Let the random vector $\begin{pmatrix} X_1 \\ X_2 \\ X_3 \end{pmatrix}$ have the positive definite dispersion matrix $\begin{pmatrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{pmatrix}$ Then, which of the following statements are true?
To determine the correct answers for this problem, we need to evaluate each option based on its statement about the dispersion matrix and principal components. The given dispersion matrix is:
| \( \begin{pmatrix} 1 & \rho & \rho \\ \rho & 1 & \rho \\ \rho & \rho & 1 \end{pmatrix} \) |
This matrix is positive definite, which means that for any non-zero vector \( \mathbf{v} \), the quadratic form \( \mathbf{v}^T \mathbf{\Sigma} \mathbf{v} > 0 \) (where \( \mathbf{\Sigma} \) is the dispersion matrix).
\(\rho\) may be \(-0.47\).
The matrix remains positive definite if its eigenvalues are positive. The eigenvalues of the given matrix can be found using the characteristic equation. The conditions for positive definiteness are satisfied for \(-\frac{1}{2} < \rho < 1\). Thus, since \(-0.47 > -\frac{1}{2}\), \(\rho\) can indeed be \(-0.47\). Therefore, this option is true.
The first principal component can only explain \(32\%\) of the total variation for some \(\rho\).
The percentage of total variation explained by a principal component is given by its corresponding eigenvalue divided by the sum of all eigenvalues. Given the matrix symmetry and structure, the maximum eigenvalue (which corresponds to the first principal component) can be greater than \(0.32\) of the total eigenvalues for typical values of \(\rho\). Thus, it is unlikely that it can only explain \(32\%\) without any other conditions specified. This option is false.
The second principal component can explain more than \(32\%\) of the total variation for any \(\rho\).
The second principal component corresponds to the second largest eigenvalue. Typically, due to symmetry and correlation effects, achieving over \(32\%\) for any \(\rho\) violates the typical eigenvalue distribution unless specifically constrained. Hence, this option is not necessarily true for any \(\rho\). This option is false.
The variance of the first principal component is \(1 + 2\rho\) for any \(\rho\).
The variance of the first principal component is indeed equal to the largest eigenvalue of the matrix, which is generally \(1 + 2\rho\). However, this depends on the range \(-\frac{1}{2} < \rho < 1\) for positive definiteness. Therefore, this statement is conditionally true but not universal for all \(\rho\).
Therefore, the correct statement is: \(\rho\) may be \(-0.47\).
Important Questions from Principle Component Analysis
- Let $(X_1, X_2, \cdots, X_p)$ be a random vector with mean $\mu$ and a positive definite dispersion matrix $\Sigma$. Then the coefficient vector $(l_1, l_2, \cdots, l_p)$ of the first principal component $\sum_{i=1}^p l_i X_i$ is
- Let $X$ be a $4 \times 1$ random vector with Multivariate normal distribution with mean $\mu$ and dispersion matrix $\Sigma$. Suppose, the eigenvalues of $\Sigma$ are $\lambda_1 = 6, \lambda_2 = 3, \lambda_3 = 2, \lambda_4 = 1$. Let $Y_1, Y_2, Y_3, Y_4$ be the four principal components. Which of the following statements are correct?
