Question

Let (X₁ X₂) be a bivariate normal distributed random variable. The joint density is f(x₁,x₂) =1/2π√det(V) exp (-x- μ)V⁻¹(-x- μ)ᵗ/2)

where x = (x₁,x₂), μ = (μ₁ μ₂), μₖ-E(Xₖ), and V is the covariance matrix. The covariance matrix is defined as V = [vkj], where vkj-cou(XₖXⱼ)
The covariance matrix V of Xi and X₂ is

V=[1 1]
[1 3]
Determine the variance σ₁²=var(X), the variance σ₂² =var(X₂), and the correlation coefficient rho between X₁ and X₃

Answer 1

The variance σ±² = 1, variance σ²² = 3, and correlation coefficient rho = 1/√3 for the bivariate normal distribution with the given covariance matrix.

Step-by-step explanation:

To determine the variance σ¹², variance σ²², and the correlation coefficient rho for the bivariate normal distribution with given covariance matrix V as [1 1; 1 3], we follow these steps:

1. Identify the diagonal elements of the covariance matrix V, which represent the variances of X₁ and X₂, respectively. Thus, σ¹² = var(X₁) = 1 and σ²² = var(X₂) = 3.
2. Notice that the off-diagonal elements represent the covariance between X₁ and X₂. Thus, cov(X₁,X₂) = 1.
3. To find the correlation coefficient rho, use the formula ρ = cov(X₁, X₂) / √(σ¹² * σ²²). Substituting the values gives us ρ = 1 / √(1*3), which simplifies to ρ = 1 / √3.

Therefore, we have the variance σ±² = 1, variance σ²² = 3, and the correlation coefficient rho = 1/√3.

To find the variances, we can use the diagonal elements of the covariance matrix V. The variance of X₁ (σ₁²) is the first element of the diagonal, which is 1. The variance of X₂ (σ₂²) is the second element of the diagonal, which is 3. So, σ₁² = 1 and σ₂² = 3.

The correlation coefficient between X₁ and X₂ (ρ) can be calculated using the formula: ρ = cov(X₁, X₂) / (√(var(X₁)) * √(var(X₂))). Here, cov(X₁, X₂) is the off-diagonal element of the covariance matrix V, which is 1. Substituting the values, we get ρ = 1 / (√(1) * √(3)) = 1 / √(3).