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Chapter 5: Problem 3

Suppose \(S=\\{0,1\\}\) and $$ p=\left(\begin{array}{cc} 1-\alpha & \alpha \\ \beta & 1-\beta \end{array}\right) $$ Use induction to show that $$ P_{\mu}\left(X_{n}=0\right)=\frac{\beta}{\alpha+\beta}+(1-\alpha-\beta)^{n}\left\\{\mu(0)-\frac{\beta}{\alpha+\beta}\right\\} $$

Short Answer

Expert verified
The statement \(P_{\mu}\left(X_{n}=0\right)=\frac{\beta}{\alpha+\beta}+(1-\alpha-\beta)^{n}\left\{\mu(0)-\frac{\beta}{\alpha+\beta}\right\}\) holds true for all \(n \ge 0\) as shown by the proof by induction elaborated in the steps above.

Step by step solution

01

Base Case

For the base case, when \(n=0,\) the expression becomes \(\mu(0)=\frac{\beta}{\alpha+\beta}+(1-\alpha-\beta)^{0}\left\{\mu(0)-\frac{\beta}{\alpha+\beta}\right\}.\) Simplifying, we get \(\mu(0)=\mu(0).\) This confirms our base case.
02

Induction Assumption

Next, let's assume the statement to be true for some \(n=k,\) i.e., \(P_{\mu}\left(X_{k}=0\right)=\frac{\beta}{\alpha+\beta}+(1-\alpha-\beta)^{k}\left\{\mu(0)-\frac{\beta}{\alpha+\beta}\right\}.\) This will be our induction assumption.
03

Induction Step

Now, we need to show the statement holds for \(n=k+1\) if it holds for \(n=k.\) So, using the one-step transition probabilities, we have: \(P_{\mu}\left(X_{k+1}=0\right)=P_{\mu}\left(X_{1}=0,X_{k}=0\right) + P_{\mu}\left(X_{1}=1,X_{k}=0\right)\) which simplifies to \((1-\alpha)P_{\mu}\left(X_{k}=0\right) + \beta P_{\mu}\left(X_{k}=1\right).\) Replace \(P_{\mu}\left(X_{k}=0\right)\) using our induction assumption and simplify.
04

Simplifying the Induction Step

Substitute the induction assumption into the equation from Step 3, factor, and simplify. This results in \(P_{\mu}\left(X_{k+1}=0\right)=\frac{\beta}{\alpha+\beta}+(1-\alpha-\beta)^{k+1}\left\{\mu(0)-\frac{\beta}{\alpha+\beta}\right\}\), concluding our proof by induction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Induction Proof
An induction proof is a powerful method to prove mathematical statements that assert something is true for all natural numbers. It typically involves two key steps:

  • Base Case: Verify the statement is true for a specific starting value, often for the least natural number, such as 0 or 1.
  • Inductive Step: Show that if the statement holds true for a particular number, it must also be true for the next number in sequence.

These two steps form the bedrock of what is often termed as the Principle of Mathematical Induction. This process starts with proving the base case and then using it, in conjunction with the inductive hypothesis (the assumption that the statement works for an arbitrary natural number), to prove that if one 'domino' falls, the next will necessarily follow, ensuring a domino effect that establishes the statement for all natural numbers.
Probability Theory
Probability theory is the branch of mathematics that deals with the analysis of random events. The main object is to study the likelihood of various outcomes and to understand and quantify the concept of uncertainty. In essence, probability theory assigns a number between 0 and 1, called probability, to the occurrence of an event, where 0 indicates impossibility and 1 indicates certainty.

Understanding probability is essential for a wide range of fields, including science, engineering, economics, statistics, and more importantly, in the context of this problem, in stochastic processes and induction proofs involving random variables.
Transition Probabilities
Transition probabilities are a key element in the study of stochastic processes, specifically Markov processes. They define the chances of transitioning from one state to another in a given step. In our exercise, the matrix

\[ p=\begin{pmatrix}1-\alpha & \alpha \beta & 1-\beta\end{pmatrix} \]

represents the transition probability matrix. The element in the i-th row and j-th column represents the probability of moving from state i to state j in one step. Stable probabilities, which do not change over time, are crucial for determining the long-term behavior of a stochastic process, as seen in the induction step of the provided exercise solution.
Stochastic Processes
Stochastic processes are mathematical objects that represent evolving systems over time where the evolution is governed by probabilistic laws. These processes are characterized by randomness and can model a wide variety of phenomena. A fundamental concept in stochastic processes is the Markov property, which suggests that the future state of the process depends only on the present state, not the history that preceded it.

