Unbiased Estimator And Variance Of Normal Distribution Sum Of Squares

JU07/15/2025 08, 2025 by THE IDEN 70 views

$Understanding Unbiased Estimators and Variance in Normal Distributions: A Deep Dive into ${ \sum_{i=1}^{n} X_i^2 / n }$$

In the realm of statistics, estimating parameters of a population based on a sample is a fundamental task. When dealing with a normal distribution, one often seeks to estimate the variance, which is a crucial measure of data dispersion. This article delves into a specific scenario: a random sample ${ X_1, X_2, ..., X_n }$ drawn from a normal distribution with a mean of zero and a variance of ${ \theta }$ , where ${ 0 < \theta < \infty }$ . Our primary focus will be on demonstrating that the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is an unbiased estimator of ${ \theta }$ and that its variance is ${ 2\theta^2 / n }$ . This exploration will not only reinforce your understanding of statistical estimation but also illuminate the properties of estimators in the context of normal distributions.

Unbiased Estimator: ${ \sum_{i=1}^{n} X_i^2 / n }$

An unbiased estimator is a statistic used to estimate a population parameter where the mean of the sampling distribution of the statistic is equal to the true value of the parameter being estimated. In simpler terms, if we were to take many samples and calculate the estimator each time, the average of these estimates would converge to the true parameter value. This property is highly desirable as it ensures that, on average, our estimator is neither overestimating nor underestimating the true value. In the context of our problem, we aim to show that the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is an unbiased estimator of the variance ${ \theta }$ . To achieve this, we need to prove that the expected value of the estimator is equal to ${ \theta }$ .

Let's delve into the mathematical proof. We start with the definition of the expected value of our estimator:

${ E\left[ \frac{\sum_{i=1}^{n} X_i^2}{n} \right] }$

Using the linearity of expectation, we can rewrite this as:

${ \frac{1}{n} E\left[ \sum_{i=1}^{n} X_i^2 \right] = \frac{1}{n} \sum_{i=1}^{n} E[X_i^2] }$

Since each ${ X_i }$ is drawn from a normal distribution with mean 0 and variance ${ \theta }$ , we know that the expected value of ${ X_i^2 }$ is equal to the variance ${ \theta }$ . This is because for any random variable ${ X }$ , ${ Var(X) = E[X^2] - (E[X])^2 }$ , and in our case, ${ E[X_i] = 0 }$ , so ${ E[X_i^2] = Var(X_i) = \theta }$ . Substituting this into our equation, we get:

${ \frac{1}{n} \sum_{i=1}^{n} \theta = \frac{1}{n} (n \theta) = \theta }$

This result clearly demonstrates that the expected value of our estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is indeed ${ \theta }$ . Therefore, we have successfully shown that ${ \sum_{i=1}^{n} X_i^2 / n }$ is an unbiased estimator of the variance ${ \theta }$ for a normal distribution with mean zero.

Detailed Breakdown of the Unbiased Estimator Proof

To solidify the understanding of why ${ \sum_{i=1}^{n} X_i^2 / n }$ serves as an unbiased estimator for the variance ${ \theta }$ , let's break down the proof into more granular steps. This detailed exposition will clarify the underlying principles and assumptions that make this estimator so valuable in statistical inference.

Starting with the Estimator: The estimator in question is ${ \hat{\theta} = \sum_{i=1}^{n} X_i^2 / n }$ . This formula represents the average of the squared values of the random sample. It's an intuitive measure because squaring the values eliminates the sign, and the average of these squared values gives us an idea of the spread or dispersion of the data around the mean (which is zero in this case).
Applying the Expectation Operator: To prove unbiasedness, we need to show that the expected value of the estimator is equal to the true parameter value, i.e., ${ E[\hat{\theta}] = \theta }$ . We start by applying the expectation operator to our estimator:

${ E[\hat{\theta}] = E\left[ \frac{\sum_{i=1}^{n} X_i^2}{n} \right] }$
Using Linearity of Expectation: The expectation operator has a property called linearity, which states that the expected value of a sum is the sum of the expected values, and the expected value of a constant times a random variable is the constant times the expected value of the random variable. Applying this property, we can rewrite the expression as:

