Chapter 4

Finite-state Markov chains

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Problem 4.1

Let [P][P] be the transition matrix for a finite state Markov chain and let state ii be recurrent. Prove that ii is aperiodic if Pii>0P_{ii} > 0.

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Problem 4.2

Show that every Markov chain with M<∞\mathsf{M} < \infty states contains at least one recurrent set of states. Explaining each of the following statements is sufficient.

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(a)

If state i1i_{1} is transient, then there is some other state i2i_{2} such that i1→i2i_{1} \rightarrow i_{2} and i2↛i1i_{2} \not\rightarrow i_{1}.

(b)

If the i2i_{2} of (a) is also transient, there is a third state i3i_{3} such that i2→i3i_{2} \rightarrow i_{3}, i3↛i2i_{3} \not\rightarrow i_{2}; that state must satisfy i3≠i2i_{3} \neq i_{2}, i3≠i1i_{3} \neq i_{1}.

(c)

Continue iteratively to repeat (b) for successive states, i1,i2,…i_{1}, i_{2}, \ldots. That is, if i1,…,iki_{1}, \ldots , i_{k} are generated as above and are all transient, generate ik+1i_{k+1} such that ik→ik+1i_{k} \rightarrow i_{k+1} and ik+1↛iki_{k+1} \not\rightarrow i_{k}. Then ik+1≠iji_{k+1} \neq i_{j} for 1≤j≤k1 \leq j \leq k.

(d)

Show that for some k≤Mk \leq \mathsf{M}, kk is not transient, i.e., it is recurrent, so a recurrent class exists.

Problem 4.3

Consider a finite-state Markov chain in which some given state, say state 1, is accessible from every other state. Show that the chain has exactly one recurrent class R\mathcal{R} of states and state 1∈R1 \in \mathcal{R}. (Note that the chain is then a unichain.)

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Problem 4.4

Show how to generalize the graph in Figure 4.4 to an arbitrary number of states M≥3\mathsf{M} \geq 3 with one cycle of M\mathsf{M} nodes and one of M−1\mathsf{M} - 1 nodes. For M=4\mathsf{M} = 4, let node 1 be the node not in the cycle of M−1\mathsf{M} - 1 nodes. List the set of states accessible from node 1 in nn steps for each n≤12n \leq 12 and show that the bound in Theorem 4.2.11 is met with equality. Explain why the same result holds for all larger M\mathsf{M}.

Theorem 4.2.11 (referenced in this problem, restated from Section 4.2): if [P][P] is the matrix of an ergodic (i.e., recurrent, aperiodic, irreducible) finite-state Markov chain with M\mathsf{M} states, then there is an integer nn such that Pijn>0P_{ij}^{n} > 0 for all i,ji, j; the smallest such nn satisfies n≤(M−1)2+1n \leq (\mathsf{M} - 1)^{2} + 1.

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Remark.
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This is the proof of Theorem 4.2.11, split across Exercises 4.4--4.7; each generalizes the previous one, culminating in the bound (M−1)2+1(\mathsf{M}-1)^2+1 above.

Problem 4.5
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(a)

Show that an ergodic Markov chain with M>1\mathsf{M} > 1 states must contain a cycle with τ<M\tau < \mathsf{M} states. Hint: Use ergodicity to show that the smallest cycle cannot contain M\mathsf{M} states.

(b)

Let ℓ\ell be a fixed state on this cycle of length τ\tau. Let T(m)\mathcal{T}(m) be the set of states accessible from ℓ\ell in mm steps. Show that for each m≥1m \geq 1, T(m)⊆T(m+τ)\mathcal{T}(m) \subseteq \mathcal{T}(m + \tau ). Hint: For any given state j∈T(m)j \in \mathcal{T}(m), show how to construct a walk of m+τm + \tau steps from ℓ\ell to jj from the assumed walk of mm steps.

(c)

Define T(0)\mathcal{T}(0) to be the singleton set {ℓ}\left\{ \ell \right\} and show that

T(0)⊆T(τ)⊆T(2τ)⊆⋯⊆T(nτ)⊆⋯ . \mathcal{T}(0) \subseteq \mathcal{T}(\tau ) \subseteq \mathcal{T}(2\tau ) \subseteq \cdots \subseteq \mathcal{T}(n\tau ) \subseteq \cdots .
(d)

Show that if one of the inclusions above is satisfied with equality, then all subsequent inclusions are satisfied with equality. Show from this that at most the first M−1\mathsf{M} - 1 inclusions can be satisfied with strict inequality and that T(nτ)=T((M−1)τ)\mathcal{T}(n\tau ) = \mathcal{T}((\mathsf{M}-1)\tau ) for all n≥M−1n \geq \mathsf{M} - 1.

(e)

Show that all states are included in T((M−1)τ)\mathcal{T}((\mathsf{M}-1)\tau ).

(f)

Show that Pij(M−1)2+1>0P_{ij}^{(\mathsf{M}-1)^{2}+1} > 0 for all i,ji, j.

Problem 4.6

Consider a Markov chain with one ergodic class of rr states, say {1,2,…,r}\left\{ 1, 2, \ldots , r\right\} and M−r\mathsf{M} - r other states that are all transient. Show that Pijn>0P_{ij}^{n} > 0 for all j≤rj \leq r and n≥(r−1)2+1+M−rn \geq (r-1)^{2} + 1 + \mathsf{M} - r.

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Problem 4.7
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(a)

Let τ\tau be the number of states in the smallest cycle of an arbitrary ergodic Markov chain of M≥3\mathsf{M} \geq 3 states. Show that Pijn>0P_{ij}^{n} > 0 for all n≥(M−2)τ+Mn \geq (\mathsf{M}-2)\tau + \mathsf{M}. Hint: Look at the proof of Theorem 4.2.11 in Exercise 4.5.

(b)

For τ=1\tau = 1, draw the graph of an ergodic Markov chain (generalized for arbitrary M≥3\mathsf{M} \geq 3) for which there is an i,ji,j for which Pijn=0P_{ij}^{n} = 0 for n=2M−3n = 2\mathsf{M}-3. Hint: Look at Figure 4.4.

(c)

For arbitrary τ<M−1\tau < \mathsf{M}-1, draw the graph of an ergodic Markov chain (generalized for arbitrary M\mathsf{M}) for which there is an i,ji,j for which Pijn=0P_{ij}^{n} = 0 for n=(M−2)τ+M−1n = (\mathsf{M}-2)\tau + \mathsf{M}-1. Assume that M\mathsf{M} and τ\tau are relatively prime.

Problem 4.8

A transition probability matrix [P][P] is said to be doubly stochastic if

∑jPij=1for all i,∑iPij=1for all j. \sum _{j} P_{ij} = 1 \quad \text{for all } i, \qquad \sum _{i} P_{ij} = 1 \quad \text{for all } j.

That is, the row sum and the column sum each equal 1. If a doubly stochastic chain has M\mathsf{M} states and is ergodic (i.e., has a single class of states and is aperiodic), calculate its steady-state probabilities.

