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Let $$ P= \begin{pmatrix} p_{0} & p_{1} & p_{2} & p_{3} & \cdots \\ 1 & 0 & 0 & 0 & \cdots\\ 0 & 1 & 0 & 0 & \cdots \\ 0 & 0 & 1 & 0 & \cdots \\ 0 & 0 & 0 & 1 & \cdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \end{pmatrix} $$

This Markov chain arises in models of a system component which failes and is replaced instantaneously with an identical component. When this identical component also fails, it too is instantaneously replaced with another identical component, and so on. The Markov chain {$X_{n}$ : n ≥ 0}, denotes the remaining lifetime of the component in use at time step n. The probability $p_{}k$ denotes the probability that a newly installed component will last for k time steps.

(a) Under what condition(s) does a stationary distribution exist for this Markov chain?

(b) Compute this distribution assuming your condition(s) hold.


My thought is that this is a renewal reward process in which a component is used for a specific period of time and then replaced with an identical one. The condition for a) would probably be $z*P=z$ where $z$ is a normalized row vector (elements sum up to 1). The problem is, $P$ has only one non-zero element per row/column so even if my condition is right, I don't know what approach to take to compute the distribution.

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Given $z = (z_0,z_1,z_2,\dots)$, we compute $$ zP = \pmatrix{p_0z_0 + z_1, p_1 z_0 + z_2, p_2z_0 + z_3, \dots} $$ So, if $z$ satisfies $z = zP$, then we have $$ z_0 = p_0z_0 + z_1\\ z_1 = p_1 z_0 + z_2\\ z_2 = p_2 z_0 + z_3\\ \vdots $$ We can rearrange these to get $$ z_1 = (1-p_0)z_0\\ z_2 = z_1 - p_1 z_0 = (1 - p_0 - p_1)z_0\\ z_3 = z_2 - p_2 z_0 = (1 - p_0 - p_1 - p_2)z_0\\ \vdots $$ With this in mind, $P$ will have a stationary distribution if and only if there exists a $z_0$ such that the above defines a distribution. In other words: $P$ will have a stationary distribution if and only if the sum $$ \sum_{k=0}^\infty \left(1 - \sum_{j=0}^k p_k\right) $$ is convergent.

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  • $\begingroup$ Thanks! one more question - how can I know use it to compute b) ? Should I just solve the above equations ? $\endgroup$
    – jreft56
    Commented Jul 28, 2019 at 19:55
  • $\begingroup$ With the “rearranged equations”, it’s enough to choose the $z_0$ that makes it so that the entries of $z$ add up to $1$. $\endgroup$ Commented Jul 28, 2019 at 21:35

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