Given the transition matrix P = [1/2 1/2 1 0] the general expression for P^k is

Question

Given the transition matrix P = [1/2 1/2 1 0] the general expression for P^k is P^k =1/3 [1 - (1-1/2)^k 2 + (-11/2)^k - 1 1 - (-1/2)^k -1] (you do not need to show this). Do the following steps: Show that the formula for P^k is valid for k = 1 and k = 2 by direct calculation. Conclude that P is regular. Calculate the stable state vector v_S for P. Show that lim_k righarrow infinity 1/3 [2 + (-1/2)^k 1 - (-1/2)^k] is equal to the stable state vector for P, hence lim k rightarrowinfinity P^k = M where each row of M is equal to the stable state vector v_S Keeping the same definitions as in the previous part, let w = [w_1, w_2] be an arbitrary state vector v_s. Show by direct calculation that wM = v S. regardless of the choice of M, hence together with lim _k rightarrow infinity P^k = M we can conclude that

Explanation / Answer

a)

for k = 1

P1 = 1/3 [ 2 + (-1/2)^1 1 - (-1/2)^1 ;  2 + (-1/2)^0 1 - (-1/2)^0];

= 1/3 [ 3/2 3/2 ; 3 0]

= [1/2 1/2 ; 1 0];

P2  = 1/3 [ 2 + (-1/2)^2 1 - (-1/2)^2 ;  2 + (-1/2)^1 1 - (-1/2)^1];

= 1/3[ 9/4 3/4 ; 3/2 3/2];

= [0.75 0.25 ; 0.55 0.5];

P*P = [1/2 1/2 ; 1 0]*[[1/2 1/2 ; 1 0]

= [1/2*1/2 +1/2*1 1/2*1/2 +1/2*0 ; 1*1/2 +0*1 1*1/2 + 0 *0]

= [0.75 0.25 ; 0.5 0.5];

hence both are equal

The matrix is "regular" so long as the sum of all values in each row is 1 and each element >= 0.

sum of first row = 1/3 ( 2 + (-1/2)^k + 1 - (-1/2)^k] = 1/3 *3 = 1

sum of 2nd row = 1/3 [2 + (-1/2)^(k-1) 1 - (-1/2)^(k-1) ] = 1

clearly each term >=0

b) stable vector

X P = X

or X(P -I) = 0

X [ -0.5 0.5 ; 1 -1] =0

let X = [x1 x2]

-x1 *0.5 +x2*1 =0

x1 = 2x2

x1 +x2 = 1

so x1 = 2/3 x2 = 1/3

hence it is [2/3 1/3];

solving we get

c) lim is 1/3 [ 2 1 ]

= [2/3 1/3] as lim k-> infinity (1/2)^k = 0

same as stable state vector

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.