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Lemma 8.1.4. Let \( g \) be a lower bounded, measurable function on \( {\mathbf{R}}^{n} \) and define, for fixed \( t \geq 0 \)\n\n\[ u\left( x\right) = {E}^{x}\left\lbrack {g\left( {X}_{t}\right) }\right\rbrack \]\n\na) If \( g \) is lower semicontinuous, then \( u \) is lower semicontinuous.\n\nb) If \( g \) is bound... | Proof. By (5.2.10) we have\n\n\[ E\left\lbrack {\left| {X}_{t}^{x} - {X}_{t}^{y}\right| }^{2}\right\rbrack \leq {\left| y - x\right| }^{2}C\left( t\right) ,\]\n\nwhere \( C\left( t\right) \) does not depend on \( x \) and \( y \) . Let \( \left\{ {y}_{n}\right\} \) be a sequence of points converging to \( x \) . Then\n... | Yes |
Theorem 8.1.5. a) If \( f \in {C}_{0}^{2}\left( {\mathbf{R}}^{n}\right) \) then \( {R}_{\alpha }\left( {\alpha - A}\right) f = f \) for all \( \alpha > 0 \) . | a) If \( f \in {C}_{0}^{2}\left( {\mathbf{R}}^{n}\right) \) then by Dynkin’s formula\n\n\[ \n{R}_{\alpha }\left( {\alpha - A}\right) f\left( x\right) = \left( {\alpha {R}_{\alpha }f - {R}_{\alpha }{Af}}\right) \left( x\right) \n\] \n\n\[ \n= \alpha {\int }_{0}^{\infty }{e}^{-{\alpha t}}{E}^{x}\left\lbrack {f\left( {X}_... | Yes |
Theorem 8.2.1 (The Feynman-Kac formula). Let \( f \in {C}_{0}^{2}\left( {\mathbf{R}}^{n}\right) \) and \( q \in C\left( {\mathbf{R}}^{n}\right) \) . Assume that \( q \) is lower bounded. a) \( {Put} \)\n\n\[ v\left( {t, x}\right) = {E}^{x}\left\lbrack {\exp \left( {-{\int }_{0}^{t}q\left( {X}_{s}\right) {ds}}\right) f\... | Proof. a) Let \( {Y}_{t} = f\left( {X}_{t}\right) ,{Z}_{t} = \exp \left( {-{\int }_{0}^{t}q\left( {X}_{s}\right) {ds}}\right) \) . Then \( d{Y}_{t} \) is given by\n\n(7.3.1) and\n\n\[ d{Z}_{t} = - {Z}_{t}q\left( {X}_{t}\right) {dt} \]\n\nSo\n\n\[ d\left( {{Y}_{t}{Z}_{t}}\right) = {Y}_{t}d{Z}_{t} + {Z}_{t}d{Y}_{t},\;\te... | Yes |
Theorem 8.3.1. If \( {X}_{t} \) is an Itô diffusion in \( {\mathbf{R}}^{n} \) with generator \( A \), then for all \( f \in {C}_{0}^{2}\left( {\mathbf{R}}^{n}\right) \) the process | \[ {M}_{t} = f\left( {X}_{t}\right) - {\int }_{0}^{t}{Af}\left( {X}_{r}\right) {dr} \] is a martingale w.r.t. \( \left\{ {\mathcal{M}}_{t}\right\} \) . | Yes |
However, as it stands this is not a stochastic differential equation of the form (5.2.3), so it is not apparent from (8.4.1) that \( R \) is an Itô diffusion. But this will follow if we can show that\n\n\[ {Y}_{t} \mathrel{\text{:=}} {\int }_{0}^{t}\mathop{\sum }\limits_{{i = 1}}^{n}\frac{{B}_{i}}{\left| B\right| }d{B}... | For then (8.4.1) can be written\n\n\[ d{R}_{t} = \frac{n - 1}{2{R}_{t}}{dt} + d\widetilde{B} \]\n\nwhich is of the form (5.2.3), thus showing by weak uniqueness (Lemma 5.3.1) that \( {R}_{t} \) is an Itô diffusion with generator\n\n\[ {Af}\left( x\right) = \frac{1}{2}{f}^{\prime \prime }\left( x\right) + \frac{n - 1}{2... | No |
An Itô process\n\n\\[ d{Y}_{t} = {vd}{B}_{t};\\;{Y}_{0} = 0\\text{ with }v\\left( {t,\\omega }\\right) \\in {\\mathcal{V}}_{\\mathcal{H}}^{n \\times m} \\]\n\ncoincides (in law) with n-dimensional Brownian motion if and only if\n\n\\[ v{v}^{T}\\left( {t,\\omega }\\right) = {I}_{n}\\;\\text{ for a.a. }\\left( {t,\\omega... | Note that in the example above we have\n\n\\[ {Y}_{t} = {\\int }_{0}^{t}{vdB} \\]\n\nwith\n\n\\[ v = \\left\\lbrack {\\frac{{B}_{1}}{\\left| B\\right| },\\ldots ,\\frac{{B}_{n}}{\\left| B\\right| }}\\right\\rbrack ,\\;B = \\left( \\begin{matrix} {B}_{1} \\\\ \\vdots \\\\ {B}_{n} \\end{matrix}\\right) \\]\n\nand since \... | No |
Lemma 8.4.4. Let \( d{Y}_{t} = u\left( {t,\omega }\right) {dt} + v\left( {t,\omega }\right) d{B}_{t},{Y}_{0} = x \) be as in Theorem 8.4.3. Then there exists an \( {\mathcal{N}}_{t} \) -adapted process \( W\left( {t,\omega }\right) \) such that\n\n\[ v{v}^{T}\left( {t,\omega }\right) = W\left( {t,\omega }\right) \;\tex... | Proof. By Itô’s formula we have (if \( {Y}_{i}\left( {t,\omega }\right) \) denotes component number \( i \) of \( Y\left( {t,\omega }\right) )\n\n\[ {Y}_{i}{Y}_{j}\left( {t,\omega }\right) = {x}_{i}{x}_{j} + {\int }_{0}^{t}{Y}_{i}d{Y}_{j}\left( s\right) + {\int }_{0}^{t}{Y}_{j}d{Y}_{i}\left( s\right) + {\int }_{0}^{t}{... | Yes |
Corollary 8.5.4. Let \( {Y}_{t},{\beta }_{s} \) be as in Corollary 8.5.3. Assume that\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{n}{v}_{i}^{2}\left( {r,\omega }\right) > 0\;\text{ for a.a. }\left( {r,\omega }\right) . \]\n\nThen there exists a Brownian motion \( {\widehat{B}}_{t} \) such that\n\n\[ {Y}_{t} = {\widehat{B}}_... | Proof. Let\n\n\[ {\widehat{B}}_{t} = {Y}_{{\alpha }_{t}} \]\n\nbe the Brownian motion from Corollary 8.5.3. By (8.5.9) \( {\beta }_{t} \) is strictly increasing and hence (8.5.4) holds, So choosing \( t = {\beta }_{s} \) in (8.5.11) we get (8.5.10). | Yes |
Lemma 8.5.6. Suppose \( s \rightarrow \alpha \left( {s,\omega }\right) \) is continuous, \( \alpha \left( {0,\omega }\right) = 0 \) for a.a. \( \omega \) . Fix \( t > 0 \) such that \( {\beta }_{t} < \infty \) a.s. and assume that \( E\left\lbrack {\alpha }_{t}\right\rbrack < \infty \) . For \( k = 1,2,\ldots \) \( {pu... | Proof. For all \( k \) we have \[ E\left\lbrack {\left( \mathop{\sum }\limits_{j}f\left( {\alpha }_{j},\omega \right) \Delta {B}_{{\alpha }_{j}} - {\int }_{0}^{{\alpha }_{t}}f\left( s,\omega \right) d{B}_{s}\right) }^{2}\right\rbrack \] \[ = E\left\lbrack {\left( \mathop{\sum }\limits_{j}{\int }_{{\alpha }_{j}}^{{\alph... | Yes |
Theorem 8.5.7 (Time change formula for Itô integrals).\n\nSuppose \( c\left( {s,\omega }\right) \) and \( \alpha \left( {s,\omega }\right) \) are \( s \) -continuous, \( \alpha \left( {0,\omega }\right) = 0 \) for a.a. \( \omega \) and that \( E\left\lbrack {\alpha }_{t}\right\rbrack < \infty \) . Let \( {B}_{s} \) be ... | Proof. The existence of the limit in (8.5.13) and the second identity in (8.5.13) follow by applying Lemma 8.5.6 to the function\n\n\[ \nf\left( {s,\omega }\right) = \sqrt{c\left( {s,\omega }\right) }.\n\]\n\nThen by Corollary 8.5.5 we have that \( {\widetilde{B}}_{t} \) is an \( {\mathcal{F}}_{{\alpha }_{t}}^{\left( m... | Yes |
Example 8.5.8 (Brownian motion on the unit sphere in \( {\mathbf{R}}^{n} \) ; \( n > 2 \) ).\n\nIn Examples 5.1.4 and 7.5.5 we constructed Brownian motion on the unit circle. It is not obvious how to extend the method used there to obtain Brownian motion on the unit sphere \( S \) of \( {\mathbf{R}}^{n};n \geq 3 \) . H... | Now perform the following time change: Define\n\n\[ {Z}_{t}\left( \omega \right) = {Y}_{\alpha \left( {t,\omega }\right) }\left( \omega \right) \]\n\nwhere\n\n\[ {\alpha }_{t} = {\beta }_{t}^{-1},\;\beta \left( {t,\omega }\right) = {\int }_{0}^{t}\frac{1}{{\left| B\right| }^{2}}{ds}. \]\n\nThen \( Z \) is again an Itô ... | Yes |
Let \( \mu \) and \( \nu \) be two probability measures on a measurable space \( \left( {\Omega ,\mathcal{G}}\right) \) such that \( {d\nu }\left( \omega \right) = f\left( \omega \right) {d\mu }\left( \omega \right) \) for some \( f \in {L}^{1}\left( \mu \right) \) . Let \( X \) be a random variable on \( \left( {\Omeg... | Proof. By the definition of conditional expectation (Appendix B) we have that if \( H \in \mathcal{H} \) then\n\n\[ \n{\int }_{H}{E}_{\nu }\left\lbrack {X \mid \mathcal{H}}\right\rbrack {fd\mu } = {\int }_{H}{E}_{\nu }\left\lbrack {X \mid \mathcal{H}}\right\rbrack {d\nu } = {\int }_{H}{Xd\nu }\n\]\n\n\[ \n= {\int }_{H}... | Yes |
