Understanding the Wishart Distribution \(W_p(n, \Sigma)\)

The question deals with a random matrix \(A = ((a_{ij}))\) that follows a Wishart distribution \(W_3(5, \Sigma)\). This means the matrix \(A\) is \(3 \times 3\) (\(p=3\)), has degrees of freedom \(n=5\), and the parameter matrix is \(\Sigma=\left(\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{array}\right)\). The Wishart distribution is a multivariate generalization of the chi-squared distribution or Gamma distribution. If \(A \sim W_p(n, \Sigma)\), it can be defined as the sum of outer products of \(n\) independent \(p\)-dimensional multivariate normal vectors with mean zero and covariance matrix \(\Sigma\).

Properties of Wishart Matrix Elements

There are known properties regarding the distributions of the elements of a Wishart matrix. For a matrix \(A \sim W_p(n, \Sigma)\), the diagonal elements \(a_{ii}\) are related to the chi-squared distribution. Specifically, the distribution of the diagonal element \(a_{ii}\) is given by \(a_{ii} \sim \sigma_{ii} \chi_n^2\), where \(\sigma_{ii}\) is the \(i\)-th diagonal element of \(\Sigma\).

In this problem, \(p=3\) and \(n=5\). The diagonal elements of the given \(\Sigma\) matrix are \(\sigma_{11} = 2\), \(\sigma_{22} = 2\), and \(\sigma_{33} = 2\). Using the property, we can state the distributions of the diagonal elements of \(A\):

• \(a_{11} \sim \sigma_{11} \chi_5^2 = 2 \chi_5^2\), which means \(\frac{1}{2} a_{11} \sim \chi_5^2\).
• \(a_{22} \sim \sigma_{22} \chi_5^2 = 2 \chi_5^2\), which means \(\frac{1}{2} a_{22} \sim \chi_5^2\).
• \(a_{33} \sim \sigma_{33} \chi_5^2 = 2 \chi_5^2\), which means \(\frac{1}{2} a_{33} \sim \chi_5^2\).

Evaluating Options Based on Diagonal Elements

Let's evaluate the first two options using the property for \(a_{22}\):

• Option 1: \(a_{22} \sim \chi_3^2\). Our finding is \(\frac{1}{2} a_{22} \sim \chi_5^2\), which implies \(a_{22} \sim 2 \chi_5^2\). A \(2 \chi_5^2\) distribution is not the same as a \(\chi_3^2\) distribution. So, option 1 is false.
• Option 2: \(\frac{1}{2} a_{22} \sim \chi_5^2\). Our finding is \(\frac{1}{2} a_{22} \sim \chi_5^2\). This matches the statement in option 2. So, option 2 is true.

Analyzing Quadratic Forms of Wishart Matrix

Options 3 and 4 involve a linear combination of elements: \((a_{11}-4 a_{13}+4 a_{33})\). This expression is a quadratic form of the matrix \(A\). For any vector \(c \in \mathbb{R}^p\), the quadratic form \(c^T A c\) follows a distribution related to the chi-squared distribution. Specifically, if \(A \sim W_p(n, \Sigma)\), then \(c^T A c \sim (c^T \Sigma c) \chi_n^2\).

Let's identify the vector \(c\) such that \(c^T A c = a_{11}-4 a_{13}+4 a_{33}\). The quadratic form \(c^T A c\) for a symmetric matrix \(A\) with \(p=3\) is given by:

\(c^T A c = c_1^2 a_{11} + c_2^2 a_{22} + c_3^2 a_{33} + 2c_1 c_2 a_{12} + 2c_1 c_3 a_{13} + 2c_2 c_3 a_{23}\).

Comparing this to \(a_{11}-4 a_{13}+4 a_{33}\), we match the coefficients:

• Coefficient of \(a_{11}\): \(c_1^2 = 1 \implies c_1 = \pm 1\). Let's choose \(c_1 = 1\).
• Coefficient of \(a_{22}\): \(c_2^2 = 0 \implies c_2 = 0\).
• Coefficient of \(a_{33}\): \(c_3^2 = 4 \implies c_3 = \pm 2\).
• Coefficient of \(a_{12}\): \(2c_1 c_2 = 2(1)(0) = 0\). This matches the expression (zero coefficient for \(a_{12}\)).
• Coefficient of \(a_{13}\): \(2c_1 c_3 = -4\). With \(c_1=1\), this gives \(2(1)c_3 = -4 \implies c_3 = -2\).
• Coefficient of \(a_{23}\): \(2c_2 c_3 = 2(0)(-2) = 0\). This matches the expression (zero coefficient for \(a_{23}\)).

Thus, the vector is \(c = \left(\begin{array}{c}1 \\ 0 \\ -2\end{array}\right)\). Now we calculate the scaling factor \(c^T \Sigma c\):

\(c^T \Sigma c = \left(\begin{array}{lll}1 & 0 & -2\end{array}\right) \left(\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{array}\right) \left(\begin{array}{c}1 \\ 0 \\ -2\end{array}\right)\)

First, calculate \(\Sigma c\):

\(\Sigma c = \left(\begin{array}{lll}2 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2\end{array}\right) \left(\begin{array}{c}1 \\ 0 \\ -2\end{array}\right) = \left(\begin{array}{c}2(1)+1(0)+1(-2) \\ 1(1)+2(0)+0(-2) \\ 1(1)+0(0)+2(-2)\end{array}\right) = \left(\begin{array}{c}2+0-2 \\ 1+0-0 \\ 1+0-4\end{array}\right) = \left(\begin{array}{c}0 \\ 1 \\ -3\end{array}\right)\)

Next, calculate \(c^T (\Sigma c)\):

\(c^T (\Sigma c) = \left(\begin{array}{lll}1 & 0 & -2\end{array}\right) \left(\begin{array}{c}0 \\ 1 \\ -3\end{array}\right) = 1(0) + 0(1) + (-2)(-3) = 0 + 0 + 6 = 6\).

So, the quadratic form \(a_{11}-4 a_{13}+4 a_{33}\) follows the distribution \((c^T \Sigma c) \chi_n^2 = 6 \chi_5^2\). This means \(\frac{1}{6}(a_{11}-4 a_{13}+4 a_{33}) \sim \chi_5^2\).

Evaluating Options Based on Quadratic Forms

Now let's evaluate options 3 and 4:

• Option 3: \(\frac{1}{33}\left(a_{11}-4 a_{13}+4 a_{33}\right) \sim \chi_3^2\). Our finding is \(\frac{1}{6}(a_{11}-4 a_{13}+4 a_{33}) \sim \chi_5^2\). The scaling factor (1/33 vs 1/6) and the degrees of freedom (3 vs 5) do not match. So, option 3 is false.
• Option 4: \(\frac{1}{9}\left(a_{11}-4 a_{13}+4 a_{33}\right) \sim \chi_5^2\). Our finding is \(\frac{1}{6}(a_{11}-4 a_{13}+4 a_{33}) \sim \chi_5^2\). The scaling factor (1/9 vs 1/6) does not match, although the degrees of freedom (5) is correct. So, option 4 is false.

Conclusion

Based on the analysis of the properties of the Wishart distribution \(W_3(5, \Sigma)\), we found that only the statement in Option 2 is true.

\(\frac{1}{2} a_{22} \sim \chi_5^2\).