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Suppose that \(y_{i}=\mu+e_{i},\) where \(i=1, \ldots ., n\) and the \(e_{i}\) are independent errors with mean zero and variance \(\sigma^{2}\). Show that \(\bar{y}\) is the least squares estimate of \(\mu\).

Short Answer

Expert verified
The sample mean \(\bar{y}\) is the least squares estimate of \(\mu\).

Step by step solution

01

Understanding the Problem

The given problem asks us to prove that the sample mean \(\bar{y}\) is the least squares estimate of \(\mu\) given the model \(y_i = \mu + e_i\). In this context, the term 'least squares estimate' refers to the value that minimizes the sum of squared deviations from the observed values.
02

Expressing the Least Squares Criterion

The least squares criterion involves finding the value of \(\mu\) that minimizes the sum of squared differences between the observed values \(y_i\) and their estimated values \(\mu\).Thus, we need to minimize the function \[S(\mu) = \sum_{i=1}^{n}(y_i - \mu)^2.\]
03

Taking the Derivative

To find the minimum of \(S(\mu)\), we take its derivative with respect to \(\mu\).The derivative is given by:\[\frac{dS}{d\mu} = -2\sum_{i=1}^{n}(y_i - \mu).\]
04

Setting the Derivative to Zero

Set the derivative to zero and solve for \(\mu\) to find the minimum.\[-2\sum_{i=1}^{n}(y_i - \mu) = 0\\Rightarrow \sum_{i=1}^{n}y_i = n\mu\]
05

Solving for \(\mu\)

Rearrange the equation \(\sum_{i=1}^{n}y_i = n\mu\) to solve for \(\mu\).\[\mu = \frac{1}{n}\sum_{i=1}^{n}y_i = \bar{y}\]This shows that \(\bar{y}\), the sample mean, is the least squares estimate of \(\mu\).
06

Verifying Minimization

After finding \(\mu = \bar{y}\), check the second derivative to ensure it represents a minimum. The second derivative is:\[\frac{d^2S}{d\mu^2} = 2n > 0\]Since the second derivative is positive, \(\bar{y}\) is indeed a minimum.

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Key Concepts

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

Independent Errors
In statistical modeling, independent errors are crucial because they imply that each measurement error in a dataset does not influence another. This means if you have an error at one data point, it won’t distort or imply any error at another point. In our example, the errors are noted as \(e_i\) and each is independent from the others.

To break this down further, consider that in a good dataset or model, each observation has its own error that is isolated from others. The idea of independence ensures that these errors are not correlated, which helps in creating accurate predictions. Independence simplifies calculations, especially when estimating parameters like the mean or variance in a dataset.
  • The independence assures no single error value has undue influence on another.
  • This helps in making the sample mean an unbiased estimator.
Sample Mean
The sample mean, denoted by \( \bar{y} \), is one of the most straightforward statistical measures and represents the average of a set of observations. Calculating the sample mean involves summing all observed values and dividing by the number of observations.
  • Formula: \( \bar{y} = \frac{1}{n} \, \sum_{i=1}^{n} y_i \)
  • The sample mean offers a central value that balances all data points.

In least squares estimation, the sample mean plays a vital role as it serves as the point around which the sum of squared deviations from individual data points is minimized. Essentially, it evenly distributes the differences from each data point, making it the best estimate of the population mean \( \mu \). The process of minimizing squared differences ensures that the mean is as representative of the entire dataset as possible.
Sum of Squared Deviations
The sum of squared deviations is a key concept in data analysis, providing insights into the variation of data points from a mean \( \mu \). It represents the total squared distance of each observation from a target value, typically the sample mean.

In our problem, the function to minimize is \( S(\mu) = \sum_{i=1}^{n} (y_i - \mu)^2 \), which aggregates these squared differences. Minimizing this function involves finding the \( \mu \) that leads to the smallest overall squared differences, thereby offering the best fit for the given data.
  • Squared deviations emphasize larger differences more than linear ones.
  • The focus on squaring helps manage positive and negative differences efficiently.

Ultimately, by selecting the \( \mu \) which minimizes this sum, we employ the least squares approach. Doing so ensures that the chosen \( \mu \), in this instance our sample mean \( \bar{y} \), represents the dataset in the most balanced manner possible.
Variance
Variance measures the spread of data points in a dataset. In our context, the variance of the error term \(e_i\), denoted as \(\sigma^2\), reflects the consistency or dispersion of errors.
  • Variance is calculated as the average of the squared differences from the mean.
  • It gives an overall picture of how much the data spreads out around the mean.

A smaller variance indicates that the data points are closely clustered around the mean, while a larger variance indicates more spread out data. In the least squares estimation scenario, understanding variance is important because it underlines the precision of our estimates.
When the errors are assumed to be independent with equal variance, as in our model where \(e_i\) has variance \(\sigma^2\), it supports the hypothesis that the sample mean \(\bar{y}\) is indeed the least squares estimate of the true mean \(\mu\). This is because the variance interacts directly with the consistency and reliability of the mean as a central tendency measure.

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