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Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\) distributions, respectively. (a) If \(\Omega \subset R^{3}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{2}, \theta_{3}\right):-\infty<\theta_{i}<\infty, i=1,2 ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\). (b) If \(\Omega \subset R^{2}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{3}\right):-\infty<\theta_{1}=\theta_{2}<\infty ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}\) and \(\theta_{3}\).

Short Answer

Expert verified
For the first condition, \(\hat{\theta_{1}} = \frac{1}{n} \sum_{i=1}^{n} X_{i}\), \(\hat{\theta_{2}} = \frac{1}{m} \sum_{i=1}^{m} Y_{i}\), and \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{2}})^{2}\). For the second condition, \(\hat{\theta_{1}} = \frac{1}{n+m} \sum_{i=1}^{n} X_{i} + \sum_{j=1}^{m} Y_{j}\) and \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{1}})^{2}\)

Step by step solution

01

Compute mles for the first condition

For \(X_{1}, X_{2}, \ldots, X_{n}\sim N(\theta_{1}, \theta_{3})\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\sim N(\theta_{2}, \theta_{3})\), we calculate the sample means as the mle estimates for \(\theta_{1}\) and \(\theta_{2}\) as these are independent normal samples. Thus, \(\hat{\theta_{1}} = \frac{1}{n} \sum_{i=1}^{n} X_{i}\) and \(\hat{\theta_{2}} = \frac{1}{m} \sum_{i=1}^{m} Y_{i}\). Compute also for \(\theta_{3}\) as \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{2}})^{2}\)
02

Compute mles for the second condition

For \(X_{1}, X_{2}, \ldots, X_{n}\sim N(\theta_{1}, \theta_{3})\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\sim N(\theta_{1}, \theta_{3})\), the mle estimate for \(\theta_{1}\) is the combined sample mean. Thus, \(\hat{\theta_{1}} = \frac{1}{n+m} \sum_{i=1}^{n} X_{i} + \sum_{j=1}^{m} Y_{j}\). Compute for \(\theta_{3}\) as \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{1}})^{2}\)

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Key Concepts

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

Independent Random Samples
When dealing with statistics and probability, it's crucial to understand independent random samples. In simple terms, independent random samples mean that each sample taken from a population has no influence on any other sample.

This independence ensures that the occurrence or measurement of one sample doesn't affect another, making statistical analysis more reliable.
  • **Independent Sampling**: Each draw or observation is made in isolation from others.
  • **Randomness**: Each sample is drawn without bias, error or external influence.
  • **Applications**: This is particularly useful in experiments and surveys where the objective is to generalize findings to a larger population.
In the context of Maximum Likelihood Estimation (MLE) for independent samples, such as those described in the exercise, the independence makes it easier to find estimates that maximize the likelihood function for the group of samples. This simplifies calculations and helps in deriving unbiased estimators for the population parameters.
Normal Distribution
The concept of normal distribution is foundational in statistics. Often visualized as a bell-shaped curve, the normal distribution is a probability distribution that is symmetric around its mean.

It is characterized by two main parameters: the mean ( \(\mu\)) and the standard deviation ( \(\sigma\)).
  • **Properties**: Approximately 68% of the data falls within one standard deviation ( \(\sigma\)) of the mean, and about 95% within two.
  • **Relevance**: Many natural phenomena follow a normal distribution, which is why it's heavily used in statistical analyses.
  • **Practical Application**: In the context of the exercise, both samples are drawn from populations assumed to be normally distributed with different means but the same variance.
Understanding normal distribution is essential, especially in tasks involving MLE because the assumption of normality makes numerous statistical techniques applicable and accurate.
Sample Mean
The sample mean is an essential concept in statistics. It represents the average value of a group of numbers from a dataset, often symbolized as \(\bar{x}\) or \(\bar{y}\) for different samples. Calculating the sample mean ensures that we have a central tendency measure of the data.

