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

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 ,P^3 ,P^4 ,... up to P^(n−1) . Show that the element of P^n in the second row and first column is a_(2,1) =((p_1 (1−(1−p_1 −p_2 )^n ))/(p_1 +p_2 )). {The columns of P^n represent probability vectors for all n∈N. Hence, a_(1,1) +a_(2,1) =1 and a_(1,2) +a_(2,2) =1 for example.}

$${Let}\:\boldsymbol{\mathrm{P}}=\begin{pmatrix}{\mathrm{1}−{p}_{\mathrm{1}} }&{{p}_{\mathrm{2}} }\\{{p}_{\mathrm{1}} }&{\mathrm{1}−{p}_{\mathrm{2}} }\end{pmatrix}=\begin{pmatrix}{{a}_{\mathrm{1},\mathrm{1}} }&{{a}_{\mathrm{1},\mathrm{2}} }\\{{a}_{\mathrm{2},\mathrm{1}} }&{{a}_{\mathrm{2},\mathrm{2}} }\end{pmatrix} \\ $$$${and}\:{that}\:\boldsymbol{\mathrm{P}}^{{n}} \:{be}\:{the}\:{nth}\:{power}\:{of}\:\boldsymbol{\mathrm{P}}\:{evaluated} \\ $$$${as}\:\boldsymbol{\mathrm{P}}^{{n}} =\boldsymbol{\mathrm{P}}×\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} ;{i}.{e}\:{by}\:{successive}\: \\ $$$${pre}−{multiplication}\:{of}\:\boldsymbol{\mathrm{P}}\:{to}\:\boldsymbol{\mathrm{P}}^{\mathrm{2}} ,\boldsymbol{\mathrm{P}}^{\mathrm{3}} ,\boldsymbol{\mathrm{P}}^{\mathrm{4}} ,... \\ $$$${up}\:{to}\:\boldsymbol{\mathrm{P}}^{{n}−\mathrm{1}} .\:{Show}\:{that}\:{the}\:{element}\:{of} \\ $$$$\boldsymbol{\mathrm{P}}^{{n}} \:{in}\:{the}\:{second}\:{row}\:{and}\:{first}\:{column} \\ $$$${is}\:{a}_{\mathrm{2},\mathrm{1}} =\frac{{p}_{\mathrm{1}} \left(\mathrm{1}−\left(\mathrm{1}−{p}_{\mathrm{1}} −{p}_{\mathrm{2}} \right)^{{n}} \right)}{{p}_{\mathrm{1}} +{p}_{\mathrm{2}} }.\: \\ $$$$\left\{{The}\:{columns}\:{of}\:\boldsymbol{\mathrm{P}}^{{n}} \:{represent}\:\right. \\ $$$${probability}\:{vectors}\:{for}\:{all}\:{n}\in\mathbb{N}.\:{Hence}, \\ $$$$\left.{a}_{\mathrm{1},\mathrm{1}} +{a}_{\mathrm{2},\mathrm{1}} =\mathrm{1}\:{and}\:{a}_{\mathrm{1},\mathrm{2}} +{a}_{\mathrm{2},\mathrm{2}} =\mathrm{1}\:{for}\:{example}.\right\} \\ $$$$ \\ $$

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