Let $\begin{pmatrix} X \\ Y \\ Z \end{pmatrix} \sim N_3 \left( \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \rho & 0 \\ \rho & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right)$, where $|\rho| < 1$. Which of the following are TRUE?
Properties of Multivariate Normal Distribution Correlations
The random vector $\begin{pmatrix} X \\ Y \\ Z \end{pmatrix}$ follows a 3-variate normal distribution $N_3 \left( \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \rho & 0 \\ \rho & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \right)$, with $|\rho| < 1$. Key implications include:
- Means: $E[X]=E[Y]=E[Z]=0$.
- Variances: $Var(X) = 1$, $Var(Y) = 1$, $Var(Z) = 1$.
- Covariances: $Cov(X, Y) = \rho$, $Cov(X, Z) = 0$, $Cov(Y, Z) = 0$.
- Independence: $Z$ is independent of $X$ and $Y$ because their covariances are zero.
Evaluating Statement 1: $Cor(X^2, Y^2) = \rho^2$.
For standard normal variables $X, Y$ with covariance $\sigma_{XY}$, the correlation of their squares is given by $Cor(X^2, Y^2) = \frac{\sigma_{XY}^2}{\sigma_X^2 \sigma_Y^2}$.
- Given $\sigma_X^2 = Var(X) = 1$, $\sigma_Y^2 = Var(Y) = 1$, and $\sigma_{XY} = Cov(X, Y) = \rho$.
- Calculation: $Cor(X^2, Y^2) = \frac{\rho^2}{1 \times 1} = \rho^2$.
- Conclusion: Statement 1 is TRUE.
Evaluating Statement 2: $Cor(X^2, Y) = 0$.
The correlation is $\frac{Cov(X^2, Y)}{\sqrt{Var(X^2)Var(Y)}}$. We calculate the covariance $Cov(X^2, Y) = E[X^2 Y] - E[X^2]E[Y]$.
- $E[X^2] = Var(X) + (E[X])^2 = 1 + 0^2 = 1$.
- $E[Y] = 0$.
- So, $Cov(X^2, Y) = E[X^2 Y]$.
- For centered bivariate normal variables $(X, Y)$, $E[X^2 Y] = 0$. This can be shown by writing $Y = \rho X + \sqrt{1-\rho^2} U$ (where $U$ is standard normal independent of $X$) and using $E[X^3]=0$ (odd moment) and $E[X^2 U]=E[X^2]E[U]=1 \times 0=0$.
- Thus, $Cov(X^2, Y) = 0$.
- Conclusion: Statement 2 is TRUE.
Evaluating Statement 3: $Cor(X^2, Z^2) = 0$.
The correlation is $\frac{Cov(X^2, Z^2)}{\sqrt{Var(X^2)Var(Z^2)}}$. We calculate the covariance $Cov(X^2, Z^2) = E[X^2 Z^2] - E[X^2]E[Z^2]$.
- $E[X^2] = 1$ and $E[Z^2] = 1$.
- From the covariance matrix, $Cov(X, Z) = 0$, implying $X$ and $Z$ are independent.
- Independence of $X$ and $Z$ implies independence of $X^2$ and $Z^2$.
- Therefore, $Cov(X^2, Z^2) = 0$.
- Conclusion: Statement 3 is TRUE.
Evaluating Statement 4: $Cor(X^2, Y^2 + Z^2) = \rho^2$.
The correlation is $\frac{Cov(X^2, Y^2 + Z^2)}{\sqrt{Var(X^2)Var(Y^2 + Z^2)}}$.
- Numerator: $Cov(X^2, Y^2 + Z^2) = Cov(X^2, Y^2) + Cov(X^2, Z^2)$. From Statement 1 and 3, this is $\rho^2 + 0 = \rho^2$.
- Denominator: $Var(X^2) = 2$ (since $X$ is standard normal, $Var(X^2)=E[X^4]-(E[X^2])^2 = 3-1^2=2$).
- $Var(Y^2 + Z^2) = Var(Y^2) + Var(Z^2) + 2 Cov(Y^2, Z^2)$. Since $Y$ and $Z$ are independent, $Y^2$ and $Z^2$ are independent, so $Cov(Y^2, Z^2)=0$.
- $Var(Y^2 + Z^2) = 2 + 2 + 0 = 4$.
- The correlation is $\frac{\rho^2}{\sqrt{Var(X^2)Var(Y^2 + Z^2)}} = \frac{\rho^2}{\sqrt{2 \times 4}} = \frac{\rho^2}{\sqrt{8}} = \frac{\rho^2}{2\sqrt{2}}$.
- This value is not equal to $\rho^2$ (unless $\rho=0$).
- Conclusion: Statement 4 is FALSE.
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.
