To show that $\frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2$ is an unbiased estimator of the population variance $\sigma^2$, let's break it down step by step.

Problem Understanding

We have a random sample $X_1, X_2, \dots, X_n$ drawn from a population with the following properties:

• The population mean is $\mu = \mathbb{E}[X_i]$, for each $i$.
• The population variance is $\sigma^2 = \mathbb{V}ar(X_i) = \mathbb{E}[(X_i - \mu)^2]$. We are tasked with proving that the estimator: $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2$ is an unbiased estimator for $\sigma^2$, meaning that: $\mathbb{E}[\hat{\sigma}^2] = \sigma^2$

Step-by-Step Proof

Step 1: Express the Estimator

The estimator of the population variance is given by: $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2$ We want to find the expected value of $\hat{\sigma}^2$, that is, $\mathbb{E}[\hat{\sigma}^2]$.

Step 2: Calculate the Expected Value

We can express $\hat{\sigma}^2$ as: [ \mathbb{E}[\hat{\sigma}^2] = \mathbb{E} \left[\frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2 \right] ] Using the linearity of expectation: [ \mathbb{E}[\hat{\sigma}^2] = \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[(X_i - \mu)^2] ]

Step 3: Use the Definition of Variance

Since $\mathbb{E}[(X_i - \mu)^2] = \text{Var}(X_i)$, and by assumption, the variance is $\sigma^2$ for each $X_i$, we have: $\mathbb{E}[(X_i - \mu)^2] = \sigma^2$ Therefore: $\mathbb{E}[\hat{\sigma}^2] = \frac{1}{n} \sum_{i=1}^{n} \sigma^2$

Step 4: Simplify

Since the sum consists of $n$ terms, each equal to $\sigma^2$, we can simplify the expression: $\mathbb{E}[\hat{\sigma}^2] = \frac{1}{n} \cdot n \cdot \sigma^2 = \sigma^2$

Conclusion

Thus, we have shown that: $\mathbb{E}[\hat{\sigma}^2] = \sigma^2$ which means that $\frac{1}{n} \sum_{i=1}^{n} (X_i - \mu)^2$ is an unbiased estimator of the population variance $\sigma^2$.

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Would you like more details or have any further questions on this topic?

Here are 5 related questions:

1. What is the difference between an unbiased estimator and a biased estimator?
2. How does the formula for the sample variance $\frac{1}{n-1} \sum (X_i - \bar{X})^2$ differ from the population variance formula?
3. Why is the sample variance formula divided by $n-1$ instead of $n$?
4. How would the proof change if we were estimating the population mean instead of the variance?
5. What is the Law of Large Numbers and how does it relate to the accuracy of estimators?

Tip: In practice, when estimating the population variance from a sample, you often use $n-1$ in the denominator to obtain an unbiased estimator. This is because dividing by $n$ tends to underestimate the variance.