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Problem 4.9
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(a)

Find the steady-state probabilities π0,…,πk−1\pi_{0}, \ldots , \pi_{k-1} for the Markov chain below. Express your answer in terms of the ratio ρ=p/q\rho = p/q, where q=1−pq = 1-p. Pay particular attention to the special case ρ=1\rho = 1.

[width=0.9figs/figs_original/ch4_ex9.png

(b)

Sketch π0,…,πk−1\pi_{0}, \ldots , \pi_{k-1}. Give one sketch for ρ=1/2\rho = 1/2, one for ρ=1\rho = 1, and one for ρ=2\rho = 2.

(c)

Find the limit of π0\pi_{0} as kk approaches ∞\infty; give separate answers for ρ<1\rho < 1, ρ=1\rho = 1, and ρ>1\rho > 1. Find limiting values of πk−1\pi_{k-1} for the same cases.

Problem 4.10
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(a)

Find the steady-state probabilities for each of the Markov chains in the figure below. Assume that all clockwise probabilities in the first graph are the same, say pp, and assume that P4,5=P4,1P_{4,5} = P_{4,1} in the second graph.

[width=0.85figs/figs_original/ch4_ex10.png

(b)

Find the matrices [P2][P^{2}] for the same chains. Draw the graphs for the Markov chains represented by [P2][P^{2}], i.e., the graph of two step transitions for the original chains. Find the steady-state probabilities for these two-step chains. Explain why your steady-state probabilities are not unique.

(c)

Find lim⁡n→∞[P2n]\lim_{n \rightarrow \infty }[P^{2n}] for each of the chains.

Problem 4.11
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(a)

Assume that ν(i)\nu^{(i)} is a right eigenvector and π(j)\pi^{(j)} is a left eigenvector of an M×M\mathsf{M} \times \mathsf{M} stochastic matrix [P][P] where λi≠λj\lambda_{i} \neq \lambda_{j}. Show that π(j)ν(i)=0\pi^{(j)}\nu^{(i)} = 0. Hint: Consider two ways of finding π(j)[P]ν(i)\pi^{(j)}[P]\nu^{(i)}.

(b)

Assume that [P][P] has M\mathsf{M} distinct eigenvalues. The right eigenvectors of [P][P] then span M\mathsf{M} space (see Section 5.2 of Strang [28]), so the matrix [U][U] with those eigenvectors as columns is non-singular. Show that U−1U^{-1} is a matrix whose rows are the M\mathsf{M} left eigenvectors of [P][P]. Hint: Use (a).

(c)

For any given integer i∈{1,M}i \in \left\{ 1, \mathsf{M}\right\}, let [A][A] be a diagonal matrix with a single non-zero element, Aii=a≠0A_{ii} = a \neq 0. Using the assumptions and results of (b), show that

[UAU−1]=aν(i)π(i). [UAU^{-1}] = a\nu ^{(i)}\pi ^{(i)}.

Hint: Visualize straightforward vector/matrix multiplication.

(d)

Verify (4.30)(4.30) (referenced here; restated: [UΛnU−1]=∑i=1Mλinν(i)π(i)[U\Lambda^{n}U^{-1}] = \sum_{i=1}^{\mathsf{M}} \lambda_{i}^{n}\nu^{(i)}\pi^{(i)}, where [Λ][\Lambda ] is the diagonal matrix of eigenvalues λi\lambda_{i} of [P][P]).

Remark.
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I added a restatement of equation (4.30), cited bare in part (d), from Section 4.4 of the source PDF, so the problem is solvable without looking it up.

Problem 4.12
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(a)

Let λk\lambda_{k} be an eigenvalue of a stochastic matrix [P][P] and let π(k)\pi^{(k)} be a left eigenvector for λk\lambda_{k}. Show that for each component πj(k)\pi_{j}^{(k)} of π(k)\pi^{(k)} and each nn that

λknπj(k)=∑iπi(k)Pijn. \lambda _{k}^{n}\pi _{j}^{(k)} = \sum _{i} \pi _{i}^{(k)} P_{ij}^{n}.
(b)

By taking magnitudes of each side and looking at the appropriate jj, show that

∣λk∣n≤M. \left|\lambda _{k}\right|^{n} \leq \mathsf{M}.
(c)

Show that ∣λk∣≤1\left|\lambda_{k}\right| \leq 1.

Problem 4.13

Consider a finite state Markov chain with matrix [P][P] which has κ\kappa aperiodic recurrent classes, R1,…,Rκ\mathcal{R}_{1}, \ldots , \mathcal{R}_{\kappa } and a set T\mathcal{T} of transient states. For any given recurrent class Rℓ\mathcal{R}_{\ell }, consider a vector ν\nu such that νi=1\nu_{i} = 1 for each i∈Rℓi \in \mathcal{R}_{\ell }, νi=lim⁡n→∞P(Xn∈Rℓ∣X0=i)\nu_{i} = \lim_{n \rightarrow \infty } \mathbb {P}\left( X_{n} \in \mathcal{R}_{\ell } \mid X_{0} = i \right) for each i∈Ti \in \mathcal{T}, and νi=0\nu_{i} = 0 otherwise. Show that ν\nu is a right eigenvector of [P][P] with eigenvalue 1. Hint: Redraw Figure 4.5 (a graph showing the block-triangular structure of [P][P] for a unichain with transient set T\mathcal{T} and recurrent class R\mathcal{R}, transitions going only from T\mathcal{T} to T\mathcal{T}, from T\mathcal{T} to R\mathcal{R}, or within R\mathcal{R}) for multiple recurrent classes and first show that ν\nu is an eigenvector of [Pn][P^{n}] in the limit.

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Remark.
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I added a description of Figure 4.5, cited bare in the hint, restated from the source PDF, so the problem is solvable without looking it up.

Problem 4.14

Answer the following questions for the following stochastic matrix [P][P]:

[P]=[1/21/2001/21/2001]. [P] = \begin{bmatrix} 1/2 & 1/2 & 0 \\ 0 & 1/2 & 1/2 \\ 0 & 0 & 1 \end{bmatrix}.
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(a)

Find [Pn][P^{n}] in closed form for arbitrary n>1n > 1.

(b)

Find all distinct eigenvalues and the multiplicity of each distinct eigenvalue for [P][P].

(c)

Find a right eigenvector for each distinct eigenvalue, and show that the eigenvalue of multiplicity 2 does not have two linearly independent eigenvectors.

(d)

Use (c) to show that there is no diagonal matrix [Λ][\Lambda ] and no invertible matrix [U][U] for which [P][U]=[U][Λ][P][U] = [U][\Lambda ].

(e)

Rederive the result of (d) using the result of (a) rather than (c).

Problem 4.15
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(a)

Let [Ji][J_{i}] be a 3 by 3 block of a Jordan form, i.e.,

[Ji]=[λi100λi100λi]. [J_{i}] = \begin{bmatrix} \lambda _{i} & 1 & 0 \\ 0 & \lambda _{i} & 1 \\ 0 & 0 & \lambda _{i} \end{bmatrix}.

