Question

57. In point estimation problems it is often unnecessary to know the distribution of a particular statistic. For example, in Exercise 56 we were able to show that s^2 is an unbiased estimator for ?^2 without mentioning the family of random variables to which s^2 belongs. However, when constructing confidence intervals or testing hypotheses it is necessary to know the distribution of the statistics involved. In this exercise we prove another result that is probably familiar to you from elementary courses. In particular, we show that if y is assumed to have a multivariate normal distribution then the random variable ((n - 1)s^2) / ?^2 = ?(i=1 to n) ((y_i - ?)^2 / ?^2) follows a chi-squared distribution with n - 1 degrees of freedom. a. Let y be an n × 1 multivariate normal random variable with mean ?' = [? ? ... ?] and variance-covariance matrix V = ?^2 I. Note that from Exercise 56 we know that ?(i=1 to n) (y_i - ?)^2 = y' [I - (zz') / n] y where z is an n × 1 matrix of ones. Show that I - (zz') / n is symmetric. b. Show that I - (zz') / n is idempotent. c. Find tr[I - (zz') / n]. d. Find r[I - (zz') / n]. (Hint: See Theorem 1.5.2.) e. Use Corollary 2.3.2 to show that ((n - 1)s^2) / ?^2 = (1 / ?^2) y' [I - zz' / n] y follows a noncentral chi-squared distribution. f. Show that the noncentrality parameter is zero, thus showing that the distribution is chi-squared with n - 1 degrees of freedom as claimed.

          57. In point estimation problems it is often unnecessary to know the distribution of a particular statistic. For example, in Exercise 56 we were able to show that s^2 is an unbiased estimator for ?^2 without mentioning the family of random variables to which s^2 belongs. However, when constructing confidence intervals or testing hypotheses it is necessary to know the distribution of the statistics involved. In this exercise we prove another result that is probably familiar to you from elementary courses. In particular, we show that if y is assumed to have a multivariate normal distribution then the random variable

((n - 1)s^2) / ?^2 = ?(i=1 to n) ((y_i - ?)^2 / ?^2)

follows a chi-squared distribution with n - 1 degrees of freedom.
a. Let y be an n × 1 multivariate normal random variable with mean ?' = [? ? ... ?] and variance-covariance matrix V = ?^2 I. Note that from Exercise 56 we know that

?(i=1 to n) (y_i - ?)^2 = y' [I - (zz') / n] y

where z is an n × 1 matrix of ones. Show that I - (zz') / n is symmetric.
b. Show that I - (zz') / n is idempotent.
c. Find tr[I - (zz') / n].
d. Find r[I - (zz') / n]. (Hint: See Theorem 1.5.2.)
e. Use Corollary 2.3.2 to show that

((n - 1)s^2) / ?^2 = (1 / ?^2) y' [I - zz' / n] y

follows a noncentral chi-squared distribution.
f. Show that the noncentrality parameter is zero, thus showing that the distribution is chi-squared with n - 1 degrees of freedom as claimed.

57. In point estimation problems it is often unnecessary to know the distribution of a particular statistic. For example, in Exercise 56 we were able to show that s^2 is an unbiased estimator for ?^2 without mentioning the family of random variables to which s^2 belongs. However, when constructing confidence intervals or testing hypotheses it is necessary to know the distribution of the statistics involved. In this exercise we prove another result that is probably familiar to you from elementary courses. In particular, we show that if y is assumed to have a multivariate normal distribution then the random variable

((n - 1)s^2) / ?^2 = ?(i=1 to n) ((yi - ?)^2 / ?^2)

follows a chi-squared distribution with n - 1 degrees of freedom.
a. Let y be an n × 1 multivariate normal random variable with mean ?' = [? ? ... ?] and variance-covariance matrix V = ?^2 I. Note that from Exercise 56 we know that

?(i=1 to n) (yi - ?)^2 = y' [I - (zz') / n] y

where z is an n × 1 matrix of ones. Show that I - (zz') / n is symmetric.
b. Show that I - (zz') / n is idempotent.
c. Find tr[I - (zz') / n].
d. Find r[I - (zz') / n]. (Hint: See Theorem 1.5.2.)
e. Use Corollary 2.3.2 to show that

((n - 1)s^2) / ?^2 = (1 / ?^2) y' [I - zz' / n] y

follows a noncentral chi-squared distribution.
f. Show that the noncentrality parameter is zero, thus showing that the distribution is chi-squared with n - 1 degrees of freedom as claimed.

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