Theorem 8.6.3 (The Girsanov theorem I).\n\nLet \( Y\left( t\right) \in {\mathbf{R}}^{n} \) be an Itô process of the form\n\n\[ \n{dY}\left( t\right) = a\left( {t,\omega }\right) {dt} + {dB}\left( t\right) ;\;t \leq T,{Y}_{0} = 0.\n\]\n\nwhere \( T \leq \infty \) is a given constant and \( B\left( t\right) \) is n-dimen... | Proof of Theorem 8.6.3. For simplicity we assume that \( a\left( {s,\omega }\right) \) is bounded. In view of Theorem 8.6.1 we have to verify that\n\n\[ \n\text{(i)}\;Y\left( t\right) = \left( {{Y}_{1}\left( t\right) ,\ldots ,{Y}_{n}\left( t\right) }\right) \text{is a martingale w.r.t.}Q\n\]\n\n\( \left( {8.6.10}\right... | Yes |
Theorem 8.6.4 (The Girsanov theorem II).\n\nLet \( Y\left( t\right) \in {\mathbf{R}}^{n} \) be an Itô process of the form\n\n\[ \n{dY}\left( t\right) = \beta \left( {t,\omega }\right) {dt} + \theta \left( {t,\omega }\right) {dB}\left( t\right) ;\;t \leq T \n\]\n\n\( \left( {8.6.16}\right) \)\n\nwhere \( B\left( t\right... | Proof. It follows from Theorem 8.6.3 that \( \widehat{B}\left( t\right) \) is a Brownian motion w.r.t. Q. So, substituting (8.6.21) in (8.6.16) we get, by (8.6.17),\n\n\[ \n{dY}\left( t\right) = \beta \left( {t,\omega }\right) {dt} + \theta \left( {t,\omega }\right) \left( {d\widehat{B}\left( t\right) - u\left( {t,\ome... | Yes |
Theorem 8.6.5 (The Girsanov theorem III).\n\nLet \( X\left( t\right) = {X}^{x}\left( t\right) \in {\mathbf{R}}^{n} \) and \( Y\left( t\right) = {Y}^{x}\left( t\right) \in {\mathbf{R}}^{n} \) be an Itô diffusion and an Itô process, respectively, of the forms\n\n\[ \n{dX}\left( t\right) = b\left( {X\left( t\right) }\righ... | Proof. The representation (8.6.28) follows by applying Theorem 8.6.4 to the case \( \theta \left( {t,\omega }\right) = \sigma \left( {Y\left( t\right) }\right) ,\beta \left( {t,\omega }\right) = \gamma \left( {t,\omega }\right) + b\left( {Y\left( t\right) }\right) ,\alpha \left( {t,\omega }\right) = b\left( {Y\left( t\... | Yes |
Let \( a : {\mathbf{R}}^{n} \rightarrow {\mathbf{R}}^{n} \) be a bounded, measurable function. Then we can construct a weak solution \( {X}_{t} = {X}_{t}^{x} \) of the stochastic differential equation\n\n\[ d{X}_{t} = a\left( {X}_{t}\right) {dt} + d{B}_{t};\;{X}_{0} = x \in {\mathbf{R}}^{n}. \] | We proceed according to the procedure above, with \( \sigma = I, b = 0 \) and\n\n\[ d{Y}_{t} = d{B}_{t};\;{Y}_{0} = x. \]\n\nChoose\n\n\[ {u}_{0} = {\sigma }^{-1} \cdot \left( {b - a}\right) = - a \]\n\nand define\n\n\[ {M}_{t} = \exp \left\{ {-{\int }_{0}^{t}{u}_{0}\left( {Y}_{s}\right) d{B}_{s} - \frac{1}{2}{\int }_{... | Yes |
Theorem 9.1.1 (Uniqueness theorem (1)).\n\nSuppose \( \phi \) is bounded and \( g \) satisfies (9.1.7). Suppose \( w \in {C}^{2}\left( D\right) \) is bounded and satisfies\n\n\( \left( {9.1.8}\right) \)\n\nand\n\n(ii)’ \( \;\mathop{\lim }\limits_{{t \uparrow {\tau }_{D}}}w\left( {X}_{t}\right) = \phi \left( {X}_{{\tau ... | Proof. Let \( {\left\{ {D}_{k}\right\} }_{k = 1}^{\infty } \) be an increasing sequence of open sets \( {D}_{k} \) such that \( {D}_{k} \subset \subset D \) and \( D = \mathop{\bigcup }\limits_{{k = 1}}^{\infty }{D}_{k} \) . Define\n\n\[ {\alpha }_{k} = k \land {\tau }_{{D}_{k}};\;k = 1,2,\ldots \]\n\nThen by the Dynki... | Yes |
a) Let \( f \) be \( X \) -harmonic in \( D \) . Then \( \mathcal{A}f = 0 \) in \( D \) .\nb) Conversely, suppose \( f \in {C}^{2}\left( D\right) \) and \( \mathcal{A}f = 0 \) in \( D \) . Then \( f \) is \( X \) -harmonic. | a) follows directly from the formula for \( \mathcal{A} \) .\nb) follows from the Dynkin formula: Choose \( U \) as in Definition 9.2.2. Then\n\n\[
{E}^{x}\left\lbrack {f\left( {X}_{{\tau }_{U}}\right) }\right\rbrack = \mathop{\lim }\limits_{{k \rightarrow \infty }}{E}^{x}\left\lbrack {f\left( {X}_{{\tau }_{U} \land k}... | Yes |
Lemma 9.2.4. Let \( \\phi \) be a bounded measurable function on \( \\partial D \) and put\n\n\[ \nu\\left( x\\right) = {E}^{x}\\left\\lbrack {\\phi \\left( {X}_{{\\tau }_{D}}\\right) }\\right\\rbrack ;\\;x \\in D.\n\]\n\nThen \( u \) is \( X \) -harmonic. Thus, in particular, \( \\mathcal{A}u = 0 \) . | Proof. From the mean value property (7.2.9) we have, if \( \\bar{V} \\subset D \)\n\n\[ u\\left( x\\right) = {\\int }_{\\partial V}u\\left( y\\right) {Q}^{x}\\left\\lbrack {{X}_{{\\tau }_{V}} \\in {dy}}\\right\\rbrack = {E}^{x}\\left\\lbrack {u\\left( {X}_{{\\tau }_{V}}\\right) }\\right\\rbrack .\n\]\n\n\( ▱ \) | Yes |
Theorem 9.2.5 (Solution of the stochastic Dirichlet problem). Let \( \phi \) be a bounded measurable function on \( \partial D \) . a) (Existence) Define \[ u\left( x\right) = {E}^{x}\left\lbrack {\phi \left( {X}_{{\tau }_{D}}\right) }\right\rbrack . \] Then \( u \) solves the stochastic Dirichlet problem (9.2.6),(9.2.... | Proof. a) It follows from Lemma 9.2.4 that \( {\left( i\right) }_{s} \) holds. Fix \( x \in D \) . Let \( \left\{ {D}_{k}\right\} \) be an increasing sequence of open sets such that \( {D}_{k} \subset \subset D \) and \( D = \mathop{\bigcup }\limits_{k}{D}_{k} \) . Put \( {\tau }_{k} = {\tau }_{{D}_{k}},\tau = {\tau }_... | Yes |
Lemma 9.2.6 (The \( 0 - 1 \) law). Let \( H \in \mathop{\bigcap }\limits_{{t > 0}}{\mathcal{M}}_{t} \) . Then either \( {Q}^{x}\left( H\right) = 0 \) or \( {Q}^{x}\left( H\right) = 1 \) . | Proof. From the strong Markov property (7.2.5) we have\n\n\[ \n{E}^{x}\left\lbrack {{\theta }_{t}\eta \mid {\mathcal{M}}_{t}}\right\rbrack = {E}^{{X}_{t}}\left\lbrack \eta \right\rbrack \n\] \n\nfor all bounded, \( {\mathcal{M}}_{\infty } \) -measurable \( \eta : \Omega \rightarrow \mathbf{R} \) . This implies that \n\... | Yes |
Corollary 9.2.7. Let \( y \in {\mathbf{R}}^{n} \) . Then\n\n\[ \text{either}{Q}^{y}\left\lbrack {{\tau }_{D} = 0}\right\rbrack = 0\;\text{or}\;{Q}^{y}\left\lbrack {{\tau }_{D} = 0}\right\rbrack = 1\text{.} \] | Proof. \( H = \left\{ {\omega ;{\tau }_{D} = 0}\right\} \in \mathop{\bigcap }\limits_{{t > 0}}{\mathcal{M}}_{t} \) | No |
Example 9.2.9. Corollary 9.2.7 may seem hard to believe at first glance. For example, if \( {X}_{t} \) is a 2-dimensional Brownian motion \( {B}_{t} \) and \( \bar{D} \) is the square \( \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \) one might think that, starting from \( \left( {\frac{1}{2},... | Symmetry considerations imply that the first alternative is impossible. Thus \( \left( {\frac{1}{2},0}\right) \), and similarly all the other points of \( \partial D \), are regular for \( D \) w.r.t. \( {B}_{t} \) . | Yes |
Let \( D = \left\lbrack {0,1}\right\rbrack \times \left\lbrack {0,1}\right\rbrack \) and let \( L \) be the parabolic differential operator \[ {Lf}\left( {t, x}\right) = \frac{\partial f}{\partial t} + \frac{1}{2} \cdot \frac{{\partial }^{2}f}{\partial {x}^{2}};\;\left( {t, x}\right) \in {\mathbf{R}}^{2}. \] | Here \[ b = \left( \begin{array}{l} 1 \\ 0 \end{array}\right) \;\text{ and }\;a = \left\lbrack {a}_{ij}\right\rbrack = \frac{1}{2}\left( \begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right) . \] So, for example, if we choose \( \sigma = \left( \begin{array}{ll} 0 & 0 \\ 1 & 0 \end{array}\right) \), we have \( \frac{1}{2... | Yes |
