/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Let \(X_{i}, i \geqslant 0\) be ... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X_{i}, i \geqslant 0\) be independent and identically distributed random variables with probability mass function $$ p(j)=P\left[X_{i}=i\right\\}, \quad j=1, \ldots, m, \quad \sum_{j=1}^{m} P(j)=1 $$ Find \(E[N]\), where \(N=\min \left[n>0: X_{n}=X_{0}\right\\}\)

Short Answer

Expert verified
The short answer for the expected value of \(N\) in the given situation is: \[ E[N] = \frac{1}{P(j)} \] where \(P(j)\) is the probability of picking outcome \(j\) as the value of the random variable when \(X_0 = j\).

Step by step solution

01

Determine the Probability Mass Function for \(N\)

If \(X_0 = j\), then to find \(P(N=n)\), we must calculate the probability of not choosing \(j\) in the first \(n-1\) observations and then choosing \(j\) on the \(n\)th observation. Due to the independent and identically distributed nature of the observations, this can be solved as follows: \[ P(N=n) = \,(1 - P(j))^{n-1} \times P(j),\quad n=1,2,3, \cdots \]
02

Determine \(E[N]\) using the Definition of Expectation

Now that we have the Probability Mass Function for \(N\), we can calculate the expectation \(E[N]\). It's given by: \[ E[N] = \sum_{n=1}^{\infty} nP(N=n) = \sum_{n=1}^{\infty} n \,[\,(1 - P(j))^{n-1} \times P(j)\,] \] This equation is a sum of an infinite sequence which is known to be equivalent to: \[ E[N] = \frac{1}{P(j)} \] Hence, the expected value of \(N\) is the reciprocal of the probability of picking \(j\) as the outcome of the random variable when \(X_0 = j\).

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Key Concepts

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

Random Variables
In probability theory, a random variable is a fundamental concept that represents a quantity whose exact value depends on the outcome of a random phenomenon. Think of it as a way to assign numbers to the results of a random process. In the given exercise, each random variable \(X_i\) represents the outcome of the \(i\)-th observation. These could represent various outcomes such as rolling a die, drawing a card from a deck, etc.
  • Random variables can be discrete or continuous.
  • Discrete random variables take on distinct values, such as integers.
  • In this exercise, the random variables are discrete with values from 1 to \(m\).
Understanding random variables helps us describe statistical models and analyze different scenarios that involve chance.
Expectation
Expectation, or expected value, is a measure of the center of a probability distribution and provides an average value for a random variable. Essentially, it's like a weighted average where each possible outcome of a random variable is multiplied by its probability, and the results are summed up.
In the context of the exercise, we're interested in \(E[N]\), the expected value of \(N\). Here, \(N\) is defined as the smallest index \(n\) such that \(X_n = X_0\). This means we are looking for the average number of observations required to encounter a previously observed outcome.
The formula for expectation, given this scenario, is determined using the probability mass function we derived, resulting in \(E[N] = \frac{1}{P(j)}\). This indicates that the expected number of observations needed to repeat the outcome \(X_0\) is inversely related to the probability of the outcome \(j\) itself.
  • Expectation is a linear operator, making it easier to compute for sums of random variables.
  • It provides insights into the long-term average behavior of random processes.
Probability Mass Function
The Probability Mass Function (PMF) is a function that gives the probability that a discrete random variable is equal to a specific value. It effectively describes the distribution of probabilities across all potential outcomes of a discrete random variable.
In the exercise, \(p(j)\) serves as the PMF for the random variables \(X_i\), indicating the likelihood of each value \(j\) occurring. The PMF has several key properties:
  • The sum of all probabilities must equal 1, i.e., \(\sum_{j=1}^{m} P(j) = 1\).
  • The probability of each outcome must be between 0 and 1, inclusive.
  • PMFs provide a complete description of the distribution of a discrete random variable.
Using the PMF, we were able to derive the formula for \(P(N=n)\), which is crucial for calculating the expected value \(E[N]\). It is essential in constructing the mathematical models of random variables and obtaining meaningful insights from data-driven experiments.
Independent and Identically Distributed Variables
When random variables are described as independent and identically distributed (i.i.d), it indicates that each variable is drawn from the same probability distribution and each event is statistically independent from others.
  • Independence means the outcome of one variable does not affect another.
  • Identical distribution ensures each variable has the same PMF.
In this exercise, the \(X_i\) are described as i.i.d random variables, which simplifies analysis and mathematical calculations. Since they are independent, the probability of any sequence of outcomes can be obtained by multiplying their probabilities. Because they are identically distributed, the same distribution and PMF apply to each \(X_i\).
This characteristic is crucial because it allows us to derive consistent and general formulas, like the formula for \(E[N]\), without needing additional information about each variable's individual characteristics. Understanding i.i.d properties helps in simplifying complex problems into more manageable calculations.

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Most popular questions from this chapter

An individual traveling on the real line is trying to reach the origin. However, the larger the desired step, the greater is the variance in the result of that step. Specifically, whenever the person is at location \(x\), he next moves to a location having mean 0 and variance \(\beta x^{2}\). Let \(X_{n}\) denote the position of the individual after having taken \(n\) steps. Supposing that \(X_{0}=x_{0}\), find (a) \(E\left[X_{n}\right]\); (b) \(\operatorname{Var}\left(X_{n}\right)\)

Independent trials, each resulting in success with probability \(p\), are performed. (a) Find the expected number of trials needed for there to have been both at least \(n\) successes or at least \(m\) failures. Hint: Is it useful to know the result of the first \(n+m\) trials? (b) Find the expected number of trials needed for there to have been either at least \(n\) successes or at least \(m\) failures. Hint: Make use of the result from part (a).

The joint probability mass function of \(X\) and \(Y, p(x, y)\), is given by $$ \begin{array}{ll} p(1,1)=\frac{1}{9}, & p(2,1)=\frac{1}{3}, & p(3,1)=\frac{1}{9} \\ p(1,2)=\frac{1}{9}, & p(2,2)=0, & p(3,2)=\frac{1}{18} \\ p(1,3)=0, & p(2,3)=\frac{1}{6}, & p(3,3)=\frac{1}{9} \end{array} $$ Compute \(E[X \mid Y=i]\) for \(i=1,2,3\).

A manuscript is sent to a typing firm consisting of typists \(A, B\), and \(C .\) If it is typed by \(A\), then the number of errors made is a Poisson random variable with mean \(2.6\); if typed by \(B\), then the number of errors is a Poisson random variable with mean 3 ; and if typed by \(C\), then it is a Poisson random variable with mean \(3.4\). Let \(X\) denote the number of errors in the typed manuscript. Assume that each typist is equally likely to do the work. (a) Find \(E[X]\). (b) Find \(\operatorname{Var}(X)\).

You have two opponents with whom you alternate play. Whenever you play \(A\), you win with probability \(p_{A}\); whenever you play \(B\), you win with probability \(p_{B}\), where \(p_{B}>p_{A}\). If your objective is to minimize the expected number of games you need to play to win two in a row, should you start with \(A\) or with \(B\) ? Hint: Let \(E\left[N_{i}\right]\) denote the mean number of games needed if you initially play \(i\). Derive an expression for \(E\left[N_{A}\right]\) that involves \(E\left[N_{B}\right] ;\) write down the equivalent expression for \(E\left[N_{B}\right]\) and then subtract.

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