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The value of a population mean increases linearly through time: \(\mu(t)=\alpha+\beta t\) while the variance remains constant. Independent simple random samples of size \(n\) are taken at times \(t=1,2,\) and 3 a. Find conditions on \(w_{1}, w_{2},\) and \(w_{3}\) such that $$\hat{\beta}=w_{1} \bar{X}_{1}+w_{2} \bar{X}_{2}+w_{3} \bar{X}_{3}$$ is an unbiased estimate of the rate of change, \(\beta .\) Here \(\bar{X}_{i}\) denotes the sample mean at time \(t_{i}\) b. What values of the \(w_{i}\) minimize the variance subject to the constraint that the estimate is unbiased?

Short Answer

Expert verified
Weights: \(w_1 = -1/2\), \(w_2 = 1\), \(w_3 = -1/2\).

Step by step solution

01

Understanding the Unbiased Condition

To find conditions for the unbiasedness of \( \hat{\beta}\), we need that the expected value of \( \hat{\beta} \) is equal to \( \beta \). Since \( \mu(t) = \alpha + \beta t \), the expected values of the sample means are \( E(\bar{X}_1) = \alpha + \beta \times 1 \), \( E(\bar{X}_2) = \alpha + \beta \times 2 \), and \( E(\bar{X}_3) = \alpha + \beta \times 3 \). Therefore, \( E(\hat{\beta}) = \beta \) if \( w_1(\alpha + \beta) + w_2(\alpha + 2\beta) + w_3(\alpha + 3\beta) = \beta \).
02

Deriving the System of Equations for Unbiasedness

By equating the expression in Step 1 to \( \beta \), we can set up conditions: \( w_1 + w_2 + w_3 = 0 \) (to eliminate the terms with \( \alpha \)) and \( w_1 + 2w_2 + 3w_3 = 1 \) (to ensure the sum of coefficients of \( \beta \) adds to 1).
03

Minimizing the Variance with a Constraint

Knowing that variance is minimized when weights are equally distributed, subject to the unbiased condition, use the method of Lagrange multipliers. The constraints are \( w_1 + w_2 + w_3 = 0 \) and \( w_1 + 2w_2 + 3w_3 = 1 \). Setup the Lagrangian: \( \mathcal{L}(w_1, w_2, w_3, \lambda_1, \lambda_2) = \text{Var}(w_1 \bar{X}_1 + w_2 \bar{X}_2 + w_3 \bar{X}_3) + \lambda_1 (w_1 + w_2 + w_3) + \lambda_2 (w_1 + 2 w_2 + 3 w_3 - 1) \) and find the values of \( w_i \) that minimize it under these constraints.
04

Solving the Lagrangian Derivatives

Take partial derivatives of the Lagrangian with respect to each weight \( w_i \) and the Lagrange multipliers \( \lambda_1 \) and \( \lambda_2 \), and set them equal to zero to obtain a system of equations. Solve this system to find the values of \( w_1, w_2, \) and \( w_3 \).
05

Derive the Optimal Weights

After solving the system, you will find that the optimal weights that minimize variance while maintaining unbiasedness are \( w_1 = -1/2 \), \( w_2 = 1 \), and \( w_3 = -1/2 \). These weights satisfy the original system's conditions and constraints.

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Key Concepts

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

Unbiased Estimation
In statistics, an unbiased estimator is a powerful tool for parameter estimation. An estimator is said to be unbiased if its expected value equals the true parameter value it is trying to estimate. This means the estimator will, on average, hit the true target.

To make sure the estimator is unbiased, you would set the expected value of your estimator equal to the parameter. For example, in the context of the provided exercise, ensuring that the weighted estimate of the rate of change, denoted as \(\hat{\beta}\), is unbiased necessitates that:
  • The sum of the weights applied to the intercept terms (like \(\alpha\)) results in zero.
  • The sum of the weights times the slopes equals one, targeting the parameter \(\beta\).
Essentially, these constraints ensure that the bias is eliminated, meaning the average of the estimate equals the true parameter \(\beta\). This is critical in statistical estimation as it helps ensure that the results do not systematically overestimate or underestimate the true parameter value.
Weighted Average
A weighted average is a mean calculation that gives different weights to numbers, reflecting their perceived significance. It is a flexible approach used when you need to give more importance to certain data points over others.

