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Suppose a certain type of fertilizer has an expected yield per acre of \(\mu_{1}\) with variance \(\sigma^{2}\), whereas the expected yield for a second type of fertilizer is \(\mu_{2}\) with the same variance \(\sigma^{2}\). Let \(S_{1}^{2}\) and \(S_{2}^{2}\) denote the sample variances of yields based on sample sizes \(n_{1}\) and \(n_{2}\), respectively, of the two fertilizers. Show that the pooled (combined) estimator $$ \hat{\sigma}^{2}=\frac{\left(n_{1}-1\right) S_{1}^{2}+\left(n_{2}-1\right) S_{2}^{2}}{n_{1}+n_{2}-2} $$ is an unbiased estimator of \(\sigma^{2}\).

Short Answer

Expert verified
The pooled estimator \( \hat{\sigma}^2 \) is unbiased for \( \sigma^2 \).

Step by step solution

01

Understand the Estimator Formula

The pooled estimator for variance, denoted as \( \hat{\sigma}^2 \), combines the sample variances \( S_1^2 \) and \( S_2^2 \) from two different samples with sizes \( n_1 \) and \( n_2 \) respectively. The formula given is: \[ \hat{\sigma}^{2}=\frac{ (n_{1}-1) S_{1}^{2} + (n_{2}-1) S_{2}^{2}}{n_{1}+n_{2}-2}.\] We need to show that this formula calculates an unbiased estimator of the common variance \( \sigma^{2} \).
02

Recall Properties of Unbiased Estimators

Recall that an estimator \( T \) for a parameter \( \theta \) is unbiased if its expected value equals the parameter: \( E[T] = \theta \). In this problem, we want to find \( E[\hat{\sigma}^2] = \sigma^2 \).
03

Compute the Expected Values of Sample Variances

The expected value of a sample variance \( S_i^2 \) is given by \( E[S_i^2] = \sigma^2 \) for an unbiased estimator. Both \( S_1^2 \) and \( S_2^2 \) are unbiased, thus: \[ E[S_1^2] = \sigma^2 \quad \text{and} \quad E[S_2^2] = \sigma^2. \]
04

Apply Linearity of Expectation

Using the linearity of expectation, apply it to the pooled estimator: \[ E[\hat{\sigma}^2] = E\left[ \frac{(n_{1}-1) S_{1}^{2} + (n_{2}-1) S_{2}^{2}}{n_{1}+n_{2}-2} \right]. \] The linearity allows separating the terms: \[ E[\hat{\sigma}^2] = \frac{(n_{1}-1)E[S_1^2] + (n_{2}-1)E[S_2^2]}{n_{1}+n_{2}-2}. \]
05

Substitute Known Expected Values

Substitute the known expected values: \[ E[\hat{\sigma}^2] = \frac{(n_{1}-1)\sigma^2 + (n_{2}-1)\sigma^2}{n_{1}+n_{2}-2}. \] Simplifying, we factor \( \sigma^2 \) out: \[ E[\hat{\sigma}^2] = \sigma^2 \cdot \frac{(n_{1}-1) + (n_{2}-1)}{n_{1}+n_{2}-2}. \]
06

Evaluate the Expression

By simplifying the expression in the last step: \[ \frac{(n_{1}-1) + (n_{2}-1)}{n_{1}+n_{2}-2} = \frac{n_{1}+n_{2}-2}{n_{1}+n_{2}-2} = 1. \] Therefore, the expected value becomes: \[ E[\hat{\sigma}^2] = \sigma^2 \cdot 1 = \sigma^2. \]
07

Conclusion

Thus, since \( E[\hat{\sigma}^2] = \sigma^2 \), the pooled estimator \( \hat{\sigma}^2 \) is an unbiased estimator of \( \sigma^2 \).

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Key Concepts

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

Unbiased Estimator
An unbiased estimator is a fascinating concept in statistics that ensures precision and accuracy when estimating unknown parameters of a population. When we say an estimator is unbiased, it means the expected value of the estimator is equal to the true value of the parameter being estimated. In mathematical terms, for an estimator \( T \) of parameter \( \theta \), it is unbiased if \( E[T] = \theta \).
  • An example of this is the sample mean as an estimator for the population mean. The expected value of the sample mean equals the population mean.
  • Likewise, our goal in the original exercise is to demonstrate that the pooled variance estimator \( \hat{\sigma}^2 \) is an unbiased estimator of the population variance \( \sigma^2 \).
  • Achieving this requires substantiating that \( E[\hat{\sigma}^2] = \sigma^2 \), as shown in the step by step solution.

Thus, unbiasedness plays a crucial role in ensuring the accuracy of statistical predictions and data analysis.
Sample Variance
Sample variance is pivotal in understanding the variability or spread within a dataset. When we talk about variance, we generally refer to how much the data points differ from the mean of the data. Sample variance, specifically, is a measure of variance calculated from a sample rather than the full population.
  • The formula for sample variance is \( S^2 = \frac{1}{n-1} \sum (X_i - \bar{X})^2 \), where \( X_i \) are sample data points and \( \bar{X} \) is the sample mean.
  • This formula uses \( n-1 \) instead of \( n \) to provide an unbiased estimate of the population variance. This is often referred to as "Bessel's correction."

