91Ó°ÊÓ

Exercise 63 from Chapter 7 introduced "regression through the origin" to relate a dependent variable \(y\) to an independent variable \(x\). The assumption there was that for any fixed \(x\) value, the dependent variable is a random variable \(Y\) with mean value \(\beta x\) and variance \(\sigma^{2}\) (so that \(Y\) has mean value zero when \(x=0\) ). The data consists of \(n\) independent \(\left(x_{i}, Y_{i}\right)\) pairs, where each \(Y_{i}\) is normally distributed with mean \(\beta x_{i}\) and variance \(\sigma^{2}\). The likelihood is then a product of normal pdf's with different mean values but the same variance. a. Show that the mle of \(\beta\) is \(\beta=\Sigma x_{i} Y_{i} / \Sigma x_{i}^{2}\). b. Verify that the mle of (a) is unbiased. c. Obtain an expression for \(V(\hat{\beta})\) and then for \(\sigma_{\bar{\beta}}\). d. For purposes of obtaining a precise estimate of \(\beta\), is it better to have the \(x_{i}{ }^{\prime}\) s all close to 0 (the origin) or quite far from 0? Explain your reasoning. e. The natural prediction of \(Y_{i}\) is \(\hat{\beta} x_{i}\). Let \(S^{2}=\Sigma\left(Y_{i}-\hat{\beta} x_{i}\right)^{2} /(n-1)\) which is analogous to our earlier sample variance \(S^{2}=\Sigma\left(X_{i}-X\right)^{2} /(n-1)\) for a univariate sample \(X_{1}, \ldots, X_{n}\) (in which case \(X\) is a natural prediction for each \(X_{i}\) ). Then it can be shown that \(T=(\hat{\beta}-\beta) /\left(S / \sqrt{\sum x_{i}^{2}}\right)\) has a \(t\) distribution based on \(n-1\) df. Use this to obtain a CI formula for estimating \(\beta\), and calculate a \(95 \%\) CI using the data from the cited exercise.

Step by step solution

01

Understand the Likelihood Function

Given that each \(Y_i\) is normally distributed with mean \(\beta x_i\) and variance \(\sigma^2\), the probability density function for \(Y_i\) is \(\frac{1}{\sqrt{2\pi\sigma^2}}\exp \left(-\frac{(Y_i - \beta x_i)^2}{2\sigma^2}\right)\). The likelihood \(L(\beta)\) is the product of these densities for all data points: \[ L(\beta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}}\exp \left(-\frac{(Y_i - \beta x_i)^2}{2\sigma^2}\right) \]This simplifies to: \[ L(\beta) = \left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^n \exp \left(-\frac{1}{2\sigma^2} \sum (Y_i - \beta x_i)^2\right) \]

02

Find the MLE of \(\beta\)

To find the maximum likelihood estimate (MLE) of \(\beta\), we want to maximize the logarithm of the likelihood function (the log-likelihood), which is given by:\[ \log L(\beta) = -\frac{n}{2} \log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum (Y_i - \beta x_i)^2 \]Taking the derivative with respect to \(\beta\) and setting it to zero gives:\[ \frac{\partial \log L}{\partial \beta} = \frac{1}{\sigma^2} \sum x_i(Y_i - \beta x_i) = 0 \]Solving this equation yields:\[ \beta = \frac{\sum x_i Y_i}{\sum x_i^2} \]

03

Verify Unbiasedness of MLE

To verify that the MLE is unbiased, we need to show that \(E[\hat{\beta}]=\beta\). Substituting, we have:\[ E[\hat{\beta}] = E\left[ \frac{\sum x_i Y_i}{\sum x_i^2} \right] = \frac{\sum x_i E[Y_i]}{\sum x_i^2} \]Since \(E[Y_i] = \beta x_i\):\[ E[\hat{\beta}] = \frac{\sum x_i (\beta x_i)}{\sum x_i^2} = \beta \frac{\sum x_i^2}{\sum x_i^2} = \beta \]Thus, the estimator is unbiased.

04

Compute Variance of \(\hat{\beta}\)

Using properties of variance, we calculate:\[ V(\hat{\beta}) = V\left( \frac{\sum x_i Y_i}{\sum x_i^2} \right) = \frac{1}{(\sum x_i^2)^2} V\left(\sum x_i Y_i \right) \]Because \(Y_i\) are independent and \(V(Y_i) = \sigma^2\),\[ V\left(\sum x_i Y_i\right) = \sum x_i^2 \sigma^2 \]So, \(V(\hat{\beta}) = \frac{\sigma^2}{\sum x_i^2}\), and the standard deviation is:\[ \sigma_{\hat{\beta}} = \frac{\sigma}{\sqrt{\sum x_i^2}} \]

05

Analyze Precision of \(x_i\) Position

For obtaining a precise estimate of \(\beta\), notice that \(V(\hat{\beta}) = \frac{\sigma^2}{\sum x_i^2}\). To minimize variance, we need the sum \(\sum x_i^2\) to be as large as possible. Therefore, it's better for \(x_i\) values to be farther from zero.