Show that the nnth power of [Ji][J_{i}] is given by

[Jin]=[λinnλin−1(n2)λin−20λinnλin−100λin]. [J_{i}^{n}] = \begin{bmatrix} \lambda _{i}^{n} & n\lambda _{i}^{n-1} & \binom {n}{2}\lambda _{i}^{n-2} \\ 0 & \lambda _{i}^{n} & n\lambda _{i}^{n-1} \\ 0 & 0 & \lambda _{i}^{n} \end{bmatrix}.

Hint: Perhaps the easiest way is to calculate [Ji2][J_{i}^{2}] and [Ji3][J_{i}^{3}] and then use iteration.

(b)

Generalize (a) to a k×kk \times k block of a Jordan form. Note that the nnth power of an entire Jordan form is composed of these blocks along the diagonal of the matrix.

(c)

Let [P][P] be a stochastic matrix represented by a Jordan form [J][J] as [P]=[U][J][U−1][P] = [U][J][U^{-1}] and consider [U−1][Pn][U]=[J][U^{-1}][P^{n}][U] = [J] (i.e., [U−1][P][U]=[J][U^{-1}][P][U] = [J]). Show that any repeated eigenvalue of [P][P] (and, in particular, any eigenvalue represented by a Jordan block of 2×22 \times 2 or more) must be strictly less than 1. Hint: Upper bound the elements of [U−1][Pn][U][U^{-1}][P^{n}][U] by taking the magnitude of the elements of [U][U] and [U−1][U^{-1}] and upper bounding each element of a stochastic matrix by 1.

(d)

Let λs\lambda_{s} be the eigenvalue of largest magnitude less than 1. Assume that the Jordan blocks for λs\lambda_{s} are at most of size kk. Show that each ergodic class of [P][P] converges at least as fast as nkλsnn^{k}\lambda_{s}^{n}.

Problem 4.16
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(a)

Let λ\lambda be an eigenvalue of a matrix [A][A], and let ν\nu and π\pi be right and left eigenvectors respectively of λ\lambda, normalized so that πν=1\pi \nu = 1. Show that

[[A]−λνπ]2=[A2]−λ2νπ. [[A] - \lambda \nu \pi ]^{2} = [A^{2}] - \lambda ^{2}\nu \pi .
(b)

Show that [[An]−λnνπ][[A]−λνπ]=[An+1]−λn+1νπ[[A^{n}] - \lambda^{n}\nu \pi ][[A] - \lambda \nu \pi ] = [A^{n+1}] - \lambda^{n+1}\nu \pi.

(c)

Use induction to show that [[A]−λνπ]n=[An]−λnνπ[[A] - \lambda \nu \pi ]^{n} = [A^{n}] - \lambda^{n}\nu \pi.

Problem 4.17

Let [P][P] be the transition matrix for an aperiodic Markov unichain with the states numbered as in Figure 4.5 (referenced here; restated: state 1 through tt are the transient states T\mathcal{T}, numbered before the recurrent states t+1t+1 through M\mathsf{M}, denoted R\mathcal{R}, so that [P][P] has the block form [[PT][Px]0[PR]]\begin{bmatrix} [P_{\mathcal{T}}] & [P_{x}] \\ 0 & [P_{\mathcal{R}}]\end{bmatrix}, where [Px][P_{x}] is the (generally non-square) block of transition probabilities from T\mathcal{T} to R\mathcal{R}).

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(a)

Show that [Pn][P^{n}] can be partitioned as

[Pn]=[[PTn][Pxn]0[PRn]]. [P^{n}] = \begin{bmatrix} [P_{\mathcal{T}}^{n}] & [P_{x}^{n}] \\ 0 & [P_{\mathcal{R}}^{n}] \end{bmatrix}.

That is, the blocks on the diagonal are simply products of the corresponding blocks of [P][P], and the upper right block is whatever it turns out to be.

(b)

Let qiq_{i} be the probability that the chain will be in a recurrent state after tt transitions, starting from state ii, i.e., qi=∑j∈RPijtq_{i} = \sum_{j \in \mathcal{R}} P_{ij}^{t}. Show that qi>0q_{i} > 0 for all transient ii.

(c)

Let qq be the minimum qiq_{i} over all transient ii and show that Pijnt≤(1−q)nP_{ij}^{nt} \leq (1-q)^{n} for all transient i,ji,j (i.e., show that [PTn][P_{\mathcal{T}}^{n}] approaches the all zero matrix [0][0] with increasing nn).

(d)

Let π=(πT,πR)\pi = (\pi_{\mathcal{T}}, \pi_{\mathcal{R}}) be a left eigenvector of [P][P] of eigenvalue 1. Show that πT=0→\pi_{\mathcal{T}} = \overrightarrow {0} and show that πR\pi_{\mathcal{R}} must be positive and be a left eigenvector of [PR][P_{\mathcal{R}}]. Thus show that π\pi exists and is unique (within a scale factor).

(e)

Show that e→\overrightarrow {e} is the unique right eigenvector of [P][P] of eigenvalue 1 (within a scale factor).

Remark.
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I added a description of Figure 4.5's block structure, cited bare in the task, restated from the source PDF's Section 4.3, so the problem is solvable without looking it up.

Problem 4.18

Generalize Exercise 4.17 to the case of a Markov chain [P][P] with κ\kappa recurrent classes and one or more transient classes. In particular, show that:

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(a)

[P][P] has exactly κ\kappa linearly independent left eigenvectors, π(1),π(2),…,π(κ)\pi^{(1)}, \pi^{(2)}, \ldots , \pi^{(\kappa )} of eigenvalue 1, and that the mmth can be taken as a probability vector that is positive on the mmth recurrent class and zero elsewhere.

(b)

[P][P] has exactly κ\kappa linearly independent right eigenvectors, ν(1),ν(2),…,ν(κ)\nu^{(1)}, \nu^{(2)}, \ldots , \nu^{(\kappa )} of eigenvalue 1, and the mmth can be taken as a vector with νi(m)\nu_{i}^{(m)} equal to the probability that recurrent class mm will ever be entered starting from state ii.

(c)

Show that

lim⁡n→∞[Pn]=∑mν(m)π(m). \lim _{n \rightarrow \infty } [P^{n}] = \sum _{m} \nu ^{(m)}\pi ^{(m)}.
Problem 4.19

Suppose a recurrent Markov chain has period dd and let Sm\mathcal{S}_{m}, 0≤m≤d−10 \leq m \leq d-1, be the mmth subset in the sense of Theorem 4.2.9. Assume the states are numbered so that the first s0s_{0} states are the states of S0\mathcal{S}_{0}, the next s1s_{1} are those of S1\mathcal{S}_{1}, and so forth. Thus the matrix [P][P] for the chain has the block form given by

[P] = \begin{bmatrix} 0 & [P_{0}] & \ddots & \ddots & 0 \\ 0 & 0 & [P_{1}] & \ddots & \ddots \\ \ddots & \ddots & \ddots & \ddots & \ddots \\ 0 & 0 & \ddots & \ddots & [P_{d-2}] \\[P_{d-1}] & 0 & \ddots & \ddots & 0 \end{bmatrix},

where [Pm][P_{m}] has dimension sm×sm+1s_{m} \times s_{m+1} for 0≤m≤d−10 \leq m \leq d-1, where the index (d−1)+1(d-1)+1 is interpreted as 0 throughout (i.e., mod dd arithmetic is used on indices). In what follows it is often more convenient to express [Pm][P_{m}] as an M×M\mathsf{M} \times \mathsf{M} matrix [Pm′][P_{m}'] whose entries are 0 except for the rows of Sm\mathcal{S}_{m} and the columns of Sm+1\mathcal{S}_{m+1}, where the entries are equal to those of [Pm][P_{m}]. In this view, [P]=∑m=0d−1[Pm′][P] = \sum_{m=0}^{d-1}[P_{m}'].

