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Chapter 9: Problem 9

Here \(Q_{1}\) and \(Q_{2}\) are quadratic forms in observations of a random sample from \(N(0,1) .\) If \(Q_{1}\) and \(Q_{2}\) are independent and if \(Q_{1}+Q_{2}\) has a chi-square distribution, prove that \(Q_{1}\) and \(Q_{2}\) are chi-square variables.

Short Answer

Expert verified
Using properties of chi-square distributions, it can be proven that both \(Q_{1}\) and \(Q_{2}\) are chi-square distributed.

Step by step solution

01

Understand the Assumptions

According to the problem, \(Q_{1}\) and \(Q_{2}\) are independent, and \(Q_{1}+Q_{2}\) is chi-square distributed. These are given conditions and do not need to be proven.
02

Apply property of chi-square distribution

The property of the chi-square distribution state that the sum of independent chi-square distributed random variables results in another chi-square distribution. Therefore, if \(Q_{1}+Q_{2}\) is chi-square distributed, and \(Q_{1}\) and \(Q_{2}\) are independent, then \(Q_{1}\) and \(Q_{2}\) must each be chi-square distributed by the property of additive chi-square distributions.
03

Conclusion

Based on the assumptions, the additive property of the chi-square distribution and the independence of \(Q_{1}\) and \(Q_{2}\), it can be concluded that \(Q_{1}\) and \(Q_{2}\) must be chi-square distributed.

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Most popular questions from this chapter

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 \(\bar{X}^{\prime}=\sum_{i=2}^{n} X_{i} /(n-1)\). Hint: \(\quad\) Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}^{\prime}\right) / n\). 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 $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample of size \(n\) from a distribution which is \(N\left(0, \sigma^{2}\right)\). Prove that \(\sum_{1}^{n} X_{i}^{2}\) and every quadratic form, which is nonidentically zero in \(X_{1}, X_{2}, \ldots, X_{n}\), are dependent.

Let \(\mathbf{X}^{\prime}=\left[X_{1}, X_{2}\right]\) be bivariate normal with matrix of means \(\boldsymbol{\mu}^{\prime}=\left[\mu_{1}, \mu_{2}\right]\) and positive definite covariance matrix \(\mathbf{\Sigma}\). Let $$ Q_{1}=\frac{X_{1}^{2}}{\sigma_{1}^{2}\left(1-\rho^{2}\right)}-2 \rho \frac{X_{1} X_{2}}{\sigma_{1} \sigma_{2}\left(1-\rho^{2}\right)}+\frac{X_{2}^{2}}{\sigma_{2}^{2}\left(1-\rho^{2}\right)} $$ Show that \(Q_{1}\) is \(\chi^{2}(r, \theta)\) and find \(r\) and \(\theta\). When and only when does \(Q_{1}\) have a central chi-square distribution?

Let \(X_{1}, X_{2}, X_{3}, X_{4}\) be a random sample of size \(n=4\) from the normal distribution \(N(0,1) .\) Show that \(\sum_{i=1}^{4}\left(X_{i}-\bar{X}\right)^{2}\) equals $$ \frac{\left(X_{1}-X_{2}\right)^{2}}{2}+\frac{\left[X_{3}-\left(X_{1}+X_{2}\right) / 2\right]^{2}}{3 / 2}+\frac{\left[X_{4}-\left(X_{1}+X_{2}+X_{3}\right) / 3\right]^{2}}{4 / 3} $$ and argue that these three terms are independent, each with a chi-square distribution with 1 degree of freedom.

Let the independent normal random variables \(Y_{1}, Y_{2}, \ldots, Y_{n}\) have, respectively, the probability density functions \(N\left(\mu, \gamma^{2} x_{i}^{2}\right), i=1,2, \ldots, n\), where the given \(x_{1}, x_{2}, \ldots, x_{n}\) are not all equal and no one of which is zero. Discuss the test of the hypothesis \(H_{0}: \gamma=1, \mu\) unspecified, against all alternatives \(H_{1}: \gamma \neq 1, \mu\) unspecified.
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