To solve this problem, we need to analyze the given bivariate normal distribution and the properties of the random vector \(\begin{pmatrix} X \\ Y \end{pmatrix}\). Let's go through the steps in detail to arrive at the correct conclusion:
1. **Understanding the Given Density Function:**
The density function is given as:
\(\frac{1}{2}\left(\phi\left(x, y; \frac{1}{2}\right) + \phi\left(x, y; -\frac{1}{2}\right)\right)\)
This means the distribution is a mixture of two bivariate normal distributions with correlation coefficients \(\frac{1}{2}\) and \(-\frac{1}{2}\). Each component has mean vector \(\begin{pmatrix} 0 \\ 0 \end{pmatrix}\) and variance-covariance matrices:
\(\begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1 \end{pmatrix}\) for \(\rho = \frac{1}{2}\)
\(\begin{pmatrix} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{pmatrix}\) for \(\rho = -\frac{1}{2}\)
2. **Checking Marginal Distributions:**
The marginal distribution of a bivariate normal distribution is normal with mean equal to the corresponding element of the mean vector and variance equal to the corresponding diagonal element of the variance-covariance matrix. For both components, the marginal distribution of \(X\) and \(Y\) is standard normal because the variances are 1, and the means are 0. So, the marginal distribution of both \(X\) and \(Y\) is indeed standard normal.
3. **Covariance Calculation:**
The covariance for the distribution is calculated as follows:
\(\text{Cov}(X, Y) = \frac{1}{2}\cdot\frac{1}{2} + \frac{1}{2}\cdot(-\frac{1}{2}) = 0\)
This shows that the covariance between \(X\) and \(Y\) is 0.
4. **Determining Independence:**
For two variables to be independent, their joint distribution should factor into their marginals. Even though the covariance is 0, which suggests uncorrelatedness, it does not imply independence in this context, as the density is a mixture and not a simple product of marginals.
After evaluating all these points, let's conclude:
- The marginal distribution of both \(X\) and \(Y\) is standard normal.
- The covariance \(Cov(X, Y) = 0\).
- X and Y are not necessarily independent due to the nature of the mixed density.
- The entire distribution is a mixture distribution, not a single bivariate normal distribution.
Therefore, the correct answers are: "Marginal distribution of both \(X\) and \(Y\) is standard normal." and "Covariance \((X, Y) = 0\)".
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.