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(a)

Show that [Pd][P^{d}] has the form

[Pd]=[[Q0]0⋱00[Q1]⋱⋱00⋱[Qd−1]], [P^{d}] = \begin{bmatrix} [Q_{0}] & 0 & \ddots & 0 \\ 0 & [Q_{1}] & \ddots & \ddots \\ 0 & 0 & \ddots & [Q_{d-1}] \end{bmatrix},

where [Qm]=[Pm][Pm+1]⋯[Pd−1][P0]⋯[Pm−1][Q_{m}] = [P_{m}][P_{m+1}] \cdots [P_{d-1}][P_{0}] \cdots [P_{m-1}]. Expressing [Qm][Q_{m}] as an M×M\mathsf{M} \times \mathsf{M} matrix [Qm′][Q_{m}'] whose entries are 0 except for the rows and columns of Sm\mathcal{S}_{m} where the entries are equal to those of [Qm][Q_{m}], this becomes [Pd]=∑m=0d−1[Qm′][P^{d}] = \sum_{m=0}^{d-1}[Q_{m}'].

(b)

Show that [Qm][Q_{m}] is the matrix of an ergodic Markov chain and let its eigenvector of eigenvalue 1 (scaled to be a probability vector) be π^m\hat{\pi }_{m} and let ν^m\hat{\nu }_{m} be the corresponding right eigenvector scaled so that π^mν^m=1\hat{\pi }_{m}\hat{\nu }_{m} = 1. Let π^m′\hat{\pi }_{m}' and ν^m′\hat{\nu }_{m}' be the corresponding M\mathsf{M}-vectors. Show that lim⁡n→∞[Pnd]=∑m=0d−1ν^m′π^m′\lim_{n \rightarrow \infty }[P^{nd}] = \sum_{m=0}^{d-1} \hat{\nu }_{m}'\hat{\pi }_{m}'.

(c)

Show that π^m′[Pm′]=π^m+1′\hat{\pi }_{m}'[P_{m}'] = \hat{\pi }_{m+1}'. Note that π^m′\hat{\pi }_{m}' is an M\mathsf{M}-tuple that is non-zero only on the components of Sm\mathcal{S}_{m}.

(d)

Let ϕ=2π−1/d\phi = 2\pi \sqrt{-1}/d and let π(k)=∑m=0d−1π^m′emkϕ\pi^{(k)} = \sum_{m=0}^{d-1} \hat{\pi }_{m}'e^{mk\phi }. Show that π(k)\pi^{(k)} is a left eigenvector of [P][P] of eigenvalue e−ϕke^{-\phi k}.

Problem 4.20

(Continuation of Exercise 4.19.)

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(a)

Show that, with the eigenvectors defined in Exercise 4.19,

lim⁡n→∞[Pnd][P]=∑i=0d−1ν(i)π(i+1),(A.56) \lim _{n \rightarrow \infty } [P^{nd}][P] = \sum _{i=0}^{d-1} \nu ^{(i)}\pi ^{(i+1)}, \tag {A.56}

where, as before, (d−1)+1(d-1)+1 is taken to be 0.

(b)

Show that, for 1≤j<d1 \leq j < d,

lim⁡n→∞[Pnd][Pj]=∑i=0d−1ν(i)π(i+j).(A.57) \lim _{n \rightarrow \infty } [P^{nd}][P^{j}] = \sum _{i=0}^{d-1} \nu ^{(i)}\pi ^{(i+j)}. \tag {A.57}
(c)

Show that

lim⁡n→∞[Pnd]{I+[P]+⋯+[Pd−1]}=(∑i=0d−1ν(i))(∑j=0d−1π(i+j)).(A.58) \lim _{n \rightarrow \infty } [P^{nd}]\left\{ I + [P] + \cdots + [P^{d-1}]\right\} = \left(\sum _{i=0}^{d-1} \nu ^{(i)}\right) \left(\sum _{j=0}^{d-1} \pi ^{(i+j)}\right). \tag {A.58}
(d)

Show that

lim⁡n→∞1d([Pn]+[Pn+1]+⋯+[Pn+d−1])=e→π,(A.59) \lim _{n \rightarrow \infty } \frac{1}{d}\left([P^{n}] + [P^{n+1}] + \cdots + [P^{n+d-1}]\right) = \overrightarrow {e}\pi , \tag {A.59}

where π\pi is the steady-state probability vector for [P][P]. Hint: Show that e→=∑mν(m)\overrightarrow {e} = \sum_{m} \nu^{(m)} and π=(1/d)∑mπ(m)\pi = (1/d)\sum_{m} \pi^{(m)}.

(e)

Show that the above result is also valid for periodic unichains.

Problem 4.21

Suppose AA and BB are each ergodic Markov chains with transition probabilities {PAi,Aj}\left\{ P_{A_{i},A_{j}}\right\} and {PBi,Bj}\left\{ P_{B_{i},B_{j}}\right\} respectively. Denote the steady-state probabilities of AA and BB by {πAi}\left\{ \pi_{A_{i}}\right\} and {πBi}\left\{ \pi_{B_{i}}\right\} respectively. The chains are now connected and modified as shown below. In particular, states A1A_{1} and B1B_{1} are connected and the new transition probabilities P′P' for the combined chain are given by

PA1,B1′=ε,PA1,Aj′=(1−ε)PA1,Ajfor all Aj;PB1,A1′=δ,PB1,Bj′=(1−δ)PB1,Bjfor all Bj. \begin{aligned} P_{A_{1},B_{1}}' & = \varepsilon , & P_{A_{1},A_{j}}' & = (1-\varepsilon )P_{A_{1},A_{j}} & \text{for all } A_{j}; \\ P_{B_{1},A_{1}}' & = \delta , & P_{B_{1},B_{j}}' & = (1-\delta )P_{B_{1},B_{j}} & \text{for all } B_{j}. \end{aligned}

All other transition probabilities remain the same. Think intuitively of ε\varepsilon and δ\delta as being small, but do not make any approximations in what follows. Give your answers to the following questions as functions of ε\varepsilon, δ\delta, {πAi}\left\{ \pi_{A_{i}}\right\} and {πBi}\left\{ \pi_{B_{i}}\right\}.

[width=0.85figs/figs_original/ch4_ex21.png

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(a)

Assume that ϵ>0\epsilon > 0, δ=0\delta = 0 (i.e., that AA is a set of transient states in the combined chain). Starting in state A1A_{1}, find the conditional expected time to return to A1A_{1} given that the first transition is to some state in chain AA.

