, increasing, strictly concave and zero at ruin. Its HJB equation makes the discounted value plus accrued utility a local supermartingale for every admissible control, and a local martingale for the stated feedback. Localization at positive lower and finite upper portfolio wealth levels gives the finite-horizon comparison. The investment value transversality condition follows from the same debt-free bound: applying the Itô formula to $w^p$ and maximizing its risky term givesThe nonnegative consumption and debt-service drifts only decrease this bound. Thus the expected terminal candidate tends to zero. The feedback has at most linear growth, including a finite limit as portfolio wealth decreases to zero; stopping it at ruin gives an admissible policy. Letting localization levels and then the horizon tend to their limits proves that the candidate is the value, rather than just a formal dual solution. If $D=0$, the absorbing-debt boundary disappears and the ordinary Merton consumption-investment problem formula is recovered.

Zero market price of risk. With $\sigma \ne 0$ and $\kappa =0$, the dual equation is first order. If $\rho >r$, the preceding formulas remain valid with $m=\rho /(\rho -r)$ for $r\ne 0$, and the logarithmic formula with $b=\rho$ for $r=0$; there is no $B$ term and ${\theta }^{\ast }=0$. The same boundary and transversality argument verifies this deterministic consumption policy.

The remaining finite-value case has $r>0$ and $pr<\rho \le r$. Put ${\gamma }_0=(\rho -pr)/R$, $w_c=h/(rR)$ andThe correct convex dual and its corresponding value areThe two value branches have the same value and derivative at $w_c$. Above $w_c$, hold no stock and consume ${\gamma }_0(w-h/r)$; the surplus over $h/r$ grows at rate $r-{\gamma }_0\ge 0$. If $\rho =r$, the lower branch is attained by zero stock holding and constant consumption $ph/R$: portfolio wealth solves $\dot{w}=r(w-w_c)$ until ruin, and direct integration gives $z_{c}w$.

If $\rho <r$ and $0<w<w_c$, the lower branch is a supremum attained in a limit of increasingly rapid fair stock lotteries between zero and $w_c$, followed by the upper-branch policy on success. The success probability tends to $w/w_c$ and the fixed service cost during the lottery tends to zero. This is possible because unrestricted dollar holdings in the nonzero-volatility stock produce a fair Brownian motion exposure even when its excess drift is zero. The supporting linear branch has optimized waiting residual $(r-\rho )z_c(w-w_c)<0$; the fast lotteries, rather than a finite feedback optimizer, supply the missing control limit. The piecewise candidate is concave, has nonpositive waiting residual everywhere, and the preceding moment bound still supplies an upper-bound verification. This degenerate case can have a supremum without an ordinary maximizing strategy.

Assume nonnegative wealth and rewards, and initially $0<p_i\le 1$. Set $F_0(x)=x$, and let $R_i$ take the values $0,2a_i$ with equal probability conditional on not being caught. For the retirement threshold with multiplicative capture risk, the dynamic programming equation is the displayed maximum; capture contributes zero. Write $t={\max }_i{a}_i{q}_i/p_i$.

Inductively, if $x\ge t$, all future successful wealth levels also lie above $t$, so $F_{s-1}(x+R_i)=x+R_i$. Continuing in town $i$ is then worth $q_i(x+a_i)\le x$, and retirement is optimal. If $x<t$, choose $i$ with $a_i{q}_i/p_i>x$. Continuing for just one night and then retiring is worth $q_i(x+a_i)>x$, so immediate retirement cannot be optimal for any $s\ge 1$. This proves an optimal policy continues exactly when $\boxed{x<t}$, with retirement chosen at the threshold.

Now suppose $q_1>q_2$ and $q_1{a}_1>q_2{a}_2$. Since also $p_1<p_2$, town $1$ has the larger threshold, so $t=a_1{q}_1/p_1$. The functions $F_s$ are nondecreasing in wealth and in $s$, and $F_s(x)\ge x$; these properties follow by induction from the dynamic programming equation.

Prove by induction on the remaining horizon that town $2$ is never optimal when continuation is worthwhile. For $s=1$, the difference between the town values is $(q_1-q_2)x+q_1{a}_1-q_2{a}_2>0$. If $a_2\le a_1$, monotonicity giveswith strictness because the first bracket is positive under the hypotheses.

