4.35. Let X1 and X2 be independent random variables that are chi-square distributed with v1 and v2 degrees of freedom, respectively, (a) Show that the moment generating function of Z X1 X2 is , thereby (b) show that Z is chi-square distributed with v1 v2 degrees of freedom. (a) The moment generating function of Z X1 X2 is using Problem 4.34. (b) It is seen from Problem 4.34 that a distribution whose moment generating function is is the chi-square distribution with v1 v2 degrees of freedom. This must be the distribution of Z, by Theorem 3-10, page 77.

By generalizing the above results, we obtain a proof of Theorem 4-4, page 115. 4.36. Let X be a normally distributed random variable having mean 0 and variance 1. Show that X2 is chi-square distributed with 1 degree of freedom. We want to find the distribution of Y X2 given a standard normal distribution for X. Since the correspondence between X and Y is not one-one, we cannot apply Theorem 2-3 as it stands but must proceed as follows.

4.37. Prove Theorem 4-3, page 115, for v 2. By Problem 4.36 we see that if X1 and X2 are normally distributed with mean 0 and variance 1, then and are chi square distributed with 1 degree of freedom each. Then, from Problem 4.35(b), we see that is chi square distributed with 1 1 2 degrees of freedom if X1 and X2 are independent. The general result for all positive integers v follows in the same manner.

4.38. The graph of the chi-square distribution with 5 degrees of freedom is shown in Fig. 4-18. (See the remarks on notation on page 115.) Find the values for which (a) the shaded area on the right 0.05, (b) the total shaded area 0.05, (c) the shaded area on the left 0.10, (d) the shaded area on the right

(a) If the shaded area on the right is 0.05, then the area to the left of is (1 0.05) 0.95, and represents the 95th percentile, . Referring to the table in Appendix E, proceed downward under column headed v until entry 5 is reached. Then proceed right to the column headed . The result, 11.1, is the required value of . (b) Since the distribution is not symmetric, there are many values for which the total shaded area 0.05. For example, the right-hand shaded area could be 0.04 while the left-hand shaded area is 0.01. It is customary, however, unless otherwise specified, to choose the two areas equal. In this case, then, each area 0.025. If the shaded area on the right is 0.025, the area to the left of is 1 0.025 0.975 and represents the 97.5th percentile, , which from Appendix E is 12.8. Similarly, if the shaded area on the left is 0.025, the area to the left of is 0.025 and represents the 2.5th percentile, , which equals 0.831. Therefore, the values are 0.831 and 12.8. (c) If the shaded area on the left is 0.10, represents the 10th percentile, , which equals 1.61. (d) If the shaded area on the right is 0.01, the area to the left of is 0.99, and represents the 99th percentile, , which equals 15.1.

4.39. Find the values of for which the area of the right-hand tail of the distribution is 0.05, if the number of degrees of freedom v is equal to (a) 15, (b) 21, (c) 50. Using the table in Appendix E, we find in the column headed the values: (a) 25.0 corresponding to v 15; (b) 32.7 corresponding to v 21; (c) 67.5 corresponding to v 50. 4.40. Find the median value of corresponding to (a) 9, (b) 28, (c) 40 degrees of freedom. Using the table in Appendix E, we find in the column headed (since the median is the 50th percentile) the values: (a) 8.34 corresponding to v 9; (b) 27.3 corresponding to v 28; (c) 39.3 corresponding to v 40. It is of interest to note that the median values are very nearly equal to the number of degrees of freedom. In fact, for v 10 the median values are equal to v 0.7, as can be seen from the table. 4.41. Find for (a) v 50, (b) v 100 degrees of freedom. For v greater than 30, we can use the fact that is very closely normally distributed with mean zero and variance one. Then if zp is the (100p)th percentile of the standardized normal distribution, we can write, to a high degree of approximation

4.43. The graph of Student's t distribution with 9 degrees of freedom is shown in Fig. 4-19. Find the value of t1 for which (a) the shaded area on the right 0.05, (b) the total shaded area 0.05, (c) the total unshaded area 0.99, (d) the shaded area on the left 0.01, (e) the area to the left of t1 is 0.90. av 1 2 b 2pva v 2b 3 x u` du (1 u2>v)(v1)>2 F(x) 1 !2pv 2v>2(v>2) ? 2(v1)>2 3 x u` c 3 ` w0 w(v1)>2ew (1 u2>v)(v1)>2 dwd du w z 2 a1 u2 v b 1 !2pv 2v>2(v>2) 3 x u` c 3 ` z0 z(v1)>2 e(z>2)[1(u2>v)] dzd du F(x) 1 !2p2v>2(v>2) 3 ` z0 3 ` u` z(v>2)1ez>2!z>v eu2z>2vdu dz y u2z> v (a) If the shaded area on the right is 0.05, then the area to the left of t1 is (1 0.05) 0.95, and t1 represents the 95th percentile, t0.95. Referring to the table in Appendix D, proceed downward under the column headed v until entry 9 is reached. Then proceed right to the column headed t0.95. The result 1.83 is the required value of t. (b) If the total shaded area is 0.05, then the shaded area on the right is 0.025 by symmetry. Therefore, the area to the left of t1 is (1 0.025) 0.975, and t1 represents the 97.5th percentile, t0.975. From Appendix D, we find 2.26 as the required value of t. (c) If the total unshaded area is 0.99, then the total shaded area is (1 0.99) 0.01, and the shaded area to the right is 0.01 2 0.005. From the table we find t0.995 3.25. (d) If the shaded area on the left is 0.01, then by symmetry the shaded area on the right is 0.01. From the table, t0.99 2.82. Therefore, the value of t for which the shaded area on the left is 0.01 is 2.82. (e) If the area to the left of t1 is 0.90, then t1 corresponds to the 90th percentile, t0.90, which from the table equals 1.38. > Fig. 4-19

3.104. Find the coefficient of (a) skewness, (b) kurtosis, for the distribution with density function Jerk x 20 f(x) = 10 Miscellaneous problems 3.105. Let X be a random variable that can take on the values 2, 1, and 3 with respective probabilities 1/3, 1/6, and 1/2. Find (a) the mean, (b) the variance, (c) the moment generating function, (d) the characteristic function, (e) the third moment about the mean. 3.106. Work Problem 3.105 if X has density function f(x) = [ )/2, thereby (b) show that Z is chi-square distributed with v, + v2 degrees of freedom. (a) The moment generating function of Z = X, + X2 is M(1) = Elenx, +x,)] = E(exx,) E(ex,) = (1 - 21)-w/2(1 - 21)-w/2 = (1 - 21)-(v,+v;1/2 using Problem 4.34. (b) It is seen from Problem 4.34 that a distribution whose moment generating function is (1 - 27)-(+2)/2 is the chi-square distribution with v, + v2 degrees of freedom. This must be the distribution of Z, by Theorem 3-10, page 77. By generalizing the above results, we obtain a proof of Theorem 4-4, page 115. 4.36. Let X be a normally distributed random variable having mean 0 and variance 1. Show that X2 is chi-square distributed with 1 degree of freedom. We want to find the distribution of Y = X2 given a standard normal distribution for X. Since the correspondence between X and Y' is not one-one, we cannot apply Theorem 2-3 as it stands but must proceed as follows