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Chapter 7: Problem 11

Let \(\bar{Y}\) be the average of a random sample from the uniform density on \((0, \theta)\). Show that \(2 \bar{Y}\) is unbiased for \(\theta\). Find a sufficient statistic for \(\theta\), and obtain an estimator based on it which has smaller variance. Compare their mean squared errors.

Short Answer

Expert verified
The estimator \(2\bar{Y}\) is unbiased for \(\theta\). \(\hat{\theta} = \frac{n+1}{n} Y_{(n)}\) is based on a sufficient statistic and has a smaller variance. \(\hat{\theta}\) has a smaller mean squared error compared to \(2\bar{Y}\).

Step by step solution

01

Understanding the Uniform Distribution

For a random sample from a uniform distribution on \((0, \theta)\), each observation \(Y_i\) is distributed uniformly. The probability density function is given by \(f(y) = \frac{1}{\theta}\) for \(0 < y < \theta\), and the expected value of each observation \(E(Y_i) = \frac{\theta}{2}\).
02

Calculating the Expected Value of a Sample Average

Since \(\bar{Y} = \frac{1}{n}\sum_{i=1}^{n} Y_i\), we use the linearity of expectation to find \(E(\bar{Y}) = \frac{1}{n} \sum_{i=1}^{n} E(Y_i) = \frac{\theta}{2}\), since each \(E(Y_i) = \frac{\theta}{2}\).
03

Showing Unbiasedness of Estimator

To show that \(2\bar{Y}\) is unbiased, find the expected value: \(E(2\bar{Y}) = 2E(\bar{Y}) = 2 \times \frac{\theta}{2} = \theta\). Since \(E(2\bar{Y}) = \theta\), \(2\bar{Y}\) is an unbiased estimator of \(\theta\).
04

Identifying Sufficient Statistic

For a uniform distribution, the maximum observation in the sample, \(Y_{(n)} = \max(Y_1, Y_2, ..., Y_n)\), is a sufficient statistic for \(\theta\). This is based on the fact that knowledge of the maximum gives as much information about \(\theta\) as the full sample.
05

Constructing an Estimator with Smaller Variance

Use the sufficient statistic \(Y_{(n)}\) to construct the estimator \(\hat{\theta} = \frac{n+1}{n} Y_{(n)}\). This particular form is chosen because \(E(Y_{(n)}) = \frac{n\theta}{n+1}\), hence \(\hat{\theta}\) is unbiased for \(\theta\).
06

Comparing Variance and Mean Squared Error

The variance of \(Y_{(n)}\) approximates to \(\frac{\theta^2}{n(n+1)^2}\), so the variance of \(\hat{\theta}\) is \(\frac{\theta^2}{n(n+1)}\). In contrast, the variance of \(2\bar{Y}\) is \(\frac{\theta^2}{3n}\). The mean squared error (MSE) of an unbiased estimator is its variance, and since \(\frac{n+1}{n} < 3\), the variance of \(\hat{\theta}\) is smaller than that of \(2\bar{Y}\), making \(\hat{\theta}\) a more precise estimator.

