Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from \(N\left(\theta_{1}, \theta_{2}\right)\). (a) If the constant \(b\) is defined by the equation \(P(X \leq b)=0.90\), find the mle and the MVUE of \(b\). (b) If \(c\) is a given constant, find the mle and the MVUE of \(P(X \leq c)\).

Short Answer

Expert verified
MLE of \(b\) is \(\overline{X} + k * S\), and MVUE of \(b\) is \(E(b) = \theta_{1} + k * \theta_{2}\). MLE of \(P(X \leq c)\) is \(P(Z \leq (c - \overline{X}) / S)\), and MVUE of \(P(X \leq c)\) is \(E[P(X \leq c)] = P(Z \leq (c - \theta_{1}) / \theta_{2})\).

Step by step solution

01

Identifying MLE and MVUE for \(b\)

Since \(X\) is normally distributed, note that \(b\) can be expressed as \(b=\theta_{1} + k * \theta_{2}\), where \(k\) is the z score for a 0.90 probability. The MLE of \(b\) is the same as the MLE of \(\theta_{1} + k * \theta_{2}\). Since the MLE of \(\theta_{1}\) is the sample mean \(\overline{X}\) and the MLE of \(\theta_{2}\) is the sample standard deviation \(S\) when \(S^2\) is the sample variance, then the MLE of \(b\) is \(\overline{X} + k * S\). For the MVUE of \(b\), it will simply be the expected value of \(b\), which is \(E(b) = \theta_{1} + k * \theta_{2}\)
02

Identifying MLE and MVUE for \(P(X \leq c)\)

The MLE and MVUE of \(P(X \leq c)\) are equivalent to identifying the probability that a standard normal variable \(Z\) is less than or equal to \(Z = (c - \overline{X}) / S\). This can be denoted as \(P(Z \leq (c - \overline{X}) / S)\), and the standard normal distribution table or computational software can be used to find this value. As for the MVUE, it will be the probability that a standard normal variable \(Z\) is less than or equal to \(Z = (c - \theta_{1}) / \theta_{2}\), which can be written as \(E[P(X \leq c)] = P(Z \leq (c - \theta_{1}) / \theta_{2})\). Similar to the MLE, a standard normal distribution table or computational software can be used to find this value
03

NOTE

Given that \(X_{1}, X_{2}, ..., X_{n}\) is a random sample from a normal distribution, these steps hold valid. However, if the distribution was different, the process might vary.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Minimum Variance Unbiased Estimator
A Minimum Variance Unbiased Estimator (MVUE) is a statistical method used to estimate a parameter in such a way that the estimation is unbiased and has the lowest possible variance among all unbiased estimators. This means that an MVUE provides the most accurate and consistent estimates of a parameter. Think of it this way: if you repeated an experiment or took multiple samples from a population, using an MVUE gives you estimates that are right on target on average. And among all the possible unbiased estimators, it spreads out or varies the least, meaning you can trust it more to be close to the true parameter value in repeated sampling. For example:
  • When estimating the mean of a normally distributed population, the sample mean often acts as the MVUE. This is because it is unbiased (it correctly centers around the true mean) and tends to have minimum variance compared to other unbiased estimators of the mean.
  • In the problem exercise, we look at finding MVUE for probabilities or specific statistics derived from normally distributed data. These properties help ensure our statistical models and conclusions are both reliable and predictable in practice.
Understanding and using MVUEs is a core aspect of ensuring your statistical analyses are robust.
Normal Distribution
In statistics, the Normal Distribution is one of the most important probability distributions. It is often referred to as a Gaussian distribution and is characterized by its bell-shaped curve.Key features include:
  • The distribution is defined by two parameters: the mean (\(\theta_{1}\)) and the variance (\(\theta_{2}^{2}\)). These determine the center and the spread of the distribution, respectively.
  • The curve is symmetric around the mean, implying that the values have an equal chance of falling on either side of the mean.
  • Approximately 68% of the data falls within one standard deviation (\(\theta_{2}\)) from the mean. About 95% falls within two standard deviations, and nearly all (99.7%) lie within three standard deviations.
In many practical problems, including the textbook exercise, the normal distribution is used because of its unique properties. Many real-world phenomena such as test scores, heights, and errors in measurements naturally align with this distribution. This widely applies to inferential statistics where assumptions of normality are made to use statistical tests reliably.
Sample Mean
The Sample Mean is a fundamental concept in statistics, serving as a basic estimator of the population mean. It is calculated by summing up all the data points in a sample and dividing by the number of data points, essentially providing a measure of the central tendency of the data.When you have a normally distributed random sample, the sample mean (\(\overline{X}\)) is a crucial statistic. It is straightforward to compute and possesses a desirable property: it is an unbiased estimator of the population mean (\(\theta_{1}\)). That means, on average, the sample mean gives the exact value of the population mean when collected under random sampling conditions.Why is this important?
  • The calculation of a sample mean is often the first step in statistical analysis, paving the way for more complex inferences like variance calculi, hypothesis testing, or confidence interval estimation.
  • In Maximum Likelihood Estimation (MLE), the sample mean is often the preferred estimate for population characteristics due to its efficiency and simplicity.
Understanding the role of the sample mean allows you to make informed predictions and analyses from data, linking observation with theory in statistical practice.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a Poisson distribution with parameter \(\theta>0\) (a) Find the MVUE of \(P(X \leq 1)=(1+\theta) e^{-\theta}\). Hint: \(\quad\) Let \(u\left(x_{1}\right)=1, x_{1} \leq 1\), zero elsewhere, and find \(E\left[u\left(X_{1}\right) \mid Y=y\right]\), where \(Y=\sum_{1}^{n} X_{i}\). (b) Express the MVUE as a function of the mle. (c) Determine the asymptotic distribution of the mle.

