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To solve this problem, we need to show that \( y' A y \) and \( y' B y \) are independent, given that \( y \) is an \( n \times 1 \) multivariate normal random vector with mean \( \mu \) and variance \( V \), and that \( A \) and \( B \) are symmetric \( n \times n \) matrices of rank \( r_1 \) and \( r_2 \), respectively. Additionally, it is given that \( AVB = 0 \), which will play a key role in demonstrating the independence.

### Steps for the solution:

1. **Multivariate normal distribution**:  
   Since \( y \sim N(\mu, V) \), the random vector \( y \) is multivariate normally distributed with covariance matrix \( V \). This means that any linear combination of the components of \( y \) will also be normally distributed.

2. **Quadratic forms in normal distributions**:  
   The terms \( y' A y \) and \( y' B y \) are quadratic forms of the random vector \( y \). The properties of quadratic forms of normally distributed random vectors will help in showing their independence.

3. **Covariance of quadratic forms**:  
   The covariance between two quadratic forms \( y' A y \) and \( y' B y \) can be expressed as:
   \[
   \text{Cov}(y' A y, y' B y) = 2 \text{tr}(AVB V)
   \]
   where \( \text{tr} \) denotes the trace operator, which sums the diagonal elements of a matrix. This formula can be derived from the structure of quadratic forms and their covariance properties.

4. **Condition \( AVB = 0 \)**:  
   The condition \( AVB = 0 \) implies that the trace \( \text{tr}(AVB V) = 0 \), which leads to:
   \[
   \text{Cov}(y' A y, y' B y) = 0
   \]
   Therefore, the covariance between \( y' A y \) and \( y' B y \) is zero.

5. **Independence of the quadratic forms**:  
   Since the covariance between \( y' A y \) and \( y' B y \) is zero and both forms are normally distributed (as they are quadratic forms of a multivariate normal distribution), we can conclude that they are independent.

Thus, \( y' A y \) and \( y' B y \) are independent under the condition \( AVB = 0 \).

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Would you like further clarification on any part of this explanation, or details about any step?

Here are 5 related questions that expand on this problem:

1. How are quadratic forms of multivariate normal variables typically used in statistics?
2. What is the geometric interpretation of matrices \( A \) and \( B \) in terms of transformations of \( y \)?
3. Can the condition \( AVB = 0 \) be relaxed and still have independence of quadratic forms?
4. How would the result change if \( A \) and \( B \) were not symmetric matrices?
5. What is the significance of the rank of matrices \( A \) and \( B \) in this problem?

**Tip**: When dealing with quadratic forms in multivariate statistics, always check for any covariance structure or matrix conditions (such as \( AVB = 0 \)) that simplify the problem.

Let y be an n x 1 multivariate normal random variable with mean μ and variance V. Let A and B be symmetric n x n matrices of rank r1 and r2 respectively. If AVB = 0, show that y' A y and y' B y are independent.

Explore the solution to a problem involving the independence of quadratic forms in a multivariate normal distribution. The solution involves calculating the covariance of quadratic forms and using the condition AVB = 0 to prove independence. Suitable for graduate-level studies in probability and linear algebra.

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