A communication system (Markov chain)

[Win_odd Dhamnekar]

A Communications System) Consider a communications system that transmits the digits 0 and 1. Each digit transmitted must pass through several stages, at each of which there is a probability p that the digit entered will be unchanged when it leaves. Letting [imath]X_n[/imath] denote the digit entering the nth stage, then [imath]\{X_n, n = 0,1 , . . .\}[/imath] is a two-state Markov chain having transition probability matrix[math]P=\begin{Vmatrix} p & 1-p \\ 1-p & p \end{Vmatrix}[/math]
Now, how to show by mathematical indiction, that [math]P^{(n)} = \begin{Vmatrix} \frac12 + \frac12 (2p-1)^n & \frac12 -\frac12(2p-1)^n \\ \frac12 - \frac12 (2p-1)^n & \frac12 + \frac12(2p-1)^n \end {Vmatrix}[/math]
How to answer this question?

 

[Deleted member 4993]

A Communications System) Consider a communications system that transmits the digits 0 and 1. Each digit transmitted must pass through several stages, at each of which there is a probability p that the digit entered will be unchanged when it leaves. Letting [imath]X_n[/imath] denote the digit entering the nth stage, then [imath]\{X_n, n = 0,1 , . . .\}[/imath] is a two-state Markov chain having transition probability matrix[math]P=\begin{Vmatrix} p & 1-p \\ 1-p & p \end{Vmatrix}[/math]
Now, how to show by mathematical indiction, that [math]P^{(n)} = \begin{Vmatrix} \frac12 + \frac12 (2p-1)^n & \frac12 -\frac12(2p-1)^n \\ \frac12 - \frac12 (2p-1)^n & \frac12 + \frac12(2p-1)^n \end {Vmatrix}[/math]
How to answer this question?

It is really disheartening that after 150 posts (in about 4 years}, we have to still remind you:

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[BigBeachBanana]

A Communications System) Consider a communications system that transmits the digits 0 and 1. Each digit transmitted must pass through several stages, at each of which there is a probability p that the digit entered will be unchanged when it leaves. Letting [imath]X_n[/imath] denote the digit entering the nth stage, then [imath]\{X_n, n = 0,1 , . . .\}[/imath] is a two-state Markov chain having transition probability matrix[math]P=\begin{Vmatrix} p & 1-p \\ 1-p & p \end{Vmatrix}[/math]
Now, how to show by mathematical indiction, that [math]P^{(n)} = \begin{Vmatrix} \frac12 + \frac12 (2p-1)^n & \frac12 -\frac12(2p-1)^n \\ \frac12 - \frac12 (2p-1)^n & \frac12 + \frac12(2p-1)^n \end {Vmatrix}[/math]
How to answer this question?

First of all, [imath]n[/imath] should exclude 0, i.e. [imath]n \in \{1,2,...\}[/imath]

Test the base case [imath](n=1)[/imath]. Is it true?
If the base is true, then assume [imath]n=k[/imath] is true and show that for [imath]n = k+1[/imath]:

[imath]P^{k+1}=P^k\times P^1 = \begin{Vmatrix} \frac12 + \frac12 (2p-1)^{k+1} & \frac12 -\frac12(2p-1)^{k+1} \\ \frac12 - \frac12 (2p-1)^{k+1} & \frac12 + \frac12(2p-1)^{k+1} \end {Vmatrix}[/imath]

 

[Win_odd Dhamnekar]

It is really disheartening that after 150 posts (in about 4 years}, we have to still remind you:

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

Sorry for ignoring to follow the rules of posting.

[math]P^{(2)} = \begin{Vmatrix} 2p^2 -2p +1 & -(2p^2-2p)\\- (2p^2-2p) & 2p^2 -2p +1 \end{Vmatrix} = \begin{Vmatrix} \frac12 + \frac12(2p-1)^2 & \frac12 - \frac12(2p-1)^2 \\ \frac12 - \frac12(2p-1)^2 & \frac12 + \frac12(2p-1)^2 \end{Vmatrix}[/math]
Hence,
[math]P^{(n)} =\begin{Vmatrix} \frac12 + \frac12 (2p-1)^n & \frac12 -\frac12(2p-1)^n \\ \frac12 -\frac12(2p-1)^n & \frac12 +\frac12(2p-1)^n \end{Vmatrix}[/math]
I solved this problem myself. Thanks to advise me to try to answer this question myself. It was easy to answer this question.

 

[BigBeachBanana]

Sorry for ignoring to follow the rules of posting.

[math]P^{(2)} = \begin{Vmatrix} 2p^2 -2p +1 & -(2p^2-2p)\\- (2p^2-2p) & 2p^2 -2p +1 \end{Vmatrix} = \begin{Vmatrix} \frac12 + \frac12(2p-1)^2 & \frac12 - \frac12(2p-1)^2 \\ \frac12 - \frac12(2p-1)^2 & \frac12 + \frac12(2p-1)^2 \end{Vmatrix}[/math]
Hence,
[math]P^{(n)} =\begin{Vmatrix} \frac12 + \frac12 (2p-1)^n & \frac12 -\frac12(2p-1)^n \\ \frac12 -\frac12(2p-1)^n & \frac12 +\frac12(2p-1)^n \end{Vmatrix}[/math]
I solved this problem myself. Thanks to advise me to try to answer this question myself. It was easy to answer this question.

You've only shown that it's true for n=2, not for all n. That's where induction proof comes in.
If I were to ask you is it true for n = 100? Are you going to multiply the matrix 100 times?
Refer to post #3.