For $a_2>a_1$, first suppose $x+2a_2\ge t$. The second town's successful large-reward branch retires, so its value is $q_2[F_{s-1}(x)+x+2a_2]/2$. Town $1$ is worth at least $q_1[F_{s-1}(x)+x+2a_1]/2$. Their difference is at leastIf instead $x+2a_2<t$, all states after either first-night reward remain below the threshold. By induction a policy starting in town $2$ then chooses town $1$ on its second night. Its value isInterchanging the first two towns gives exactly the same value, because capture probabilities multiply and independent rewards add. After starting in town $1$, however, the induction hypothesis says town $2$ is strictly inferior on the second night at every possible surviving wealth level. Replacing that second choice by the optimal one strictly improves the value. This proves town $1$ dominates throughout every finite horizon.

If a town has $p_i=0$ and $a_i>0$, its threshold is interpreted as infinity and retirement is not optimal before the horizon ends. A town with $q_i{a}_i=0$ offers no gain; the natural threshold is zero. These conventions handle the degenerate capture probabilities rather than dividing by zero.

Treat the white noise as innovations with conditional mean zero and conditional second moment $v$, independent of the chosen past control. This is the convention needed for the dynamic programming calculation. The printed stage cost is $\frac{1}{2}{x}^2+u^2$, not one-half of both terms.

With terminal value $V_0(x)=x^2$, induction in the Bellman equation givesIf $V_{h-1}(z)=z^2+(h-1)v$, completing the square yieldsThus $\boxed{V_h(x)=x^2+hv,u^{\ast }(x)=-x/2}$ for every horizon, in particular $V_6(x_0)=x_0^2+6v$. The same completion gives the average-cost Bellman equation solution $\boxed{\varphi (x)=x^2,\lambda =v}$, up to an additive constant in $\varphi$.

For every policy in $P$, conditional second moments satisfy $\mathbb{E}[x_{t+1}^2\mid {\mathcal{F}}_t]=(x_t+u_t{)}^2+v\le 0.9x_t^2+v$. Iterating givesThe Bellman inequality implies ${\mathbb{E}}_{\pi }{\sum }_{t<h}(\frac{1}{2}{x}_t^2+u_t^2)\ge hv+x_0^2-{\mathbb{E}}_{\pi }{x}_h^2$. Dividing by $h$ and using the bound proves that every admissible policy has lower limiting average cost at least $v$, even if its ordinary limit does not exist. The feedback $u=-x/2$ belongs to $P$ because $(x+u{)}^2=x^2/4\le 0.9x^2$. It makes the Bellman inequality an equality, and its bounded terminal second moment proves its average cost tends to $\boxed{v}$. Therefore it is optimal over $P$.

Merely uncorrelated noise with the displayed unconditional moments would not ensure the conditional identities for arbitrary adapted controls. The innovation interpretation specifies the hypothesis used in the calculation.

Fix the first tape of the group multiplier automaton $M_x$ to be $u=a_1\cdots a_m$. Construct its possible runs by dynamic programming, one column at a time. A record consists of a state and a flag indicating whether the output tape has ended. At column $i\le m$, the first symbol is $a_i$ and the second symbol may be any letter of the alphabet, or $\#$; after the second tape first uses $\#$ it must continue to use $\#$. At later columns the first symbol is $\#$, and an additional column must have a genuine letter on the second tape. Never add a $(\#,\#)$ column.

Keep one predecessor record and its output symbol for each reachable state-and-flag pair in each layer. Two runs reaching the same record have exactly the same possible future completions, so this merging loses no accepting output. The merged dead state of a finite automaton can be discarded. At each layer $i\ge m$, test whether a reached state is accepting; the initial layer is included when $m=0$. Upon success, follow predecessor records backwards and omit padding symbols to recover $v$.

The automatic structure for a group supplies a representative in $L$ of $\overline{u}x$, so an accepted padded convolution of words exists and this search terminates. The output is the required representative:The search uses only the finite transition table; it does not presuppose a solution to the word problem for a group.

