We are tasked with showing that the statistic

$\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$

is an unbiased estimator of the population variance $\sigma^2$, given a random sample of size $n$ from a population with mean $\mu$ and variance $\sigma^2$.

Step-by-step Solution:

1. Definition of Unbiased Estimator

An estimator $\hat{\theta}$ of a parameter $\theta$ is unbiased if:

$\mathbb{E}[\hat{\theta}] = \theta$

In this case, we want to show that:

$\mathbb{E}\left[\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \right] = \sigma^2$

where $X_i$ is the $i$-th observation in the sample.

2. Expanding the Sum

We start by expanding the expression inside the expectation:

$\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$

Since $(X_i - \mu)^2$ is a squared deviation from the population mean $\mu$, we can expand this as:

$(X_i - \mu)^2 = X_i^2 - 2X_i\mu + \mu^2$

Thus, the sum becomes:

$\frac{1}{n} \sum_{i=1}^n (X_i^2 - 2X_i \mu + \mu^2)$

We can now split the sum:

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

3. Taking the Expectation

Now, we compute the expectation of each term:

1. For $\sum_{i=1}^n X_i^2$: Since $X_i$ is a random sample from the population, each $X_i$ has the same distribution. The expectation of $X_i^2$ is given by:

   [ \mathbb{E}[X_i^2] = \text{Var}(X_i) + (\mathbb{E}[X_i])^2 = \sigma^2 + \mu^2 ]

   Therefore:

   $\mathbb{E}\left[\sum_{i=1}^n X_i^2\right] = n(\sigma^2 + \mu^2)$

2. For $\sum_{i=1}^n X_i$: The expectation of the sum is:

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

3. For $n\mu^2$: Since this is a constant, its expectation is simply $n\mu^2$.

4. Substituting Expectations

Substituting these expectations back into the expression:

$\mathbb{E}\left[\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \right] = \frac{1}{n} \left( n(\sigma^2 + \mu^2) - 2\mu n \mu + n\mu^2 \right)$

Simplifying:

$= \frac{1}{n} \left( n\sigma^2 + n\mu^2 - 2n\mu^2 + n\mu^2 \right)$

$= \frac{1}{n} \left( n\sigma^2 \right)$

$= \sigma^2$

Thus, we have shown that:

$\mathbb{E}\left[\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \right] = \sigma^2$

Therefore, $\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$ is an unbiased estimator of $\sigma^2$.

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Summary

We demonstrated that the statistic $\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$ is an unbiased estimator of the population variance $\sigma^2$, as its expectation equals $\sigma^2$.

Would you like further details on any part of this derivation, or have any specific questions?

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Related Questions:

1. How do we prove that the sample variance is an unbiased estimator of the population variance?
2. What is the difference between biased and unbiased estimators?
3. How can we estimate the population mean $\mu$ from a sample?
4. What is the formula for the variance of a sample (as opposed to the population)?
5. Can we modify the estimator $\frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$ to make it unbiased when we use the sample mean instead of $\mu$?

Tip:

When working with sample statistics, remember that unbiased estimators provide an accurate estimate of the population parameter on average, but individual samples can still vary.