The following Markov matrix describes a stochastic process involving 3 states 81, 82, 83. The number given by element i, j is the probability of transition from state j to state i; i.e. the column corresponds to the current state and the row corresponds to the future state. future state 9/10( A pictorial representation of the transition probabilities is below. 1/10 s1 s2 s3 s1 9/10 3/5 0 s2 1/10 3/10 1/2 0 1/10 1/2 s3 3/5 3/10 S2 1/10 1/2 S3 (i) Find the characteristic polynomial. (ii) Use the characteristic equation to solve for the second and third eigenvalues, given A₁ = 1. [Recall, a Markov matrix has max(X) = 1] ¹1/2 (iii) Find the first eigenvector, 2₁. (iv) Show that the eigenvectors ₁, ₂ = [−2,1,1], 22 and v V3 by confirming that c₁ = C₂ = c3 = 0 is the unique solution to c₁v₁ = 0. = [3,-4, 1] are linearly independent 3 i=1

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Chapter2: Second-order Linear Odes
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The following Markov matrix describes a stochastic process involving 3 states 81, 82, 83. The number
given by element i, j is the probability of transition from state j to state i; i.e. the column corresponds
to the current state and the row corresponds to the future state.
future state
9/10
A pictorial representation of the transition probabilities is below.
0
1/10
s1
s2 s3
s1 9/10 3/5 0
s2
1/10 3/10 1/2
s3
0 1/10 1/2
3/5
3/10
1/10
1/2
(i) Find the characteristic polynomial.
(ii) Use the characteristic equation to solve for the second and third eigenvalues, given A₁ = 1.
[Recall, a Markov matrix has max(X) = 1]
(iii) Find the first eigenvector, 2₁.
(iv) Show that the eigenvectors 2₁, 22
=
[3,-4, 1] are linearly independent
3
by confirming that c₁ = C₂ = c3 = 0 is the unique solution to ci v ₁ = 0.
[-2, 1, 1], and v
=
i=1
(v) In terms of a linear combination of eigenvectors, give the general solution to the stochastic pro-
cess at the nth iteration.
(vi) With a stable population of 1000, the initial (n=1) distribution is [291, 312, 397] T. Compute the
values of the coefficients attached to each eigenvector.
(vii) What is the long-term trend for this system? That is, what is the stable state?
[Note: you should extract the stable state from the vector solution but it can be solved inde-
pendently by recognising that the input and output is identical when the system is stable. This
is referred to as a fixed point. Then MX = X ⇒ (MI)X = 0. This is equivalent to the
homogeneous equation that was solved for 21
with λ = 1.]
Transcribed Image Text:The following Markov matrix describes a stochastic process involving 3 states 81, 82, 83. The number given by element i, j is the probability of transition from state j to state i; i.e. the column corresponds to the current state and the row corresponds to the future state. future state 9/10 A pictorial representation of the transition probabilities is below. 0 1/10 s1 s2 s3 s1 9/10 3/5 0 s2 1/10 3/10 1/2 s3 0 1/10 1/2 3/5 3/10 1/10 1/2 (i) Find the characteristic polynomial. (ii) Use the characteristic equation to solve for the second and third eigenvalues, given A₁ = 1. [Recall, a Markov matrix has max(X) = 1] (iii) Find the first eigenvector, 2₁. (iv) Show that the eigenvectors 2₁, 22 = [3,-4, 1] are linearly independent 3 by confirming that c₁ = C₂ = c3 = 0 is the unique solution to ci v ₁ = 0. [-2, 1, 1], and v = i=1 (v) In terms of a linear combination of eigenvectors, give the general solution to the stochastic pro- cess at the nth iteration. (vi) With a stable population of 1000, the initial (n=1) distribution is [291, 312, 397] T. Compute the values of the coefficients attached to each eigenvector. (vii) What is the long-term trend for this system? That is, what is the stable state? [Note: you should extract the stable state from the vector solution but it can be solved inde- pendently by recognising that the input and output is identical when the system is stable. This is referred to as a fixed point. Then MX = X ⇒ (MI)X = 0. This is equivalent to the homogeneous equation that was solved for 21 with λ = 1.]
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