"A machine has 4 components and the machine cannot operate when any one of these components fail. At the beginning of each day, the machine starts running. During any day component $i$ fails with probability $p_i = \frac{1}{2^{i+1}}$ . Once a component fails, the machine stops and the failing component is replaced, and the machine starts operating the next day.\
Model ${X_n, n ∈ \mathbb{Z}}$ as a discrete time markov chain, where\
$$X_n = i \; \text{if component} \; i = 1, . . . , 4\; \; \text{fails on day n and} \; 0 \; \text{if the machine runs without any failure} $$
So I thought that because the machine starts running again the next day with component $i$ retaining it's probability of failure that this model would have the memoryless property. So I had that $$p_{ij} = P(X_{n+1} = j|X_n = i) = P(X_{n+1} = j) = \frac{1}{2^{j+1}}$$
However then the probability $p_{i0} = P(X_{n+1} = 0) =$ Probability no component fails. Now from what I have learnt so far I believed this to then be $(1- \frac{1}{2^2})(1- \frac{1}{2^3})(1- \frac{1}{2^4})(1- \frac{1}{2^5})$. This would then give my probability matrix 5 rows consisting of $<\frac{9765}{16384} \; \frac{1}{4} \; \frac{1}{8} \; \frac{1}{16} \; \frac{1}{32}>$. The sum of the rows is then not 1, so I have gone wrong somewhere but not sure where.
Can someone help please. Thanks
Calculating the probability of no components failing as $(1- \frac{1}{2^2})(1- \frac{1}{2^3})(1- \frac{1}{2^4})(1- \frac{1}{2^5})$ relies on each component failing (or not failing) independently of all others but this is not true here.
It's easy to see they are not independent because only one component can fail on any given day.
For $i=1,2,3,4$, let $F_i$ be the event $\text{"the $i^{th}$ component fails"}$. Then, for example, $P(F_1\cap F_2) = 0 \neq p_1p_2 = P(F_1)P(F_2)$, which shows $F_1,F_2$ are dependent events.
In fact, the only possible outcomes for any given day are:
\begin{eqnarray*} &&\text{No components fail} \\ &&\text{Component $1$ fails} \\ &&\text{Component $2$ fails} \\ &&\text{Component $3$ fails} \\ &&\text{Component $4$ fails.} \\ \end{eqnarray*}
So,
$$P(\text{No components fail}) = 1 - P(F_1) - P(F_2) - P(F_3) - P(F_4) = 1-p_1-p_2-p_3-p_4.$$