Question

Let 0 be an absorbing state in a success runs Markov chain \(\left\\{X_{n}\right\\}\) having Irunsition probabilities \(P_{00}=1\) and \(P_{i, i+1}=p_{i}=1-P_{l, 0}\) for \(i=1,2, \ldots . .\) ??uppose \(p_{i} \geq p_{i+1} \geq \ldots\), and let \(a\) be the unique value for which \(a p_{a-1} /(a-1)>\) \(\mathrm{I} \cdot(a+1) p_{a} / a .\) Define $$ f(i)= \begin{cases}0, & \text { for } \quad i=0 \\ a p_{i} p_{i+1} \cdots p_{a-1}, & \text { for } \quad 1 \leq i

Short answer

The unique value \(a\) can be determined from the given conditions: \(a p_{a-1} / (a - 1) > I \cdot (a + 1) p_a / a\) and \(p_{i} \geq p_{i+1} \geq \ldots\), given that the success-run Markov chain has an absorbing state at 0 and certain transition probabilities. Using this value of \(a\), we can define the function \(f(i)\) as: $$ f(i)= \begin{cases}0, & \text { for } \quad i=0 \\\ a p_{i} p_{i+1} \cdots p_{a-1}, & \text { for } \quad 1 \leq i<a \\\ i, & \text { for } \quad i \geq a\end{cases} $$ This function describes the state of the Markov chain based on the given probability inequalities and the unique value \(a\).

Step-by-step solution

To find the unique value \(a\), we need to use the inequalities provided: \(a p_{a-1} / (a - 1) > I \cdot (a + 1) p_a / a\). Since \(p_{i} \geq p_{i+1} \geq \ldots\), there exists a unique value \(a\) for which this inequality holds. #Step 2: Define the function f(i)#

Now that we have the unique value \(a\), we can define the function \(f(i)\) as given in the exercise. $$ f(i)= \begin{cases}0, & \text { for } \quad i=0 \\\ a p_{i} p_{i+1} \cdots p_{a-1}, & \text { for } \quad 1 \leq i<a \\\ i, & \text { for } \quad i \geq a\end{cases} $$ We have constructed the function \(f(i)\) based on the given transition probabilities and the unique value \(a\). This function will describe the state of the Markov chain depending on the probability inequalities and the unique value \(a\).