Stochastic Process with Stochastic matrix

I have an exercise with an answer which I don't completely understand:

Here's the exercise: Suppose $$p$$ is the transition (stochastic) matrix defined by $$p= \begin{pmatrix} 1-\alpha & \alpha \\ \beta & 1- \beta \\ \end{pmatrix}$$

What I need to find is $$\Bbb P(X_n=1 | X_0=1)$$.

So here's the solution:

Noting that $$p^{n+1}=p^np$$ this gives:

$$p^{n+1}_{11}= p^{n}_{12}\beta + p^n_{11}(1-\alpha)$$

and since p is a stochastic matrix $$p^n_{11}+p^n_{12}=1$$ for all $$n$$.

Then we suub the second into the first and solve for $$p^n_{11}$$ $$\Rightarrow$$ $$p^{n+1}_{11}=(1-\alpha-\beta)p^n_{11}+\beta$$

So for some reason they also write that $$p^0_{11}=1.$$

So then the solution is :

$$p^n_{11}= \begin{cases} \frac{\alpha}{\alpha+\beta} + \frac{\alpha}{\alpha+\beta}(1-\alpha-\beta)^n , & \text{for }\alpha+\beta \gt 0\\ 1, & \text{for } \alpha+\beta=0. \end{cases}$$

So how did they deduce the solution?

Is this even a correct solution?

There are two questions.

1. Why is $$p_{11}^0 = 1$$? This concerns powers of the matrix (which I will write as $$P$$). Because matrix powers may be unfamiliar, here is a quick sketch of the theory. For powers to be useful, they need to have some basic properties, one of which is that $$P^1=P$$ and another is that for any natural number $$n,$$ $$P^{n+1} = P\,P^n = P^n\, P.$$ When $$n=0$$ this means $$P = P^1 = P^{0+1} = P^0\,P = P\, P^0.$$ There is an obvious universal solution $$P^0 = \mathbb{I} = \pmatrix{1&0\\0&1}.$$ In particular, its upper left hand entry is $$(P^0)_{11} = 1.$$

2. How does one solve a recursive sequence of equations $$\begin{cases}x_0 &= 1 \\ x_{n+1} &= \rho\, x_n + \lambda, & n \gt 0\end{cases}$$ Here $$x_n$$ refers to the $$1,1$$ entry of $$P^n;$$ that is, to $$(P^n)_{11}.$$ The argument recounted in the question establishes that $$\rho = 1-\alpha-\beta$$ and $$\lambda=\beta.$$

One way is to simplify this to a "homogeneous" relation by setting $$y_n + \mu = x_n$$ for some number $$\mu$$ to be found. Substituting in the recursion gives $$y_{n+1} + \mu = x_{n+1} = \rho\, x_n + \lambda = \rho(y_n + \mu) + \lambda.$$ Equivalently, we may isolate $$y_{n+1}$$ to give $$y_{n+1} = \rho(y_n + \mu) + \lambda - \mu = \rho\,y_n + (\lambda + (\rho-1)\mu).$$ Thus if we set $$\mu = \frac{\lambda}{1-\rho}$$ (assuming $$\rho\ne 1$$), this simplifies to $$y_{n+1} = \rho\, y_n.$$ This is a geometric series with the well-known (unique) solution $$y_{n} = \rho^n\, y_0 = \rho^n(x_0 - \mu),\quad n=0, 1, 2, \ldots.$$ Adding $$\mu$$ back to both sides yields $$x_{n} = y_{n} + \mu = \rho^n(x_0 - \mu) + \mu.$$ In the present application $$\mu = \frac{\lambda}{1-\rho} = \frac{\beta}{\alpha+\beta}.$$ Coupled with the initial condition $$x_0 = (P^0)_{11} = 1$$ this gives the stated solution when $$\rho \ne 1;$$ that is, when $$\alpha+\beta\ne 0.$$ In this last case necessarily $$\alpha=\beta=0$$ (because these are both transition probabilities), whence the recursion is simply $$(P^{n+1})_{11} = (P^{n})_{11}$$ whose solution obviously is $$(P^{n})_{11}=(P^0)_{11} = 1$$ for all $$n.$$

• Thank you so much for clarifying! And thanks for editing! – user6969 Jan 14 at 18:27
• so $x_n$ would be equal to $\rho^{n-1}(x_0-\mu)+\mu$? Because you defined $x_{n+1}$ to be $p^{n+1}_{11}$ – user6969 Jan 14 at 18:41
• As I wrote at the outset, "$x_n$" is shorthand for "$(P^n)_{11}.$" – whuber Jan 14 at 18:44
• I understand this, but in the end you wrote how much is $x_{n+1}$ equal to but not $x_n$. So that's why I asked, not because I was confused by your notation. – user6969 Jan 14 at 18:48
• oh, okay, I see now. Yes, thank you! – user6969 Jan 14 at 18:52