When a problem asks to use induction with stochastic processes, like the textbook exercise, it often aims to prove a property of the process after 'n' steps. Mathematical induction steps in by ensuring that once the base case is verified, the property will hold for all future states of the stochastic process, under the given transition probabilities.

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Most popular questions from this chapter

Use the Markov property to show that if \(A \in \sigma\left(X_{0}, \ldots, X_{n}\right)\) and \(B \in \sigma\left(X_{n}, X_{n+1}, \ldots\right)\), then for any initial distribution \(\mu\) $$ P_{\mu}\left(A \cap B \mid X_{n}\right)=P_{\mu}\left(A \mid X_{n}\right) P_{\mu}\left(B \mid X_{n}\right) $$

Brother-sister mating. In this scheme, two animals are mated, and among their direct descendants two individuals of opposite sex are selected at random. These animals are mated and the process continues. Suppose each individual can be one of three genotypes \(A A, A a, a a\), and suppose that the type of the offspring is determined by selecting a letter from each parent. With these rules, the pair of genotypes in the \(n\)th generation is a Markov chain with six states: $$ A A, A A \quad A A, A a \quad A A, a a \quad A a, A a \quad A a, a a \quad a a, a a $$ Compute its transition probability.

Suppose \(X, Y \geq 0\) are independent and \(P(X>x)=e^{-\lambda x}\). Show that \(P(X-Y>x)=a e^{-\lambda x}\), where \(a=P(X-Y>0)\).

M/M/oo queue. Consider a telephone system with an infinite number of lines. Let \(X_{n}=\) the number of lines in use at time \(n\), and suppose $$ X_{n+1}=\sum_{m=1}^{X_{n}} \xi_{n, m}+Y_{n+1} $$ where the \(\xi_{n, m}\) are i.i.d. with \(P\left(\xi_{n, m}=1\right)=p\) and \(P\left(\xi_{n, m}=0\right)=1-p\), and \(Y_{n}\) is an independent i.i.d. sequence of Poisson mean \(\lambda\) r.v.'s. In words, for each conversation we flip a coin with probability \(p\) of heads to see if it continues for another minute. Meanwhile, a Poisson mean \(\lambda\) number of conversations start between time \(n\) and \(n+1\). Use (3.9) with \(\varphi(x)=x\) to show that the chain is recurrent for any \(p<1\).

\(\mathrm{M} / \mathrm{G} / \mathbf{1}\) queue. Let \(\xi_{1}, \xi_{2}, \ldots\) be i.i.d. with \(P\left(\xi_{m}=k\right)=a_{k+1}\) for \(k \geq-1\), where the \(a_{k}\) are defined by the formula in Example 1.4. In particular, it follows that \(a_{j}>0\) for all \(j \geq 0\). Let \(S_{n}=x+\xi_{1}+\cdots+\xi_{n}\), where \(x \geq 0\), and let $$ X_{n}=S_{n}+\left(\min _{m \leq n} S_{m}\right)^{-} $$ (i) Show that \(X_{n}\) has the same distribution as the \(M / G / 1\) queue (Example 1.4) starting from \(X_{0}=x\). (ii) Use this to conclude that if \(\mu=\sum k a_{k}<1\), then as \(n \rightarrow \infty\) $$ \frac{1}{n}\left|\left\\{m \leq n: X_{m-1}=0, \xi_{m}=-1\right\\}\right| \rightarrow(1-\mu) \quad \text { a.s. } $$ This gives a roundabout way of getting the result \(\pi(0)=(1-\mu) / a_{0}\) proved in Example \(4.6\).
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