Let \( \Delta = \left\{ {\left( {x, y}\right) ;{x}^{2} + {y}^{2} < 1}\right\} \subset {\mathbf{R}}^{2} \) and let \( \left\{ {\Delta }_{n}\right\} \) be a sequence of disjoint open discs in \( \Delta \) centered at \( \left( {{2}^{-n},0}\right) \), respectively, \( n = 1,2,\ldots \) . Put \[ D = \Delta \smallsetminus \... | This is a consequence of the famous Wiener criterion. See Port and Stone (1979), p. 225. | No |
Theorem 9.2.13. Suppose \( {X}_{t} \) satisfies Hunt’s condition (H). Let \( \phi \) be a bounded continuous function on \( \partial D \) . Suppose there exists a bounded \( u \in \) \( {C}^{2}\left( D\right) \) such that\n\n(i) \( {Lu} = 0 \) in \( D \)\n\n(ii) \( \mathop{\lim }\limits_{\substack{{x \rightarrow y} \\ ... | Proof. Let \( \left\{ {D}_{k}\right\} \) be as in the proof Theorem 9.1.1. By Lemma 9.2.3 b) \( u \) is \( X \) -harmonic and therefore\n\n\[ u\left( x\right) = {E}^{x}\left\lbrack {u\left( {X}_{{\tau }_{k}}\right) }\right\rbrack \;\text{ for all }x \in {D}_{k}\text{ and all }k. \]\n\nIf \( k \rightarrow \infty \) then... | Yes |
This example shows that it need not hold even when \( L \) is elliptic: Consider Example 9.2.11 again, in the case when the point 0 is not regular. Choose \( \phi \in C\left( {\partial D}\right) \) such that \[ \phi \left( 0\right) = 1,0 \leq \phi \left( y\right) < 1\;\text{ for }y \in \partial D \smallsetminus \{ 0\} ... | Since \( \{ 0\} \) is polar for \( {B}_{t} \) (see Exercise 9.7 a) we have \( {B}_{{\tau }_{D}}^{0} \neq 0 \) a.s and therefore \[ u\left( 0\right) = {E}^{0}\left\lbrack {\phi \left( {B}_{{\tau }_{D}}\right) }\right\rbrack < 1. \] By a slight extension of the mean value property (7.2.9) (see Exercise 9.4) we get \[ {E}... | Yes |
Assume that \( \phi \) is a bounded continuous function on \( \partial D = \{ \left( {t, R}\right) ;t \in \mathbf{R}\} \) . Then by Theorem 9.2.5 the function \( u\left( {s, x}\right) = {E}^{s, x}\left\lbrack {\phi \left( {X}_{{\tau }_{D}}\right) }\right\rbrack \) is the solution of the stochastic Dirichlet problem (9.... | Using the Laplace transform it is possible to find the distribution of the first exit point on \( \partial D \) for \( X \), i.e. to find the distribution of the first time \( t = \widehat{\tau } \) that \( {B}_{t} \) reaches the value \( R \) . (See Karlin and Taylor (1975), p. 363. See also Exercise 7.19.) The result... | Yes |
Theorem 9.3.1 (Solution of the stochastic Poisson problem). Assume that\n\n\[ \n{E}^{x}\left\lbrack {{\int }_{0}^{{\tau }_{D}}\left| {g\left( {X}_{s}\right) }\right| {ds}}\right\rbrack < \infty \;\text{ for all }x \in D.\n\]\n\n(This occurs, for example, if \( g \) is bounded and \( {E}^{x}\left\lbrack {\tau }_{D}\righ... | Proof. Choose \( U \) open, \( x \in U \subset \subset D \) . Put \( \eta = {\int }_{0}^{{\tau }_{D}}g\left( {X}_{s}\right) {ds},\tau = {\tau }_{U} \).\n\nThen by the strong Markov property (7.2.5)\n\n\[ \n\frac{{E}^{x}\left\lbrack {v\left( {X}_{\tau }\right) }\right\rbrack - v\left( x\right) }{{E}^{x}\left\lbrack \tau... | Yes |
Theorem 9.3.2 (Uniqueness theorem for the Poisson equation).\n\nAssume that \( {X}_{t} \) satisfies Hunt’s condition (H) ((9.2.15)). Assume that (9.3.3) holds and that there exists a function \( v \in {C}^{2}\left( D\right) \) and a constant \( C \) such that\n\n\[ \left| {v\left( x\right) }\right| \leq C\left( {1 + {E... | Proof. Let \( {D}_{k},{\tau }_{k} \) be as in the proof of Theorem 9.2.5. Then by Dynkin’s formula\n\n\[ {E}^{x}\left\lbrack {v\left( {X}_{{\tau }_{k}}\right) }\right\rbrack - v\left( x\right) = {E}^{x}\left\lbrack {{\int }_{0}^{{\tau }_{k}}\left( {Lv}\right) \left( {X}_{s}\right) {ds}}\right\rbrack = - {E}^{x}\left\lb... | Yes |
Corollary 9.3.5 (The Green formula). Let \( {E}^{x}\left\lbrack {\tau }_{D}\right\rbrack < \infty \) for all \( x \in D \) and assume that \( f \in {C}_{0}^{2}\left( {\mathbf{R}}^{n}\right) \) . Then\n\n\[ f\left( x\right) = {E}^{x}\left\lbrack {f\left( {X}_{{\tau }_{D}}\right) }\right\rbrack - {\int }_{D}\left( {{L}_{... | Proof. By Dynkin's formula and (9.3.24) we have\n\n\[ {E}^{x}\left\lbrack {f\left( {X}_{{\tau }_{D}}\right) }\right\rbrack = f\left( x\right) + {E}^{x}\left\lbrack {{\int }_{0}^{{\tau }_{D}}\left( {{L}_{X}f}\right) \left( {X}_{s}\right) {ds}}\right\rbrack = f\left( x\right) + {\int }_{D}\left( {{L}_{X}f}\right) \left( ... | Yes |
If \( {X}_{t} = {B}_{t} \) is 1-dimensional Brownian motion in a bounded interval \( \left( {a, b}\right) \subset \mathbf{R} \) then we can compute the Green function \( G\left( {x, y}\right) \) explicitly. | To this end, choose a bounded continuous function \( g : \left( {a, b}\right) \rightarrow \mathbf{R} \) and let us compute\n\n\[ v\left( x\right) \mathrel{\text{:=}} {E}^{x}\left\lbrack {{\int }_{0}^{{\tau }_{D}}g\left( {B}_{t}\right) {dt}}\right\rbrack .\n\]\n\nBy Corollary 9.1.2 we know that \( v \) is the solution o... | Yes |
a) If \( f \) is superharmonic (supermeanvalued) and \( \alpha > 0 \) , then \( {\alpha f} \) is superharmonic (supermeanvalued). | a) and b) are straightforward. | No |
Theorem 10.1.6. Let \( f : {\mathbf{R}}^{n} \rightarrow \left\lbrack {0,\infty }\right\rbrack \) . Then \( f \) is excessive w.r.t. \( {X}_{t} \) if and only if \( f \) is superharmonic w.r.t. \( {X}_{t} \) . | Proof in a special case. Let \( L \) be the differential operator associated to \( X \) (given by the right hand side of (7.3.3)), so that \( L \) coincides with the generator \( A \) of \( X \) on \( {C}_{0}^{2} \) . We only prove the theorem in the special case when \( f \in {C}^{2}\left( {\mathbf{R}}^{n}\right) \) a... | Yes |
Theorem 10.1.7. (Construction of the least superharmonic majorant).\n\nLet \( g = {g}_{0} \) be a nonnegative, lower semicontinuous function on \( {\mathbf{R}}^{n} \) and define inductively\n\n\[ \n{g}_{n}\left( x\right) = \mathop{\sup }\limits_{{t \in {S}_{n}}}{E}^{x}\left\lbrack {{g}_{n - 1}\left( {X}_{t}\right) }\ri... | Proof. Note that \( \left\{ {g}_{n}\right\} \) is increasing. Define \( \check{g}\left( x\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}{g}_{n}\left( x\right) \) . Then\n\n\[ \n\check{g}\left( x\right) \geq {g}_{n}\left( x\right) \geq {E}^{x}\left\lbrack {{g}_{n - 1}\left( {X}_{t}\right) }\right\rbrack \;\tex... | Yes |
Corollary 10.1.8. Define \( {h}_{0} = g \) and inductively\n\n\[ \n{h}_{n}\left( x\right) = \mathop{\sup }\limits_{{t \geq 0}}{E}^{x}\left\lbrack {{h}_{n - 1}\left( {X}_{t}\right) }\right\rbrack ;\;n = 1,2,\ldots \n\]\n\nThen \( {h}_{n} \uparrow \widehat{g} \) . | Proof. Let \( h = \lim {h}_{n} \) . Then clearly \( h \geq \check{g} = \widehat{g} \) . On the other hand, since \( \widehat{g} \) is excessive we have\n\n\[ \n\widehat{g}\left( x\right) \geq \mathop{\sup }\limits_{{t \geq 0}}{E}^{x}\left\lbrack {\widehat{g}\left( {X}_{t}\right) }\right\rbrack \n\]\n\nSo by induction\n... | Yes |