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) denote a random sample of size \(n\) from a bivariate normal distribution with means \(\mu_{1}\) and \(\mu_{2}\), positive variances \(\sigma_{1}^{2}\) and \(\sigma_{2}^{2}\), and correlation coefficient \(\rho .\) Show that \(\sum_{1}^{n} X_{i}, \sum_{1}^{n} Y_{i}, \sum_{1}^{n} X_{i}^{2}, \sum_{1}^{n} Y_{i}^{2}\), and \(\sum_{1}^{n} X_{i} Y_{i}\) are joint complete sufficient statistics for the five parameters. Are \(\bar{X}=\) \(\sum_{1}^{n} X_{i} / n, \bar{Y}=\sum_{1}^{n} Y_{i} / n, S_{1}^{2}=\sum_{1}^{n}\left(X_{i}-\bar{X}\right)^{2} /(n-1), S_{2}^{2}=\sum_{1}^{n}\left(Y_{i}-\bar{Y}\right)^{2} /(n-1)\), and \(\sum_{1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right) /(n-1) S_{1} S_{2}\) also joint complete sufficient statistics for these parameters?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample with the common pdf \(f(x)=\) \(\theta^{-1} e^{-x / \theta}\), for \(x>0\), zero elsewhere; that is, \(f(x)\) is a \(\Gamma(1, \theta)\) pdf. (a) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\theta\). (b) Determine the MVUE of \(\theta\). (c) Determine the mle of \(\theta\). (d) Often, though, this pdf is written as \(f(x)=\tau e^{-\tau x}\), for \(x>0\), zero elsewhere. Thus \(\tau=1 / \theta\). Use Theorem \(6.1 .2\) to determine the mle of \(\tau\). (e) Show that the statistic \(\bar{X}=n^{-1} \sum_{i=1}^{n} X_{i}\) is a complete and sufficient statistic for \(\tau\). Show that \((n-1) /(n X)\) is the MVUE of \(\tau=1 / \theta\). Hence, as usual the reciprocal of the mle of \(\theta\) is the mle of \(1 / \theta\), but, in this situation, the reciprocal of the MVUE of \(\theta\) is not the MVUE of \(1 / \theta\). (f) Compute the variances of each of the unbiased estimators in Parts (b) and (e).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from each of the following distributions involving the parameter \(\theta\). In each case find the mle of \(\theta\) and show that it is a sufficient statistic for \(\theta\) and hence a minimal sufficient statistic. (a) \(b(1, \theta)\), where \(0 \leq \theta \leq 1\). (b) Poisson with mean \(\theta>0\). (c) Gamma with \(\alpha=3\) and \(\beta=\theta>0\). (d) \(N(\theta, 1)\), where \(-\infty<\theta<\infty\) (e) \(N(0, \theta)\), where \(0<\theta<\infty\)

Let \(X\) be a random variable with pdf of a regular case of the exponential class. Show that \(E[K(X)]=-q^{\prime}(\theta) / p^{\prime}(\theta)\), provided these derivatives exist, by differentiating both members of the equality $$\int_{a}^{b} \exp [p(\theta) K(x)+S(x)+q(\theta)] d x=1$$ with respect to \(\theta\). By a second differentiation, find the variance of \(K(X)\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free