$\Sigma = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & c \\ 1 & c & 2 \end{pmatrix}$. The value of $c$ such that
$X_2$ and $-X_1 + X_2 - X_3$ are independent is
Multivariate Normal Independence Condition
For a random vector $X \sim N_p(\underline{\mu}, \Sigma)$, two linear combinations, $Y_1 = \mathbf{a}^T X$ and $Y_2 = \mathbf{b}^T X$, are independent if their covariance is zero, specifically $\mathbf{a}^T \Sigma \mathbf{b} = 0$, provided $X$ follows a multivariate normal distribution.
Defining Linear Combinations
Let $X = (X_1, X_2, X_3)^T$. We are interested in the independence of $V_1 = X_2$ and $V_2 = -X_1 + X_2 - X_3$. We represent these in vector form:
- $V_1 = X_2 = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} X$. Thus, $\mathbf{a} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$.
- $V_2 = -X_1 + X_2 - X_3 = \begin{pmatrix} -1 & 1 & -1 \end{pmatrix} X$. Thus, $\mathbf{b} = \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix}$.
Calculating the Covariance Term $\mathbf{a}^T \Sigma \mathbf{b}$
The given covariance matrix is $\Sigma = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & c \\ 1 & c & 2 \end{pmatrix}$. We compute $\mathbf{a}^T \Sigma \mathbf{b}$:
- Calculate $\mathbf{a}^T \Sigma$: $ \mathbf{a}^T \Sigma = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & c \\ 1 & c & 2 \end{pmatrix} = \begin{pmatrix} 1 & 3 & c \end{pmatrix} $
- Multiply the result by $\mathbf{b}$: $ (\mathbf{a}^T \Sigma) \mathbf{b} = \begin{pmatrix} 1 & 3 & c \end{pmatrix} \begin{pmatrix} -1 \\ 1 \\ -1 \end{pmatrix} = (1)(-1) + (3)(1) + (c)(-1) = -1 + 3 - c = 2 - c $
Solving for $c$
For independence, the covariance term must be zero:
$ \mathbf{a}^T \Sigma \mathbf{b} = 0 $ $ 2 - c = 0 $Solving for $c$, we get:
$ c = 2 $The value $c=2$ ensures that $X_2$ and $-X_1 + X_2 - X_3$ are independent.
Important Questions from Multivariate Normal Distribution
- Suppose $(X_1, X_2)$ follows a bivariate normal distribution with $E(X_1) = E(X_2) = 0$, $V(X_1) = V(X_2) = 2$ and $Cov(X_1, X_2) = -1$. If $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-y^2/2} dy$, then $P[X_1 - X_2 > 6]$ is equal to
Let $Y = (Y_1, \dots, Y_n)'$ have the multivariate normal distribution $N_n(0, I)$. Which of the following is the covariance matrix of the conditional distribution of $Y$ given $$\sum_{i=1}^n Y_i ?$$ ($1$ denotes the $n \times 1$ vector with all elements 1.)
Let $\Sigma = ((\sigma_{ij}))$ be an $n \times n$ symmetric and positive definite matrix such that $\sigma_{ij} \neq 0$ for all $i, j$. Which of the following matrices will always be the covariance matrix of a multivariate normal random vector?
- Suppose $X_1 \sim N_{p_1}(0, \Sigma_1), \ X_2 \sim N_{p_2}(0, \Sigma_2)$, where $X_1$ and $X_2$ are independently distributed. If $p_1 > p_2$ and $\Sigma_1, \Sigma_2$ are positive definite then which of the following statements is necessarily true?
- Let $X_1 \sim N(0, 1)$ and let
$$X_2 = \begin{cases} -X_1, & -2 \le X_1 \le 2 \\ X_1, & \text{otherwise}. \end{cases}$$
Then identify the correct statement.
