To address part 1 of the problem, we need to **prove that** the quantity \( (X - \mu)^T \Sigma^{-1} (X - \mu) \) is distributed as a chi-squared distribution with \( p \) degrees of freedom, given that \( X \) is distributed as \( N_p(\mu, \Sigma) \), where \( \mu \) is the mean vector and \( \Sigma \) is the covariance matrix.

### Proof:

#### Step 1: Definition and Setup
- Let \( X \) be a random vector following a multivariate normal distribution:
  \[
  X \sim N_p(\mu, \Sigma)
  \]
  This means:
  \[
  X \text{ has mean vector } \mu \text{ and covariance matrix } \Sigma.
  \]
  
#### Step 2: Standardization
- Define the standardized random vector \( Z \) as:
  \[
  Z = \Sigma^{-\frac{1}{2}} (X - \mu)
  \]
  Here, \( Z \) is a random vector with a standard normal distribution \( N_p(0, I) \), where \( I \) is the identity matrix of size \( p \). This is because:
  \[
  Z \sim N_p(0, I)
  \]
  where \( Z \) has mean \( 0 \) and covariance matrix \( I \).

#### Step 3: Expression of the Quadratic Form
- Consider the quadratic form:
  \[
  (X - \mu)^T \Sigma^{-1} (X - \mu)
  \]
  Substitute \( Z \) into the expression:
  \[
  (X - \mu)^T \Sigma^{-1} (X - \mu) = \left[\Sigma^{\frac{1}{2}}Z\right]^T \Sigma^{-1} \left[\Sigma^{\frac{1}{2}}Z\right]
  \]
  Simplifying:
  \[
  = Z^T \Sigma^{\frac{1}{2}T} \Sigma^{-1} \Sigma^{\frac{1}{2}} Z = Z^T Z
  \]

#### Step 4: Distribution of the Quadratic Form
- Since \( Z \sim N_p(0, I) \), the sum of the squares of \( p \) independent standard normal variables \( Z_1^2 + Z_2^2 + \cdots + Z_p^2 \) follows a chi-squared distribution with \( p \) degrees of freedom:
  \[
  Z^T Z = Z_1^2 + Z_2^2 + \cdots + Z_p^2 \sim \chi^2_p
  \]

#### Conclusion:
- Therefore, \( (X - \mu)^T \Sigma^{-1} (X - \mu) \) follows a chi-squared distribution with \( p \) degrees of freedom:
  \[
  (X - \mu)^T \Sigma^{-1} (X - \mu) \sim \chi^2_p
  \]

This completes the proof.

### Do you have any questions on this proof? Would you like to proceed to part 2 or any other section?

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### Related Questions:
1. How does the interpretation of \( (X - \mu)^T \Sigma^{-1} (X - \mu) \) change when the covariance matrix \( \Sigma \) is not diagonal?
2. What role does the covariance matrix \( \Sigma \) play in shaping the distribution of the quadratic form?
3. Can you provide a practical example where this chi-squared distribution would be useful in real-world data analysis?
4. How would you modify this proof if \( X \) followed a non-normal distribution?
5. Why is it important that the random vector \( Z \) is standardized in this proof?

### Tip:
When proving distribution properties, it's often useful to standardize the variables involved. This reduces the problem to working with standard distributions, making it easier to apply known results.

Understand the proof that (X - \mu)^T \Sigma^{-1} (X - \mu) follows a chi-squared distribution with p degrees of freedom in multivariate normal distribution theory. This detailed explanation covers key concepts such as multivariate normal distribution, quadratic forms, and chi-squared distributions.

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