We are tasked with showing that the following expression is an unbiased estimator of the population variance $\sigma^2$:

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

Here, $X_1, X_2, \dots, X_n$ are random samples drawn from a population with known mean $\mu$ and variance $\sigma^2$. To prove that $\hat{\sigma}^2$ is an unbiased estimator, we need to show that the expected value of $\hat{\sigma}^2$ is equal to $\sigma^2$. In other words, we need to prove that:

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

Step 1: Break down the sum

We begin by analyzing the sum inside the expectation:

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

Step 2: Expand the squared term

First, expand the square:

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

Now, take the sum over $i$:

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

Step 3: Take the expectation

Next, we compute the expectation of the entire sum. We can separate each term because the expectation operator is linear:

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

Now, evaluate the expectations for each term:

• $E[X_i^2]$ is the second moment of $X_i$, and since $X_i$ is drawn from a population with variance $\sigma^2$ and mean $\mu$, we know that:

  $E[X_i^2] = \sigma^2 + \mu^2$

• $E[X_i]$ is the mean of the random variable $X_i$, which is $\mu$, so:

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

Thus, the expectation becomes:

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

Simplifying this expression:

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

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

Step 4: Final Calculation

Now, divide by $n$ to compute $\hat{\sigma}^2$:

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

Conclusion

We have shown that:

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

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

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Would you like more details on any of the steps or concepts? Here are a few related questions you might find interesting:

1. How does this result change if we estimate the population mean $\mu$ from the sample data rather than assuming it's known?
2. What would the expression for an unbiased estimator of variance be if we used $n-1$ in the denominator instead of $n$?
3. How do unbiased estimators of population parameters differ from biased estimators in terms of properties like consistency and efficiency?
4. Can you derive the variance of the estimator $\hat{\sigma}^2$?
5. How does the sample size $n$ influence the reliability of this estimator?

Tip: The choice of using $n$ or $n-1$ in the denominator when estimating the variance depends on whether you're working with the full population or a sample. For sample variance, $n-1$ corrects the bias.