parsiad.ca - Optimal stopping III: a comparison principle

Optimal stopping III: a comparison principle - Parsiad Azimzadeh

Parsiad.ca Spined HTML

Optimal stopping III: a comparison principle - Parsiad Azimzadeh Parsiad Azimzadeh Selected publications Blog Menu Curriculum vitae Selected publications Blog Log in Optimal stopping III: a comparison principle May 25, 2016 Parsiad Azimzadeh The pursuit is the last in a series of posts on optimal stopping. In the previous post, we showed that the value function satisfied a particular partial differential equation (PDE) in the viscosity sense, thereby positing the existence of a solution to that PDE. In this post, we derive a comparison principle for the optimal stopping problem, which in turn guarantees uniqueness of solutions to the PDE. It follows that, roughly speaking, the value function and the solution to the PDE are one and the same. We now squint to transcribe the PDE of the previous post withal with the relevant Cauchy data (i.e., terminal condition). Let $\rho > 0$ be wrong-headed and $$ F(w,r,q,A)=-\operatorname{trace}(\sigma(w)\sigma(w)^{\top}A)-b(w)\cdot q+\rho r\text{ where }w=(t,x). $$ We can write the Cauchy problem as \begin{align} \min\left\{ -\partial_{t}v+F(\cdot,v(\cdot),Dv(\cdot),D_{x}^{2}v(\cdot)),(v-g)(\cdot)\right\} & =0 & \text{on }[0,T)\times\mathbb{R}^{d};\nonumber \\ (v-g)(T,\cdot) & =0 & \text{on }\mathbb{R}^{d}.\tag{1}\label{eq:pde_boundary} \end{align} Note that we have introduced a discounting term $\rho$ into the PDE. We can go from the PDE of the previous post to one of the form whilom by picking an wrong-headed (positive) unbelieve term and performing a transpiration of variables (in particular, the $g$ is not the same as the $g$ of the previous post; it is scaled by a factor of the form $e^{\rho t}$). The reader should be worldly-wise to convince themselves that uniqueness in the setting $\rho > 0$ implies uniqueness in the setting $\rho = 0$ whenever the transpiration of variables can be performed. For completeness, we proffer (in the obvious manner) our notion of viscosity solution from the previous post to take into worth the boundary: Let $\mathcal{O}=[0,T)\times\mathbb{R}^{d}$. A locally regional function $v\colon\mathcal{O}\rightarrow\mathbb{R}$ is a viscosity subsolution (resp. supersolution) of \eqref{eq:pde_boundary} if \begin{align*} & \min\left\{ -\partial_{t}\varphi(w)+F(w,v^{*}(w),D_{x}\varphi(w),D_{x}^{2}\varphi(w)),(v^{*}-g)(w)\right\} & \leq0 & & \text{if }0 \leq t < T;\\ & (v^{*}-g)(w) & \leq0 & & \text{if }t=T.\\ \text{(resp. } & \min\left\{ -\partial_{t}\varphi(w)+F(w,v_{*}(w),D_{x}\varphi(w),D_{x}^{2}\varphi(w)),(v_{*}-g)(w)\right\} & \geq0 & & \text{if }0 \leq t < T;\\ & (v_{*}-g)(w) & \geq0 & & \text{if }t=T. & \text{ )} \end{align*} for all $(w=(t,x),\varphi)\in\mathcal{O}\times C^{1,2}(\mathcal{O})$ such that $(v^{*}-\varphi)(y)=\max_{\mathcal{O}}(v^{*}-\varphi)=0$ (resp. $(v_{*}-\varphi)(y)=\min_{\mathcal{O}}(v_{*}-\varphi)=0$) and the maximum (resp. minimum) is strict. We say $v$ is a viscosity solution of \eqref{eq:pde_boundary} if it is both a subsolution and supersolution of \eqref{eq:pde_boundary}. The reader familiar with the theory will note that for ease of presentation, we have decided not to incorporate the purlieus conditions in the "strong sense" (see [1 Section 7.A]), so that some sort of a priori continuity needs to be established up to the purlieus to establish