(b)

Assume that ϵ>0\epsilon > 0, δ=0\delta = 0. Find TA,BT_{A,B}, the expected time to first reach state B1B_{1} starting from state A1A_{1}. Your answer should be a function of ϵ\epsilon and the original steady-state probabilities {πAi}\left\{ \pi_{A_{i}}\right\} in chain AA.

(c)

Assume ε>0\varepsilon > 0, δ>0\delta > 0. Find TB,AT_{B,A}, the expected time to first reach state A1A_{1}, starting in state B1B_{1}. Your answer should depend only on δ\delta and {πBi}\left\{ \pi_{B_{i}}\right\}.

(d)

Assume ε>0\varepsilon > 0 and δ>0\delta > 0. Find P′(A)P'(A), the steady-state probability that the combined chain is in one of the states {Aj}\left\{ A_{j}\right\} of the original chain AA.

(e)

Assume ε>0\varepsilon > 0, δ=0\delta = 0. For each state Aj≠A1A_{j} \neq A_{1} in AA, find vAjv_{A_{j}}, the expected number of visits to state AjA_{j}, starting in state A1A_{1}, before reaching B1B_{1}. Your answer should depend only on ε\varepsilon and {πAi}\left\{ \pi_{A_{i}}\right\}.

(f)

Assume ε>0\varepsilon > 0, δ>0\delta > 0. For each state AjA_{j} in AA, find πAj′\pi_{A_{j}}', the steady-state probability of being in state AjA_{j} in the combined chain. Hint: Be careful in your treatment of state A1A_{1}.

Problem 4.22

Section 4.5.1 showed how to find the expected first-passage times to a fixed state, say 1, from all other states. It is often desirable to include the expected first recurrence time from state 1 to return to state 1. This can be done by splitting state 1 into two states, first an initial state with no transitions coming into it but the original transitions going out, and second, a final trapping state with the original transitions coming in.

?
(a)

For the chain on the left side of Figure 4.6 (a four-state recurrent Markov chain, restated below), draw the graph for the modified chain with five states where state 1 has been split into two states.

[width=0.5figs/figs_original/ch4_fig46.png

(b)

Suppose one has found the expected first-passage times vjv_{j} for states j=2,…,4j = 2, \ldots , 4 (or in general from 2 to M\mathsf{M}). Find an expression for v1v_{1}, the expected first recurrence time for state 1 in terms of v2,v3,…vMv_{2}, v_{3}, \ldots v_{\mathsf{M}} and P12,…,P1MP_{12}, \ldots , P_{1\mathsf{M}}.

Remark.
?

I added the left-hand chain of Figure 4.6, cited bare in part (a), restated from the source PDF's Section 4.5.1, so the problem is solvable without looking it up.

Problem 4.23
?
(a)

Assume throughout that [P][P] is the transition matrix of a unichain (and thus the eigenvalue 1 has multiplicity 1). Show that a solution to the equation [P]w→−w→=r→−ge→[P]\overrightarrow {w} - \overrightarrow {w} = \overrightarrow {r} - g\overrightarrow {e} exists if and only if r→−ge→\overrightarrow {r} - g\overrightarrow {e} lies in the column space of [P−I][P-I], where [I][I] is the identity matrix.

(b)

Show that this column space is the set of vectors x→\overrightarrow {x} for which πx→=0\pi \overrightarrow {x} = 0. Then show that r→−ge→\overrightarrow {r} - g\overrightarrow {e} lies in this column space.

(c)

Show that, with the extra constraint that πw→=0\pi \overrightarrow {w} = 0, the equation [P]w→−w→=r→−ge→[P]\overrightarrow {w} - \overrightarrow {w} = \overrightarrow {r} - g\overrightarrow {e} has a unique solution.

Problem 4.24

For the Markov chain with rewards shown below (states 1 and 2, transition probabilities P12=P21=0.01P_{12}=P_{21}=0.01, P11=P22=0.99P_{11}=P_{22}=0.99, rewards r1=0r_{1}=0, r2=1r_{2}=1):

[width=0.45figs/figs_original/ch4_fig48.png

?
(a)

Find the solution to (4.37)(4.37) (referenced here; restated: the relative-gain equation w→+ge→=[P]w→+r→\overrightarrow {w} + g\overrightarrow {e} = [P]\overrightarrow {w} + \overrightarrow {r}, together with the normalization πw→=0\pi \overrightarrow {w}=0) and find the gain gg.

(b)

Modify the figure above by letting P12P_{12} be an arbitrary probability. Find gg and w→\overrightarrow {w} again and give an intuitive explanation of why P12P_{12} affects w2w_{2}.

Remark.
?

I added a restatement of equation (4.37), cited bare in part (a), from Section 4.5 of the source PDF, so the problem is solvable without looking it up.

Problem 4.25
?
(a)

Show that the gain per stage gg is 0. Hint: Show that r→\overrightarrow {r} is zero where the steady-state vector π\pi is non-zero.

(b)

Let [PR][P_{\mathcal{R}}] be the transition matrix for the recurrent states and let r→R=0\overrightarrow {r}_{\mathcal{R}} = 0 be the reward vector and w→R\overrightarrow {w}_{\mathcal{R}} the relative-gain vector for [PR][P_{\mathcal{R}}]. Show that w→R=0\overrightarrow {w}_{\mathcal{R}} = 0. Hint: Use Theorem 4.5.4.

(c)

Show that wi=0w_{i} = 0 for all i∈Ri \in \mathcal{R}. Hint: Compare the relative-gain equations for [P][P] to those for [PR][P_{\mathcal{R}}].

(d)

Show that for each n≥0n \geq 0, [Pn]w→=[Pn+1]w→+[Pn]r→[P^{n}]\overrightarrow {w} = [P^{n+1}]\overrightarrow {w} + [P^{n}]\overrightarrow {r}. Hint: Start with the relative-gain equation for [P][P].

(e)

Show that w→=[Pn+1]w→+∑m=0n[Pm]r→\overrightarrow {w} = [P^{n+1}]\overrightarrow {w} + \sum_{m=0}^{n}[P^{m}]\overrightarrow {r}. Hint: Sum the result in (d).

(f)

Show that lim⁡n→∞[Pn+1]w→=0\lim_{n \rightarrow \infty }[P^{n+1}]\overrightarrow {w} = 0 and that lim⁡n→∞∑m=0n[Pm]r→\lim_{n \rightarrow \infty }\sum_{m=0}^{n}[P^{m}]\overrightarrow {r} is finite, non-negative, and has positive components for ri>0r_{i} > 0. Hint: Use Lemma 4.3.6.

(g)

Demonstrate the final result of the corollary by using the previous results on r→=r→′−r→′′\overrightarrow {r} = \overrightarrow {r}' - \overrightarrow {r}''.