Part (a)(i), applied to $w=u$ and $g=\overline{u}x$, guarantees an accepted output of length at most $m+N$. Its padded convolution of words has at most $m+N$ columns. Thus the dynamic programming search in part (b) reaches an accepting layer by this bound.

Each layer has at most twice the fixed number of states of $M_x$, and each record has at most $|A|+1$ outgoing choices. These are constants for the fixed automatic structure for a group. Storing a pointer and a symbol takes constant time; copying whole partial output strings at every step would unnecessarily spoil this bound. Reconstruction also takes at most $m+N$ steps. Therefore automatic multiplication has linear time and bounded length increase:For nonempty inputs this is the requested $O(|u|)$ time complexity. The additive $1$ merely includes the constant work on the empty word.

For the intended random walk model, take $S_0$ to be deterministic, or independent of all the independent and identically distributed random variables $Y_j=S_j-S_{j-1}$. If $\nu$ is their common probability distribution, then $Y_{t+1}$ is independent of ${\mathcal{F}}_t=\sigma (S_0,Y_1,\dots ,Y_t)$. This supplies the Markov property that the backward dynamic programming argument needs. Define the transition operatorand the deterministic optimal stopping value function recursively byAt any state where the conditional law is evaluated, independence givesBackward induction, starting with $U_T=f(S_T)$, proves the intended representationIntegrability along the actual random walk makes the recursion finite at the states used by that process, up to null sets. If finite real-valued value functions on all of $\mathbb{R}$ are intended, a convenient sufficient convention is statewise integrability: $\mathbb{E}|f(s+Y_1+\cdots +Y_k)|<\infty$ for every $s$ and $0\le k\le T$. Indeed each stopping value is bounded in absolute value by the expectation of the sum of the absolute rewards. This convention makes the displayed recursion finite and measurable everywhere; the source only assumes integrability from the given starting process.

The printed independence of the increments alone does not suffice if $S_0$ can reveal future increments. Here is a bounded finite-state counterexample, also with a convex reward. Let $T=2$, let ${\epsilon }_1,{\epsilon }_2$ be independent fair signs, and setThe two increments are exactly ${\epsilon }_1,{\epsilon }_2$, hence independent and identically distributed. But ${\mathcal{F}}_1$ already knows both signs, so $U_1=\max \{S_1^{+},S_2^{+}\}$. On the two positive-probability histories with $S_1=0$, its values are respectively $0$ and $1$. Thus no deterministic $V(1,0)$ can work. Adding independence of $S_0$ from the increments repairs this missing Markov property hypothesis.

For the plug-in Neyman allocation, the estimated Bernoulli distribution variances are $6/25$ and $3/16$, soThis uses the target ratio directly as the allocation odds; a procedure that actively corrects previous allocation imbalances would need an additional rule.

For the randomized play-the-winner rule RPW$(1,1)$, return the drawn ball and add one ball of the same treatment after a success or of the opposite treatment after a failure. Treatment 0 has three successes and two failures, and treatment 1 has one success and three failures. The urn therefore has $1+3+3=7$ treatment-0 balls and $1+1+2=4$ treatment-1 balls. HenceFor dynamic programming with only the final patient left, there is no future value from learning. Independent uniform prior distributions and Beta-binomial conjugacy give posteriors $p_0\mid \mathcal{D}\sim \mathrm{Beta}(4,3)$ and $p_1\mid \mathcal{D}\sim \mathrm{Beta}(2,4)$. Their posterior predictive probabilities of success are $4/7$ and $2/6=1/3$. The optimal terminal action therefore givesThis is optimal for expected successes with no imposed lower bound on the randomization probabilities.

The dynamic programming principle over a short interval $[t,t+\Delta t]$ givesA first-order Taylor expansion of the value function givesAfter subtracting $F(x,t)$, dividing by $\Delta t$ and taking the limit, the Hamilton-Jacobi-Bellman equation is

A stochastic control problem chooses an admissible control input using information available in a filtration to optimize an expected objective for a random dynamical system. Dynamic programming often characterizes its value function through a Hamilton-Jacobi-Bellman equation.