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Chapter 2: Problem 6

If binomial sampling is continued until \(m\) successes have been obtained, let \(X_{i}(i=\) \(1, \ldots, m)\) be the number of failures between the \((i-1) \mathrm{st}\) and \(i\) th success. (a) The \(X_{i}\) are iid according to the geometric distribution \(P\left(X_{i}=x\right)=p q^{x}, x=\) \(0,1, \ldots\) (b) The statistic \(Y=\Sigma X_{i}\) is sufficient for \(\left(X_{1}, \ldots, X_{m}\right)\) and has the distribution (3.3).

Short Answer

Expert verified
\( X_i \) are geometric; \( Y \) is negative binomial and sufficient.

Step by step solution

01

Understanding the setup

We are dealing with a situation known as negative binomial sampling, where we stop sampling once a certain number of successes \( m \) have been achieved. In this context, each \( X_i \) represents the number of failures before the \( i^{th} \) success. Our task is to show the distribution properties of \( X_i \) and \( Y \).
02

Showing \(X_i\) follows a geometric distribution

The given probability \( P(X_i = x) = pq^x \) is the probability mass function of a geometric distribution. This means that the number of failures before a single success in a sequence of Bernoulli trials follows a geometric distribution, with \( p \) being the probability of success and \( q = 1-p \) being the probability of failure. Therefore, each \( X_i \) is independent and identically distributed (iid) according to the geometric distribution.
03

Defining the statistic \( Y \)

The statistic \( Y = \sum_{i=1}^{m} X_i \) represents the total number of failures before achieving \( m \) successes. This is a random variable of interest which combines the individual \( X_i \). Given that each \( X_i \) is iid geometric, \( Y \) is the sum of \( m \) independent geometric variables.
04

Distribution of \( Y \) and Sufficient Statistic

The sum \( Y \) of \( m \) geometric random variables corresponds to a negative binomial distribution, which models the number of failures before a specified number \( m \) of successes is reached in a series of independent and identically distributed Bernoulli trials. Since \( Y \) aggregates all information about the trials, it acts as a sufficient statistic. This distribution is given by: \[ P(Y = k) = \binom{k + m - 1}{k} p^m q^k \] where \( k \) is the total number of failures.

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Key Concepts

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

Geometric Distribution
In probability theory, the geometric distribution is a crucial building block.The geometric distribution specifically deals with finding the probability of the first success on the nth trial in a series of Bernoulli trials. A Bernoulli trial is an experiment in which there are only two possible outcomes: success or failure. For example, flipping a coin (heads or tails) or passing a certain test (pass or fail). The geometric distribution asks us how many trials until we hit our first success.
Let's break it down further:
  • The probability of success in each trial is denoted by \( p \).
  • The probability of failure (not succeeding) in each trial is \( q = 1 - p \).
The probability mass function (PMF) for a geometric distribution is given by: \[ P(X = x) = pq^x \]Where \( X \) is a random variable representing the number of trials until the first success, \( p \) is the success probability, \( q \) is the failure probability, and \( x \) is the number of failures before the first success. This distribution is memoryless, meaning past failures do not affect the probability of future success.
Sufficient Statistic
A sufficient statistic is an essential idea in statistics because it summarizes all the necessary information needed from a sample to make inferences about the population or process being studied.
  • To be sufficient, the statistic must capture all the data's information about the parameter.
  • Sufficient statistics help in simplifying calculations and estimating parameters efficiently.
In our context of negative binomial distribution, we use the sum \( Y = \sum_{i=1}^m X_i \) to represent the total failures before hitting \( m \) successes. This sum serves as a sufficient statistic for:\[(X_1, X_2, ..., X_m)\]No other statistic derived from the same sample could provide any additional insight into the process. The use of sufficient statistics helps to minimize the complexity of the data without losing essential information needed for prediction and inference.
Bernoulli Trials
A Bernoulli trial is one of the simplest forms of a random experiment. As mentioned earlier, it involves an experiment where there are only two possible outcomes: a success or a failure. Understanding Bernoulli trials is fundamental when studying more complex probability distributions, such as the negative binomial distribution.
  • Each trial is independent, meaning that the outcome of one trial does not affect others.
  • The trials are identically distributed, with each having the same probability \( p \) of success and \( q = 1 - p \) of failure.
When performing negative binomial sampling, we use a series of Bernoulli trials to determine when a certain number of successes is achieved. It's important to note that in Bernoulli trials, the number of trials is not fixed; instead, trials are repeated until the desired number of successes is obtained. This framework forms the basis of more complex distributions, such as the geometric and negative binomial distributions, which extend upon the concept of independent Bernoulli trials.