${ E[\hat{\theta}] = \frac{1}{n} E\left[ \sum_{i=1}^{n} X_i^2 \right] = \frac{1}{n} \sum_{i=1}^{n} E[X_i^2] }$

This step is crucial because it allows us to deal with each term in the sum individually.
Relating Expected Value to Variance: Here's where the properties of the normal distribution come into play. We know that for any random variable ${ X }$ , the variance is defined as ${ Var(X) = E[X^2] - (E[X])^2 }$ . In our case, each ${ X_i }$ has a mean of zero, i.e., ${ E[X_i] = 0 }$ . Therefore, the variance simplifies to:

${ Var(X_i) = E[X_i^2] - 0^2 = E[X_i^2] }$

We also know that each ${ X_i }$ is drawn from a normal distribution with variance ${ \theta }$ , so ${ Var(X_i) = \theta }$ . This gives us a direct relationship:

${ E[X_i^2] = \theta }$
Substituting and Simplifying: Now we substitute this result back into our expression:

${ E[\hat{\theta}] = \frac{1}{n} \sum_{i=1}^{n} E[X_i^2] = \frac{1}{n} \sum_{i=1}^{n} \theta }$

Since we are summing ${ \theta }$ ${ n }$ times, we get:

${ E[\hat{\theta}] = \frac{1}{n} (n \theta) }$

Finally, simplifying the expression, we arrive at:

${ E[\hat{\theta}] = \theta }$
Conclusion: This final result is the cornerstone of our proof. It shows that the expected value of our estimator ${ \hat{\theta} = \sum_{i=1}^{n} X_i^2 / n }$ is equal to the true variance ${ \theta }$ . By definition, this means that the estimator is unbiased. The estimator does not systematically overestimate or underestimate the true variance. It is a consistent and reliable way to estimate the variance from a sample drawn from a normal distribution with a mean of zero.

By dissecting the proof into these detailed steps, we gain a deeper appreciation for the properties of the normal distribution, the linearity of expectation, and the crucial relationship between variance and the expected value of squared deviations from the mean. This understanding empowers us to make informed decisions when estimating population parameters from sample data.

Variance of the Estimator: ${ 2\theta^2 / n }$

Now that we've established the unbiasedness of our estimator, the next critical step is to determine its variance. The variance of an estimator quantifies how much the estimator's values vary across different samples. A lower variance indicates that the estimator's values are more tightly clustered around its mean (which, for an unbiased estimator, is the true parameter value), making it a more precise estimator. In our case, we aim to show that the variance of the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is ${ 2\theta^2 / n }$ . This result will provide insights into the estimator's precision and how it scales with the sample size ${ n }$ .

To calculate the variance, we start with the definition of variance:

${ Var\left( \frac{\sum_{i=1}^{n} X_i^2}{n} \right) }$

We can rewrite this using the properties of variance. First, we factor out the constant ${ 1/n }$ :

${ Var\left( \frac{\sum_{i=1}^{n} X_i^2}{n} \right) = \frac{1}{n^2} Var\left( \sum_{i=1}^{n} X_i^2 \right) }$

Since the ${ X_i }$ 's are independent (as they are drawn from a random sample), the variance of the sum is the sum of the variances:

${ \frac{1}{n^2} Var\left( \sum_{i=1}^{n} X_i^2 \right) = \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i^2) }$

To find ${ Var(X_i^2) }$ , we recall that ${ X_i }$ follows a normal distribution with mean 0 and variance ${ \theta }$ . This implies that ${ X_i / \sqrt{\theta} }$ follows a standard normal distribution, denoted as ${ Z \sim N(0, 1) }$ . Consequently, ${ X_i^2 / \theta }$ follows a chi-squared distribution with 1 degree of freedom, denoted as ${ \chi^2(1) }$ . The variance of a chi-squared distribution with ${ k }$ degrees of freedom is ${ 2k }$ . Therefore, the variance of ${ X_i^2 / \theta }$ is ${ 2 }$ .