Here’s why it matters:
  • **Calculation**: Sum all the values in a dataset, then divide by the number of observations to get the sample mean.
  • **Significance in MLE**: In the exercise, the sample mean of random variables \(X\) and \(Y\) are used as the maximum likelihood estimates for the normal distribution means \(\theta_1\) and \(\theta_2\).
  • **Central Tendency**: It reflects the approximate center of the data, allowing for inferences about the population mean.
By using the sample mean, especially for independent and normally distributed data, you can establish a point estimate for the population mean. This is crucial when determining MLE, as it provides an unbiased estimation of the parameters.

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Most popular questions from this chapter

Let the table $$ \begin{array}{c|cccccc} x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & 6 & 10 & 14 & 13 & 6 & 1 \end{array} $$ represent a summary of a sample of size 50 from a binomial distribution having \(n=5\). Find the mle of \(P(X \geq 3)\). For the data in the table, using the \(\mathrm{R}\) function pbinom determine the realization of the mle.

Rao (page 368,1973 ) considers a problem in the estimation of linkages in genetics. McLachlan and Krishnan (1997) also discuss this problem and we present their model. For our purposes, it can be described as a multinomial model with the four categories \(C_{1}, C_{2}, C_{3}\), and \(C_{4}\). For a sample of size \(n\), let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}, X_{4}\right)^{\prime}\) denote the observed frequencies of the four categories. Hence, \(n=\sum_{i=1}^{4} X_{i} .\) The probability model is $$ \begin{array}{|c|c|c|c|} \hline C_{1} & C_{2} & C_{3} & C_{4} \\ \hline \frac{1}{2}+\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4}-\frac{1}{4} \theta & \frac{1}{4} \theta \\ \hline \end{array} $$ where the parameter \(\theta\) satisfies \(0 \leq \theta \leq 1\). In this exercise, we obtain the mle of \(\theta\). (a) Show that likelihood function is given by $$ L(\theta \mid \mathbf{x})=\frac{n !}{x_{1} ! x_{2} ! x_{3} ! x_{4} !}\left[\frac{1}{2}+\frac{1}{4} \theta\right]^{x_{1}}\left[\frac{1}{4}-\frac{1}{4} \theta\right]^{x_{2}+x_{3}}\left[\frac{1}{4} \theta\right]^{x_{4}} $$ (b) Show that the log of the likelihood function can be expressed as a constant (not involving parameters) plus the term $$ x_{1} \log [2+\theta]+\left[x_{2}+x_{3}\right] \log [1-\theta]+x_{4} \log \theta $$ (c) Obtain the partial derivative with respect to \(\theta\) of the last expression, set the result to 0 , and solve for the mle. (This will result in a quadratic equation that has one positive and one negative root.)

Given \(f(x ; \theta)=1 / \theta, 00\), formally compute the reciprocal of $$ n E\left\\{\left[\frac{\partial \log f(X: \theta)}{\partial \theta}\right]^{2}\right\\} $$ Compare this with the variance of \((n+1) Y_{n} / n\), where \(Y_{n}\) is the largest observation of a random sample of size \(n\) from this distribution. Comment.

Let \(\left(X_{1}, Y_{1}\right),\left(X_{2}, Y_{2}\right), \ldots,\left(X_{n}, Y_{n}\right)\) be a random sample from a bivariate normal distribution with \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2}, \rho=\frac{1}{2}\), where \(\mu_{1}, \mu_{2}\), and \(\sigma^{2}>0\) are unknown real numbers. Find the likelihood ratio \(\Lambda\) for testing \(H_{0}: \mu_{1}=\mu_{2}=0, \sigma^{2}\) unknown against all alternatives. The likelihood ratio \(\Lambda\) is a function of what statistic that has a well- known distribution?

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from the two normal distributions \(N\left(0, \theta_{1}\right)\) and \(N\left(0, \theta_{2}\right)\). (a) Find the likelihood ratio \(\Lambda\) for testing the composite hypothesis \(H_{0}: \theta_{1}=\theta_{2}\) against the composite alternative \(H_{1}: \theta_{1} \neq \theta_{2}\). (b) This \(\Lambda\) is a function of what \(F\) -statistic that would actually be used in this test?

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