Let $X_1$ and $X_2$ be independent random variables each having $N(\mu, \sigma^2)$ distribution, where $\mu \in \mathbb{R}, \sigma^2 > 0$. Let $0 \leq \theta < 2\pi$ and $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$. Put $Y = (Y_1, Y_2)^t = A X$ with $X = (X_1, X_2)^t$. Which of the following statements are correct?
Evaluating Statements about Transformed Gaussian Variables
We are given two independent random variables, $X_1$ and $X_2$, both following a normal distribution $N(\mu, \sigma^2)$. Their vector form is $X = (X_1, X_2)^t$. A rotation matrix $A$ is defined as $A = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix}$. The transformed vector is $Y = (Y_1, Y_2)^t = AX$. We need to determine the correctness of the given statements.
Analysis of Statement 3: $Y_1$ and $Y_2$ are Gaussian.
- The vector $X = (X_1, X_2)^t$ follows a bivariate normal distribution because $X_1$ and $X_2$ are independent Gaussian random variables.
- The transformation $Y = AX$ is a linear transformation of the random vector $X$.
- A key property of Gaussian distributions is that a linear transformation of a Gaussian random vector results in another Gaussian random vector.
- Therefore, $Y = (Y_1, Y_2)^t$ is a bivariate Gaussian random vector, which implies that both $Y_1$ and $Y_2$ are individually Gaussian random variables.
- Conclusion: Statement 3 is correct.
Analysis of Statement 4: $Y_1$ and $Y_2$ may be correlated.
- The covariance matrix of $X$ is $Cov(X) = \sigma^2 I = \begin{pmatrix} \sigma^2 & 0 \\ 0 & \sigma^2 \end{pmatrix}$, since $X_1$ and $X_2$ are independent and have the same variance $\sigma^2$.
- The covariance matrix of $Y$ is given by $Cov(Y) = A Cov(X) A^t$.
- Since $A$ is a rotation matrix, $A A^t = I$ (the identity matrix).
- Thus, $Cov(Y) = A (\sigma^2 I) A^t = \sigma^2 A A^t = \sigma^2 I = \begin{pmatrix} \sigma^2 & 0 \\ 0 & \sigma^2 \end{pmatrix}$.
- The covariance between $Y_1$ and $Y_2$ is the off-diagonal element of $Cov(Y)$, which is $0$.
- This means $Y_1$ and $Y_2$ are uncorrelated. Since $Y$ is a bivariate Gaussian vector, uncorrelated components are also independent.
- Therefore, $Y_1$ and $Y_2$ are not correlated.
- Conclusion: Statement 4 is incorrect.
Analysis of Statements 1 and 2: $Y = X$ in distribution.
- For $Y = X$ in distribution, we require $E[Y] = E[X]$ and $Cov(Y) = Cov(X)$.
- We already established $Cov(Y) = Cov(X) = \sigma^2 I$.
- The expected value of $X$ is $E[X] = (\mu, \mu)^t$.
- The expected value of $Y$ is $E[Y] = A E[X] = A (\mu, \mu)^t = \begin{pmatrix} \mu(\cos\theta + \sin\theta) \\ \mu(\cos\theta - \sin\theta) \end{pmatrix}$.
- For $E[Y] = E[X]$, we need $\mu(\cos\theta + \sin\theta) = \mu$ and $\mu(\cos\theta - \sin\theta) = \mu$.
- Case 1: $\mu = 0$. Then $E[X] = (0, 0)^t$ and $E[Y] = (0, 0)^t$. So $E[Y] = E[X]$ holds. In this case, $Y=X$ in distribution.
- Case 2: $\mu \neq 0$. We need $\cos\theta + \sin\theta = 1$ and $\cos\theta - \sin\theta = 1$. Adding gives $2\cos\theta = 2 \implies \cos\theta = 1$. Subtracting gives $2\sin\theta = 0 \implies \sin\theta = 0$. This implies $\theta = 0$. If $\theta = 0$, then $A=I$, and $Y=AX=X$ always, regardless of $\mu$.
- Statement 1 claims $Y=X$ iff $\mu=0$. This is false because if $\theta=0$, $Y=X$ for any $\mu$.
- Statement 2 claims $Y=X$ iff $\mu=0, \theta=0$. This is false because if $\theta=0$, $Y=X$ regardless of $\mu$.
- Conclusion: Statements 1 and 2 are incorrect.
Final Conclusion
Based on the analysis, only Statement 3 is correct.
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.
