/*! 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 16 Suppose the true average growth ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose the true average growth \(\mu\) of one type of plant during a 1-year period is identical to that of a second type, but the variance of growth for the first type is \(\sigma^{2}\), whereas for the second type the variance is \(4 \sigma^{2}\). Let \(X_{1}, \ldots, X_{m}\) be \(m\) independent growth observations on the first type [so \(\left.E\left(X_{i}\right)=\mu, V\left(X_{i}\right)=\sigma^{2}\right]\), and let \(Y_{1}, \ldots, Y_{n}\) be \(n\) independent growth observations on the second type \(\left[E\left(Y_{i}\right)=\mu\right.\), \(\left.V\left(Y_{i}\right)=4 \sigma^{2}\right]\) a. Show that for any \(\delta\) between 0 and 1, the estimator \(\hat{\mu}=\delta \bar{X}+(1-\delta) \bar{Y}\) is unbiased for \(\mu\). b. For fixed \(m\) and \(n\), compute \(V(\hat{\mu})\), and then find the value of \(\delta\) that minimizes \(V(\hat{\mu})\). [Hint: Differentiate \(V(\hat{\mu})\) with respect to \(\delta\).]

Short Answer

Expert verified
a. \( \hat{\mu} \) is unbiased. b. Optimal \( \delta = \frac{4n}{4n + m} \).

Step by step solution

01

Define the Estimator

We are given the estimator \( \hat{\mu} = \delta \bar{X} + (1 - \delta) \bar{Y} \), where \( \bar{X} \) is the average of the observations of the first plant type, and \( \bar{Y} \) is the average of the observations of the second plant type.
02

Determine Unbiasedness

An estimator is unbiased if its expected value equals the parameter it estimates. The expectation of \( \hat{\mu} \) is \( E(\hat{\mu}) = E(\delta \bar{X} + (1 - \delta) \bar{Y}) = \delta E(\bar{X}) + (1 - \delta) E(\bar{Y}) = \delta \mu + (1 - \delta) \mu = \mu \). Thus, \( \hat{\mu} \) is unbiased for \( \mu \).
03

Compute the Variance

The variance of \( \hat{\mu} \) is given by \( V(\hat{\mu}) = V(\delta \bar{X} + (1 - \delta) \bar{Y}) = \delta^2 V(\bar{X}) + (1 - \delta)^2 V(\bar{Y}) \) since \( \bar{X} \) and \( \bar{Y} \) are independent. Given \( V(\bar{X}) = \frac{\sigma^2}{m} \) and \( V(\bar{Y}) = \frac{4\sigma^2}{n} \), substitute to get \( V(\hat{\mu}) = \delta^2 \frac{\sigma^2}{m} + (1 - \delta)^2 \frac{4\sigma^2}{n} \).
04

Differentiate to Minimize Variance

To find the \( \delta \) that minimizes \( V(\hat{\mu}) \), differentiate \( V(\hat{\mu}) \) with respect to \( \delta \). The derivative is \( \frac{d}{d\delta} \left( \delta^2 \frac{\sigma^2}{m} + (1 - \delta)^2 \frac{4\sigma^2}{n} \right) = 2\delta \frac{\sigma^2}{m} - 2(1-\delta) \frac{4\sigma^2}{n} \).
05

Solve for Optimal \\\delta

Set the derivative to zero to find the minimum: \( 2\delta \frac{\sigma^2}{m} = 2(1-\delta) \frac{4\sigma^2}{n} \). Simplify to get \( \delta \frac{1}{m} = \frac{4(1-\delta)}{n} \), which results in \( \delta = \frac{4n}{4n + m} \) when solved for \( \delta \).

