Question

X=\binom{X^{(1)}}{X_{(2)}}\sim N_4\left ( \mu ,\sum \right ),  where \begin{pmatrix} -4\\ \frac{1}{4} \\ 0 \end{pmatrix} and \sum=\begin{pmatrix} \begin{matrix} 2&2 \\ 2& 1\end{matrix} & \begin{matrix} 3&0 \\ 2& 0\end{matrix} \\ \begin{matrix} 3&2 \\ 0& 0\end{matrix} & \begin{matrix} 2&1 \\ 1& 1\end{matrix} \end{pmatrix}. Find the 

E(X^{(2)}|X^{(1)}=x^{(1)}) and Cov \left ( X^{(2)}|X^{(1)}=x^{(1)} \right ).

02 Apr 2024
Answer :
Word Count : 808

To find \( E(X^{(2)}|X^{(1)}=x^{(1)}) \) and \( \text{Cov} \left ( X^{(2)}|X^{(1)}=x^{(1)} \right ) \), we'll use the properties of conditional expectation and covariance.

Given that \( X^{(1)} = x^{(1)} \), we know that \( X^{(2)} \) follows a conditional normal distribution. The conditional expectation \( E(X^{(2)}|X^{(1)}=x^{(1)}) \) is the mean of this conditional normal distribution, and the conditional covariance \( \text{Cov} \left ( X^{(2)}|X^{(1)}=x^{(1)} \right ) \) is the covariance matrix.

First, let's write down the properties of conditional normal distributions:

1. Conditional Mean:
\[ E(X^{(2)}|X^{(1)}=x^{(1)}) = \mu_2 + \Sigma_{21}\Sigma_{11}^{-1}(x^{(1)} - \mu_1) \]

2. Conditional Covariance:
\[ \text{Cov} \left ( X^{(2)}|X^{(1)}=x^{(1)} \right ) = \Sigma_{22} - \Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12} \]

Given:
\[ \mu = \begin{pmatrix} -4\\ \frac{1}{4} \\ 0 \end{pmatrix} \]
and
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