In our exercise, the sample variances \( S_1^2 \) and \( S_2^2 \) from two different samples are combined using the pooled variance estimator. Both are individually unbiased estimators of the true variance \( \sigma^2 \), an important property leveraged to demonstrate the unbiasedness of the pooled variance estimator.
Linearity of Expectation
The linearity of expectation is a powerful property used in probability and statistics to simplify complex calculations. It states that the expected value of a sum of random variables is equal to the sum of their expected values, regardless of whether the individual random variables are independent. Mathematically, it can be expressed as:
  • \( E[X + Y] = E[X] + E[Y] \)

In the context of the pooled variance estimator problem, this property allows us to handle the combined expectation of the weighted sample variances \( (n_1 - 1)S_1^2 + (n_2 - 1)S_2^2 \). By applying the linearity of expectation:
  • The expected value \( E[\hat{\sigma}^2] \) can be broken down into \( \frac{(n_1 - 1)E[S_1^2] + (n_2 - 1)E[S_2^2]}{n_1 + n_2 - 2} \).
  • This simplifies the task of showing \( E[\hat{\sigma}^2] = \sigma^2 \), as it allows straightforward substitution of the unbiased expected values \( E[S_1^2] = \sigma^2 \) and \( E[S_2^2] = \sigma^2 \).

Through this property, the calculations become tractable and illustrate why the pooled estimator is effective.
Mathematical Statistics
Mathematical statistics is a branch of mathematics focused on the theoretical underpinnings and fundamental principles of statistical methods. It provides a framework for designing statistical experiments, analyzing data, and making inferences about populations based on samples. Some key elements of mathematical statistics include:
  • Estimation theory: Concerned with deriving estimators and properties like unbiasedness to accurately infer population parameters.
  • Hypothesis testing: Involves making decisions about population characteristics based on sample data.
  • Probability theory: Provides the foundation for all statistical methods, as it models randomness and uncertainty inherent in data.

In the context of the pooled variance estimator, mathematical statistics offers a structured method for understanding and proving why certain estimations are preferable. It aids in defining terms rigorously such as unbiasedness and utilizing concepts like the linearity of expectation to justify the unbiased nature of \( \hat{\sigma}^2 \). This rigorous approach helps us confidently apply statistical techniques to analyze real-world scenarios, like comparing different fertilizer treatments in agricultural studies as seen in the exercise.

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Most popular questions from this chapter

Here is a result that allows for easy identification of a minimal sufficient statistic: Suppose there is a function \(t\left(x_{1}, \ldots, x_{n}\right)\) such that for any two sets of observations \(x_{1}, \ldots, x_{n}\) and \(y_{1}, \ldots, y_{n}\), the likelihood ratio \(f\left(x_{1}, \ldots, x_{n} ; \theta\right) / f\left(y_{1}, \ldots, y_{n} ; \theta\right)\) doesn't depend on \(\theta\) if and only if \(t\left(x_{1}, \ldots, x_{n}\right)\) \(=t\left(y_{1}, \ldots, y_{n}\right)\). Then \(T=t\left(X_{1}, \ldots, X_{n}\right)\) is a minimal sufficient statistic. The result is also valid if \(\theta\) is replaced by \(\theta_{1}, \ldots, \theta_{k}\), in which case there will typically be several jointly minimal sufficient statistics. For example, if the underlying pdf is exponential with parameter \(\lambda\), then the likelihood ratio is \(\lambda^{\sum x_{i}-\Sigma y_{i}}\), which will not depend on \(\lambda\) if and only if \(\sum x_{i}=\sum y_{i}\), so \(T=\sum x_{i}\) is a minimal sufficient statistic for \(\lambda\) (and so is the sample mean). a. Identify a minimal sufficient statistic when the \(X_{i}\) 's are a random sample from a Poisson distribution. b. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and variance \(\theta\). c. Identify a minimal sufficient statistic or jointly minimal sufficient statistics when the \(X_{i}\) 's are a random sample from a normal distribution with mean \(\theta\) and standard deviation \(\theta\).

The principle of unbiasedness (prefer an unbiased estimator to any other) has been criticized on the grounds that in some situations the only unbiased estimator is patently ridiculous. Here is one such example. Suppose that the number of major defects \(X\) on a randomly selected vehicle has a Poisson distribution with parameter \(\lambda\). You are going to purchase two such vehicles and wish to estimate \(\theta=P\left(X_{1}=0, \quad X_{2}=0\right)=e^{-2 \lambda}\), the probability that neither of these vehicles has any major defects. Your estimate is based on observing the value of \(X\) for a single vehicle. Denote this estimator by \(\hat{\theta}=\delta(X)\). Write the equation implied by the condition of unbiasedness, \(E[\delta(X)]=e^{-2 \lambda}\), cancel \(e^{-\lambda}\) from both sides, then expand what remains on the right-hand side in an infinite series,and compare the two sides to determine \(\delta(X)\). If \(X=200\), what is the estimate? Does this seem reasonable? What is the estimate if \(X=199 ?\) Is this reasonable?

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). [Hint: For any rv \(Y, E\left(Y^{2}\right)=\) \(V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X}\). b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ?

Suppose a measurement is made on some physical characteristic whose value is known, and let \(X\) denote the resulting measurement error. For an unbiased measuring instrument or technique, the mean value of \(X\) is 0 . Assume that any particular measurement error is normally distributed with variance \(\sigma^{2}\). Let \(X_{1}, \ldots . X_{n}\) be a random sample of measurement errors. a. Obtain the method of moments estimator of \(\sigma^{2}\). b. Obtain the maximum likelihood estimator of \(\sigma^{2}\).

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cl} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, \quad x_{2}=.79, \quad x_{3}=.90\), \(x_{4}=.65, x_{5}=.86, x_{6}=.47, x_{7}=.73, x_{8}=.97\), \(x_{9}=.94, x_{10}=.77\). a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

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