Theorem 10.1.9 (Existence theorem for optimal stopping).\n\nLet \( {g}^{ * } \) denote the optimal reward and \( \widehat{g} \) the least superharmonic majorant of a continuous reward function \( g \geq 0 \) .\n\na) Then\n\n\[ \n{g}^{ * }\left( x\right) = \widehat{g}\left( x\right) .\n\]\n\n\( \left( {10.1.17}\right) \... | Proof. First assume that \( g \) is bounded and define\n\n\[ \n{\widetilde{g}}_{\epsilon }\left( x\right) = {E}^{x}\left\lbrack {\widehat{g}\left( {X}_{{\tau }_{\epsilon }}\right) }\right\rbrack \;\text{ for }\epsilon > 0.\n\]\n\n\( \left( {10.1.22}\right) \)\n\nThen \( {\widetilde{g}}_{\epsilon } \) is supermeanvalued... | Yes |
Corollary 10.1.10. Suppose there exists a Borel set \( H \) such that\n\n\[ \n{\widetilde{g}}_{H}\left( x\right) \mathrel{\text{:=}} {E}^{x}\left\lbrack {g\left( {X}_{{\tau }_{H}}\right) }\right\rbrack \n\]\n\nis a supermeanvalued majorant of \( g \) . Then\n\n\[ \n{g}^{ * }\left( x\right) = {\widetilde{g}}_{H}\left( x... | Proof. If \( {\widetilde{g}}_{H} \) is a supermeanvalued majorant of \( g \) then clearly\n\n\[ \n\bar{g}\left( x\right) \leq {\widetilde{g}}_{H}\left( x\right) \n\]\n\nOn the other hand we of course have\n\n\[ \n{\widetilde{g}}_{H}\left( x\right) \leq \mathop{\sup }\limits_{\tau }{E}^{x}\left\lbrack {g\left( {X}_{\tau... | Yes |
Corollary 10.1.11. Let\n\n\[ D = \{ x;g\left( x\right) < \widehat{g}\left( x\right) \} \]\n\nand put\n\n\[ \widetilde{g}\left( x\right) = {\widetilde{g}}_{D}\left( x\right) = {E}^{x}\left\lbrack {g\left( {X}_{{\tau }_{D}}\right) }\right\rbrack . \]\n\nIf \( \widetilde{g} \geq g \) then \( \widetilde{g} = {g}^{ * } \) . | Proof. Since \( {X}_{{\tau }_{D}} \notin D \) we have \( g\left( {X}_{{\tau }_{D}}\right) \geq \widehat{g}\left( {X}_{{\tau }_{D}}\right) \) and therefore \( g\left( {X}_{{\tau }_{D}}\right) = \widehat{g}\left( {X}_{{\tau }_{D}}\right) \), a.s. \( {Q}^{x} \). So \( \widetilde{g}\left( x\right) = {E}^{x}\left\lbrack {\w... | Yes |
Let \( {X}_{t} = {B}_{t} \) be a Brownian motion in \( {\mathbf{R}}^{2} \). Using that \( {B}_{t} \) is recurrent in \( {\mathbf{R}}^{2} \) (Example 7.4.2) one can show that the only (nonnegative) superharmonic functions in \( {\mathbf{R}}^{2} \) are the constants (Exercise 10.2). Therefore | \[ {g}^{ * }\left( x\right) = \parallel g{\parallel }_{\infty } \mathrel{\text{:=}} \sup \left\{ {g\left( y\right) ;y \in {\mathbf{R}}^{2}}\right\} \;\text{ for all }x. \] So if \( g \) is unbounded then \( {g}^{ * } = \infty \) and no optimal stopping time exists. Assume therefore that \( g \) is bounded. The continua... | Yes |
The situation is different in \( {\mathbf{R}}^{n} \) for \( n \geq 3 \). a) To illustrate this let \( {X}_{t} = {B}_{t} \) be Brownian motion in \( {\mathbf{R}}^{3} \) and let the reward function be \[ g\left( \xi \right) = \left\{ {\begin{array}{ll} {\left| \xi \right| }^{-1} & \text{ for }\left| \xi \right| \geq 1 \\... | b) Let us change \( g \) to \[ h\left( x\right) = \left\{ \begin{array}{ll} {\left| x\right| }^{-\alpha } & \text{ for }\left| x\right| \geq 1 \\ 1 & \text{ for }\left| x\right| < 1 \end{array}\right. \] for some \( \alpha > 1 \) . Let \( H = \{ x;\left| x\right| > 1\} \) and define \[ \widetilde{h}\left( x\right) = {E... | Yes |
Let \( {X}_{t} = {B}_{t} \) be 1-dimensional Brownian motion and let the reward function be\n\n\[ g\left( {t,\xi }\right) = {e}^{-{\alpha t} + {\beta \xi }};\;\xi \in \mathbf{R} \]\n\nwhere \( \alpha ,\beta \geq 0 \) are constants. The characteristic operator \( \widehat{\mathcal{A}} \) of \( {Y}_{t}^{s, x} = \left\lbr... | Thus\n\n\[ \mathcal{A}g = \left( {-\alpha + \frac{1}{2}{\beta }^{2}}\right) g \]\n\nso if \( {\beta }^{2} \leq {2\alpha } \) then \( {g}^{ * } = g \) and the best policy is to stop immediately. If \( {\beta }^{2} > {2\alpha } \) we have\n\n\[ U \mathrel{\text{:=}} \{ \left( {s, x}\right) ;\widehat{\mathcal{A}}g\left( {... | Yes |
Example 10.2.2. (When is the right time to sell the stocks?) We now return to a specified version of Problem 5 in the introduction:\n\nSuppose the price \( {X}_{t} \) at time \( t \) of a person’s assets (e.g. a house, stocks, oil ...) varies according to a stochastic differential equation of the form\n\n\[ d{X}_{t} = ... | The characteristic operator \( \widehat{\mathcal{A}} \) of the process \( {Y}_{t} = \left( {s + t,{X}_{t}}\right) \) is given by\n\n\[ \widehat{\mathcal{A}}f\left( {s, x\right) = \frac{\partial f}{\partial s} + {rx}\frac{\partial f}{\partial x} + \frac{1}{2}{\alpha }^{2}{x}^{2}\frac{{\partial }^{2}f}{\partial {x}^{2}};... | No |
Consider the optimal stopping problem\n\n\\[ \n\\gamma \\left( x\\right) = \\mathop{\\sup }\\limits_{\\tau }\\mathbb{E}^{x}\\left\\lbrack {{\\int }_{0}^{\\tau }\\theta {e}^{-{\\rho t}}{X}_{t}{dt} + {e}^{-{\\rho \\tau }}{X}_{\\tau }}\\right\\rbrack \n\\]\n\nwhere\n\n\\[ \nd{X}_{t} = \\alpha {X}_{t}{dt} + \\beta {X}_{t}d... | Then with\n\n\\[ \nf\\left( y\\right) = f\\left( {s, x}\\right) = \\theta {e}^{-{\\rho s}}x,\\;g\\left( y\\right) = {e}^{-{\\rho s}}x \n\\]\n\nand\n\n\\[ \nG\\left( {s, x, w}\\right) = g\\left( {s, x}\\right) + w = {e}^{-{\\rho s}}x + w \n\\]\n\nwe have\n\n\\[ \n{\\mathcal{A}}_{Z}G = \\frac{\\partial G}{\\partial s} + ... | No |
Theorem 10.4.1 (Variational inequalities for optimal stopping).\n\nSuppose we can find a function \( \phi : \bar{V} \rightarrow \mathbf{R} \) such that\n\n(i) \( \;\phi \in {C}^{1}\left( V\right) \cap C\left( \bar{V}\right) \)\n\n(ii) \( \phi \geq g \) on \( V \) and \( \phi = g \) on \( \partial V \) .\n\nDefine\n\n\[... | Proof. By (i), (iv) and (v) we can find a sequence of functions \( {\phi }_{j} \in {C}^{2}\left( V\right) \cap C\left( \bar{V}\right), j = 1,2,\ldots \), such that\n\n(a) \( {\phi }_{j} \rightarrow \phi \) uniformly on compact subsets of \( \bar{V} \), as \( j \rightarrow \infty \)\n\n(b) \( L{\phi }_{j} \rightarrow {L... | Yes |
To illustrate Theorem 10.4.1 let us apply it to reconsider Example 10.2.2: Rather than proving (10.2.8) and the following properties of \( D \), we now simply guess/assume that \( D \) has the form \[ D = \left\{ {\left( {s, x}\right) ;0 < x < {x}_{0}}\right\} \] for some \( {x}_{0} > 0 \], which is intuitively reasona... | This is easily done by direct calculation (assuming \( r < \rho \) ). We conclude that \( \phi = {g}^{ * } \) and \( {\tau }^{ * } = {\tau }_{D} \) is optimal (with the value (10.2.13) for \( {x}_{0} \) ). | Yes |
Theorem 11.2.1 (The Hamilton-Jacobi-Bellman (HJB) equation (I)). Define\n\n\[ \Phi \left( y\right) = \sup \left\{ {{J}^{u}\left( y\right) ;u = u\left( Y\right) }\right. \text{Markov control}\} \text{.} \]\n\nSuppose that \( \Phi \in {C}^{2}\left( G\right) \cap C\left( \bar{G}\right) \) satisfies\n\n\[ {E}^{y}\left\lbra... | Proof. The last two statements are easy to prove: Since \( {u}^{ * } = {u}^{ * }\left( y\right) \) is optimal we have\n\n\[ \Phi \left( y\right) = {J}^{{u}^{ * }}\left( y\right) = {E}^{y}\left\lbrack {{\int }_{0}^{T}F\left( {{Y}_{s},{u}^{ * }\left( {Y}_{s}\right) }\right) {ds} + K\left( {Y}_{T}\right) }\right\rbrack . ... | Yes |