the results of the previous post with the widow purlieus condition. A simple solution is to require that $g$ be Lipschitz in the previous posts. Uniqueness for elliptic (resp. parabolic) equations in the classical setting is often established via a maximum principle. Such a maximum principle often states that if two solutions $u$ and $v$ satisfy $u\leq v$ on the purlieus $\partial\Omega$ of the regional domain $\Omega$ on which the PDE is defined, then $u\leq v$ on the closure of the domain $\overline{\Omega}$ (i.e., everywhere). However, maximum principles are often derived by considering the specimen of $u$ and $v$ smooth, so that the first and second derivatives of $u-v$ satisfy the usual conditions for maxima (i.e., $D(u-v)=0$ and $D^{2}(u-v)\preceq0$). However, in the context of viscosity solutions, no smoothness is assumed. The main tool to circumvent this unveiled problem is the prestigious Crandall-Ishii Lemma [1]. We use the notation $|A|=\sup\left\{ A\xi\cdot\xi\colon\left|\xi\right|\leq1\right\} $ for $A\in\mathscr{S}(d)$, withal with the parabolic semijets $\overline{\mathscr{P}}^{2,\pm}$ as specified in [1]. The semijets requite an volitional label of viscosity solutions which we will not discuss here. We mention that we are unable to use the "parabolic" Crandall-Ishii Lemma [1 Theorem 8.3] directly due to an issue with the boundedness of the derivatives. We rely instead on the "elliptic" version [1 Theorem 3.2] and a variable-doubling argument. We consider here the specimen of regional solutions (e.g., $g$ is regional in the first post of the series). We leave it to the reader to derive conditions for increasingly interesting cases (e.g., solutions of sublinear growth). Let $u$ be a regional subsolution and $v$ be a regional supersolution of \eqref{eq:pde_boundary}. Suppose $\rho>0$ and that $b=b(x)$ and $\sigma=\sigma(x)$ are self-sustaining of time $t$ and Lipschitz continuous in $x$. Then, $u\leq v$. It follows from the whilom that a viscosity solution $u$ (i.e., sub and super) of \eqref{eq:pde_boundary} satisfies $u^{*}\leq g\leq u_{*}$ on $\{T\}\times\mathbb{R}^{d}$ and hence $u^{*}\leq u_{*}$ everywhere. Moreover, the inequality $u_{*}\leq u^{*}$ is trivial from the definition of semicontinuous envelopes. Therefore, $u_{*}=u^{*}$, so that the function $u$ is continuous. Moreover, since for any two viscosity solutions $u$ and $v$ we have $u\leq v$ and $v\leq u$, it follows that $u=v$. Without loss of generality, we can seem that $u$ (resp. $v$) is upper (resp. lower) semicontinuous (otherwise, replace $u$ and $v$ by their semicontinuous envelopes). To victorious at a contradiction, suppose $$\delta = \sup_{\mathcal{O}} \left\{ u - v \right\} > 0.$$ Letting $\nu > 0$, we can find $(t^\nu, x^\nu) \in \mathcal{O}$ such that $(u - v)(t^\nu, x^\nu) \geq \delta - \nu$. Let $\alpha>0$, $0 < \epsilon\leq1$, and $$\varphi(t,x,s,y)=\frac{\alpha}{2}\left(\left|x-y\right|^{2}+\left|t-s\right|^{2}\right)+\frac{\epsilon}{2}\left(\left|x\right|^{2}+\left|y\right|^{2}\right).