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To solve the problem, we'll follow these steps:

### Problem Overview:
- We have a random sample \( X_1, X_2, \dots, X_n \) from a normal distribution \( N(\mu, \sigma^2) \).
- The goal is to show that the maximum likelihood estimator (MLE) of the variance \( \sigma^2 \) is given by:
  \[
  \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2
  \]
  where \( \bar{X} \) is the sample mean:
  \[
  \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i
  \]

### Step-by-Step Solution:

1. **Log-Likelihood Function:**
   The likelihood function for the normal distribution \( N(\mu, \sigma^2) \) based on the sample \( X_1, X_2, \dots, X_n \) is:
   \[
   L(\mu, \sigma^2) = \prod_{i=1}^n \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left(-\frac{(X_i - \mu)^2}{2\sigma^2}\right)
   \]
   Taking the natural logarithm of the likelihood function (the log-likelihood), we have:
   \[
   \ell(\mu, \sigma^2) = -\frac{n}{2} \log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \mu)^2
   \]

2. **Without Differentiation:**
   To find the MLE for \( \sigma^2 \) without differentiating the log-likelihood function:
   - Notice that the log-likelihood function is maximized when the sum of squared differences \( \sum_{i=1}^n (X_i - \mu)^2 \) is minimized.
   - The minimum value of this sum occurs when \( \mu = \bar{X} \), the sample mean. Thus, the MLE for \( \mu \) is \( \hat{\mu} = \bar{X} \).

3. **Substituting \( \mu = \bar{X} \):**
   Substitute \( \hat{\mu} = \bar{X} \) into the expression for the log-likelihood:
   \[
   \ell(\bar{X}, \sigma^2) = -\frac{n}{2} \log(2\pi\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (X_i - \bar{X})^2
   \]
   The term \( \sum_{i=1}^n (X_i - \bar{X})^2 \) represents the sum of squared deviations from the mean, also known as the total variance.

4. **MLE for \( \sigma^2 \):**
   The MLE for \( \sigma^2 \) maximizes the log-likelihood. It is obtained by minimizing the denominator in the second term, which happens when:
   \[
   \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2
   \]
   This is the desired result.

Thus, the MLE for \( \sigma^2 \) is \( \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2 \).

Would you like a more detailed explanation of any part of this solution? Here are 8 related questions to deepen your understanding:

1. How is the log-likelihood function for a normal distribution derived?
2. Why is the sample mean \( \bar{X} \) the MLE for \( \mu \)?
3. What is the significance of the sum of squared deviations in statistics?
4. Can the MLE for \( \sigma^2 \) be biased or unbiased? How can it be adjusted if biased?
5. How would the MLE change if the distribution were not normal?
6. What role does the Central Limit Theorem play in this context?
7. How would the estimation change for small sample sizes?
8. What assumptions are critical for the validity of this MLE?

**Tip:** When working with MLE, understanding the properties of the likelihood function is key to finding estimators that maximize the likelihood without necessarily requiring complex calculus.

Learn how to find the maximum likelihood estimator (MLE) for the variance in a normal distribution with a detailed step-by-step solution. This problem covers key statistical concepts, including maximum likelihood estimation, normal distribution, and sample mean calculations. Suitable for advanced undergraduate students and anyone interested in statistical inference.

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