Section 7.2 (p.385)

 

1.         .

If n=1, log f(x) = const. - ½ log x – x/2. It doesn’t have a mode because that as x ® 0, log f(x) increases without a bound.

If n=2, log f(x) = const. - x/2. The mode is 0.

If n ³ 3, let . We have that x = n-2 maximizes log f(x). So the mode is n-2.

 

2.         Try to draw them yourself.

 

3.         We want to find the smallest radius r such that Pr(X2 + Y2 £ r2) ³ 0.99. Since X2 + Y2 has a c2 distribution with 2 degrees of freedom (Theorem 2), r2 ³ 9.210 based on table on p.690.

 

4.         X2 + Y2 + Z2 has a c2 distribution with 3 degrees of freedom. Pr(X2 + Y2 + Z2 £ 1) » 0.20 based on the Chi-square table.

 

6.         Fi (Xi) has a uniform distribution on (0,1) (see p.154 for details).

Let Zi = -2 log Fi (Xi), we can show that Zi has pdf  f(z) = ½ e-z/2, for z > 0 (You should be able to figure this out by yourself. If not, review Sec. 3.8). In other words, Zi has a Gamma distribution with parameters a = 1 and b = ½. From Theorem 3. of Sec. 5.9, we know that Y = -2 S log Fi (Xi) = S Zi has a Gamma distribution with a = n and b = 1/2., or a c2 distribution with 2n degrees of freedom

 

8.         has a standard normal distribution. Therefore  has a c2 distribution with 1 degrees of freedom

 

9.         Both (X1 + X2 + X3) and (X4 + X5 + X6) have normal distributions with mean 0 and variance 3. Therefore (X1 + X2 + X3)2/3 and (X4 + X5 + X6)2/3 both have c2 distributions with 1 degrees of freedom. Let c = 1/3, Y has a c2 distributions with 2 degrees of freedom.