Problem 4.26

Consider the Markov chain below:

[width=0.4figs/figs_original/ch4_ex26.png

?
(a)

Suppose the chain is started in state ii and goes through nn transitions; let vi(n)v_{i}(n) be the expected number of transitions (out of the total of nn) until the chain enters the trapping state, state 1. Find an expression for v→(n)=(v1(n),v2(n),v3(n))T\overrightarrow {v}(n) = (v_{1}(n), v_{2}(n), v_{3}(n))^{\mathsf{T}} in terms of v→(n−1)\overrightarrow {v}(n-1) (take v1(n)=0v_{1}(n) = 0 for all nn). Hint: View the system as a Markov reward system; what is the value of r→\overrightarrow {r}?

(b)

Find the numerical value of lim⁡n→∞v→(n)\lim_{n \rightarrow \infty } \overrightarrow {v}(n). Interpret the meaning of the elements viv_{i} in the solution of (4.32)(4.32).

(c)

Give a direct argument why (4.32)(4.32) provides the solution directly to the expected time from each state to enter the trapping state.

Problem 4.27
?
(a)

Show that (4.48)(4.48) (referenced here; restated: the dynamic-programming recursion vi∗(n,u→)=max⁡k→(ri(k)+∑jPij(k)vj∗(n−1,u→))v_{i}^{*}(n, \overrightarrow {u}) = \max_{\overrightarrow {k}} \left(r_{i}^{(k)} + \sum_{j} P_{ij}^{(k)}v_{j}^{*}(n-1,\overrightarrow {u})\right), in vector form v→∗(n,u→)=max⁡k→(r→k→+[P(k→)]v→∗(n−1,u→))\overrightarrow {v}^{*}(n,\overrightarrow {u}) = \max_{\overrightarrow {k}}\left(\overrightarrow {r}^{\overrightarrow {k}} + [P^{(\overrightarrow {k})}]\overrightarrow {v}^{*}(n-1,\overrightarrow {u})\right)) can be rewritten in the more compact form

v→∗(n,u→)=v→∗(1,v→∗(n−1,u→)). \overrightarrow {v}^{*}(n, \overrightarrow {u}) = \overrightarrow {v}^{*}(1, \overrightarrow {v}^{*}(n-1, \overrightarrow {u})).
(b)

Explain why it is also true that

v→∗(2n,u→)=v→∗(n,v→∗(n,u→)).(A.63) \overrightarrow {v}^{*}(2n, \overrightarrow {u}) = \overrightarrow {v}^{*}(n, \overrightarrow {v}^{*}(n, \overrightarrow {u})). \tag {A.63}
(c)

One might guess that (A.63) could be used iteratively, finding v→∗(2n+1,u→)\overrightarrow {v}^{*}(2^{n+1}, \overrightarrow {u}) from v→∗(2n,u→)\overrightarrow {v}^{*}(2^{n}, \overrightarrow {u}). Explain why this is not possible in any straightforward way. Hint: Think through explicitly how one might calculate v→∗(n,v→∗(n,u→))\overrightarrow {v}^{*}(n, \overrightarrow {v}^{*}(n, \overrightarrow {u})) from v→∗(n,u→)\overrightarrow {v}^{*}(n, \overrightarrow {u}).

Remark.
?

I added a restatement of equation (4.48), cited bare in part (a), from Section 4.6 of the source PDF, so the problem is solvable without looking it up.

Problem 4.28

Consider finding the expected time until a given string appears in an IID binary sequence with P(Xi=1)=p1\mathbb {P}\left(X_{i}=1\right) = p_{1}, P(Xi=0)=p0=1−p1\mathbb {P}\left(X_{i}=0\right) = p_{0} = 1-p_{1}.

?
(a)

Following the procedure in Example 4.5.1, draw the 3 state Markov chain for the string (0,1)(0,1). Find the expected number of trials until the first occurrence of that string.

(b)

For (b) and (c), let (a1,a2,a3,…,ak)=(0,1,1,…,1)(a_{1}, a_{2}, a_{3}, \ldots , a_{k}) = (0,1,1,\ldots ,1), i.e., zero followed by k−1k-1 ones. Draw the corresponding Markov chain for k=4k = 4.

(c)

Let viv_{i}, 1≤i≤k1 \leq i \leq k be the expected first-passage time from state ii to state kk. Note that vk=0v_{k} = 0. For each ii, 1≤i<k1 \leq i < k, show that vi=αi+vi+1v_{i} = \alpha_{i} + v_{i+1} and v0=βi+vi+1v_{0} = \beta_{i} + v_{i+1} where αi\alpha_{i} and βi\beta_{i} are each expressed as a product of powers of p0p_{0} and p1p_{1}. Hint: use induction on ii using i=1i=1 as the base. For the inductive step, first find βi+1\beta_{i+1} as a function of βi\beta_{i} starting with i=1i=1 and using the equation v0=1/p0+v1v_{0} = 1/p_{0} + v_{1}.

(d)

Let a→=(0,1,0)\overrightarrow {a} = (0,1,0). Draw the corresponding Markov chain for this string. Evaluate v0v_{0}, the expected time for (0,1,0)(0,1,0) to occur.

Problem 4.29
?
(a)

Find lim⁡n→∞[Pn]\lim_{n \rightarrow \infty }[P^{n}] for the Markov chain below. Hint: Think in terms of the long term transition probabilities. Recall that the edges in the graph for a Markov chain correspond to the positive transition probabilities.

[width=0.35figs/figs_original/ch4_ex29.png

(b)

Let π(1)\pi^{(1)} and π(2)\pi^{(2)} denote the first two rows of lim⁡n→∞[Pn]\lim_{n \rightarrow \infty }[P^{n}] and let ν(1)\nu^{(1)} and ν(2)\nu^{(2)} denote the first two columns of lim⁡n→∞[Pn]\lim_{n \rightarrow \infty }[P^{n}]. Show that π(1)\pi^{(1)} and π(2)\pi^{(2)} are independent left eigenvectors of [P][P], and that ν(1)\nu^{(1)} and ν(2)\nu^{(2)} are independent right eigenvectors of [P][P]. Find the eigenvalue for each eigenvector.

(c)

Let r→\overrightarrow {r} be an arbitrary reward vector and consider the equation

w→+g(1)ν(1)+g(2)ν(2)=r→+[P]w→.(A.64) \overrightarrow {w} + g^{(1)}\nu ^{(1)} + g^{(2)}\nu ^{(2)} = \overrightarrow {r} + [P]\overrightarrow {w}. \tag {A.64}

Determine what values g(1)g^{(1)} and g(2)g^{(2)} must have in order for (A.64) to have a solution. Argue that with the additional constraints w1=w2=0w_{1} = w_{2} = 0, (A.64) has a unique solution for w→\overrightarrow {w} and find that w→\overrightarrow {w}.

Problem 4.30

Let u→\overrightarrow {u} and u→′\overrightarrow {u}' be arbitrary final reward vectors with u→≤u→′\overrightarrow {u} \leq \overrightarrow {u}'.

?
(a)

Let k→\overrightarrow {k} be an arbitrary stationary policy and prove that v→k→(n,u→)≤v→k→(n,u→′)\overrightarrow {v}^{\overrightarrow {k}}(n, \overrightarrow {u}) \leq \overrightarrow {v}^{\overrightarrow {k}}(n, \overrightarrow {u}') for each n≥1n \geq 1.

