Suppose that $\begin{pmatrix} X \\ Y \end{pmatrix}$ has a bivariate density $f = \frac{1}{2} f_1 + \frac{1}{2} f_2$, where $f_1$ and $f_2$ are respectively, the densities of bivariate normal distributions $N(\mu_1, \Sigma)$ and $N(\mu_2, \Sigma)$, with $\mu_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$, $\mu_2 = \begin{pmatrix} -1 \\ -1 \end{pmatrix}$ and $\Sigma = I_2$, the $2 \times 2$ identity matrix. Then which of the following is correct?
The given question involves analyzing a bivariate distribution that is a mixture of two bivariate normal distributions. Let's break down the problem step by step:
The bivariate random vector \(\begin{pmatrix} X \\ Y \end{pmatrix}\) has the density:
\(f = \frac{1}{2} f_1 + \frac{1}{2} f_2\)
where \(f_1\) and \(f_2\) are the densities of the bivariate normal distributions:
- \(N\left(\mu_1, \Sigma\right)\) with \(\mu_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}\) and \(\Sigma = I_2\)
- \(N\left(\mu_2, \Sigma\right)\) with \(\mu_2 = \begin{pmatrix} -1 \\ -1 \end{pmatrix}\) and \(\Sigma = I_2\)
The covariance matrix \(\Sigma\) is given as \(I_2\), the identity matrix, indicating that within each component distribution, \(X\) and \(Y\) are independent. However, because \(f\) is a mixture, we must analyze the correlation created by the mixture to determine the relationship between \(X\) and \(Y\).
The mixture distribution has mean (since both components are equally weighted):
\(\mu = \frac{1}{2} \mu_1 + \frac{1}{2} \mu_2 = \frac{1}{2} \begin{pmatrix} 1 \\ 1 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} -1 \\ -1 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}\)
To calculate the covariance matrix of the mixture, we use the law of total covariance:
\(\text{Cov}(X, Y) = \frac{1}{2} \Sigma + \frac{1}{2} \Sigma + \frac{1}{2} (\mu_1 - \mu)(\mu_1 - \mu)^T + \frac{1}{2} (\mu_2 - \mu)(\mu_2 - \mu)^T\)
Since the identity matrix \(\Sigma\) is involved, the first part simplifies to just \(\Sigma\). Also calculate the additional covariance term:
\((\mu_1 - \mu)(\mu_1 - \mu)^T = \begin{pmatrix} 1 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\)
\((\mu_2 - \mu)(\mu_2 - \mu)^T = \begin{pmatrix} -1 \\ -1 \end{pmatrix} \begin{pmatrix} -1 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\)
Thus, the covariance contributions sum to:
\(\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = 2 \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\)
So, the total covariance matrix becomes:
\(\Sigma + \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\)
From this, it is evident that \(\text{Cov}(X, Y) = 1\), which implies that \(X\) and \(Y\) are positively correlated.
Therefore, the correct answer is: $X$ and $Y$ are positively correlated.
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.
