Let-P-1-p-1-p-2-p-1-1-p-2-a-1-1-a-1-2-a-2-1-a-2-2-and-that-P-n-be-the-nth-power-of-P-evaluated-as-P-n-P-P-n-1-i-e-by-successive-pre-multiplication-of-P-to-P-2

Question Number 4638 by Yozzii last updated on 17/Feb/16

L e t P = (\begin{matrix} 1 - p_{1} & p_{2} \\ p_{1} & 1 - p_{2} \end{matrix}) = (\begin{matrix} a_{1, 1} & a_{1, 2} \\ a_{2, 1} & a_{2, 2} \end{matrix})

a n d t h a t P^{n} b e t h e n t h p o w e r o f P e v a l u a t e d

a s P^{n} = P \times P^{n - 1}; i . e b y s u c c e s s i v e

p r e - m u l t i p l i c a t i o n o f P t o P^{2}, P^{3}, P^{4}, \dots

u p t o P^{n - 1} . S h o w t h a t t h e e l e m e n t o f

P^{n} i n t h e s e c o n d r o w a n d f i r s t c o l u m n

i s a_{2, 1} = \frac{p_{1} (1 - {(1 - p_{1} - p_{2})}^{n})}{p_{1} + p_{2}} .

{T h e c o l u m n s o f P^{n} r e p r e s e n t

p r o b a b i l i t y v e c t o r s f o r a l l n \in N . H e n c e,

a_{1, 1} + a_{2, 1} = 1 a n d a_{1, 2} + a_{2, 2} = 1 f o r e x a m p l e .}