Theorem 11.2.2 (The HJB (II) equation - a converse of HJB (I)).\n\nLet \( \phi \) be a function in \( {C}^{2}\left( G\right) \cap C\left( \bar{G}\right) \) such that, for all \( v \in U \) ,\n\n\[ \n{F}^{v}\left( y\right) + \left( {{L}^{v}\phi }\right) \left( y\right) \leq 0;\;y \in G \n\]\n\n\( \left( {11.2.9}\right) ... | Proof. Assume that \( \phi \) satisfies (11.2.9) and (11.2.10) above. Let \( u \) be a Markov control. Since \( {L}^{u}\phi \leq - {F}^{u} \) in \( G \) we have by Dynkin’s formula\n\n\[ \n{E}^{y}\left\lbrack {\phi \left( {Y}_{{T}_{R}}\right) }\right\rbrack = \phi \left( y\right) + {E}^{y}\left\lbrack {{\int }_{0}^{{T}... | Yes |
Theorem 11.2.3. Let\n\n\[ \n{\Phi }_{M}\left( y\right) = \sup \left\{ {{J}^{u}\left( y\right) ;u = u\left( Y\right) }\right. \text{Markov control}\} \n\] \n\nand \n\n\[ \n{\Phi }_{a}\left( y\right) = \sup \left\{ {{J}^{u}\left( y\right) ;u = u\left( {t,\omega }\right) {\mathcal{F}}_{t}^{\left( m\right) }\text{-adapted ... | Proof. Let \( \phi \) be a function in \( {C}^{2}\left( G\right) \cap C\left( \bar{G}\right) \) satisfying (11.2.15) and \n\n\[ \n{F}^{v}\left( y\right) + \left( {{L}^{v}\phi }\right) \left( y\right) \leq 0\;\text{ for all }y \in G, v \in U \n\] \n\n(11.2.16) \n\nand \n\n\[ \n\phi \left( y\right) = K\left( y\right) \;\... | Yes |
Example 11.2.4 (The linear stochastic regulator problem).\n\nSuppose that the state \( {X}_{t} \) of the system at time \( t \) is given by a linear stochastic differential equation:\n\n\[ d{X}_{t} = \left( {{H}_{t}{X}_{t} + {M}_{t}{u}_{t}}\right) {dt} + {\sigma }_{t}d{B}_{t},\;t \geq s;\;{X}_{s} = x \]\n\n\( \left( {1... | In this case the HJB-equation for \( \Psi \left( {s, x}\right) = \mathop{\inf }\limits_{u}{J}^{u}\left( {s, x}\right) \) becomes\n\n\[ 0 = \mathop{\inf }\limits_{v}\left\{ {{F}^{v}\left( {s, x}\right) + \left( {{L}^{v}\Psi }\right) \left( {s, x}\right) }\right\} \]\n\n\[ = \frac{\partial \Psi }{\partial s} + \mathop{\i... | Yes |
Example 11.2.5 (An optimal portfolio selection problem). Let \( {X}_{t} \) denote the wealth of a person at time \( t \) . Suppose that the person has the choice of two different investments. The price \( {p}_{1}\left( t\right) \) at time \( t \) of one of the assets is assumed to satisfy the equation \[ \frac{d{p}_{1}... | If \( {\Phi }_{x} \mathrel{\text{:=}} \frac{\partial \Phi }{\partial x} > 0 \) and \( {\Phi }_{xx} \mathrel{\text{:=}} \frac{{\partial }^{2}\Phi }{\partial {x}^{2}} < 0 \), the solution is \[ v = u\left( {t, | No |
Suppose the system is a 1-dimensional Itô integral\n\n\[ d{X}_{t} = d{X}_{t}^{u} = u\left( {t,\omega }\right) d{B}_{t},\;t \geq s;{X}_{s} = x > 0 \]\n\n\( \left( {11.2.56}\right) \)\n\nand consider the stochastic control problem\n\n\[ \Phi \left( {t, x}\right) = \mathop{\sup }\limits_{u}{E}^{t, x}\left\lbrack {K\left( ... | Assuming that \( \Phi \in {C}^{2} \) and that \( {u}^{ * } \) exists we get by the HJB (I) equation\n\n\[ \mathop{\sup }\limits_{{v \in \mathbf{R}}}\left\{ {\frac{\partial \Phi }{\partial t} + \frac{1}{2}{v}^{2}\frac{{\partial }^{2}\Phi }{\partial {x}^{2}}}\right\} = 0\;\text{ for }t < {t}_{1},\Phi \left( {{t}_{1}, x}\... | Yes |
Theorem 11.3.1. Suppose that we for all \( \lambda \in \Lambda \subset {\mathbf{R}}^{l} \) can find \( {\Phi }_{\lambda }\left( y\right) \) and \( {u}_{\lambda }^{ * } \) solving the (unconstrained) stochastic control problem (11.3.5)-(11.3.6). Moreover, suppose that there exists \( {\lambda }_{0} \in \Lambda \) such t... | Proof. Let \( u \) be a Markov control, \( \lambda \in \Lambda \) . Then by the definition of \( {u}_{\lambda }^{ * } \) we have\n\n\[ \n{E}^{y}\left\lbrack {{\int }_{0}^{T}{F}^{{u}_{\lambda }^{ * }}\left( {Y}_{t}^{{u}_{\lambda }^{ * }}\right) {dt} + K\left( {Y}_{T}^{{u}_{\lambda }^{ * }}\right) + \lambda \cdot M\left(... | Yes |
Consider the following market \[ d{X}_{0}\left( t\right) = 0,\;d{X}_{1}\left( t\right) = {dB}\left( t\right) ,\;0 \leq t \leq T = 1. \] Let \[ Y\left( t\right) = {\int }_{0}^{t}\frac{{dB}\left( s\right) }{\sqrt{1 - s}}\;\text{ for }0 \leq t < 1. \] By Corollary 8.5 .5 there exists a Brownian motion \( \widehat{B}\left(... | This example illustrates that with portfolios only required to be self-financing and satisfy (12.1.5) one can virtually generate any terminal value \( V\left( {T,\omega }\right) \) from \( {V}_{0} = 0 \), even when the risky price process \( {X}_{1}\left( t\right) \) is Brownian motion. This clearly contradicts the rea... | No |
Theorem 12.1.5. Let \( F \) be an \( {\mathcal{F}}_{T}^{\left( m\right) } \) -measurable random variable and let \( B\left( t\right) \) be \( m \) -dimensional Brownian motion. Then there exists \( \phi \in {\mathcal{W}}^{m} \) such that\n\n\[ F\left( \omega \right) = {\int }_{0}^{T}\phi \left( {t,\omega }\right) {dB}\... | Note that \( \phi \) is not unique. See Exercise 3.4.22 in Karatzas and Shreve (1991). See also Exercise 12.4. | No |
Lemma 12.1.6. Suppose there exists a measure \( Q \) on \( {\mathcal{F}}_{T}^{\left( m\right) } \) such that \( P \sim Q \) and such that the normalized price process \( \{ \bar{X}\left( t\right) {\} }_{t \in \left\lbrack {0, T}\right\rbrack } \) is a local martingale w.r.t. \( Q \) . Then the market \( \{ X\left( t\ri... | Proof. Suppose \( \theta \left( t\right) \) is an arbitrage for \( \{ \bar{X}\left( t\right) {\} }_{t \in \left\lbrack {0, T}\right\rbrack } \) . Let \( {\bar{V}}^{\theta }\left( t\right) \) be the corresponding value process for the normalized market with \( {\bar{V}}^{\theta }\left( 0\right) = 0 \) . Then \( {\bar{V}... | No |
Theorem 12.1.8. a) Suppose there exists a process \( u\left( {t,\omega }\right) \in {\mathcal{V}}^{m}\left( {0, T}\right) \) such that, with \( \widehat{X}\left( {t,\omega }\right) = \left( {{X}_{1}\left( {t,\omega }\right) ,\ldots ,{X}_{n}\left( {t,\omega }\right) }\right) \), \[ \sigma \left( {t,\omega }\right) u\lef... | Proof. a) We may assume that \( \{ X\left( t\right) \} \) is normalized, i.e. that \( \rho = 0 \) (Exercise 12.1). Define the measure \( Q = {Q}_{u} \) on \( {\mathcal{F}}_{T}^{\left( m\right) } \) by \[ {dQ}\left( \omega \right) = \exp \left( {-{\int }_{0}^{T}u\left( {t,\omega }\right) {dB}\left( t\right) - \frac{1}{2... | No |
Consider the price process \( X\left( t\right) \) given by\n\n\[ d{X}_{0}\left( t\right) = 0,\;d{X}_{1}\left( t\right) = {2dt} + d{B}_{1}\left( t\right) ,\;d{X}_{2}\left( t\right) = - {dt} + d{B}_{1}\left( t\right) + d{B}_{2}\left( t\right) . \]\n\nIn this case we have\n\[ \mu = \left\lbrack \begin{matrix} 2 \\ - 1 \en... | From Theorem 12.1.8a) we conclude that \( X\left( t\right) \) has no arbitrage. | Yes |
Lemma 12.2.2. Let \( \bar{X}\left( t\right) = \xi \left( t\right) X\left( t\right) \) be the normalized price process, as in (12.1.8)-(12.1.11). Suppose \( \theta \left( t\right) \) is an admissible portfolio for the market \( \{ X\left( t\right) \} \) with value process\n\n\[ \n{V}^{\theta }\left( t\right) = \theta \l... | Proof. Note that \( {\bar{V}}^{\theta }\left( t\right) \) is lower bounded if and only if \( {V}^{\theta }\left( t\right) \) is lower bounded (since \( \rho \left( t\right) \) is bounded). Consider first the market consisting of the price process \( X\left( t\right) \) . Let \( \theta \left( t\right) \) be an admissibl... | Yes |
