Problem 5 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from normal distributions with variances \(\sigma^{2}\). It can be shown that $$ \operatorname{Var}\left(S_{X}^{2}\right)=\frac{2 \sigma^{4}}{n-1} \text { and } \operatorname{Var}\left(S_{Y}^{2}\right)=\frac{2 \sigma^{4}}{m-1} $$ Consider linear combinations \(a S_{X}^{2}+b S_{Y}^{2}\) that are unbiased estimators for \(\sigma^{2}\). a. Show that \(a\) and \(b\) must satisfy \(a+b=1\). b. Show that \(\operatorname{Var}\left(a S_{X}^{2}+(1-a) S_{Y}^{2}\right)\) is minimized for \(a=(n-1) /(n+m-2)\) (and hence \(b=(m-1) /(n+m-2))\).

Step-by-Step Solution

Answer

a and b must satisfy a+b=1. a=(n-1)/(n+m-2) minimizes the variance.

Step 1: Unbiased Estimator Constraint

To find unbiased estimators, the linear combination \( a S_X^2 + b S_Y^2 \) must satisfy the condition that its expected value equals \( \sigma^2 \). This gives us the equation \( a E(S_X^2) + b E(S_Y^2) = \sigma^2 \). Since both \( E(S_X^2) \) and \( E(S_Y^2) \) are \( \sigma^2 \), the constraint simplifies to \( a + b = 1 \). Thus, \( a \) and \( b \) must satisfy \( a + b = 1 \).

Step 2: Express Variance of Linear Combination

The variance of the linear combination \( a S_X^2 + (1-a) S_Y^2 \) is \[ \operatorname{Var}\left(a S_X^2 + (1-a) S_Y^2\right) = a^2 \operatorname{Var}(S_X^2) + (1-a)^2 \operatorname{Var}(S_Y^2) \].

Step 3: Substitute Variances

Substitute the given variances: \( \operatorname{Var}(S_X^2) = \frac{2\sigma^4}{n-1} \) and \( \operatorname{Var}(S_Y^2) = \frac{2\sigma^4}{m-1} \) into the equation from Step 2. This gives: \[ \operatorname{Var}\left(a S_X^2 + (1-a) S_Y^2\right) = a^2 \frac{2\sigma^4}{n-1} + (1-a)^2 \frac{2\sigma^4}{m-1} \].

Step 4: Simplify the Expression

Factor out \( 2\sigma^4 \) for simplicity, giving: \[ 2\sigma^4 \left( \frac{a^2}{n-1} + \frac{(1-a)^2}{m-1} \right) \].

Step 5: Find Minimum of the Variance

To minimize this expression, complete the square or take the derivative with respect to \( a \), set it to zero, and solve for \( a \). The solution by setting the derivative to zero or completing the square results in \[ a = \frac{n-1}{n+m-2} \], and hence \( b = 1-a = \frac{m-1}{n+m-2} \). This minimizes the variance.

Key Concepts

Normal Distribution

Understanding the concept of a normal distribution is crucial for statistical estimation. When a variable follows a normal distribution, its data is symmetrically distributed around the mean, creating a bell-shaped curve. This characteristic makes it remarkably useful in real-world data analysis, as many natural phenomena approach this type of distribution.

In statistics, the normal distribution is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). These parameters help determine the shape and spread of the distribution.

• The mean describes the center of the distribution.
• The standard deviation describes the spread or variability around the mean.

For the given problem, two independent samples are taken from normal distributions which share the same variance (\( \sigma^{2} \)). This says that while the mean may differ, the spread of values around the mean is consistent across both samples.

Unbiased Estimator

An unbiased estimator is a statistical tool that accurately predicts the true value of a parameter. The idea is that the expected value of the estimator equals the parameter it estimates. In this case, the noise or systemic error in prediction is minimized. This makes the concept invaluable for drawing accurate conclusions from data.

For example, if we want to create an unbiased estimate for the variance (\( \sigma^{2} \)), any linear combination like \( a S_{X}^{2} + b S_{Y}^{2} \) must satisfy

\[ a \times E(S_{X}^{2}) + b \times E(S_{Y}^{2}) = \sigma^{2} \]
Given that both expected values are equal to \( \sigma^{2} \), this simplifies to \( a + b = 1 \). This constraint ensures no bias in estimating the true variance of the samples.

Variance Minimization

Variance minimization plays a critical role in improving the precision of the estimator. In this context, variance refers to the spread of the estimator's outcomes from the expected value. A smaller variance indicates that your estimator is more likely to be close to the actual parameter.

The problem involves finding values of coefficients \( a \)and \( b \) that not only satisfy \( a + b = 1 \)but also minimize the variance of the linear combination \( aS_{X}^{2} + (1-a)S_{Y}^{2} \). We factor in the variance for both \( S_{X}^{2} \)and \( S_{Y}^{2} \) from the samples.

After substitution and simplification, we derive:
\[2 \sigma^{4} \left( \frac{a^{2}}{n-1} + \frac{(1-a)^{2}}{m-1} \right)\]
Minimizing this expression leads us to find that
\( a = \frac{n-1}{n+m-2} \)and consequently
\( b = \frac{m-1}{n+m-2} \),
which results in minimized variance. This approach ensures the estimator is as stable and reliable as possible.

Random Samples

Random samples are foundational in creating unbiased and reliable estimators in statistics. When we draw a random sample from a population, each member has an equal chance of being selected. This randomness reduces biases that might distort statistical analyses.

In the problem context, the samples \( X_{1}, X_{2}, \ldots, X_{n} \)and \( Y_{1}, Y_{2}, \ldots, Y_{m} \) are independently drawn from normal distributions. This independence ensures that the characteristics of one group do not influence the other. This is crucial as any pattern that might emerge is purely due to chance, reinforcing the integrity of statistical predictions,

With such randomness, the assumptions on variance and mean used in calculations hold true, making analyses robust. Hence, using random samples provides a genuine reflection of the underlying population's characteristics, allowing statisticians to make valid inferences.