Question

Let \(\left\\{X_{n}\right\\}\) be a success runs Markov chain having transition probabilities \(P_{i, i+1}=p_{i}=1-P_{i, 0}\), for \(i=0,1, \ldots\) Suppose \(01 \geq(a+1) \beta_{P a} / a .\) Define $$ f(i)=\left\\{\begin{array}{lll} a \beta^{a-t} p_{i} \cdot p_{i+1} \cdots p_{a-1}, & \text { for } \quad i

Short answer

#tag_title# Part (b) - Prove \(f(i) \geq \beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right]\) #tag_content# To show that \(f(i) \geq \beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right]\), note that \(\beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right] = \beta \left[ p_i f(i+1) + (1-p_i) f(0)\right]\), since the Markov chain can either transition to state \(i+1\) with probability \(p_i\) or reset to state 0 with probability \(1-p_i\). Now consider two cases: Case 1: \(i < a\) \(\beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right] = \beta \left[ p_i a \beta^{a-i-1} p_{i+1} \cdot ... \cdot p_{a-1} + (1-p_i) f(0)\right] \leq \beta\left[p_i a \beta^{a-i-1} p_i + (1-p_i) a \beta^{a-i} p_i\right] = f(i)\), because \(f(0) \leq a \) and \(\beta^{a-i-1} \geq \beta^{a-i}\), as \(0 < \beta < 1\). Case 2: \(i \geq a\) \(\beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right] = \beta \left[ p_i (i+1) + (1-p_i) f(0)\right] \leq \beta \left[p_i (i+1) + (1-p_i) i\right] = \beta i\), since \(f(0) \leq i\) in this case. As \(f(i) = i \geq \beta i\), the inequality holds. In both cases, we have shown that \(f(i) \geq \beta E\left[f\left(X_{n}\right) \mid X_{n-1}=i\right]\), proving the desired result.

Step-by-step solution

Part (a) - Prove \(f(i) \geq i\) for all \(i\)

To show that \(f(i) \geq i\) for all \(i\), we will break it down into two cases: when \(i < a\) and when \(i \geq a\). Case 1: \(i < a\) \(f(i) = a \beta^{a-i} p_i \cdot p_{i+1} \cdot ... \cdot p_{a-1}\) Since \(p_i \geq p_{i+1} \geq ... \geq p_{a-1} > 0\), we have \(f(i) \geq a \beta^{a-i} p_i = (a-i) \cdot (\frac{a \beta^{a-i}}{a-i} \cdot p_i) \geq (a-i) \cdot (\frac{a+1}{a} \cdot \frac{a \beta^{a-i}}{a-1} \cdot p_{a-1}) \geq (a-i)\), as defined. Since \(i < a\), \(a-i > 0\) and thus \(f(i) \geq i\). Case 2: \(i \geq a\) \(f(i) = i\) In this case, \(f(i) = i\) by definition, so the inequality holds. Thus, we have shown that \(f(i) \geq i\) for all \(i\).