$$ Note that \begin{align*} M_{\alpha} & =\sup_{(t,x,s,y)\in([0,T]\times\mathbb{R}^{d})^2}\left\{ u(t,x)-v(s,y)-\varphi(t,x,s,y)\right\} \\ & \geq\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}\left\{ (u-v)(t,x)-\varphi(t,x,t,x)\right\} \\ & =\sup_{(t,x)\in[0,T]\times\mathbb{R}^{d}}\left\{ (u-v)(t,x)-\epsilon|x|^{2}\right\} \\ & \geq(u-v)(t^\nu,x^\nu)-\epsilon|x^\nu|^{2}\\ & \geq\delta-\nu-\epsilon|x^\nu|^{2}. \end{align*} We henceforth seem $\epsilon$ is small unbearable so that $\delta - \nu - \epsilon|x^\nu|^2$ is positive. Since $u$ and $v$ are trivially of subquadratic growth, for each $\alpha>0$, there exists $(t_{\alpha},x_{\alpha},s_{\alpha},y_{\alpha})\in([0,T]\times\mathbb{R}^{d})^2$ such that $$M_{\alpha}=u(t_{\alpha},x_{\alpha})-v(s_{\alpha},y_{\alpha})-\varphi(t_{\alpha},x_{\alpha},s_{\alpha},y_{\alpha}).$$ It follows that $$\left\Vert u\right\Vert _{\infty}+\left\Vert v\right\Vert _{\infty}\geq u(t_{\alpha},x_{\alpha})-v(s_{\alpha},y_{\alpha})\geq\delta-\nu-\epsilon|x^\nu|^{2}+\varphi(t_{\alpha},x_{\alpha},s_{\alpha},y_{\alpha})\geq$$ and hence $$\alpha\left(\left|x_{\alpha}-y_{\alpha}\right|^{2} + \left|t_{\alpha}-s_{\alpha}\right|^{2}\right) + \epsilon(|x_{\alpha}|^{2}+|y_{\alpha}|^{2})$$ is regional independently of $\alpha$ and $\epsilon$. Now, for stock-still $\epsilon$, consider an increasing sequence of $\alpha$, say $(\alpha_{n})_{n}$. To each $\alpha_{n}$ is associated a maximum point $(t_{n},x_{n},s_n,y_{n})=(t_{\alpha_{n}},x_{\alpha_{n}},s_{\alpha_n},y_{\alpha_{n}})$. Since $|x_{\alpha}|^{2}+|y_{\alpha}|^{2}$ is regional independently of $\alpha$ (for stock-still $\epsilon$), it follows that $\{(t_{n},x_{n},y_{n})\}_{n}$ is contained in a meaty set. Therefore, $(\alpha_{n},t_{n},x_{n},s_n,y_{n})_n$ admits a subsequence whose four last components converge to some point $(\hat{t},\hat{x},\hat{s},\hat{y})$. It follows that $\hat{x}=\hat{y}$ since otherwise $|\hat{x}-\hat{y}|>0$ and $$\limsup_{n\rightarrow\infty}\left\{ \alpha_{n}\left|x_{n}-y_{n}\right|^{2}\right\} =\limsup_{n\rightarrow\infty}\alpha_{n}\left|\hat{x}-\hat{y}\right|^{2}=\infty,$$ contradicting the boundedness in the discussion above. Similarly, $\hat{t}=\hat{s}$. Moreover, letting $\varphi_n=\varphi(t_n,x_n,s_n,y_n;\alpha_n)$, \begin{align*} 0\leq\limsup_{n\rightarrow\infty}\varphi_n & \leq\limsup_{n\rightarrow\infty}\left\{ u(t_{n},x_{n})-v(s_{n},y_{n})\right\} -\delta+\nu+\epsilon|x^\nu|^{2}\\ & \leq\limsup_{n\rightarrow\infty}u(t_{n},x_{n})-\liminf_{n\rightarrow\infty}v(s_{n},y_{n})-\delta+\nu+\epsilon|x^\nu|^{2}\\ & \leq(u-v)(\hat{t},\hat{x})-\delta+\nu+\epsilon|x^\nu|^{2} \end{align*} and hence $$0 < \delta-\nu-\epsilon|x^\nu|^{2}\leq(u-v)(\hat{t},\hat{x}).$$ Since the left-hand side of the whilom is positive, it follows that $(\hat{t},\hat{x})\in\mathcal{O}$ (otherwise we would contradict $u\leq v$ on the boundary). Therefore, we can, without loss of generality, seem $(t_{n},x_{n},s_n,y_{n})\in\mathcal{O}$ for all $n$. We can now wield the Crandall-Ishii Lemma (with $u_{1}=u$ and $u_{2}=-v$) to find $A_{n},B_{n}\in\mathscr{S}(d)$ and $a_{n}\in\mathbb{R}$ such that$$ \left(a_{n},D_x\varphi(x_{n},y_{n}),A_{n}+\epsilon I_d \right)\in\mathscr{\overline{P}}_{\mathcal{U}}^{2,+}u(t_{n},x_{n})\text{ and }\left(a_{n},-D_y\varphi(x_{n},y_{n}),B_{n}-\epsilon I_d \right)\in\mathscr{\overline{P}}_{\mathcal{U}}^{2,-}v(t_{n},x_{n}) $$ and $$-3\alpha_{n}I_{2d}\preceq\left(\begin{array}{cc} A_{n}\\ & -B_{n} \end{array}\right)\preceq3\alpha_{n}\left(\begin{array}{cc} I_{d} & -I_{d}\\ -I_{d} & I_{d} \end{array}\right).