Understanding p-Variate Distribution Properties
We are given a p-variate random vector X with mean E(X) = μ and variance Var(X) = Σ, where p > 2. We examine the truthfulness of the provided statements regarding its properties.
Deriving Expected Value Properties
Statement 1: $E(XX') = Σ + μμ'$
- The variance of a random vector X is defined as Var(X) = E(XX') - E(X)E(X').
- Given Var(X) = Σ, E(X) = μ, and E(X') = (E(X))' = μ'.
- Substituting these values: E(XX') = Var(X) + E(X)E(X') = Σ + μμ'.
- Therefore, Statement 1 is true.
Statement 2: $E(X'X) = μ'μ + trace(Σ)$
- X'X is a scalar value, and E(X'X) = E(trace(XX')).
- Using linearity of expectation and trace: E(X'X) = trace(E(XX')).
- From Statement 1, E(XX') = Σ + μμ'.
- So, E(X'X) = trace(Σ + μμ') = trace(Σ) + trace(μμ').
- We know that for a vector μ, trace(μμ') = μ'μ.
- Combining these results: E(X'X) = trace(Σ) + μ'μ.
- Therefore, Statement 2 is true.
Analyzing Quadratic Forms
Statement 3: $E((X - μ)'Σ-1(X - μ)) = p$
- Let Y = X - μ. Then E(Y) = E(X) - μ = 0 and Var(Y) = Var(X) = Σ.
- The expression becomes E(Y'Σ-1Y).
- Using trace properties: E(Y'Σ-1Y) = E(trace(Y'Σ-1Y)) = trace(E(Σ-1YY')).
- Since Var(Y) = E(YY') = Σ, we have E(Y'Σ-1Y) = trace(Σ-1E(YY')) = trace(Σ-1Σ).
- trace(Σ-1Σ) = trace(Ip), where Ip is the p x p identity matrix.
- The trace of Ip is p.
- Therefore, Statement 3 is true.
Applying Probability Inequalities
Statement 4: $P[(X - μ)'Σ-1(X - μ) ≥ t2] ≤ p / t2$, for t ≠ 0
- Let Q = (X - μ)'Σ-1(X - μ). From Statement 3, we know E(Q) = p.
- Q is a non-negative random variable because Σ is positive definite.
- Applying Markov's inequality (P(Z ≥ a) ≤ E(Z) / a for non-negative Z and a > 0):
- P(Q ≥ t2) ≤ E(Q) / t2 (assuming t2 > 0).
- Substituting E(Q) = p: P(Q ≥ t2) ≤ p / t2.
- Therefore, Statement 4 is true.
Conclusion: Statements 1, 2, 3, and 4 are all true properties related to the given p-variate distribution.
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.
