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if u 0, w 0 and 0 otherwise. The (marginal) density function of U can now be found by integrating with respect to w from 0 to ` , i.e., h(u) if u 0 and 0 if u 2(v1 v2)>2
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` (v1 >v2)v1>2u(v1>2) 1 [(v 30 w 1 (v1 >2) (v2 >2) v2)>2] 1e [1 (v1u>v2)](w>2)
0. But from 15, Appendix A,
30 w Therefore, we have
p 1e aw
( p) ap
(v1 >v2)v1>2u(v1>2) h(u) 2(v1 a
v2)>2
v1 2
v2)>2
(v1 >2) (v2 >2) c v2 b
1 a1 2
v1u (v1 v2 b d
v1 2
v1 v2 a b a b 2 2 if u 0 and 0 if u
v v v11>2v22>2u(v1>2) 1(v2
v1u)
(v1 v2)>2
0, which is the required result.
CHAPTER 4 Special Probability Distributions
4.46. Prove that the F distribution is unimodal at the value a v1 v1 2 ba v2 v2 2 b if v1
The mode locates the maximum value of the density function. Apart from a constant, the density function of the F distribution is u(v1>2) 1(v2 a v1 2 v1u)
(v1 v2)>2
If this has a relative maximum, it will occur where the derivative is zero, i.e., 1bu(v1>2) 2(v2 v1u)
(v1 v2)>2
u(v1>2) 1v1 a
v1 2
b(v2
v1u)
[(v1 v2)>2] 1
Dividing by u(v1>2) 2(v2 a v1 2 1b(v2
v1u)
v1u)
[(v1 v2)>2] 1,
u 2 0, we find
v2 2 b 0 or u a v1 v1 2 ba v2 v2 2 b
uv1 a
Using the second-derivative test, we can show that this actually gives the maximum.
4.47. Using the table for the F distribution in Appendix F, find (a) F0.95,10,15, (b) F0.99,15,9, (c) F0.05,8,30, (d) F0.01,15,9.
(a) From Appendix F, where v1 (b) From Appendix F, where v1 10, v2 15, v2 15, we find F0.95,10,15 9, we find F0.99,15,9 2.54. 4.96.
(c) By Theorem 4-8, page 117, F0.05,8,30 (d) By Theorem 4-8, page 117, F0.01,15,9
1 F0.95,30,8 1 F0.99,9,15
1 3.08 1 3.89
0.325. 0.257.
Relationships among F, x2, and t distributions 4.48. Verify that (a) F0.95 t2 , (b) F0.99 t2 . 0.975 0.995
(a) Compare the entries in the first column of the F0.95 table in Appendix F with those in the t distribution under t0.975. We see that 161 (12.71)2, 18.5 (4.30)2, 10.1 (3.18)2, 7.71 (2.78)2, etc.
(b) Compare the entries in the first column of the F0.99 table in Appendix F with those in the t distribution under t0.995. We see that 4050 (63.66)2, 98.5 (9.92)2, 34.1 (5.84)2, 21.2 (4.60)2, etc.
4.49. Prove Theorem 4-9, page 117, which can be briefly stated as F1
Let v1 1, v2
t2 1
(p>2)
and therefore generalize the results of Problem 4.48. v in the density function for the F distribution [(45), page 116]. Then a f (u) v 2 1 b vv>2u
1>2(v
1 v a b a b 2 2 a v 2 1 b
(v 1)>2
v !p a b 2 a v 2 1 b
vv>2u
1>2v (v 1)>2 a1
u vb
(v 1)>2
v !vp a b 2
1>2 a1
u vb
(v 1)>2
CHAPTER 4 Special Probability Distributions
for u P(U
0, and f (u) 0 for u 0. Now, by the definition of a percentile value, F1 F1 p) 1 p. Therefore, a v 2 1 b 3
1>2 a1
is the number such that
v 0 !vp a b 2 In the integral make the change of variable t a 2 v 2 1 b 3
u vb
(v 1)>2
v !vp a b 2
t2 vb
(v 1)>2
Comparing with (42), page 115, we see that the left-hand side of the last equation equals 2 P(0 T !F1 p)
where T is a random variable having Student s t distribution with v degrees of freedom. Therefore, 1 p !F1 p) P(0 T 2 P(T P(T !F1 p) !F1 p) p 2 p 2 P(T
where we have used the symmetry of the t distribution. Solving, we have P(T But, by definition, t1
(p/2)
!F1 p)
is the number such that P(T t1
(p>2))
and this number is uniquely determined, since the density function of the t distribution is strictly positive. Therefore,
which was to be proved.
(p>2)
t2 1
(p>2)
4.50. Verify Theorem 4-10, page 117, for (a) p
0.95, (b) p
` ) with the
(a) Compare the entries in the last row of the F0.95 table in Appendix F (corresponding to v2 entries under x2 in Appendix E. Then we see that 0.95 3.84 3.84 , 1 3.00 5.99 , 2 2.60 7.81 , 3 2.37 9.49 , 4 2.21 11.1 , 5
etc.
which provides the required verification. (b) Compare the entries in the last row of the F0.99 table in Appendix F (corresponding to v2 entries under x2 in Appendix E. Then we see that 0.99 6.63 6.63 , 1 4.61 9.21 , 2 3.78 11.3 , 3 3.32 13.3 , 4 3.02 15.1 , 5
` ) with the
etc.
which provides the required verification. The general proof of Theorem 4-10 follows by letting v2 S ` in the F distribution on page 116.
The bivariate normal distribution 4.51. Suppose that X and Y are random variables whose joint density function is the bivariate normal distribution. Show that X and Y are independent if and only if their correlation coefficient is zero.
CHAPTER 4 Special Probability Distributions
If the correlation coefficient r f (x, y) 0, then the bivariate normal density function (49), page 117, becomes c 1 e s1 !2p
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