(b)

For the optimal dynamic policy, prove that v→∗(n,u→)≤v→∗(n,u→′)\overrightarrow {v}^{*}(n, \overrightarrow {u}) \leq \overrightarrow {v}^{*}(n, \overrightarrow {u}') for each n≥1n \geq 1. This is known as the monotonicity theorem.

(c)

Now let u→\overrightarrow {u} and u→′\overrightarrow {u}' be arbitrary. Let α=max⁡i(ui−ui′)\alpha = \max_{i}(u_{i}-u_{i}'). Show that

v→∗(n,u→)≤v→∗(n,u→′)+αe→.(A.65) \overrightarrow {v}^{*}(n, \overrightarrow {u}) \leq \overrightarrow {v}^{*}(n, \overrightarrow {u}') + \alpha \overrightarrow {e}. \tag {A.65}
Problem 4.31

Consider a Markov decision problem with M\mathsf{M} states in which some state, say state 1, is inherently reachable from each other state.

?
(a)

Show that there must be some other state, say state 2, and some decision, k2k_{2}, such that P21(k2)>0P_{21}^{(k_{2})} > 0.

(b)

Show that there must be some other state, say state 3, and some decision, k3k_{3}, such that either P31(k3)>0P_{31}^{(k_{3})} > 0 or P32(k3)>0P_{32}^{(k_{3})} > 0.

(c)

Assume, for some ii, and some set of decisions k2,…,kik_{2}, \ldots , k_{i} that, for each jj, 2≤j≤i2 \leq j \leq i, Pjℓ(kj)>0P_{j\ell }^{(k_{j})} > 0 for some ℓ<j\ell < j (i.e., that each state from 2 to jj has a non-zero transition to a lower numbered state). Show that there is some state (other than 1 to ii), say i+1i+1 and some decision ki+1k_{i+1} such that Pi+1,ℓ(ki+1)>0P_{i+1,\ell }^{(k_{i+1})} > 0 for some ℓ≤i\ell \leq i.

(d)

Use (a), (b), and (c) to observe that there is a stationary policy k→=k1,…,kM\overrightarrow {k} = k_{1}, \ldots , k_{\mathsf{M}} for which state 1 is accessible from each other state.

Problem 4.32

George drives his car to the theater, which is at the end of a one-way street. There are parking places along the side of the street and a parking garage that costs $5 at the theater. Each parking place is independently occupied or unoccupied with probability 1/21/2. If George parks nn parking places away from the theater, it costs him nn cents (in time and shoe leather) to walk the rest of the way. George is myopic and can only see the parking place he is currently passing. If George has not already parked by the time he reaches the nnth place, he first decides whether or not he will park if the place is unoccupied, and then observes the place and acts according to his decision. George can never go back and must park in the parking garage if he has not parked before.

?
(a)

Model the above problem as a 2 state dynamic programming problem. In the `driving' state, state 2, there are two possible decisions: park if the current place is unoccupied or drive on whether or not the current place is unoccupied.

(b)

Find vi∗(n,u→)v_{i}^{*}(n, \overrightarrow {u}), the minimum expected aggregate cost for nn stages (i.e., immediately before observation of the nnth parking place) starting in state i=1i = 1 or 2; it is sufficient to express vi∗(n,u→)v_{i}^{*}(n, \overrightarrow {u}) in terms of vi∗(n−1)v_{i}^{*}(n-1). The final costs, in cents, at stage 0 should be v2(0)=500v_{2}(0) = 500, v1(0)=0v_{1}(0) = 0.

(c)

For what values of nn is the optimal decision the decision to drive on?

(d)

What is the probability that George will park in the garage, assuming that he follows the optimal policy?

Problem 4.33
?
(a)

Show that if two stationary policies k→′\overrightarrow {k}' and k→\overrightarrow {k} have the same recurrent class R′\mathcal{R}' and if ki′=kik_{i}' = k_{i} for all i∈R′i \in \mathcal{R}', then wi′=wiw_{i}' = w_{i} for all i∈R′i \in \mathcal{R}'. Hint: See the first part of the proof of Lemma 4.6.7.

(b)

Assume that k→′\overrightarrow {k}' satisfies (4.50)(4.50) (i.e., that it satisfies the termination condition of the policy improvement algorithm) and that k→\overrightarrow {k} satisfies the conditions of (a). Show that (4.64)(4.64) is satisfied for all states ℓ\ell.

(c)

Show that w→≤w→′\overrightarrow {w} \leq \overrightarrow {w}'. Hint: Follow the reasoning at the end of the proof of Lemma 4.6.7.

Problem 4.34

Consider the dynamic programming problem below with two states and two possible policies, denoted k→\overrightarrow {k} and k→′\overrightarrow {k}'. The policies differ only in state 2.

[width=0.9figs/figs_original/ch4_ex34.png

?
(a)

Find the steady-state gain per stage, gg and g′g', for stationary policies k→\overrightarrow {k} and k→′\overrightarrow {k}'. Show that g=g′g = g'.

(b)

Find the relative-gain vectors, w→\overrightarrow {w} and w→′\overrightarrow {w}', for stationary policies k→\overrightarrow {k} and k→′\overrightarrow {k}'.

(c)

Suppose the final reward, at stage 0, is u1=0u_{1} = 0, u2=uu_{2} = u. For what range of uu does the dynamic programming algorithm use decision k→\overrightarrow {k} in state 2 at stage 1?

(d)

For what range of uu does the dynamic programming algorithm use decision k→\overrightarrow {k} in state 2 at stage 2? At stage nn? You should find that (for this example) the dynamic programming algorithm uses the same decision at each stage nn as it uses in stage 1.

(e)

Find the optimal gain v2∗(n,u→)v_{2}^{*}(n, \overrightarrow {u}) and v1∗(n,u→)v_{1}^{*}(n, \overrightarrow {u}) as a function of stage nn assuming u=10u = 10.

(f)

Find lim⁡n→∞[v∗(n,u→−nge)]\lim_{n \rightarrow \infty }[v^{*}(n, \overrightarrow {u}-nge)] and show how it depends on u→\overrightarrow {u}.

Problem 4.35

Consider a Markov decision problem in which the stationary policies k→\overrightarrow {k} and k→′\overrightarrow {k}' each satisfy (4.50)(4.50) and each correspond to ergodic Markov chains.

?
(a)

Show that if r→k→′+[Pk→′]w→′≥r→k→+[Pk→]w→′\overrightarrow {r}^{\overrightarrow {k}'} + [P^{\overrightarrow {k}'}]\overrightarrow {w}' \geq \overrightarrow {r}^{\overrightarrow {k}} + [P^{\overrightarrow {k}}]\overrightarrow {w}' is not satisfied with equality, then g′>gg' > g.

(b)

Show that r→k→′+[Pk→′]w→′=r→k→+[Pk→]w→′\overrightarrow {r}^{\overrightarrow {k}'} + [P^{\overrightarrow {k}'}]\overrightarrow {w}' = \overrightarrow {r}^{\overrightarrow {k}} + [P^{\overrightarrow {k}}]\overrightarrow {w}'. Hint: Use (a).

