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An American call permits purchase of the underlying at strike $X$ at a chosen stopping time no later than maturity. Under standard pricing and integrability hypotheses its value is the displayed optimal stopping value. A European call option restricts exercise to the final time. Dividends and interest on the deferred strike affect the incentive to exercise early.

An optimal stopping problem in which observations update a posterior probability distribution over an unknown parameter. The decision state includes that posterior.

An obstacle constraint requires a solution to lie above a given function. In an optimal stopping problem, the value dominates the payoff, satisfies a generator equation where continuation is optimal, and a generator inequality where stopping is optimal. The complementary conditions can be expressed as a maximum equal to zero.

Determining the value of a European option or American option from a specified financial market model and its permitted exercise rules. Risk-neutral pricing, replicating strategies and optimal stopping address different parts of this problem.

Use $\kappa =(\mu -r)/\sigma$ and ${\gamma }_M$ from Question 1, and take the finite-value case ${\gamma }_M>0$. After retirement the value is $V_M(w)={\gamma }_M^{-R}{w}^{1-R}/(1-R)$. Working for an infinitesimal extra interval while using the retired portfolio and consumption produces a gain $\epsilon V_M^{\prime }(w)-\lambda$. Its zero isThis proves the instantaneous indifference level appearing in the question. For the stated irreversible optimal stopping problem it is generally a lower bound on the retirement boundary, rather than the boundary itself: retaining the opportunity to keep working has value. The following explicit wealth-variable Legendre dual calculation exhibits the correction. In particular, imposing continuity of $V^{\prime \prime }$ at retirement would discard that option value; smooth fit requires continuity of $V^{\prime }$.

Assume the usual borrowing-against-income solvency condition: $w>-\epsilon /r$ while working, with nonnegative wealth after retirement, and take $r>0$, $\rho >0$, $\kappa \ne 0$. The question does not specify a tighter borrowing constraint. Let $J(z)={\sup }_w\{V(w)-zw\}$, so $w=-J^{\prime }(z)$ and $V=J-z{J}^{\prime }$. The working Hamilton-Jacobi-Bellman equation becomes the linear Cauchy-Euler equationThe retired dual is $J_M(z)=R{z}^{1-1/R}/[(1-R){\gamma }_M]$. Let $m<0$ be the negative root ofThe other root exceeds one. Its growing term is excluded by the solvency asymptotic as $z\to \infty$. Therefore the working dual has the formRetirement corresponds to $z\le z_{\ast }$, where $J=J_M$. Value matching and smooth fit giveConsequently the retirement boundary with an income option isUsing the root equation, $z_{\ast }/(\lambda /\epsilon )=1+{\kappa }^{2}m/(2\rho )$, which lies strictly between zero and one. Thus $\overline{w}>w_{\mathrm{myopic}}$ for a nonzero risk premium. For example, $r=\rho =\epsilon =\lambda =\kappa =1$ and $R=2$ give $m=-1$, ${\gamma }_M=9/8$, andThese parameters give an explicit counterexample to identifying the printed level with the irreversible retirement boundary.

For $-\epsilon /r<w<\overline{w}$, find the unique $z>z_{\ast }$ fromThis right side decreases strictly from $\overline{w}$ to $-\epsilon /r$, since $J^{\prime \prime }>0$. Recover the value and controls byFor $w\ge \overline{w}$, retire immediately and use the Merton controls. Along the working strategy, $z_t$ satisfies $d{z}_t=(\rho -r)z_{t}dt-\kappa z_{t}d{W}_t$, and the optimal retirement time is its first hitting time of $z_{\ast }$, equivalently wealth's first hitting time of $\overline{w}$.

To check the optimal stopping inequalities, write $h=J-J_M$. It satisfies $h(z_{\ast })=h^{\prime }(z_{\ast })=0$ and $h^{\prime \prime }(z)=Dm(m-1)z^{m-2}>0$ for $z>z_{\ast }$. Hence continuation has positive option value there. On $z\le z_{\ast }$ the stopping candidate has working residual $\epsilon z-\lambda \le 0$. The matched dual is continuously differentiable, so the Itô formula introduces no boundary local-time term. Localization, the state-price budget, and the appropriate infinite-horizon transversality justify verification.

If nonnegative financial wealth is imposed during employment as an additional constraint, the working dual also has an upper boundary, and that extra constraint must be included in the free-boundary problem. It cannot be recovered by assuming the printed myopic equality.

Write $\alpha =2r/{\sigma }^2\in (0,1)$. In a continuation interval, the Euler differential equation has power solutions $s$ and $s^{-\alpha }$: substitution of $s^{\beta }$ gives $(\beta -1)(\beta +\alpha )=0$. A decreasing bounded value uses the second solution.

Let $b$ be the exercise boundary. Value matching and smooth fit require $A{b}^{-\alpha }=(1+b{)}^{-1}$ and $-\alpha A{b}^{-\alpha -1}=-(1+b{)}^{-2}$. Dividing gives $b/(1+b)=\alpha$. ConsequentlyThis is a perpetual reciprocal-payoff American option.

To verify the obstacle inequality, observe that $s^{\alpha }/(1+s)$ increases up to $b$ and decreases afterwards, because its logarithmic derivative is $\alpha /s-1/(1+s)$. Therefore $V(s)\ge (1+s{)}^{-1}$ for $s>b$. In the continuation region $({\mathcal{L}}_r-r)V=0$. In the exercise region,The quadratic is convex. At the two endpoints of $[0,b]$ it equals $-\alpha /2$ and $-\alpha /[2(1-\alpha )]$, respectively, so it is negative throughout that interval. Thus the obstacle problem is satisfied on both regions, with value matching and smooth fit at $b$. The second derivative has a jump at $b$; the equation is understood piecewise and in the generalized Itô sense described in part (a).