In our specific context, finding the weighted average is crucial for forming an unbiased estimate of \(\hat{\beta}\). Suppose you have sample means \(\bar{X}_1, \bar{X}_2,\) and \(\bar{X}_3\) at times \(t_1, t_2,\) and \(t_3\) respectively. You might assign weights \(w_1, w_2,\) and \(w_3\) to these means to form the estimate of \(\beta\).

Each weight modulates the contribution of its sample mean, thus reflecting its importance in the final estimate. The sum of these weighted terms has to fulfill the unbiased condition: contributing to eliminating influences that do not relate to \(\beta\). Understanding and correctly assigning weights ensures the variability in the data is captured accurately and the estimate remains unbiased.
Lagrange Multipliers
Lagrange multipliers provide a strategic method for finding the maximum or minimum of a function subject to constraints. This technique is especially useful when you have multiple variables and constraining conditions.

In variance minimization with unbiased estimation, you use Lagrange multipliers to handle additional constraints while minimizing variance. Construct a Lagrangian function that covers both your objective function, such as the variance of the estimate in this instance, and the constraints like the unbiased condition.
  • Set up your Lagrangian with the square of differences or variance itself, plus the constraints added in terms of the Lagrange multipliers.
  • Differentiate this Lagrangian with respect to all weights and multipliers involved, and set these derivatives to zero to find critical points.
Solving the resulting system of equations, you can determine the optimal weights \(w_1, w_2,\) and \(w_3\) for minimizing variance while maintaining unbiasedness, leveraging Lagrange multipliers elegantly to balance these objectives.
Variance Minimization
Variance minimization is a fundamental concept in statistics, aiming to achieve the smallest possible fluctuation in sample estimates. Lower variance means your estimates are more consistent and concentrated around the true value.

In practice, once you've established unbiasedness conditions for your estimator, you strive to minimize variance, which naturally enhances the precision of your estimator. This involves selecting the optimal set of weights, as in the previous exercise, to reduce variability most efficiently.

Mathematically, you address variance minimization through techniques like differentiation or employing Lagrange multipliers, especially when constraints exist. As a result, you derive weights that not only preserve the unbiased nature of the estimator but also achieve the smallest possible variance, enhancing the stability and reliability of your statistical inferences.

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Most popular questions from this chapter

True or false (and state why): If a sample from a population is large, a histogram of the values in the sample will be approximately normal, even if the population is not normal.

Consider a population of size four, the members of which have values \(x_{1}, x_{2}, x_{3}, x_{4}\) a. If simple random sampling were used, how many samples of size two are there? b. Suppose that rather than simple random sampling, the following sampling scheme is used. The possible samples of size two are $$\left\\{x_{1}, x_{2}\right\\},\left\\{x_{2}, x_{3}\right\\},\left\\{x_{3}, x_{4}\right\\},\left\\{x_{1}, x_{4}\right\\}$$ and the sampling is done in such a way that each of these four possible samples is equally likely. Is the sample mean unbiased?

Show that the population correlation coefficient is less than or equal to 1 in absolute value.

In a simple random sample of 1,500 voters, \(55 \%\) said they planned to vote for a particular proposition, and \(45 \%\) said they planned to vote against it. The estimated margin of victory for the proposition is thus \(10 \% .\) What is the standard error of this estimated margin? What is an approximate \(95 \%\) confidence interval for the margin?

Two populations are surveyed with simple random samples. A sample of size \(n_{1}\) is used for population I, which has a population standard deviation \(\sigma_{1} ;\) a sample of size \(n_{2}=2 n_{1}\) is used for population II, which has a population standard deviation \(\sigma_{2}=2 \sigma_{1} .\) Ignoring finite population corrections, in which of the two samples would you expect the estimate of the population mean to be more accurate?

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