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Key Concepts

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

Variance Minimization
When estimating parameters, one goal is often to minimize the variance of an estimator. Variance refers to the spread or dispersion of a set of values, and it is crucial because it tells us how much the values differ from their expected value. In this problem, we are focused on minimizing the variance of the estimator \( \hat{\mu} = \delta \bar{X} + (1 - \delta) \bar{Y} \). The challenge is to find the optimal \( \delta \) that leads to the smallest variance.The variance of the estimator is given by:- \( V(\hat{\mu}) = \delta^2 V(\bar{X}) + (1 - \delta)^2 V(\bar{Y}) \).This calculation assumes that \( \bar{X} \) and \( \bar{Y} \) are independent, which allows us to compute the variance of a sum this way.To find the minimum variance, we take the derivative of \( V(\hat{\mu}) \) with respect to \( \delta \) and set it to zero.- This step helps identify the point where the variance is at its lowest, effectively minimizing the unwanted spread in our estimator’s predictions.Calculus, specifically differentiation, becomes a powerful tool for finding this optimal \( \delta \). Through this optimization, the goal is to achieve an estimator that is both unbiased and as precise as possible.
Independent Observations
The concept of independent observations is essential to understanding the behavior of statistical estimators. Independence means that the value of one observation does not influence or predict the value of another. This independence is a fundamental assumption in many statistical models because it simplifies the calculations of expected values and variances.In this exercise, we consider two types of plants, each providing a sample of observations:
  • \( X_1, X_2, \ldots, X_m \) for the first plant type.
  • \( Y_1, Y_2, \ldots, Y_n \) for the second plant type.
These observations are stated to be independent:- This means the growth observed for one type doesn’t affect or correlate with the observations from another type.- Such independence simplifies our assumptions and means we can easily calculate combined variances for estimates like \( V(\hat{\mu}) \).Independence ensures that when we estimate \( \mu \) by combining data from both plant types, the variances you calculate from each set of data do not overlap or interact, making our statistical procedures valid and reliable.
Parameter Estimation
Parameter estimation involves taking known data samples and using them to infer or predict the values of parameters in a statistical model. A parameter, like \( \mu \) in this exercise, represents a true value that describes the population from which the sample is drawn.Here, we estimate \( \mu \) using the estimator \( \hat{\mu} = \delta \bar{X} + (1 - \delta) \bar{Y} \). This composite estimator combines data from two independent samples to create a more accurate reflection of the true parameter.Key elements in parameter estimation include:
  • **Unbiasedness**: An estimator is unbiased if its expectation equals the true parameter. The exercise shows the estimator \( \hat{\mu} \) is unbiased by demonstrating that \( E(\hat{\mu}) = \mu \).
  • **Precision**: Apart from unbiasedness, low variance is desired to ensure precision. A precise estimator will consistently return values close to the true parameter.
Efficient parameter estimation, therefore, involves finding the sweet spot between these characteristics. It aims to produce an estimator that represents the population distribution accurately, with minimal error and variability, leading to robust and trustworthy statistical analysis.

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Most popular questions from this chapter

In a random sample of 80 components of a certain type, 12 are found to be defective. a. Give a point estimate of the proportion of all such components that are not defective. b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here. The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. [Hint: If \(p\) denotes the probability that a component works properly, how can \(P\) (system works) be expressed in terms of \(p\) ?]

When the sample standard deviation \(S\) is based on a random sample from a normal population distribution, it can be shown that $$ E(S)=\sqrt{2 /(n-1)} \Gamma(n / 2) \sigma / \Gamma((n-1) / 2) $$ Use this to obtain an unbiased estimator for \(\sigma\) of the form \(c S\). What is \(c\) when \(n=20\) ?

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\).

At time \(t=0\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \(\lambda\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \(\lambda\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential \((\lambda)\) variables, which is exponential with parameter \(2 \lambda\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential rv with parameter \(3 \lambda\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \(25.2\), \(41.7,51.2,55.5,59.5,61.8\) (from which you should calculate the times between successive births). Derive the mle of \(\lambda\).

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a gamma distribution with parameters \(\alpha\) and \(\beta\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \(\alpha\) and \(\beta\). Do you think they can be solved explicitly? b. Show that the mle of \(\mu=\alpha \beta\) is \(\hat{\mu}=\bar{X}\).

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