06

Construct Confidence Interval for \(\beta\)

Use the result \[ T = \frac{\hat{\beta} - \beta}{S/\sqrt{\sum x_i^2}} \] which follows a t-distribution with \(n-1\) degrees of freedom. A CI for \(\beta\) can be calculated as:\[ \hat{\beta} \pm t_{\alpha/2, n-1} \frac{S}{\sqrt{\sum x_i^2}} \]Where \(t_{\alpha/2, n-1}\) is the critical value from a t-distribution table, and \(S\) is the estimate of variance from the given formula. Use data to find a specific CI for \(\beta\).

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Key Concepts

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

Maximum Likelihood Estimation

Maximum Likelihood Estimation, often abbreviated as MLE, is a fundamental method in statistical inference used to find parameter estimates that maximize the likelihood of observed data. In simple terms, MLE is about choosing the parameter values that make the observed data most probable. For our regression through the origin problem, we had to estimate the slope \( \beta \) that best fits our data points.
The likelihood function for the regression model is built based on probabilities given by the normal distribution. We use the log-likelihood, which is more manageable mathematically, to find the MLE. By taking the derivative of the log-likelihood with respect to the parameter \( \beta \), and setting it to zero, we obtain the MLE for \( \beta \):

• \[ \beta_{\text{MLE}} = \frac{\sum x_i Y_i}{\sum x_i^2} \]

This expression synthesizes information from all data points, rooted in the assumption that our observations are normally distributed with the same variance but potentially different means. MLE provides a robust framework, particularly because it yields estimates with desirable properties, such as consistency and efficiency, when certain regularity conditions are met.

Unbiased Estimator

An estimator is said to be unbiased if its expected value equals the true parameter value it aims to estimate. In simpler terms, an unbiased estimator hits the bullseye on average — although any given estimate might be off, if we repeatedly drew samples and re-estimated, the average of these estimates would equal the true parameter.
For our problem, once we obtained the MLE for \( \beta \), we verified its unbiasedness by calculating its expectation:

• \[ E[\hat{\beta}] = \frac{\sum x_i E[Y_i]}{\sum x_i^2} = \beta \]

This calculation shows that our MLE is unbiased because, fundamentally, the expected value of the outcome variable \( Y_i \) is correctly modeled by \( \beta x_i \). The fact that the MLE is unbiased adds confidence that on average, across many samples, our estimate of \( \beta \) will be correct.

Variance and Standard Deviation

Understanding the precision of an estimator involves examining its variance. Variance tells us how much the estimator might fluctuate from its expected value. A lower variance indicates more reliable estimates. For the MLE of \( \beta \), the variance is derived as follows:

• \[ V(\hat{\beta}) = \frac{\sigma^2}{\sum x_i^2} \]

This formula highlights the importance of the distribution of \( x_i \) values. If \( x_i \) values vary more (leading to a higher \( \sum x_i^2 \)), the variance of \( \hat{\beta} \) decreases, implying higher precision in our estimates.
The standard deviation, \( \sigma_{\hat{\beta}} \), simply is the square root of the variance and provides a measure of average deviation of \( \hat{\beta} \) from the expected value:

• \[ \sigma_{\hat{\beta}} = \frac{\sigma}{\sqrt{\sum x_i^2}} \]

This measure is crucial, especially when constructing confidence intervals, as it sets the stage for how tightly we can expect our estimator to cluster around the true value of \( \beta \).

Confidence Interval

A Confidence Interval (CI) gives a range within which we expect our parameter to lie, with a certain level of confidence — usually 95%. It reflects the variability in our estimate based on the data. For \( \beta \), the CI is constructed using the t-distribution, given the finite sample size:

• \[ \hat{\beta} \pm t_{\alpha/2, n-1} \frac{S}{\sqrt{\sum x_i^2}} \]

Here:\- \( t_{\alpha/2, n-1} \) is the critical value from the t-distribution, accounting for the sample size \( n \),\- \( S \) represents the sample standard deviation, providing an estimate of \( \sigma^2 \).
The confidence interval is crucial in statistical inference. It tells us, "We are 95% confident that the true value of \( \beta \) lies within this range." By considering both the variability in the data and the sample size, CIs provide a meaningful context to interpret how reliable or uncertain our MLE of \( \beta \) is. This combination of estimation and uncertainty quantification underlies the strength of statistical analysis in decision-making.