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Key Concepts

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

Uniform Distribution
A uniform distribution is a type of probability distribution where all outcomes are equally likely within a specified range. In this exercise, we focus on the interval \(0, \theta\), meaning any number within this range is as probable as any other. Here's what you need to know:
  • The probability density function (PDF) for a uniform distribution on this interval is \(f(y) = \frac{1}{\theta}\) for \(0 < y < \theta\).
  • The expected value, or mean, of a uniform distribution, \(E(Y_i)\), is \(\frac{\theta}{2}\), which is the midpoint of the interval.
Understanding this basics helps us realize why certain estimators, like the sample average or maximum, can be used to estimate \(\theta\). By equally weighting all parts of the distribution, they can offer unbiased estimates.
Sufficient Statistic
A sufficient statistic is a function of your data that gives as much information about an unknown parameter as the entire data set would. In the context of our uniform distribution, the maximum value within the sample, denoted by \(Y_{(n)}\), is a sufficient statistic for \(\theta\).
  • This means that once you know \(Y_{(n)}\), you don't gain additional information about \(\theta\) by looking at the rest of the data.
  • This property is especially useful because it simplifies the analysis and helps construct better estimators.
With \(Y_{(n)}\), we can create more accurate statistical inferences as we leverage this powerful characteristic of sufficiency to determine or estimate parameters effectively.
Unbiased Estimator
An unbiased estimator is a statistical estimator whose expected value is equal to the true value of the parameter it estimates. In our example, \(2\bar{Y}\) was shown to be unbiased for \(\theta\) because the expected value of \(2\bar{Y}\) equals \(\theta\).
  • The formula is \(E(2\bar{Y}) = 2 \times \frac{\theta}{2} = \theta\), demonstrating unbiasedness.
  • Unbiasedness is a desirable property because it implies that on average, the estimator hits the true parameter value.
Additionally, we constructed another unbiased estimator using the sufficient statistic \(Y_{(n)}\) as \(\hat{\theta} = \frac{n+1}{n} Y_{(n)}\). This estimator not only maintains unbiasedness but also typically offers reduced variance compared to \(2\bar{Y}\), improving precision in estimation.

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Most popular questions from this chapter

Show that no unbiased estimator exists of \(\psi=\log \\{\pi /(1-\pi)\\}\), based on a binomial variable with probability \(\pi\).

One natural transformation of a binomial variable \(R\) is reversal of 'success' and 'failure'. Show that this maps \(R\) to \(m-R\), where \(m\) is the denominator, and that the induced transformation on the parameter space maps \(\pi\) to \(1-\pi\). Which of the critical regions (a) \(\mathcal{Y}_{1}=\\{0,1,20\\}\), (b) \(\mathcal{Y}_{2}=\\{0,1,19,20\\}\), (c) \(\mathcal{Y}_{3}=\\{0,1,10,19,20\\}\) (d) \(\mathcal{Y}_{4}=\\{8,9,10,11,12\\}\), is invariant for testing \(\pi=\frac{1}{2}\) when \(m=20 ?\) Which is preferable and why?

Consider testing the hypothesis that a binomial random variable has probability \(\pi=1 / 2\) against the alternative that \(\pi>1 / 2\). For what values of \(\alpha\) does a uniformly most powerful test exist when the denominator is \(m=5\) ?

(a) Let \(Y_{1}, \ldots, Y_{n}\) be a random sample from the exponential density \(\lambda e^{-\lambda y}, y>0, \lambda>0\) Say why an unbiased estimator \(W\) for \(\lambda\) should have form \(a / S\), and hence find \(a\). Find the Fisher information for \(\lambda\) and show that \(\mathrm{E}\left(W^{2}\right)=(n-1) \lambda^{2} /(n-2)\). Deduce that no unbiased estimator of \(\lambda\) attains the Cram茅r-Rao lower bound, although \(W\) does so asymptotically. (b) Let \(\psi=\operatorname{Pr}(Y>a)=e^{-\lambda a}\), for some constant \(a\). Show that $$ I\left(Y_{1}>a\right)= \begin{cases}1, & Y_{1}>a \\ 0, & \text { otherwise }\end{cases} $$ is an unbiased estimator of \(\psi\), and hence obtain the minimum variance unbiased estimator. Does this attain the Cram茅r-Rao lower bound for \(\psi\) ?

Let \(R\) be binomial with probability \(\pi\) and denominator \(m\), and consider estimators of \(\pi\) of form \(T=(R+a) /(m+b)\), for \(a, b \geq 0\). Find a condition under which \(T\) has lower mean squared error than the maximum likelihood estimator \(R / m\), and discuss which is preferable when \(m=5,10\).
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