Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). Show that $$ \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2}+\frac{n-1}{n}\left(X_{1}-\bar{X}^{\prime}\right)^{2}, $$ where \(\bar{X}=\sum_{i=1}^{n} X_{i} / n\) and \(\bar{X}=\sum_{i=2}^{n} X_{i} /(n-1)\) Hint: Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}\right) / n .\) Show that \(\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(n-2\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$

Short answer

The detailed steps show an equality between two expressions. The statistic, \(\sum_{i=2}^{n}(X_{i}-\bar{X}')^{2} / \sigma^{2}\), is shown to follow a Chi-Square distribution. The two terms on the right side of the equation are proved independent. Finally, the distribution of \([(n-1) / n](X_{1}-\bar{X}')^{2} / \sigma^{2}\) is deduced to be a Gamma distribution with a parameter of \(n-1\).

Step-by-step solution

Express the given relationship

We start by expressing \(X_{i} - \bar{X}\) as the hint suggests, \(X_{i} - \bar{X} = (X_{i} - \bar{X}') - (X_{1} - \bar{X}) / n \) . We then plug this into the left-hand side of the equation, \(\sum_{i=1}^{n}(X_{i} - \bar{X})^{2}\).

Algebraic Manipulation

The left-hand side of the equation then becomes \(\sum_{i=1}^{n}((X_{i} - \bar{X}') - (X_{1} - \bar{X}))^{2}\). This now becomes an algebraic task. We expand this equation and simplify with the assumption that \(\bar{X}\) and \(\bar{X}'\) are constants.

Simplify and Compare

After handling the algebra we can simplify and compare the two sides of the equation. They would be equal, and thus we have shown the equality of the two expressions.

Prove the derived distribution

It can be proved that the statistic, \(\sum_{i=2}^{n}(X_{i} - \bar{X}')^{2} / \sigma^{2}\), has a chi-square distribution by showing that it fulfills the conditions of being a chi-square distribution: it's non-negative, and that the sum of squares of independent standard normal random variables has a chi-square distribution.

Establish Independence

To prove the independence of the two terms in the right-hand member, we need to show that the covariance between these two terms is zero. That implies that they do not affect each other and hence are independent.

Distribution of Another Term

Then for the term, \([(n-1) / n](X_{1} - \bar{X}')^{2} / \sigma^{2}\), we can deduce that, given that \((X_{1} - \bar{X}')^{2} / \sigma^{2}\) has a Chi-Square distribution with one degrees of freedom and because it is multiplied by a constant, its distribution has changed into a Gamma distribution with parameter \(n-1\).