Question

Let \(\mathbf{X}=\left(X_{1}, X_{2}, X_{3}\right)\) have a multivariate normal distribution with mean vector 0 and variance-covariance matrix $$ \boldsymbol{\Sigma}=\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right] $$ Find \(P\left(X_{1}>X_{2}+X_{3}+2\right)\). Hint: Find the vector a so that \(\mathbf{a X}=X_{1}-X_{2}-X_{3}\) and make use of Theorem \(3.5 .2\).

Short answer

The probability \(P\left(X_{1}>X_{2}+X_{3}+2\right)\) is approximately 0.1587.

Step-by-step solution

Identify vector \(a\)

In the equation \(\mathbf{a X}=X_{1}-X_{2}-X_{3}\), to convert it into the form, set the vector \(a\) as \(\mathbf{a}=\left[1, -1, -1\right]\). We have \(\mathbf{a X}=X_{1}-X_{2}-X_{3}\). We are interested in the distribution of \(\mathbf{a X}\) so that we can solve for \(P(\mathbf{a X}>2)\).

Obtain mean and variance of the linear combination

The mean of the normal distribution of \(\mathbf{a X}\) can be obtained by multiplying \(a\) by the mean vector 0, which gives us 0. The variance is given by \(\mathbf{a^{T} \Sigma a}\). Substituting \(\mathbf{\Sigma}\) and \(a\), we get \(\mathbf{a^{T} \Sigma a}=\left[1,-1,-1\right] \left[\begin{array}{lll}1 & 0 & 0 \0 & 2 & 1 \0 & 1 & 2\end{array}\right]\left[\begin{array}{l}1 \-1 \-1\end{array}\right]=4.\)

Convert the event

Now that we have the mean and variance of \(\mathbf{a X}\), we can convert the event of \(P(\mathbf{a X}>2)\) into probability distribution function. Standardizing the random variable, we get \(P(\mathbf{a X}-0 > 2-0) = P(\frac{\mathbf{a X}}{\sqrt{4}} > \frac{2}{\sqrt{4}}) = P(Z > 1)\), where \(Z\) is a standard normal random variable.

Calculate the probability

As we know that standard normal distribution is symmetrical about 0 and the area under its curve equals 1. We aim to find the probability for \(Z > 1\). It is given by \(1 - P(Z \leq 1)\). However, the value for \(P(Z \leq 1)\) is found using Z-tables or software, which provides the value 0.8413. Therefore, the final probability is \(1 - 0.8413 = 0.1587\).