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Paper 1 – Markov Chain
Example
The matrix
�0.2 0.75�
T � � � is a transition matrix for a Markov chain . �0.8 0.25�
13
( a ) Find the characteristic polynomial of T . [ 2 ]
( b ) Hence , write down the values of �
1
and �
2
, the eigenvalues of T , where �1 � �2. [ 2 ]
( c ) Find v , the steady state probability vector for this Markov chain . [ 2 ]
Solution
( a ) The characteristic polynomial of T �det ( T��I
) 0.2 � � 0.75
�
0.8 0.25 � � � ( 0.2 ��)( 0.25 ��) � ( 0.75 )( 0.8 ) � � � � �
2
0.05 0.2� 0.25� � 0.6
2
� � �0.45�� 0.55
A1
11
( b ) �1
�� , �2 � 1 20
( M1 ) for valid approach
( c ) v is the eigenvector of T corresponding to �2 � 1. ( R1 ) for correct reasoning
A2
�15
� � 31
� �v � � �
A1 16 � � � 31 �
[ 2 ]
[ 2 ]
[ 2 ] www . seprodstore . com
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