Multivariate Normal Quadratic Forms
We are given two independent random vectors:
- $X_1 \sim N_{p_1}(0, \Sigma_1)$
- $X_2 \sim N_{p_2}(0, \Sigma_2)$
where $\Sigma_1$ and $\Sigma_2$ are positive definite covariance matrices.
Key Statistical Theorem
A fundamental result in multivariate statistics states that if a random vector $Y \sim N_p(0, \Sigma)$ where $\Sigma$ is a positive definite matrix, then the quadratic form $Y^T \Sigma^{-1} Y$ follows a chi-squared distribution with $p$ degrees of freedom:
$ Y^T \Sigma^{-1} Y \sim \chi^2_p $
Applying Theorem to $X_1$ and $X_2$
Applying this theorem to our specific random vectors:
- For $X_1$: Since $X_1 \sim N_{p_1}(0, \Sigma_1)$ and $\Sigma_1$ is positive definite, we have $X_1^T \Sigma_1^{-1} X_1 \sim \chi^2_{p_1}$.
- For $X_2$: Similarly, since $X_2 \sim N_{p_2}(0, \Sigma_2)$ and $\Sigma_2$ is positive definite, we have $X_2^T \Sigma_2^{-1} X_2 \sim \chi^2_{p_2}$.
Independence and Summation Property
The problem states that $X_1$ and $X_2$ are independently distributed. This implies that the derived random variables $X_1^T \Sigma_1^{-1} X_1$ and $X_2^T \Sigma_2^{-1} X_2$ are also independent.
A property of the chi-squared distribution is that the sum of two independent chi-squared variables follows a chi-squared distribution whose degrees of freedom is the sum of the individual degrees of freedom.
Therefore, combining the results:
$ (X_1^T \Sigma_1^{-1} X_1) + (X_2^T \Sigma_2^{-1} X_2) \sim \chi^2_{p_1} + \chi^2_{p_2} $Since the two components are independent, their sum is:
$ X_1^T \Sigma_1^{-1} X_1 + X_2^T \Sigma_2^{-1} X_2 \sim \chi^2_{p_1 + p_2} $Conclusion
Based on the properties of multivariate normal distributions and chi-squared distributions, the statement $X_1^T \Sigma_1^{-1} X_1 + X_2^T \Sigma_2^{-1} X_2 \sim \chi^2_{p_1 + p_2}$ is necessarily true.
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?
- 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. - 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