Lemma 12.2.3. Suppose there exists an \( m \) -dimensional process \( u\left( {t,\omega }\right) \in \) \( {\mathcal{V}}^{m}\left( {0, T}\right) \) such that, with \( \widehat{X}\left( {t,\omega }\right) = \left( {{X}_{1}\left( {t,\omega }\right) ,\ldots ,{X}_{n}\left( {t,\omega }\right) }\right) \), \[ \sigma \left( {... | Proof. The first statement follows from the Girsanov theorem. To prove the representation (12.2.15) we compute \[ d{\bar{X}}_{i}\left( t\right) = d\left( {\xi \left( t\right) {X}_{i}\left( t\right) }\right) = \xi \left( t\right) d{X}_{i}\left( t\right) + {X}_{i}\left( t\right) {d\xi }\left( t\right) \] \[ = \xi \left( ... | Yes |
Corollary 12.2.6. (a) If \( n = m \) then the market is complete if and only if \( \sigma \left( {t,\omega }\right) \) is invertible for a.a. \( \left( {t,\omega }\right) \) . | Proof. (a) is a direct consequence of Theorem 12.2.5, since the existence of a left inverse implies invertibility when \( n = m \) . | Yes |
Define \( {X}_{0}\left( t\right) \equiv 1 \) and\n\n\[ \n\\left\\lbrack \\begin{array}{l} d{X}_{1}\left( t\right) \\\\ d{X}_{2}\left( t\right) \\\\ d{X}_{3}\left( t\right) \\end{array}\\right\\rbrack = \\left\\lbrack \\begin{array}{l} 1 \\\\ 2 \\\\ 3 \\end{array}\\right\\rbrack {dt} + \\left\\lbrack \\begin{array}{ll} ... | Then \( \\rho = 0 \) and the equation (12.2.12) gets the form\n\n\[ \n{\\sigma u} = \\left\\lbrack \\begin{array}{ll} 1 & 0 \\\\ 0 & 1 \\\\ 1 & 1 \\end{array}\\right\\rbrack \\left\\lbrack \\begin{array}{l} {u}_{1} \\\\ {u}_{2} \\end{array}\\right\\rbrack = \\left\\lbrack \\begin{array}{l} 1 \\\\ 2 \\\\ 3 \\end{array}\... | Yes |
Can we find such a \( T \)-claim? | Let \( \theta \left( t\right) = \left( {{\theta }_{0}\left( t\right) ,{\theta }_{1}\left( t\right) }\right) \) be an admissible portfolio. Then the corresponding value process \( {V}_{z}^{\theta }\left( t\right) \) is given by (see (12.2.20))\n\n\[ {V}_{z}^{\theta }\left( t\right) = z + {\int }_{0}^{t}{\theta }_{1}\lef... | Yes |
Theorem 12.3.2. a) Suppose (12.2.12) and (12.2.13) hold and let \( Q \) be as in (12.2.2). Let \( F \) be a (European) \( T \) -claim such that \( {E}_{Q}\left\lbrack {\xi \left( T\right) F}\right\rbrack < \infty \) . Then\n\n\[ \n\text{ess}\inf F\left( \omega \right) \leq p\left( F\right) \leq {E}_{Q}\left\lbrack {\xi... | Proof. a) Suppose \( y \in \mathbf{R} \) and there exists an admissible portfolio \( \theta \) such that\n\n\[ \n{V}_{-y}^{\theta }\left( {T,\omega }\right) = - y + {\int }_{0}^{T}\theta \left( s\right) {dX}\left( s\right) \geq - F\left( \omega \right) \;\text{ a.s. } \n\]\ni.e., using (12.2.7) and Lemma 12.2.4,\n\n\[ ... | Yes |
Theorem 12.3.3. Let \( Y\\left( t\\right) \) and \( Z\\left( t\\right) \) be as in (12.3.10) and (12.3.12), respectively, and assume that \( h : {\\mathbf{R}}^{n} \\rightarrow \\mathbf{R} \) is as in (12.3.13). Assume that (12.3.11) and (12.3.16) hold and define \( Q \) and \( \\widetilde{B}\\left( t\\right) \) by (12.... | \[ h\\left( {Y\\left( T\\right) }\\right) = {E}_{Q}^{y}\\left\\lbrack {h\\left( {Y\\left( T\\right) }\\right) }\\right\\rbrack + {\\int }_{0}^{T}\\phi \\left( {t,\\omega }\\right) d\\widetilde{B}\\left( t\\right) ,\] where \( \\phi = \\left( {{\\phi }_{1},\\ldots ,{\\phi }_{m}}\\right) \), with \[ {\\phi }_{j}\\left( {... | Yes |
Theorem 12.3.4. Let \( \{ X\left( t\right) {\} }_{t \in \left\lbrack {0, T}\right\rbrack } \) be a complete market. Suppose (12.2.12) and (12.2.13) hold and let \( Q,\widetilde{B} \) be as in (12.2.2),(12.2.3). Let \( F \) be a European \( T \) -claim such that \( {E}_{Q}\left\lbrack {\xi \left( T\right) F}\right\rbrac... | Proof. (12.3.26) is just part b) of Theorem 12.3.2. The relation (12.3.28) follows from (12.3.8). Note that the equation (12.3.28) has the solution\n\n\[ \widehat{\theta }\left( {t,\omega }\right) = {X}_{0}\left( t\right) \phi \left( {t,\omega }\right) \Lambda \left( {t,\omega }\right) \] | No |
Theorem 12.3.6 (The generalized Black & Scholes formula). Suppose \( X\left( t\right) = \left( {{X}_{0}\left( t\right) ,{X}_{1}\left( t\right) }\right) \) is given by\n\n\[ d{X}_{0}\left( t\right) = \rho \left( t\right) {X}_{0}\left( t\right) {dt};\;{X}_{0}\left( 0\right) = 1 \]\n\n(12.3.36)\n\n\[ d{X}_{1}\left( t\righ... | Proof. Part a) is already proved and part b) follows from Theorem 12.3.3 and Theorem 12.3.4: (Strictly speaking condition (12.3.11) is not satisfied for the process \( {X}_{1} \) (unless \( x \) is bounded away from 0 ), but in this case it can be verified directly that \( u\left( {t, z}\right) \) given by (12.3.13) be... | Yes |
Theorem 12.3.9. The price \( {p}_{\mathrm{A}}\left( F\right) \) of an American contingent \( T \) -claim \( F \) of the Markovian form (12.3.56) is the solution of the optimal stopping problem (12.3.60), with Itô diffusion \( Y\left( t\right) \) given by (12.3.59). | We recognize (12.3.60) as a special case of the optimal stopping problem considered in Theorem 10.4.1. We can therefore use the method there to evaluate \( {p}_{\mathrm{A}}\left( F\right) \) in special cases. | No |
Example 1.4 (Social mobility). Let \( {X}_{n} \) be a family’s social class in the \( n \) th generation, which we assume is either \( 1 = \) lower, \( 2 = \) middle, or \( 3 = \) upper. In our simple version of sociology, changes of status are a Markov chain with the following transition probability\n\n\[ \n\begin{mat... | ## Q. Do the fractions of people in the three classes approach a limit? | No |
Example 1.7 (Repair chain). A machine has three critical parts that are subject to failure, but can function as long as two of these parts are working. When two are broken, they are replaced and the machine is back to working order the next day. To formulate a Markov chain model we declare its state space to be the par... | \[ \begin{matrix} & 0 & 1 & 2 & 3 & {12} & {13} & {23} \\ 0 & {0.93} & {0.01} & {0.02} & {0.04} & 0 & 0 & 0 \\ \\mathbf{1} & 0 & {0.94} & 0 & 0 & {0.02} & {0.04} & 0 \\ \\mathbf{2} & 0 & 0 & {0.95} & 0 & {0.01} & 0 & {0.04} \\ \\mathbf{3} & 0 & 0 & 0 & {0.97} & 0 & {0.01} & {0.02} \\ \\mathbf{{12}} & 1 & 0 & 0 & 0 & 0 ... | Yes |
In a Markov chain the distribution of \( {X}_{n + 1} \) only depends on \( {X}_{n} \). This can easily be generalized to case in which the distribution of \( {X}_{n + 1} \) only depends on \( \left( {{X}_{n},{X}_{n - 1}}\right) \). For a concrete example consider a basketball player who makes a shot with the following ... | <table><thead><tr><th></th><th>HH</th><th>HM</th><th>MH</th><th>MM</th></tr></thead><tr><td>HH</td><td>3/4</td><td>\( 1/4 \)</td><td>0</td><td>0</td></tr><tr><td>HM</td><td>0</td><td>0</td><td>\( 2/3 \)</td><td>\( 1/3 \)</td></tr><tr><td>MH</td><td>\( 2/3 \)</td><td>\( 1/3 \)</td><td>0</td><td>0</td></tr><tr><td>MM</td... | Yes |
Theorem 1.1. The \( m \) step transition probability \( P\left( {{X}_{n + m} = j \mid {X}_{n} = i}\right) \) is the mth power of the transition matrix \( p \) . | The key ingredient in proving this is the Chapman-Kolmogorov equation\n\n\[ \n{p}^{m + n}\left( {i, j}\right) = \mathop{\sum }\limits_{k}{p}^{m}\left( {i, k}\right) {p}^{n}\left( {k, j}\right) \]\n\n(1.2)\n\nOnce this is proved, Theorem 1.1 follows, since taking \( n = 1 \) in (1.2), we see that\n\n\[ \n{p}^{m + 1}\lef... | Yes |