For optimality, use the Risk-neutral measure for the Black-Scholes model, under which $d{S}_t=r{S}_{t}dt+\sigma S_{t}d{W}_t^{\mathbb{Q}}$. The Itô formula makes $e^{-rt}V(S_t)$ a nonnegative supermartingale, so every exercise time $\tau$ satisfiesBefore hitting the exercise region, its drift vanishes. Since $0\le V\le 1$, the stopped process is a bounded martingale. Apply the optional sampling theorem at ${\tau }_{\ast }\wedge T$ and let $T\to \infty$. The contribution from $\{{\tau }_{\ast }>T\}$ is at most $e^{-rT}$, while the boundary value equals the payoff. Hence equality holds forIf $S_0\le b$, exercise immediately; otherwise wait for the first down-crossing. If that time is infinite, the discounted payout is zero. This proves both the value and the optimal stopping policy, and part (a) supplies its superhedge.

Backward induction makes every $U_t$ adapted and integrable: the conditional expectation of an integrable random variable is integrable, and $|\max (a,b)|\le |a|+|b|$. The Snell envelope recursion directly gives $U_t\ge \mathbb{E}[U_{t+1}\mid {\mathcal{F}}_t]$, so $U$ is a supermartingale.

If $Z$ is a submartingale, iterating its defining inequality gives $\mathbb{E}[Z_T\mid {\mathcal{F}}_t]\ge Z_t$. Starting at $U_T=Z_T$ and applying the tower property of conditional expectation backwards showsThis last conditional expectation is a true martingale. The conclusions areIn the second case waiting until $T$ attains the optimal stopping value. It is not generally true that $U_t=Z_t$: that further equality holds when $Z$ itself is a martingale.

An American call option with no expiry. For geometric Brownian motion with drift $r-y$ under a risk-neutral measure, positive dividend yield $y$, $r>0$, and volatility $\sigma >0$, let $p>1$ solve ${\sigma }^{2}p(p-1)/2+(r-y)p-r=0$. Its value is $(S_{\ast }-K)(S/S_{\ast }{)}^p$ below $S_{\ast }=pK/(p-1)$ and $S-K$ above. The power solves the stationary Black-Scholes equation with continuous stock dividends; value matching and smooth fit determine the threshold.

The candidate dominates the positive part $(S-K{)}^{+}$ because its continuation branch is convex and tangent to $S-K$ at $S_{\ast }$. In the exercise region its discounted-generator residual is $rK-yS\le 0$: the root equation gives $y{S}_{\ast }=K[r+{\sigma }^{2}p/2]>rK$. Applying Itô's formula to the discounted value, localized if necessary, gives an upper bound on every exercise policy. Stopping at the first hit of $S_{\ast }$ attains it; on paths not yet hitting, the bounded discounted continuation value tends to zero. This supplies the optimal stopping verification. As $y\downarrow 0$, $p\downarrow 1$, the threshold diverges and the value tends to $S$.

For a dividend-free Black-Scholes model with $r>0$, $\sigma >0$, $K>0$ and positive integer $n$, the optimal stopping value of the power put option isIn the continuation region the stationary Black-Scholes equation has powers $S$ and $S^{-\gamma }$. Boundedness at infinity removes $S$. Value matching gives the coefficient $(K-b{)}^n{b}^{\gamma }$, maximized at the displayed $b$; equivalently this is the smooth fit boundary.

For a verification beyond constant-trigger policies, let $g(S)=(K-S{)}_{+}^n$. The maximum property of $b^{\gamma }(K-b{)}^n$ gives $V\ge g$. For the infinitesimal generator $\mathcal{L}=rS{\partial }_S+{\sigma }^2{S}^2{\partial }_{SS}/2$, $(\mathcal{L}-r)V=0$ above $b$. Below $b$, set $y=S/(K-S)\le \gamma /n$; thenThe candidate is continuously differentiable across $b$, so the generalized Itô formula has no derivative-jump local time of a semimartingale term. The discounted candidate is a nonnegative supermartingale, bounding every stopping reward by optional stopping. First hitting of $b$ attains the bound; the remainder at infinite time vanishes since $V\le K^n$ and $r>0$. For $n=1$ this recovers the perpetual put option. At $r=0$ the value is $K^n$ for $S>0$, approached but not attained at a finite exercise time. At $r<0$ the perpetual value is infinite, because exercise at deterministic times tending to infinity has unbounded discounted expected value.

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 strictly increasing sequence $s_i$ for which $s(X_t)$ is a martingale along the birth-death chain before hitting its boundary. With upward and downward probabilities $p_i,q_i$, harmonicity means $p_i(s_{i+1}-s_i)=q_i(s_i-s_{i-1})$. This coordinate is useful for hitting probabilities and optimal stopping.

At a regular exercise boundary for a diffusion optimal stopping problem, value matching is commonly accompanied by matching first derivatives. This makes the value $C^1$ across the boundary and eliminates a derivative-jump local time of a semimartingale term in the generalized Itô formula. It does not assert that second derivatives agree, or that smooth fit holds for every possible stopping problem.