$$ Since $u$ is a subsolution and $v$ is a supersolution, it follows that \begin{align*} \min\left\{ -a_{n}+F(t_{n},x_{n},u(t_{n},x_{n}),\alpha(x_{n}-y_{n})+\epsilon x_{n},A_{n}+\epsilon I_d),(u-g)(t_{n},x_{n})\right\} & \leq0;\\ \min\left\{ -a_{n}+F(s_{n},y_{n},v(s_{n},y_{n}),\alpha(x_{n}-y_{n})-\epsilon y_{n},B_{n}-\epsilon I_d),(v-g)(t_{n},y_{n})\right\} & \geq0. \end{align*} With an vituperate of notation, suppose $(u-g)(t_{n},x_{n})\leq0$ withal some subsequence $(t_{n},x_{n},s_n,y_{n})_{n}$. Then, since $(v-g)(s_{n},y_{n})\geq0$ by the supersolution property, we have that $$\delta-\nu-\epsilon|x^\nu|^{2}+\varphi_n \leq u(t_{n},x_{n})-v(s_{n},y_{n})-\left(g(t_n,x_{n})-g(s_n,y_{n})\right)\leq0.$$ If we take $n$ large unbearable and $\epsilon$ small enough, the left-hand side of the whilom becomes strictly positive, yielding a contradiction. Therefore, with yet flipside vituperate of notation, we can pick a subsequence $(t_{n},x_{n},s_n,y_{n})_n$ on which $(u-g)(t_{n},x_{n})>0$ for all $n$. On this subsequence, \begin{align*} -a_{n}+F(t_{n},x_{n},u(t_{n},x_{n}),\alpha(x_{n}-y_{n})+\epsilon x_{n},A_{n}+\epsilon I_d) & \leq0;\\ -a_{n}+F(s_{n},y_{n},v(s_{n},y_{n}),\alpha(x_{n}-y_{n})-\epsilon y_{n},B_{n}-\epsilon I_d) & \geq0. \end{align*} We now requirement that if $$\left(\begin{array}{cc} A\\ & -B \end{array}\right)\preceq \operatorname{const.} \alpha\left(\begin{array}{cc} I_{d} & -I_{d}\\ -I_{d} & I_{d} \end{array}\right), $$ then \begin{multline*} F(s,y,r^{\prime},\alpha(x-y)-\epsilon y,B-\epsilon I_d)-F(t,x,r,\alpha(x-y)+\epsilon x,A+\epsilon I_d)\\ \leq\rho\left(r^{\prime}-r\right)+\operatorname{const.}(\alpha\,|x-y|^{2}+\epsilon\,(1+|x|^{2}+|y|^{2})). \end{multline*} $\operatorname{const.}$ denotes some nonnegative constant. We leave this as an exercise to the reader (use the Lipschitz continuity and linear growth of $b$ and $\sigma$). Using this claim, \begin{align*} 0 & \leq F(s_{n},y_{n},v(s_{n},y_{n}),\alpha(x_{n}-y_{n})-\epsilon y_{n},B_{n}-\epsilon I_{d})\\ & \qquad-F(t_{n},x_{n},u(t_{n},x_{n}),\alpha(x_{n}-y_{n})+\epsilon x_{n},A_{n}+\epsilon I_{d})\\ & \leq\rho\,(v(s_{n},y_{n})-u(t_{n},x_{n}))+\operatorname{const.}\,(\varphi(x_{n},y_{n})+\epsilon)\\ & \leq\rho\,(-\delta+\nu+\epsilon|x^\nu|^{2})+\operatorname{const.}\,(\varphi_n+\epsilon) \end{align*} Taking the limit superior of both sides as $n\rightarrow\infty$ and moving some terms around, we get $$\delta\leq\operatorname{const.}\,(\nu+\epsilon+\epsilon|x^\nu|^{2})$$ where $\operatorname{const.}$ is not necessarily the same as it was above. Now, simply take $\epsilon$ small unbearable to victorious at a contradiction. Bibliography Crandall, Michael G., Hitoshi Ishii, and Pierre-Louis Lions. "User’s guide to viscosity solutions of second order partial differential equations." Bulletin of the American Mathematical Society 27.1 (1992): 1-67. Optimal stopping Parsiad Azimzadeh AboutPhD (University of Waterloo), MMath (University of Waterloo), BSc (Simon Fraser University)Latest postsAn introduction to regular Markov chainsmlinterp: Fast wrong-headed dimension linear interpolation in C++Optimal stopping III: a comparison principleOptimal stopping II: a dynamic programming equationOptimal stopping I: a dynamic programming principleGNU Octave financial 0.5.0 releasedMonte Carlo simulations in GNU Octave financial packageIntroductory group theoryClosed-form expressions for perpetual and finite-maturity American binary optionsFast Fourier Transform with examples in GNU Octave/MATLABPagesHomeSelected publicationsWelcomeTagsMarkov villenage (1)Optimal stopping (3)GNU Octave (2)Notes (2)Mathematical finance (1) RSS | Design: HTML5 UP Please enable JavaScript to view the comments powered by Disqus.