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Suppose that \(\Theta_{1}, \Theta_{2},\) and \(\Theta_{3}\) are estimators of \(\theta .\) We know that \(E\left(\Theta_{1}\right)=E\left(\Theta_{2}\right)=0, E\left(\Theta_{3}\right) \neq \theta, V\left(\Theta_{1}\right)=12, V\left(\Theta_{2}\right)=10\) and \(E\left(\Theta_{3}-\theta\right)^{2}=6 .\) Compare these three estimators. Which do you prefer? Why?

Short Answer

Expert verified
\(\Theta_3\) is preferred because it has the lowest mean squared error.

Step by step solution

01

Identify Properties of Estimators

First, analyze the given properties: \(\Theta_1\) and \(\Theta_2\) are unbiased because \(E(\Theta_1) = E(\Theta_2) = 0\) and these equal the parameter \(\theta\) (if \(\theta = 0\)). \(\Theta_3\) is biased because \(E(\Theta_3) eq \theta\).
02

Evaluate Variances of Unbiased Estimators

For unbiased estimators, \(\Theta_1\) and \(\Theta_2\), compare their variances. Since \(V(\Theta_1) = 12\) and \(V(\Theta_2) = 10\), \(\Theta_2\) is more efficient because it has a lower variance.
03

Analyze Mean Squared Error (MSE) of Biased Estimator

Mean Squared Error (MSE) can be used to evaluate biased estimators. Since the exercise gives \(E((\Theta_3 - \theta)^2) = 6\), this is the MSE of \(\Theta_3\).
04

Compare Efficiency Using MSE

For \(\Theta_1\) and \(\Theta_2\), the MSE equals the variance because they are unbiased. Thus, \(MSE(\Theta_1) = 12\) and \(MSE(\Theta_2) = 10\). Since \(MSE(\Theta_3) = 6\), \(\Theta_3\) is actually preferred due to its lower MSE, despite its bias.
05

Conclusion on Preferred Estimator

Even though \(\Theta_3\) is biased, it has the lowest MSE of the three estimators. Therefore, it is the preferred estimator for \(\theta\) due to its efficiency in estimation.

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Key Concepts

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

Unbiased Estimators
Unbiased estimators are a fundamental concept in statistics, providing a way to estimate a parameter by ensuring that, on average, the estimates are accurate. An estimator is considered unbiased if its expected value equals the true value of the parameter it is estimating. Essentially, this means that if you were to calculate the estimator many times, the average of these calculations would be equal to the actual parameter.For example, consider two estimators, say \( \Theta_1 \) and \( \Theta_2 \), both aiming to estimate a parameter \( \theta \). If \( E(\Theta_1) = E(\Theta_2) = \theta \), then both estimators are unbiased. In our original exercise, both \( \Theta_1 \) and \( \Theta_2 \) were identified as unbiased because \( E(\Theta_1) = E(\Theta_2) = 0 \), assuming that the parameter \( \theta = 0 \). This property is crucial because unbiasedness ensures that an estimator does not systematically overestimate or underestimate the parameter, making it a reliable choice in the estimation process. However, unbiasedness alone does not account for the efficiency of the estimator, which brings us to consider variance in the analysis.
Variance Analysis
The variance of an estimator measures the spread of its estimates around the expected value. It is a critical factor in determining the reliability of an estimator. Essentially, the variance tells us how much the estimated values could deviate from the expected value, or, in simpler terms, how consistent the estimator is.In statistical analysis, low variance is generally preferred because it implies greater consistency and precision. For two unbiased estimators, the one with the lower variance is deemed more efficient. In our analysis, we found that estimator \( \Theta_2 \) has a variance of 10, whereas \( \Theta_1 \) has a variance of 12. Since \( \Theta_2 \) has a lower variance than \( \Theta_1 \), it is considered more efficient among the unbiased estimators.When comparing estimators, we acknowledge that unbiasedness and low variance are ideal features. However, sometimes we encounter biased estimators that, though not ideal in a traditional sense, offer other advantages. To assess these, we often turn to the Mean Squared Error (MSE).
Mean Squared Error (MSE)
Mean Squared Error (MSE) is a comprehensive measure used to evaluate both biased and unbiased estimators. It combines both the variance of the estimator and the square of the bias. The formula for MSE is expressed as:\[ MSE(\theta_i) = Var(\Theta_i) + (Bias(\Theta_i))^2 \]This means that MSE accounts for the total error in an estimation due to both variance and any bias present. In essence, MSE helps us quantify the trade-off between bias and variance, providing a single statistic that can be used to compare different estimators.For the estimators in the exercise, we found that both \( \Theta_1 \) and \( \Theta_2 \) have MSEs equivalent to their variances because they are unbiased. This means their MSE values are 12 and 10, respectively. On the other hand, the biased estimator \( \Theta_3 \) had an MSE of 6, indicating lower overall error than either unbiased estimator.Despite its bias, \( \Theta_3 \) is preferred in terms of efficiency because it results in a lower MSE, showing that sometimes a small amount of bias is acceptable if it significantly reduces variance, leading to more accurate and reliable estimations in practice. Thus, MSE serves as a critical criterion for decision-making, especially in scenarios where trade-offs between bias and variance need to be evaluated.

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Most popular questions from this chapter

Suppose that \(\Theta_{i}\) and \(\Theta_{2}\) are unbiased estimators of the parameter \(\theta .\) We know that \(V\left(\Theta_{1}\right)=10\) and \(V\left(\Theta_{2}\right)=4\). Which estimator is better and in what sense is it better? Calculate the relative efficiency of the two estimators.

Consider a Weibull distribution with shape parameter 1.5 and scale parameter \(2.0 .\) Generate a graph of the probability distribution. Does it look very much like a normal distribution? Construct a table similar to Table \(7-1\) by drawing 20 random samples of size \(n=10\) from this distribution. Compute the sample average from each sample and construct a normal probability plot of the sample averages. Do the sample averages seem to be normally distributed?

Suppose that \(\hat{\Theta}_{1}\) and \(\hat{\Theta}_{2}\) are estimators of the parameter \(\theta\) We know that \(E\left(\Theta_{1}\right)=\theta, E\left(\Theta_{2}\right)=\theta / 2, V\left(\Theta_{1}\right)=10, V\left(\Theta_{2}\right)=4 .\) Which estimator is better? In what sense is it better?

Scientists at the Hopkins Memorial Forest in westem Massachusetts have been collecting meteorological and environmental data in the forest data for more than 100 years. In the past few years, sulfate content in water samples from Birch Brook has averaged \(7.48 \mathrm{mg} / \mathrm{L}\) with a standard deviation of \(1.60 \mathrm{mg} / \mathrm{L}\) (a) What is the standard error of the sulfate in a collection of 10 water samples? (b) If 10 students measure the sulfate in their samples, what is the probability that their average sulfate will be between 6.49 and \(8.47 \mathrm{mg} / \mathrm{L} ?\) (c) What do you need to assume for the probability calculated in (b) to be accurate?

(a) Show that \(\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2} / n\) is a biased estimator of \(\sigma^{2}\) (b) Find the amount of bias in the estimator. (c) What happens to the bias as the sample size \(n\) increases?

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