Example 1.11 (Gambler’s ruin). Suppose for simplicity that \( N = 4 \) in Example 1.1, so that the transition probability is\n\n\[\n\\begin{matrix} & 0 & 1 & 2 & 3 & 4 \\\\ 0 & {1.0} & 0 & 0 & 0 & 0 \\\\ 1 & {0.6} & 0 & {0.4} & 0 & 0 \\\\ 2 & 0 & {0.6} & 0 & {0.4} & 0 \\\\ 3 & 0 & 0 & {0.6} & 0 & {0.4} \\\\ 4 & 0 & 0 &... | To compute \( {p}^{2} \) one row at a time we note:\n\n\( {p}^{2}\\left( {0,0}\\right) = 1 \) and \( {p}^{2}\\left( {4,4}\\right) = 1 \), since these are absorbing states.\n\n\( {p}^{2}\\left( {1,3}\\right) = {\\left( {0.4}\\right) }^{2} = {0.16} \), since the chain has to go up twice.\n\n\( {p}^{2}\\left( {1,1}\\right... | No |
Theorem 1.2 (Strong Markov property). Suppose \( T \) is a stopping time. Given that \( T = n \) and \( {X}_{T} = y \), any other information about \( {X}_{0},\ldots {X}_{T} \) is irrelevant for predicting the future, and \( {X}_{T + k}, k \geq 0 \) behaves like the Markov chain with initial state \( y \) . | Why is this true? To keep things as simple as possible we will show only that\n\n\[ P\left( {{X}_{T + 1} = z \mid {X}_{T} = y, T = n}\right) = p\left( {y, z}\right) \]\n\nLet \( {V}_{n} \) be the set of vectors \( \left( {{x}_{0},\ldots ,{x}_{n}}\right) \) so that if \( {X}_{0} = {x}_{0},\ldots ,{X}_{n} = {x}_{n} \), t... | Yes |
Example 1.12 (Gambler’s ruin). Consider, for concreteness, the case \( N = 4 \) .\n\n\[ \n\begin{matrix} & 0 & 1 & 2 & 3 & 4 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & {0.6} & 0 & {0.4} & 0 & 0 \\ 2 & 0 & {0.6} & 0 & {0.4} & 0 \\ 3 & 0 & 0 & {0.6} & 0 & {0.4} \\ 4 & 0 & 0 & 0 & 0 & 1 \end{matrix} \n\]\n\nWe will show that eventua... | It is easy to check that 0 and 4 are recurrent. Since \( p\left( {0,0}\right) = 1 \), the chain comes back on the next step with probability one, i.e.,\n\n\[ \n{P}_{0}\left( {{T}_{0} = 1}\right) = 1 \n\]\n\nand hence \( {\rho }_{00} = 1 \) . A similar argument shows that 4 is recurrent. In general if \( y \) is an abso... | Yes |
Example 1.13 (Social mobility). Recall that the transition probability is\n\n\\[ \n\\begin{matrix} & 1 & 2 & 3 \\\\ 1 & {0.7} & {0.2} & {0.1} \\\\ 2 & {0.3} & {0.5} & {0.2} \\\\ 3 & {0.2} & {0.4} & {0.4} \\end{matrix} \n\\]\n\nTo begin we note that no matter where \\( {X}_{n} \\) is, there is a probability of at least ... | i.e., we will return to 3 with probability 1 . The last argument applies even more strongly to states 1 and 2 , since the probability of jumping to them on the next step is always at least 0.2 . Thus all three states are recurrent. | Yes |
Lemma 1.3. Suppose \( {P}_{x}\left( {{T}_{y} \leq k}\right) \geq \alpha > 0 \) for all \( x \) in the state space \( S \) . Then | \[ {P}_{x}\left( {{T}_{y} > {nk}}\right) \leq {\left( 1 - \alpha \right) }^{n} \] | Yes |
Lemma 1.4. If \( x \rightarrow y \) and \( y \rightarrow z \), then \( x \rightarrow z \) . | Proof. Since \( x \rightarrow y \) there is an \( m \) so that \( {p}^{m}\left( {x, y}\right) > 0 \) . Similarly there is an \( n \) so that \( {p}^{n}\left( {y, z}\right) > 0 \) . Since \( {p}^{m + n}\left( {x, z}\right) \geq {p}^{m}\left( {x, y}\right) {p}^{n}\left( {y, z}\right) \) it follows that \( x \rightarrow z... | Yes |
Theorem 1.5. If \( {\rho }_{xy} > 0 \), but \( {\rho }_{yx} < 1 \), then \( x \) is transient. | Proof. Let \( K = \min \left\{ {k : {p}^{k}\left( {x, y}\right) > 0}\right\} \) be the smallest number of steps we can take to get from \( x \) to \( y \) . Since \( {p}^{K}\left( {x, y}\right) > 0 \) there must be a sequence \( {y}_{1},\ldots {y}_{K - 1} \) so that\n\n\[ p\left( {x,{y}_{1}}\right) p\left( {{y}_{1},{y}... | Yes |
Lemma 1.6. If \( x \) is recurrent and \( {\rho }_{xy} > 0 \) then \( {\rho }_{yx} = 1 \) . | Proof. If \( {\rho }_{yx} < 1 \) then Lemma 1.5 would imply \( x \) is transient. | No |
Example 1.14 (A Seven-state chain). Consider the transition probability: | To identify the states that are recurrent and those that are transient, we begin by drawing a graph that will contain an arc from \( i \) to \( j \) if \( p\left( {i, j}\right) > 0 \) and \( i \neq j \) . We do not worry about drawing the self-loops corresponding to states with \( p\left( {i, i}\right) > 0 \) since suc... | No |
Theorem 1.8. If the state space \( S \) is finite, then \( S \) can be written as a disjoint union \( T \cup {R}_{1} \cup \cdots \cup {R}_{k} \), where \( T \) is a set of transient states and the \( {R}_{i},1 \leq i \leq k \), are closed irreducible sets of recurrent states. | Proof. Let \( T \) be the set of \( x \) for which there is a \( y \) so that \( x \rightarrow y \) but \( y \nrightarrow x \) . The states in \( T \) are transient by Theorem 1.5. Our next step is to show that all the remaining states, \( S - T \), are recurrent.\n\nPick an \( x \in S - T \) and let \( {C}_{x} = \{ y ... | Yes |
Lemma 1.10. In a finite closed set there has to be at least one recurrent state. | To prove these results we need to introduce a little more theory. Recall the time of the \( k \) th visit to \( y \) defined by\n\n\[ \n{T}_{y}^{k} = \min \left\{ {n > {T}_{y}^{k - 1} : {X}_{n} = y}\right\} \n\]\n\nand \( {\rho }_{xy} = {P}_{x}\left( {{T}_{y} < \infty }\right) \) the probability we ever visit \( y \) a... | No |
Lemma 1.11. \( {E}_{x}N\\left( y\\right) = {\\rho }_{xy}/\\left( {1 - {\\rho }_{yy}}\\right) \) | Proof. Accept for the moment the fact that for any nonnegative integer valued random variable \( X \), the expected value of \( X \) can be computed by\n\n\[ \n{EX} = \\mathop{\\sum }\\limits_{{k = 1}}^{\\infty }P\\left( {X \geq k}\\right) \]\n\n(1.6)\n\nWe will prove this after we complete the proof of Lemma 1.11. Now... | Yes |
Lemma 1.12. \( {E}_{x}N\left( y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {x, y}\right) \) . | Proof. Let \( {1}_{\left\{ {X}_{n} = y\right\} } \) denote the random variable that is 1 if \( {X}_{n} = y,0 \) otherwise. Clearly\n\n\[ N\left( y\right) = \mathop{\sum }\limits_{{n = 1}}^{\infty }{1}_{\left\{ {X}_{n} = y\right\} } \]\n\nTaking expected values now gives\n\n\[ {E}_{x}N\left( y\right) = \mathop{\sum }\li... | Yes |
Theorem 1.13. \( y \) is recurrent if and only if\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{p}^{n}\left( {y, y}\right) = {E}_{y}N\left( y\right) = \infty \] | Proof. The first equality is Lemma 1.12. From Lemma 1.11 we see that \( {E}_{y}N\left( y\right) = \) \( \infty \) if and only if \( {\rho }_{yy} = 1 \), which is the definition of recurrence. | Yes |
Consider the weather chain (Example 1.3) and suppose that the initial distribution is \( q\left( 1\right) = {0.3} \) and \( q\left( 2\right) = {0.7} \) . | \[ \left( \begin{array}{ll} {0.3} & {0.7} \end{array}\right) \left( \begin{array}{ll} {0.6} & {0.4} \\ {0.2} & {0.8} \end{array}\right) = \left( \begin{array}{ll} {0.32} & {0.68} \end{array}\right) \] \[ \text{since}\;{0.3}\left( {0.6}\right) + {0.7}\left( {0.2}\right) = {0.32} \] \[ {0.3}\left( {0.4}\right) + {0.7}\le... | Yes |
