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If \(X_{1}, X_{2}, \ldots, X_{n}\) is a random sample from a distribution with pdf $$ f(x ; \theta)=\left\\{\begin{array}{ll} \frac{3 \theta^{3}}{(x+\theta)^{2}} & 0

Short Answer

Expert verified
The estimator \(Y = 2 \bar{X}\) is an unbiased estimator of \(\theta\) if its expected value is indeed \(\theta\). It is efficient if its variance is equal to the Cramer-Rao lower bound.

Step by step solution

01

Verify Unbiasedness

First, calculate the expected value of \(Y = 2 \bar{X}\). This is done by taking the integral of the product of \(Y\) and the given pdf over the integration intervals defined by the pdf, from 0 to \(+\infty\). In this case, the expected value of \(Y\) should equal \(\theta\), demonstrating unbiasedness.
02

Calculate Variance of Y

Now, calculate the variance of \(Y = 2 \bar{X}\). This is done by taking the expected value of \((Y - E(Y))^2\), again by integrating the square deviation from the expected value of \(Y\) over the defined intervals. This gives us the variance of \(Y\), which is an important measure of its efficiency.
03

Calculate and Compare with Cramer-Rao Lower Bound

Then, calculate the Cramer-Rao lower bound, a theoretical minimum variance for any unbiased estimator, using the formula / formulae involving the pdf and its derivative. By comparing the variance of Y with this lower bound, we can determine if \(Y\) is an efficient estimator of \(\theta\). If the variance of \(Y\) equals the Cramer-Rao lower bound, then \(Y\) is an efficient estimator of \(\theta\).

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Key Concepts

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

Understanding Expected Value
The concept of expected value is a fundamental aspect of statistics and probability theory. It represents the average outcome you would expect to see if you could repeat an experiment or random trial an infinite number of times.

For a continuous random variable, the expected value, often denoted as E(X), is calculated by integrating the product of the variable's value and its probability density function (pdf) across all possible values. In practical terms, it's a form of weighted average, with probabilities providing the weights.

When an estimator, like \(Y = 2 \bar{X}\) in our exercise, is called 'unbiased', it means that its expected value equals the true parameter it estimates, \(\theta\). Showing this involves demonstrating that E(Y) is equal to \(\theta\), which lays the foundation for trusting the estimator as a reliable measure in statistical analysis.
Variance Calculation in Estimation
The variance of an estimator is a measure of its dispersion or spread; it gives us an idea of how much the values of the estimator can fluctuate around its expected value.

To calculate the variance, denoted as Var(Y), you find the expected value of the squared difference between the estimator and its expected value: Var(Y) = E[(Y - E(Y))^2]. This calculation involves integrating the square of this difference multiplied by the pdf over all possible values. A low variance means that the estimated values are tightly clustered around the expected value, which is desirable as it indicates more precise estimates.

For our estimator \(Y = 2 \bar{X}\), computing the variance and comparing it with other estimators or bounds can help us understand its reliability and precision in estimating the parameter \(\theta\).
The Cramer-Rao Lower Bound Explained
The Cramer-Rao lower bound (CRLB) is a theoretical minimum for the variance of unbiased estimators. It's derived from the Fisher information, which measures the amount of information that a random variable contains about an unknown parameter.

The CRLB provides a benchmark to evaluate estimators; it tells us what the best possible variance is given our data. An estimator is efficient if its variance equals the CRLB for all values of the parameter being estimated.

To calculate the CRLB, one usually needs to find the Fisher information, which involves taking the derivative of the log of the pdf, squaring it, and then taking its expected value. This process requires a bit of calculus but offers a powerful tool to assess the quality of an estimator through comparison with the Cramer-Rao bound.
Efficiency of Estimators and Their Importance
An efficient estimator in statistics is one that achieves the lowest possible variance among all unbiased estimators of a parameter. The closer the variance of an estimator comes to the Cramer-Rao lower bound, the more efficient the estimator is considered to be.

Efficiency matters because it impacts the precision and reliability of the estimator. In practical applications, having an efficient estimator means that we can make more confident decisions based on the data, as there is less variability in the estimation process.

In the context of our exercise, determining the efficiency of \(Y = 2 \bar{X}\) as an estimator for \(\theta\) requires computing its variance and comparing it against the CRLB. If these values match, it implies that \(Y\) is among the best possible estimators for \(\theta\) under the given circumstances, which is a significant result in statistical inference.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(\Gamma(\alpha=4, \beta=\theta)\) distribution, where \(0 \leq \theta<\infty\). (a) Show that the likelihood ratio test of \(H_{0}: \theta=\theta_{0}\) versus \(H_{1}: \theta \neq \theta_{0}\) is based upon the statistic \(W=\sum_{i=1}^{n} X_{i} .\) Obtain the null distribution of \(2 W / \theta_{0}\). (b) For \(\theta_{0}=3\) and \(n=5\), find \(c_{1}\) and \(c_{2}\) so that the test that rejects \(H_{0}\) when \(W \leq c_{1}\) or \(W \geq c_{2}\) has significance level \(0.05 .\)

Recall that \(\widehat{\theta}=-n / \sum_{i=1}^{n} \log X_{i}\) is the mle of \(\theta\) for a beta \((\theta, 1)\) distribution. Also, \(W=-\sum_{i=1}^{n} \log X_{i}\) has the gamma distribution \(\Gamma(n, 1 / \theta)\). (a) Show that \(2 \theta W\) has a \(\chi^{2}(2 n)\) distribution. (b) Using part (a), find \(c_{1}\) and \(c_{2}\) so that $$ P\left(c_{1}<\frac{2 \theta n}{\hat{\theta}}

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the Poisson distribution with \(0<\theta \leq 2\). Show that the mle of \(\theta\) is \(\widehat{\theta}=\min \\{\bar{X}, 2\\}\).

Suppose the pdf of \(X\) is of a location and scale family as defined in Example 6.4.4. Show that if \(f(z)=f(-z)\), then the entry \(I_{12}\) of the information matrix is 0 . Then argue that in this case the mles of \(a\) and \(b\) are asymptotically independent.

Prove that \(\bar{X}\), the mean of a random sample of size \(n\) from a distribution that is \(N\left(\theta, \sigma^{2}\right),-\infty<\theta<\infty\), is, for every known \(\sigma^{2}>0\), an efficient estimator of \(\theta\).

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