Question

Let \(Q=X_{1} X_{2}-X_{3} X_{4}\), where \(X_{1}, X_{2}, X_{3}, X_{4}\) is a random sample of size 4 from a distribution that is \(N\left(0, \sigma^{2}\right)\). Show that \(Q / \sigma^{2}\) does not have a chi-square distribution. Find the mgf of \(Q / \sigma^{2}\).

Short answer

Q does not have a chi-square distribution as it cannot be expressed as a sum of squares. The moment generating function (mgf) of \(Q / \sigma^{2}\) is \((1 - 2t)^{-1}\).

Step-by-step solution

Showing Q does not have a Chi-Square distribution

We start by expressing Q as a sum of squares, a form necessary for a Chi-square distribution, but Q cannot be expressed as a sum of squares, demonstrating that Q does not follow a Chi-Square distribution.

Finding the Moment Generating Function of Q

The moment generating function (mgf) of a random variable is defined by the formula \(M(t) = E[e^{tX}]\). Therefore to find the mgf of \(Q / \sigma^{2}\), we pair \(X_1\) with \(X_4\) and \(X_2\) with \(X_3\) since they are independently distributed. Therefore, mgf of \(Q / \sigma^{2}\) = \(E[e^{t(Q / \sigma^{2})}]\) = \(E[e^{\frac{t(X_1 X_2 - X_3 X_4)}{\sigma^2}}]\) = \(E[e^{t(X_1 X_4 / \sigma^2)}]\) * \(E[e^{t(-X_2 X_3 / \sigma^2)}]\). After substituting the values of \(E[e^{t(X_1 X_4 / \sigma^2)}]\) and \(E[e^{t(-X_2 X_3 / \sigma^2)}]\), we can calculate the final form of the mgf.

Identifying the distributions

Notice that \(X_1 X_4 / \sigma^2\) and \(-X_2 X_3 / \sigma^2\) both are bi-variate normal distributions, with their covariance being 0, and their expected value being 0. Hence, both follow a distribution of a normal random variable squared, i.e., a Chi-Squared distribution with 1 degree of freedom, making their mgf \(E[e^{t(X_1 X_4 / \sigma^2)}] = E[e^{t(-X_2 X_3 / \sigma^2)}] = (1 - 2t)^{-1/2}\). Substituting this back in our mgf calculated in Step 2, we find the mgf of \(Q / \sigma^{2}\).