Question

Assuming population to be \(N\left(\mu, \sigma^{2}\right)\), show that sample variance is a consistent estimator for population variance \(\sigma^{2}\).

Short answer

The sample variance is a consistent estimator for population variance, as demonstrated by the fact that the expected value of the sample variance converges to the population variance as the sample size increases.

Step-by-step solution

Define the sample variance

Sample variance \(s^{2}\) of a sample of size \(n\) from a population with variance \(\sigma^{2}\) is \(s^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (X_{i} - \bar{X})^{2}\), where \(X_{i}\) are the sample data points and \(\bar{X}\) is the sample mean.

Expand and simplify

For simplicity, let's denote \(E[X_{i}] = \mu\) and \(E[X_{i}^{2}] = \sigma^{2} + \mu^{2}\). We expand \(s^{2}\) as \(s^{2} = \frac{1}{n-1} \sum_{i=1}^{n} (X_{i}^{2} - 2X_{i}\bar{X} + \bar{X}^{2})\) and then taking expectations on both sides: \(E[s^{2}] = \frac{1}{n-1} E[\sum_{i=1}^{n} X_{i}^{2} - 2n\bar{X}^{2} + n\bar{X}^{2})\]. Simplify this to get \(E[s^{2}] = \sigma^{2} + \mu^{2} - \frac{2\mu^{2}}{n} + \frac{\mu^{2}}{n}\).

Show that the variance is consistent

The limit of an estimator as \(n \rightarrow \infty\) shows its consistency. In this case, we have \(\lim_{n \rightarrow \infty} E[s^{2}] = \lim_{n \rightarrow \infty} (\sigma^{2} + \mu^{2} - \frac{2\mu^{2}}{n} + \frac{\mu^{2}}{n}) = \sigma^{2} + \mu^{2} - 2\mu^{2} + \mu^{2} = \sigma^{2}\). Hence, the sample variance is a consistent estimator for the population variance.

Key concepts

Sample Variance

Sample variance is a statistical measure used to determine how far each value in a sample is from the sample mean. It helps understand the variability within a sample as opposed to the entire population. Mathematically, the formula for sample variance, denoted as \(s^2\), is given by:\[ s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \bar{X})^2 \]Here, \(n\) is the number of observations in your sample, \(X_i\) represents each individual observation, and \(\bar{X}\) is the average of all observations in the sample.**Why divide by \(n-1\)?**- This division by \(n-1\) instead of \(n\) corrects the bias in the estimation of the population variance. It's called Bessel's correction.- Without this correction, you might underestimate the variance because samples tend to reflect less variability than the population.

Population Variance

Population variance is a measure of how data points in a whole population are distributed around the mean of that population. It differs from sample variance in terms of scope and calculation. The population variance is denoted as \(\sigma^2\) and calculated as:\[ \sigma^2 = \frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2 \]In this equation, \(N\) represents the total number of observations in the population, \(X_i\) is each observation, and \(\mu\) is the mean of the population. **Understanding the Importance**- Population variance provides a complete picture of the variability within the entire group.- It is crucial for making informed decisions and predictions about the population. - Unlike sample variance, which uses \(n-1\) in its calculation, population variance uses \(N\) because it accounts for every observation without needing to adjust for bias.

Consistent Estimator

A consistent estimator in statistics is one that produces values closer to the parameter being estimated as the sample size increases. For sample variance to be a consistent estimator of population variance, the calculated value of \(s^2\) should approach the population variance \(\sigma^2\) as the sample size \(n\) tends towards infinity.**Key Characteristics of Consistency**- As sample size increases, the sample variance provides a better approximation of the population variance.- The probability that the estimator deviates from the true parameter value by more than a given margin approaches zero as sample size increases.**Mathematical Evidence**The consistency of sample variance as a measure of population variance can be shown mathematically by evaluating:\[ \lim_{n \to \infty} E[s^2] = \sigma^2 \]This means as the sample size \(n\) becomes very large, the expected value of the sample variance \(s^2\) converges to the true population variance \(\sigma^2\), establishing it as a consistent estimator.