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Most popular questions from this chapter

In this problem, we establish some facts about eigenvalues and eigenvectors of square matrices. (For a more general treatment, see, for example, Marshall and Olkin 1979, Chapter 20.) We use the facts that a scalar \(\lambda>0\) is an eigenvalue of the \(n \times n\) symmetric matrix \(A\) if there exists an \(n \times 1\) vector \(p\), the corresponding eigenvector, satisfying \(A p=\lambda p\). If \(\mathrm{A}\) is nonsingular, there are \(n\) eigenvalues with corresponding linearly independent eigenvectors. (a) Show that \(A=P^{\prime} D_{\lambda} P\), where \(D_{\lambda}\) is a diagonal matrix of eigenvalues of \(A\) and \(P\) is and \(n \times n\) matrix whose rows are the corresponding eigenvalues that satisfies \(P^{\prime} P=P P^{\prime}=I\), the identity matrix. (b) Show that \(\max _{x} \frac{x^{\prime} A x}{x^{\prime} x}=\) largest eigenvalue of \(A\). (c) If \(B\) is a nonsingular symmetric matrix with eigenvector-eigenvalue representation \(B=Q^{\prime} D_{\beta} Q\), then \(\max _{x} \frac{x^{\prime} A x}{x^{\prime} B x}=\) largest eigenvalue of \(A^{*}\), where \(A^{*}=\) \(D_{\beta}^{-1 / 2} Q A Q^{\prime} D_{\beta}^{-1 / 2}\) and \(D_{\beta}^{-1 / 2}\) is a diagonal matrix whose elements are the reciprocals of the square roots of the eigenvalues of \(B\). (d) For any square matrices \(C\) and \(D\), show that the eigenvalues of the matrix \(C D\) are the same as the eigenvalues of the matrix \(D C\), and hence that \(\max _{x} \frac{x^{\prime} A x}{x^{\prime} B x}=\) largest eigenvalue of \(A B^{-1}\). (e) If \(A=a a^{\prime}\), where \(a\) is a \(n \times 1\) vector ( \(A\) is thus a rank-one matrix), then \(\max _{x} \frac{x^{\prime} a a^{\prime} x}{x^{\prime} B x}=\) \(a^{\prime} B^{-1} a\).

(Liu and Brown 1993) Let \(X\) be an observation from the normal mixture density $$ p_{\theta}(x)=\frac{1}{2 \sqrt{2 \pi}}\left\\{e^{-(1 / 2)(x-\theta)^{2}}+e^{-(1 / 2)(x+\theta)^{2}}\right\\}, \theta \in \Omega $$ where \(\Omega\) is any neighborhood of zero. Thus, the random variable \(X\) is either \(N(\theta, 1)\) or \(N(-\theta, 1)\), each with probability \(1 / 2\). Show that \(\theta=0\) is a singular point, that is, if there exists an unbiased estimator of \(\theta\) it will have infinite variance at \(\theta=0\).

If \(X\) is a single observation from \(N\left(\xi, \sigma^{2}\right)\), show that no unbiased estimator \(\delta\) of \(\sigma^{2}\) exists when \(\xi\) is unknown. [Hint: For fixed \(\sigma=a, X\) is a complete sufficient statistic for \(\xi\), and \(E[\delta(X)]=a^{2}\) for all \(\xi\) implies \(\delta(x)=a^{2}\) a.e.]

If \(X_{1}, \ldots, X_{n}\) are iid as \(N\left(\xi, \sigma^{2}\right)\) with \(\sigma^{2}\) known, find the UMVU estimator of (a) \(\xi^{2}\), (b) \(\xi^{3}\), and (c) \(\xi^{4}\). [Hint: To evaluate the expectation of \(\bar{X}^{k}\), write \(\bar{X}=Y+\xi\), where \(Y\) is \(N\left(0, \sigma^{2} / n\right)\) and expand \(E(Y+\xi)^{k}\).]

If \(X\) has the Poisson distribution \(P(\theta)\), show that \(1 / \theta\) does not have an unbiased estimator.
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