$$X_2 = \begin{cases} -X_1, & -2 \le X_1 \le 2 \\ X_1, & \text{otherwise}. \end{cases}$$
Then identify the correct statement.
Bivariate Normal Distribution Analysis for Transformed Variables
The question involves two random variables, $X_1$ and $X_2$. $X_1$ follows a standard normal distribution, $X_1 \sim N(0, 1)$. $X_2$ is defined based on $X_1$ using a piecewise function: $X_2 = -X_1$ if $-2 \le X_1 \le 2$, and $X_2 = X_1$ otherwise. We need to identify the correct statement about their joint distribution.
Checking Bivariate Normality Conditions
A key requirement for a pair of random variables $(X_1, X_2)$ to have a bivariate normal distribution is that any linear combination of them, $aX_1 + bX_2$, must also be normally distributed. Let's test this condition with the linear combination $Y = X_1 - X_2$.
- Case 1: When $|X_1| > 2$, $X_2$ is equal to $X_1$. Therefore, $Y = X_1 - X_1 = 0$.
- Case 2: When $|X_1| \le 2$, $X_2$ is equal to $-X_1$. Therefore, $Y = X_1 - (-X_1) = 2X_1$.
We know $X_1 \sim N(0, 1)$. Consequently, $2X_1$ follows a normal distribution $N(0, 2^2) = N(0, 4)$. However, the random variable $Y$ has a mixed distribution. It equals 0 with probability $P(|X_1| > 2)$ and follows $N(0, 4)$ with probability $P(|X_1| \le 2)$. A distribution composed of a discrete value (0) and a continuous normal distribution is not itself a normal distribution.
Since the linear combination $X_1 - X_2$ is not normally distributed, the pair $(X_1, X_2)$ cannot have a bivariate normal distribution.
Evaluating the Options
- Option 1: $\text{corr}(X_1, X_2) = 1$. This is false. $X_2$ is not always equal to $X_1$ (e.g., when $X_1 = 1$, $X_2 = -1$). The actual correlation is approximately -0.9088.
- Option 2: $X_2$ does not have $N(0, 1)$ distribution. This is false. Calculation shows that the probability density function of $X_2$ is identical to that of $X_1$, meaning $X_2 \sim N(0, 1)$.
- Option 3: $(X_1, X_2)$ has a bivariate normal distribution. This is false, as proven above by examining the linear combination $X_1 - X_2$.
- Option 4: $(X_1, X_2)$ does not have a bivariate normal distribution. This is true, because the linear combination $X_1 - X_2$ fails the normality test required for bivariate normal distributions.
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 $X \sim N_3 (\underline{\mu}, \Sigma)$ where $\underline{\mu} = (1, 1, 1)$ and
$\Sigma = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & c \\ 1 & c & 2 \end{pmatrix}$. The value of $c$ such that
$X_2$ and $-X_1 + X_2 - X_3$ are independent is
