Question

Suppose that \( (X_1, X_2, X_3) \) has a \( N_3(\mu, \Sigma) \) distribution with \[ \mu = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \quad \text{and} \quad \Sigma = \begin{pmatrix} 2 & 2 & 1 \\ 2 & 5 & 1 \\ 1 & 1 & 1 \end{pmatrix}. \] Given that \( \Phi(-0.5) = 0.3085 \), where \( \Phi(.) \) denotes the cumulative distribution function of a standard normal random variable, \[ P\left( (X_1 - 2X_2 + 2X_3)^2 < \frac{7}{2} \right) \text{ (rounded off to two decimal places) equals ...............} \]

Hint:

- For quadratic forms of normal variables, the result is typically a chi-square distribution or a transformation of a normal distribution.
- Use the standard normal CDF for computations involving standardization of normal variables.

Answer 1

1) Understanding the question:
The given distribution is a multivariate normal distribution, and we need to find the probability of the quadratic form \( (X_1 - 2X_2 + 2X_3)^2 \) being less than \( \frac{7}{2} \). We are also provided with the value of \( \Phi(-0.5) \), the cumulative distribution function (CDF) of the standard normal distribution.
2) Transformation of the quadratic form:
Let the new random variable \( Z = X_1 - 2X_2 + 2X_3 \). The distribution of \( Z \) is normal since it is a linear combination of normally distributed variables. We need to find the probability \( P(Z^2<\frac{7}{2}) \). This simplifies to: \[ P(-\sqrt{\frac{7}{2}}<Z<\sqrt{\frac{7}{2}}) \] Using the properties of normal distribution, we compute the standardization of \( Z \) and use the given value of \( \Phi(-0.5) \) to find the probability. The final value lies between 0.35 and 0.40.
The correct answer is \( 0.35 \) to \( 0.40 \).