Now, we need to find the variance of ${ X_i^2 }$ . Using the property ${ Var(aX) = a^2 Var(X) }$ , we have:

${ Var(X_i^2) = Var(\theta \cdot (X_i^2 / \theta)) = \theta^2 Var(X_i^2 / \theta) = \theta^2 \cdot 2 = 2\theta^2 }$

Substituting this back into our equation, we get:

${ \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i^2) = \frac{1}{n^2} \sum_{i=1}^{n} 2\theta^2 = \frac{1}{n^2} (n \cdot 2\theta^2) = \frac{2\theta^2}{n} }$

Thus, we have shown that the variance of the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is indeed ${ 2\theta^2 / n }$ . This result is significant because it tells us that as the sample size ${ n }$ increases, the variance of the estimator decreases. This is a desirable property, as it indicates that with larger samples, our estimator becomes more precise and provides a more accurate estimate of the true variance ${ \theta }$ .

Detailed Exploration of the Variance Derivation

To fully grasp the variance of the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ , it is essential to dissect the derivation process into smaller, comprehensible segments. This detailed exploration will illuminate the statistical principles and transformations that underlie the result, providing a deeper appreciation for the estimator's behavior and properties.

Starting with the Variance Expression: We begin by expressing the variance of the estimator:

${ Var\left( \frac{\sum_{i=1}^{n} X_i^2}{n} \right) }$
Applying Variance Properties – Constant Factor: The variance of a constant times a random variable is the constant squared times the variance of the random variable. In our case, the constant is ${ 1/n }$ . Applying this property, we get:

${ Var\left( \frac{\sum_{i=1}^{n} X_i^2}{n} \right) = \frac{1}{n^2} Var\left( \sum_{i=1}^{n} X_i^2 \right) }$

This step simplifies the expression by moving the constant outside the variance operator.
Variance of a Sum of Independent Random Variables: Since the ${ X_i }$ 's are drawn from a random sample, they are independent. For independent random variables, the variance of their sum is the sum of their variances. Thus, we can rewrite the expression as:

${ \frac{1}{n^2} Var\left( \sum_{i=1}^{n} X_i^2 \right) = \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i^2) }$

This step is crucial because it breaks down the problem into calculating the variance of individual ${ X_i^2 }$ terms.
Connecting to the Chi-Squared Distribution: Here, we leverage the properties of the normal distribution. Since each ${ X_i }$ follows a normal distribution with mean 0 and variance ${ \theta }$ , we can standardize it by dividing by its standard deviation ${ \sqrt{\theta} }$ . The resulting variable, ${ Z_i = X_i / \sqrt{\theta} }$ , follows a standard normal distribution (mean 0, variance 1). Squaring this standardized variable, we get ${ Z_i^2 = X_i^2 / \theta }$ , which follows a chi-squared distribution with 1 degree of freedom, denoted as ${ \chi^2(1) }$ .

The chi-squared distribution is a well-known distribution in statistics, and its properties are extensively documented. One crucial property is its variance: the variance of a chi-squared distribution with ${ k }$ degrees of freedom is ${ 2k }$ . In our case, with 1 degree of freedom, the variance of ${ X_i^2 / \theta }$ is 2.
Calculating Variance of ${ X_i^2 }$ : We now need to find the variance of ${ X_i^2 }$ itself. To do this, we use the property that ${ Var(aX) = a^2 Var(X) }$ , where ${ a }$ is a constant. In our case, we can express ${ X_i^2 }$ as ${ \theta \cdot (X_i^2 / \theta) }$ . Applying the property, we get:

${ Var(X_i^2) = Var(\theta \cdot (X_i^2 / \theta)) = \theta^2 Var(X_i^2 / \theta) }$

Since we know that ${ Var(X_i^2 / \theta) = 2 }$ (from the chi-squared distribution), we have:

${ Var(X_i^2) = \theta^2 \cdot 2 = 2\theta^2 }$
Substituting Back into the Sum: We now substitute this result back into our expression for the variance of the estimator:

${ \frac{1}{n^2} \sum_{i=1}^{n} Var(X_i^2) = \frac{1}{n^2} \sum_{i=1}^{n} 2\theta^2 }$

Since we are summing ${ 2\theta^2 }$ ${ n }$ times, we get:

${ \frac{1}{n^2} \sum_{i=1}^{n} 2\theta^2 = \frac{1}{n^2} (n \cdot 2\theta^2) }$
Simplifying the Expression: Finally, we simplify the expression to obtain the variance of the estimator:

${ \frac{1}{n^2} (n \cdot 2\theta^2) = \frac{2\theta^2}{n} }$
Conclusion: This final result, ${ 2\theta^2 / n }$ , is the variance of the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ . It shows that the variance decreases as the sample size ${ n }$ increases. This is a desirable property, as it indicates that larger samples lead to more precise estimates of the variance ${ \theta }$ . The detailed derivation highlights the interplay between the normal distribution, the chi-squared distribution, and the properties of variance, providing a comprehensive understanding of the estimator's behavior.

Implications and Significance

The findings that the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is an unbiased estimator of ${ \theta }$ with a variance of ${ 2\theta^2 / n }$ have significant implications in statistical inference and practice. These properties are crucial for understanding the reliability and precision of our estimates, especially when dealing with data that is assumed to follow a normal distribution.

Unbiasedness and Accuracy

The unbiasedness of the estimator means that, on average, the estimator will give us the true value of the variance ${ \theta }$ . This is a fundamental property that ensures our estimator is not systematically overestimating or underestimating the parameter. In practical terms, if we were to repeatedly draw samples from the same population and calculate the estimator for each sample, the average of these estimates would converge to the true variance. This makes the estimator a reliable tool for inferring the population variance from sample data.

Variance and Precision

The variance of the estimator, ${ 2\theta^2 / n }$ , provides insights into the precision of our estimate. A lower variance indicates that the estimator's values are more tightly clustered around its mean, which, in this case, is the true variance ${ \theta }$ . This means that our estimate is likely to be closer to the true value. The formula ${ 2\theta^2 / n }$ reveals a critical relationship: the variance is inversely proportional to the sample size ${ n }$ . As the sample size increases, the variance decreases, indicating that larger samples lead to more precise estimates. This underscores the importance of collecting sufficient data to obtain reliable estimates of population parameters.

Practical Applications

These properties are particularly useful in various practical applications. For example, in quality control, we might use this estimator to assess the variability of a manufacturing process. A lower variance in the process indicates better consistency and quality. In financial analysis, understanding the variance of returns is crucial for risk management. An unbiased and precise estimator of variance allows for more accurate assessment of risk and informed investment decisions. In scientific research, estimating the variance is essential for hypothesis testing and confidence interval construction. An unbiased estimator with a known variance helps researchers draw valid conclusions from their data.

Limitations and Considerations

While the estimator ${ \sum_{i=1}^{n} X_i^2 / n }$ is unbiased and has a well-defined variance, it's essential to acknowledge its limitations. This estimator is specifically designed for normal distributions with a mean of zero. If the mean is non-zero or the data does not follow a normal distribution, the properties of unbiasedness and the variance formula may not hold. In such cases, alternative estimators and methods may be more appropriate.

Impact on Statistical Inference

The properties of this estimator significantly impact statistical inference. Knowing that the estimator is unbiased and has a variance of ${ 2\theta^2 / n }$ allows us to construct confidence intervals and perform hypothesis tests about the population variance ${ \theta }$ . For instance, we can use the central limit theorem to approximate the sampling distribution of the estimator and build confidence intervals that provide a range of plausible values for ${ \theta }$ . Similarly, we can use the estimator to test hypotheses about the variance, such as whether it exceeds a certain threshold. These inference procedures are fundamental tools in statistical analysis and decision-making.

Conclusion

In conclusion, we have rigorously demonstrated that ${ \sum_{i=1}^{n} X_i^2 / n }$ is an unbiased estimator of the variance ${ \theta }$ for a normal distribution with mean zero. We have also shown that its variance is ${ 2\theta^2 / n }$ , which decreases as the sample size ${ n }$ increases. These properties are critical for ensuring the accuracy and precision of our estimates. This exploration underscores the importance of understanding the properties of estimators in statistical inference and provides a solid foundation for more advanced statistical analysis. The unbiasedness and variance results are not just theoretical curiosities; they are practical tools that enable us to make informed decisions based on data, whether in quality control, finance, research, or any other field that relies on statistical analysis.