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.)
$I - \frac{11'}{n}$
Multivariate Normal Distribution Setup
We are given a random vector $Y = (Y_1, \dots, Y_n)'$ following a multivariate normal distribution $N_n(0, I)$. This specifies the mean vector $E[Y] = 0$ and the covariance matrix $Cov(Y) = \Sigma = I$, where $I$ is the $n \times n$ identity matrix.
The objective is to determine the covariance matrix of the conditional distribution of $Y$ given the sum $S = \sum_{i=1}^n Y_i$. This sum can be represented using vector notation as $S = 1'Y$, where $1$ is the $n \times 1$ vector consisting of all ones.
Conditional Covariance Formula Application
For a multivariate normal random vector $Y \sim N_n(\mu, \Sigma)$, the covariance matrix of the conditional distribution of $Y$ given a linear constraint $a'Y = c$ is provided by the formula:
$ Cov(Y | a'Y) = \Sigma - \frac{\Sigma a a' \Sigma}{a' \Sigma a} $
In this context, $\mu = 0$, $\Sigma = I$, and the linear combination for conditioning is $1'Y$, which means we use $a = 1$.
Calculation Steps for Covariance Matrix
We substitute $\Sigma = I$ and $a = 1$ into the conditional covariance formula:
- First, calculate the denominator: $a' \Sigma a = 1' I 1 = 1' 1 = n$.
- Next, calculate the matrix fraction term: $\frac{\Sigma a a' \Sigma}{a' \Sigma a} = \frac{I (1) (1') I}{n} = \frac{1 1'}{n}$. Here, $1 1'$ denotes the outer product of the vector $1$ with itself, resulting in an $n \times n$ matrix of ones.
By substituting these calculated terms back into the formula, we obtain the conditional covariance matrix:
$ Cov(Y | 1'Y) = I - \frac{1 1'}{n} $
Final Conditional Covariance Result
Therefore, the covariance matrix of the conditional distribution of $Y$ given $\sum_{i=1}^n Y_i$ is $I - \frac{11'}{n}$. This corresponds to Option 3.
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 $\Sigma = ((\sigma_{ij}))$ 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. - 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
