Question

Mathematical Statistics with Applications Homework Help

Answer 1

7.37:

a. W follows a chi-squared distribution with 5 degrees of freedom. See theorem 7.2 from the same chapter, which says

$\displaystyle \sum_{i=1}^n\left(\frac{Y_i-\mu}{\sigma}\right)^2$

is chi-squared distributed with n d.f.. Here we have $\mu=0$ and $\sigma=1$.

b. U follows a chi-squared distribution with 4 degrees of freedom. See theorem 7.3:

$\displaystyle \frac1{\sigma^2}\sum_{i=1}^n (Y_i-\overline Y)^2$

is chi-squared distributed with n - 1 d.f..

c. Y₆² is chi-square distributed for the same reason as W, but with d.f. = 1. The sum of chi-squared distributed random variables is itself chi-squared distributed, with d.f. equal to the sum of the individual random variables' d.f.s. Then U + Y₆² is chi-squared distributed with 5 + 1 = 6 degrees of freedom.

7.38:

a. Notice that

$\dfrac{\sqrt 5 Y_6}{\sqrt W} = \dfrac{Y_6}{\sqrt{\frac W5}}$

and see definition 7.2 for the t distribution. Since Y₆ is normally distributed with mean 0 and s.d. 1, it follows that this random variable is t distributed with 5 degrees of freedom.

b. Similar manipulation gives

$\dfrac{2Y_6}{\sqrt U} = \dfrac{\sqrt4 Y_6}{\sqrt U} = \dfrac{Y_6}{\sqrt{\frac U4}}$

so this r.v. is t distributed with 4 degrees of freedom.