Consider the social mobility chain (Example 1.4) and suppose that the initial distribution: \( q\left( 1\right) = {0.5}, q\left( 2\right) = {0.2} \), and \( q\left( 3\right) = {0.3} \). Multiplying the vector \( q \) by the transition probability gives the vector of probabilities at time 1 . | \[ \left( \begin{array}{lll} {0.5} & {0.2} & {0.3} \end{array}\right) \left( \begin{array}{lll} {0.7} & {0.2} & {0.1} \\ {0.3} & {0.5} & {0.2} \\ {0.2} & {0.4} & {0.4} \end{array}\right) = \left( \begin{array}{lll} {0.47} & {0.32} & {0.21} \end{array}\right) \] To check the arithmetic note that the three entries on the... | Yes |
Example 1.17 (Weather chain). To compute the stationary distribution we want to solve\n\n\[ \left( \begin{array}{ll} {\pi }_{1} & {\pi }_{2} \end{array}\right) \left( \begin{array}{ll} {0.6} & {0.4} \\ {0.2} & {0.8} \end{array}\right) = \left( \begin{array}{ll} {\pi }_{1} & {\pi }_{2} \end{array}\right) \] | Multiplying gives two equations:\n\n\[ {0.6}{\pi }_{1} + {0.2}{\pi }_{2} = {\pi }_{1} \]\n\n\[ {0.4}{\pi }_{1} + {0.8}{\pi }_{2} = {\pi }_{2} \]\n\nBoth equations reduce to \( {0.4}{\pi }_{1} = {0.2}{\pi }_{2} \) . Since we want \( {\pi }_{1} + {\pi }_{2} = 1 \), we must have \( {0.4}{\pi }_{1} = {0.2} - {0.2}{\pi }_{1... | Yes |
Example 1.18 (Social Mobility (continuation of 1.4)).\n\n\\[ \n\\begin{matrix} & 1 & 2 & 3 \\ 1 & {0.7} & {0.2} & {0.1} \\ 2 & {0.3} & {0.5} & {0.2} \\ 3 & {0.2} & {0.4} & {0.4} \\ \\end{matrix} \n\\]\n\nThe equation \\( {\\pi p} = \\pi \\) says\n\n\\[ \n\\left( \\begin{array}{lll} {\\pi }_{1} & {\\pi }_{2} & {\\pi }_{... | which translates into three equations\n\n\\[ \n{0.7}{\\pi }_{1} + {0.3}{\\pi }_{2} + {0.2}{\\pi }_{3} = {\\pi }_{1} \n\\]\n\n\\[ \n{0.2}{\\pi }_{1} + {0.5}{\\pi }_{2} + {0.4}{\\pi }_{3} = {\\pi }_{2} \n\\]\n\n\\[ \n{0.1}{\\pi }_{1} + {0.2}{\\pi }_{2} + {0.4}{\\pi }_{3} = {\\pi }_{3} \n\\]\n\nNote that the columns of th... | Yes |
Example 1.19 (Brand Preference (continuation of 1.5)).\n\n\\[ \n\\begin{array}{rrrr} & 1 & 2 & 3 \\\\ 1 & {0.8} & {0.1} & {0.1} \\\\ 2 & {0.2} & {0.6} & {0.2} \\\\ 3 & {0.3} & {0.3} & {0.4} \\end{array} \n\\]\n\nUsing the first two equations and the fact that the sum of the \\( \\pi \\) ’s is 1\n\n\\[ \n{0.8}{\\pi }_{1... | \nSubtracting \\( {\\pi }_{1} \\) from both sides of the first equation and \\( {\\pi }_{2} \\) from both sides of the second, this translates into \\( {\\pi A} = \\left( {0,0,1}\\right) \\) with\n\n\\[ \nA = \\left( \\begin{matrix} - {0.2} & {0.1} & 1 \\\\ {0.2} & - {0.4} & 1 \\\\ {0.3} & {0.3} & 1 \\end{matrix}\\righ... | Yes |
Theorem 1.14. Suppose that the \( k \times k \) transition matrix \( p \) is irreducible. Then there is a unique solution to \( {\pi p} = \pi \) with \( \mathop{\sum }\limits_{x}{\pi }_{x} = 1 \) and we have \( {\pi }_{x} > 0 \) for all \( x \) . | Proof. Let \( I \) be the identity matrix. Since the rows of \( p - I \) add to 0, the rank of the matrix is \( \leq k - 1 \) and there is a vector \( v \) so that \( {vp} = v \) . Let \( q = \left( {I + p}\right) /2 \) be the lazy chain that stays put with probability \( 1/2 \) and otherwise takes a step according to ... | Yes |
Example 1.21 (Ehrenfest chain (continuation of 1.2)). For concreteness, suppose there are three balls. In this case the transition probability is\n\n\[ \n\begin{matrix} & 0 & 1 & 2 & 3 \\ 0 & 0 & 3/3 & 0 & 0 \\ 1 & 1/3 & 0 & 2/3 & 0 \\ 2 & 0 & 2/3 & 0 & 1/3 \\ 3 & 0 & 0 & 3/3 & 0 \end{matrix} \n\] | In the second power of \( p \) the zero pattern is shifted:\n\n\[ \n\begin{matrix} & 0 & 1 & 2 & 3 \\ 0 & 1/3 & 0 & 2/3 & 0 \\ 1 & 0 & 7/9 & 0 & 2/9 \\ 2 & 2/9 & 0 & 7/9 & 0 \\ 3 & 0 & 2/3 & 0 & 1/3 \end{matrix} \n\]\n\nTo see that the zeros will persist, note that if we have an odd number of balls in the left urn, the... | Yes |
Lemma 1.15. \( {I}_{x} \) is closed under addition. That is, if \( i, j \in {I}_{x} \), then \( i + j \in {I}_{x} \) . | Proof. If \( i, j \in {I}_{x} \) then \( {p}^{i}\left( {x, x}\right) > 0 \) and \( {p}^{j}\left( {x, x}\right) > 0 \) so\n\n\[ \n{p}^{i + j}\left( {x, x}\right) \geq {p}^{i}\left( {x, x}\right) {p}^{j}\left( {x, x}\right) > 0 \n\]\n\nand hence \( i + j \in {I}_{x} \) . | Yes |
Lemma 1.16. If \( x \) has period 1, i.e., the greatest common divisor \( {I}_{x} \) is 1, then there is a number \( {n}_{0} \) so that if \( n \geq {n}_{0} \), then \( n \in {I}_{x} \) . In words, \( {I}_{x} \) contains all of the integers after some value \( {n}_{0} \) . | Proof. We begin by observing that it enough to show that \( {I}_{x} \) will contain two consecutive integers: \( k \) and \( k + 1 \) . For then it will contain \( {2k},{2k} + 1,{2k} + 2 \), and \( {3k},{3k} + 1,{3k} + 2,{3k} + 3 \), or in general \( {jk},{jk} + 1,\ldots {jk} + j \) . For \( j \geq k - 1 \) these block... | Yes |
Lemma 1.17. If \( p\left( {x, x}\right) > 0 \), then \( x \) has period 1 . | Proof. If \( p\left( {x, x}\right) > 0 \), then \( 1 \in {I}_{x} \), so the greatest common divisor is 1 . | Yes |
Lemma 1.18. If \( {\rho }_{xy} > 0 \) and \( {\rho }_{yx} > 0 \) then \( x \) and \( y \) have the same period. | Proof. Suppose that the period of \( x \) is \( c \), while the period of \( y \) is \( d < c \) . Let \( k \) be such that \( {p}^{k}\left( {x, y}\right) > 0 \) and let \( m \) be such that \( {p}^{m}\left( {y, x}\right) > 0 \) . Since\n\n\[ \n{p}^{k + m}\left( {x, x}\right) \geq {p}^{k}\left( {x, y}\right) {p}^{m}\le... | Yes |
If we are going to operate the machine for 1,800 days (about 5 years) then how many parts of types 1, 2, and 3 will we use? | To find the stationary distribution we look at the last row of\n\n\[ \n{\left( \begin{matrix} - {0.07} & {0.01} & {0.02} & {0.04} & 0 & 0 & 1 \\ 0 & - {0.06} & 0 & 0 & {0.02} & {0.04} & 1 \\ 0 & 0 & - {0.05} & 0 & {0.01} & 0 & 1 \\ 0 & 0 & 0 & - {0.03} & 0 & {0.01} & 1 \\ 1 & 0 & 0 & 0 & - 1 & 0 & 1 \\ 1 & 0 & 0 & 0 & ... | Yes |
Example 1.24 (Inventory chain (continuation of 1.6)). We have an electronics store that sells a videogame system, with the ptential for sales of \( 0,1,2 \), or 3 of these units each day with probabilities \( {0.3},{0.4},{0.2} \), and 0.1 . Each night at the close of business new units can be ordered which will be avai... | Suppose we use a 2,3 inventory policy. That is, we order if there are \( \leq 2 \) units and we order enough stock so that we have 3 units at the beginning of the next day.\n\nIn this case we always start the day with 3 units, so the transition probability has constant rows\n\n\[ \begin{matrix} & 0 & 1 & 2 & 3 \\ 0 & {... | Yes |
Theorem 1.24. If \( p \) is a doubly stochastic transition probability for a Markov chain with \( N \) states, then the uniform distribution, \( \pi \left( x\right) = 1/N \) for all \( x \), is a stationary distribution. | Proof. To check this claim we note that if \( \pi \left( x\right) = 1/N \) then\n\n\[ \mathop{\sum }\limits_{x}\pi \left( x\right) p\left( {x, y}\right) = \frac{1}{N}\mathop{\sum }\limits_{x}p\left( {x, y}\right) = \frac{1}{N} = \pi \left( y\right) \]\n\nLooking at the second equality we see that conversely, if \( \pi ... | Yes |
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