Question

Suppose \(\mathbf{X}\) is distributed \(N_{3}(\mathbf{0}, \mathbf{\Sigma})\), where $$ \boldsymbol{\Sigma}=\left[\begin{array}{lll} 3 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{array}\right] $$ Find \(P\left(\left(X_{1}-2 X_{2}+X_{3}\right)^{2}>15.36\right)\).

Short answer

The probability that \( P((X_{1}-2X_{2}+X_{3})^{2}>15.36) \) is extremely small, essentially 0.

Step-by-step solution

Definition of Variables

The random variable \( \mathbf{X} \) is distributed as a multivariate normal distribution with mean vector \( \mathbf{0} \) and covariance matrix \( \mathbf{\Sigma} \). The elements of the random vector \( \mathbf{X} \) are denoted as \( X_{1} \), \( X_{2} \), and \( X_{3} \). Our task is to find the probability that \( (X_{1}-2X_{2}+X_{3})^{2} > 15.36 \).

Reformulating the Problem

Let \( Y = X_{1} - 2X_{2} + X_{3} \), where \( Y \) is also a normal random variable, and so \( Y^{2} \) will follow a non-central chi-square distribution \( \chi_{1}^{2}(\lambda) \) with 1 degree of freedom, where \( \lambda \) denotes the non-centrality parameter. We can write the problem as \( P(Y^{2} > 15.36) \).

Find Mean and Variance of Y

Firstly, we need to find the mean and variance of \( Y = X_{1} - 2X_{2} + X_{3} \). Since \( E[X_{i}] = 0, E[Y] = 0 \). Secondly, \( Var(Y) = E[Y^{2}] = Var(X_{1}) + 4Var(X_{2}) + Var(X_{3}) - 4Cov(X_{1},X_{2}) + 2Cov(X_{1},X_{3}) - 4Cov(X_{2},X_{3}) \). Considering the given \( \Sigma \) matrix, the variance will be 2.

Standardize Y

The standardized version of the variable \( Y \), denoted as \( Z \) is defined by \( Z = (Y - E[Y])/\sqrt{Var(Y)} \). Here, \( Z = Y/\sqrt{2} \), which will be standard normally distributed, \( i.e., Z \sim N(0,1) \).

Replicate Problem in terms of Z

The task can be reformulated as \( P(Z^{2} > 15.36/2) \). The right hand side simplifies to \( P(Z^{2} > 7.68) \). And \( Z^{2} \sim \chi_{1}^{2} \), which has a chi-square distribution with 1 degree of freedom.

Compute the Probability

The probability can be found using chi-square distribution tables or software that includes statistical functions. However, most tables do not go as far as 7.68, and software returns probability close to 0, which is too small to measure accurately.