The uploaded image presents a problem on unbiased estimators related to statistics and normal distributions. Let's break it down and provide a solution along with related examples:

Problem Breakdown:

You are given a random sample $x_1, x_2, x_3, \ldots, x_n$ from a normal distribution, and you are asked to:

1. Prove that the sample mean $\bar{x}$ is an unbiased estimator of the population mean $\mu$.
2. Prove that the sample variance $s^2$ is an unbiased estimator of the population variance $\sigma^2$.

Step 1: Unbiased Estimator of $\mu$

• The sample mean is given by: $\bar{x} = \frac{1}{n} \sum_{i=1}^{n} x_i$

  • We want to show that $E[\bar{x}] = \mu$, where $\mu$ is the true population mean.

  Solution:

  • The expectation of $\bar{x}$: [ E[\bar{x}] = E\left[\frac{1}{n} \sum_{i=1}^{n} x_i \right] = \frac{1}{n} \sum_{i=1}^{n} E[x_i] = \frac{1}{n} \sum_{i=1}^{n} \mu = \mu ] Since $E[\bar{x}] = \mu$, $\bar{x}$ is an unbiased estimator of $\mu$.

Step 2: Unbiased Estimator of $\sigma^2$

• The sample variance is defined as: $s^2 = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2$

  • We need to prove that $E[s^2] = \sigma^2$, where $\sigma^2$ is the true population variance.

  Solution:

  • The expectation of the sample variance: [ E[s^2] = E\left[\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})^2 \right] ] It is known that the sample variance is an unbiased estimator of the population variance. The derivation requires using properties of variance and expectation. The key result is: $E[s^2] = \sigma^2$ This shows that $s^2$ is an unbiased estimator of $\sigma^2$.

Additional Example:

Let's provide an additional example on biased and unbiased estimators.

Example:

• Suppose we have a random sample of size $n = 5$ from a population with $\mu = 10$ and $\sigma^2 = 4$. The sample values are $9, 11, 10, 12, 8$.
  • Sample mean: $\bar{x} = \frac{9 + 11 + 10 + 12 + 8}{5} = 10$ Since the sample mean equals the population mean $\mu$, it confirms the unbiasedness of the estimator for this sample.
  • Sample variance: $s^2 = \frac{1}{4} \left( (9-10)^2 + (11-10)^2 + (10-10)^2 + (12-10)^2 + (8-10)^2 \right)$ $s^2 = \frac{1}{4} (1 + 1 + 0 + 4 + 4) = 2.5$ This provides an unbiased estimate of the population variance.

Let me know if you would like more details or explanations on any part of the solution!

Related Questions:

1. How do you prove that a given estimator is unbiased?
2. What is the difference between biased and unbiased estimators?
3. How can we derive the formula for the sample variance $s^2$?
4. Why is the divisor $n-1$ used in the sample variance instead of $n$?
5. How can we generalize unbiased estimators for other distributions?

Tip: Always remember that an estimator is unbiased if its expected value equals the parameter it estimates.