(c)

Find the relationship between the relative gain vector w→k→\overrightarrow {w}^{\overrightarrow {k}} for policy k→\overrightarrow {k} and the relative-gain vector w→′\overrightarrow {w}' for policy k→′\overrightarrow {k}'. Hint: Show that r→k→+[Pk→]w→′=ge→+w→′\overrightarrow {r}^{\overrightarrow {k}} + [P^{\overrightarrow {k}}]\overrightarrow {w}' = g\overrightarrow {e} + \overrightarrow {w}'; what does this say about w→\overrightarrow {w} and w→′\overrightarrow {w}'?

(d)

Suppose that policy k→\overrightarrow {k} uses decision 1 in state 1 and policy k→′\overrightarrow {k}' uses decision 2 in state 1 (i.e., k1=1k_{1} = 1 for policy k→\overrightarrow {k} and k1=2k_{1} = 2 for policy k→′\overrightarrow {k}'). What is the relationship between r1(k),P11(k),P12(k),…P1J(k)r_{1}^{(k)}, P_{11}^{(k)}, P_{12}^{(k)}, \ldots P_{1J}^{(k)} for kk equal to 1 and 2?

(e)

Now suppose that policy k→\overrightarrow {k} uses decision 1 in each state and policy k→′\overrightarrow {k}' uses decision 2 in each state. Is it possible that ri(1)>ri(2)r_{i}^{(1)} > r_{i}^{(2)} for all ii? Explain carefully.

(f)

Now assume that ri(1)r_{i}^{(1)} is the same for all ii. Does this change your answer to (e)? Explain.

Problem 4.36

Consider a Markov decision problem with three states. Assume that each stationary policy corresponds to an ergodic Markov chain. It is known that a particular policy k→′=(k1,k2,k3)=(2,4,1)\overrightarrow {k}' = (k_{1}, k_{2}, k_{3}) = (2,4,1) is the unique optimal stationary policy (i.e., the gain per stage in steady state is maximized by always using decision 2 in state 1, decision 4 in state 2, and decision 1 in state 3). As usual, ri(k)r_{i}^{(k)} denotes the reward in state ii under decision kk, and Pij(k)P_{ij}^{(k)} denotes the probability of a transition to state jj given state ii and given the use of decision kk in state ii. Consider the effect of changing the Markov decision problem in each of the following ways (the changes in each part are to be considered in the absence of the changes in the other parts):

?
(a)

r1(1)r_{1}^{(1)} is replaced by r1(1)−1r_{1}^{(1)} - 1.

(b)

r1(2)r_{1}^{(2)} is replaced by r1(2)+1r_{1}^{(2)} + 1.

(c)

r1(k)r_{1}^{(k)} is replaced by r1(k)+1r_{1}^{(k)} + 1 for all state 1 decisions kk.

(d)

For all ii, ri(ki)r_{i}^{(k_{i})} is replaced by ri(ki)+1r_{i}^{(k_{i})} + 1 for the decision kik_{i} of policy k→′\overrightarrow {k}'.

For each of the above changes, answer the following questions; give explanations:

  1. Is the gain per stage, g′g', increased, decreased, or unchanged by the given change?

  2. Is it possible that another policy, k→≠k→′\overrightarrow {k} \neq \overrightarrow {k}', is optimal after the given change?

Problem 4.37

Let k→′\overrightarrow {k}' be the optimal stationary policy for a Markov decision problem and let g′g' and π′\pi ' be the corresponding gain and steady-state probability respectively. Let vi∗(n,u→)v_{i}^{*}(n, \overrightarrow {u}) be the optimal dynamic expected reward for starting in state ii at stage nn with final reward vector u→\overrightarrow {u}.

?
(a)

Show that min⁡i[vi∗(n,u→)−vi∗(n−1,u→)]≤g′≤max⁡i[vi∗(n,u→)−vi∗(n−1,u→)]\min_{i}[v_{i}^{*}(n,\overrightarrow {u}) - v_{i}^{*}(n-1,\overrightarrow {u})] \leq g' \leq \max_{i}[v_{i}^{*}(n,\overrightarrow {u}) - v_{i}^{*}(n-1,\overrightarrow {u})]; n≥1n \geq 1. Hint: Consider premultiplying v→∗(n,u→)−v→∗(n−1,u→)\overrightarrow {v}^{*}(n,\overrightarrow {u}) - \overrightarrow {v}^{*}(n-1,\overrightarrow {u}) by π′\pi ' or π\pi where k→\overrightarrow {k} is the optimal dynamic policy at stage nn.

(b)

Show that the lower bound is non-decreasing in nn and the upper bound is non-increasing in nn.

Problem 4.38

Consider an integer-time queueing system with a finite buffer of size 2. At the beginning of the nthn^{\text{th}} time interval, the queue contains at most two customers. There is a cost of one unit for each customer in queue (i.e., the cost of delaying that customer). If there is one customer in queue, that customer is served. If there are two customers, an extra server is hired at a cost of 3 units and both customers are served. Thus the total immediate cost for two customers in queue is 5, the cost for one customer is 1, and the cost for 0 customers is 0. At the end of the nnth time interval, either 0, 1, or 2 new customers arrive (each with probability 1/31/3).

?
(a)

Assume that the system starts with 0≤i≤20 \leq i \leq 2 customers in queue at time −1-1 (i.e., in stage 1) and terminates at time 0 (stage 0) with a final cost u→\overrightarrow {u} of 5 units for each customer in queue (at the beginning of interval 0). Find the expected aggregate cost vi(1,u→)v_{i}(1,\overrightarrow {u}) for 0≤i≤20 \leq i \leq 2.

(b)

Assume now that the system starts with ii customers in queue at time −2-2 with the same final cost at time 0. Find the expected aggregate cost vi(2,u→)v_{i}(2,\overrightarrow {u}) for 0≤i≤20 \leq i \leq 2.

(c)

For an arbitrary starting time −n-n, find the expected aggregate cost vi(n,u→)v_{i}(n,\overrightarrow {u}) for 0≤i≤20 \leq i \leq 2.

(d)

Find the cost per stage and find the relative cost (gain) vector.

(e)

Now assume that there is a decision maker who can choose whether or not to hire the extra server when there are two customers in queue. If the extra server is not hired, the three-unit fee is saved, but only one of the customers is served. If there are two arrivals in this case, assume that one is turned away at a cost of 5 units. Find the minimum dynamic aggregate expected cost vi∗(1)v_{i}^{*}(1), 0≤i≤20 \leq i \leq 2, for stage 1 with the same final cost as before.

(f)

Find the minimum dynamic aggregate expected cost vi∗(n,u→)v_{i}^{*}(n,\overrightarrow {u}) for stage nn, 0≤i≤20 \leq i \leq 2.

(g)

Now assume a final cost u→\overrightarrow {u} of 1 unit per customer rather than 5, and find the new minimum dynamic aggregate expected cost vi∗(n,u→)v_{i}^{*}(n,\overrightarrow {u}), 0≤i≤20 \leq i \leq 2.