Schaum's Outline Probability and Statistics (3rd Edition)

CHAPTER 4 Special Probability Distributions 137

Letting y ϭ u 2z> v in the bracketed integral, we find

1 ``
2v>2⌫(v>2)
F(x) ϭ !2p 33 z(v>2)Ϫ1eϪz>2 !z>v eϪu2z>2v du dz

zϭ0 uϭϪ`

ϭ !2pv 1 x c ` z(vϪ1)>2 eϪ(z>2)[1ϩ(u2>v)] dz d du
2v>2⌫(v>2)
3 3
u ϭ Ϫ` zϭ0

Letting w ϭ z a1 ϩ u2 b , this can then be written
2 v

1 x` w(vϪ1)>2eϪw
!2pv 2v>2⌫(v>2) ϩ u2>v)(vϩ1)>2
F(x) ϭ ? 2(vϩ1)>2 3 c 3 (1 dw d du
u w
ϭ Ϫ` ϭ 0

⌫av ϩ 1b x du
2 u2>v)(vϩ1)>2
ϭ
v 3 (1 ϩ
2pv⌫ a 2 b u ϭ Ϫ`

as required.

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 ϭ 0.05,
(b) the total shaded area ϭ 0.05,
(c) the total unshaded area ϭ 0.99,
(d) the shaded area on the left ϭ 0.01,
(e) the area to the left of t1 is 0.90.

Fig. 4-19

(a) If the shaded area on the right is 0.05, then the area to the left of t1 is (1 Ϫ 0.05) ϭ 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 Ϫ 0.025) ϭ 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 Ϫ 0.99) ϭ 0.01, and the shaded area to the
right is 0.01>2 ϭ 0.005. From the table we find t0.995 ϭ 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 ϭ 2.82. Therefore, the value of t for which the shaded area on the left is 0.01 is Ϫ2.82.

(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.

138 CHAPTER 4 Special Probability Distributions

4.44. Find the values of t for which the area of the right-hand tail of the t distribution is 0.05 if the number of
degrees of freedom v is equal to (a) 16, (b) 27, (c) 200.

Referring to Appendix D, we find in the column headed t0.95 the values: (a) 1.75 corresponding to v ϭ 16;
(b) 1.70 corresponding to v ϭ 27; (c) 1.645 corresponding to v ϭ 200. (The latter is the value that would be
obtained by using the normal curve. In Appendix D this value corresponds to the entry in the last row marked ϱ.)

The F distribution
4.45. Prove Theorem 4-7.

The joint density function of V1 and V2 is given by

f (v1, v2) ϭ a 1 > 2) v e(n1>2)Ϫ1 Ϫv1>2 b a 1 > 2) n e(n2>2)Ϫ1 Ϫv2>2 b
2v1>2⌫(n1 1 2n2>2⌫(n2 2

ϭ 1 v v e(n1>2)Ϫ1 (n2>2)Ϫ1 Ϫ(v1ϩv2)>2
2(n1ϩn2)>2⌫(n1>2)⌫(n2>2) 12

if v1 Ͼ 0, v2 Ͼ 0 and 0 otherwise. Make the transformation

u ϭ v1>v1 ϭ v2v1 , w ϭ v2 or v1 ϭ v1uw v2 ϭ w
v2>v2 v1v2 v2

Then the Jacobian is

'(v1, v2) ϭ 2 'v1>'u 'v1>'w 2 ϭ 2 v1w>v2 v1u>v2 2 ϭ v1w
'(u, w) 'v2>'u 'v2>'w 0 1 v2

Denoting the density as a function of u and w by g(u, w), we thus have

g(u, w) ϭ 1 a v1uw b (v1>2)Ϫ1 w e(v2>2)Ϫ1 Ϫ[1ϩ(v1u>v2)](w>2) v1w
2(v1ϩv2)>2⌫(v1>2)⌫(v2>2) v2 v2

if u Ͼ 0, w Ͼ 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) ϭ (v1>v2)v1>2u(v1>2)Ϫ1 2) ` dw
2(v1ϩv2)>2⌫(v1>2)⌫(v2>
3 w e[(v1ϩv2)>2]Ϫ1 Ϫ[1ϩ(v1u>v2)](w>2)
0

if u Ͼ 0 and 0 if u Յ 0. But from 15, Appendix A,

` ⌫( p)
3 wpϪ1eϪaw dw ϭ
0 ap

Therefore, we have

(v1>v2)v1>2u(v1>2)Ϫ1⌫a v1 ϩ v2 b
2
h(u) ϭ
1 v1u (v1ϩv2)>2
2(v1ϩv2)>2⌫(v1 > 2)⌫(v2 > 2) c 2 a1 ϩ v2 b d

⌫a v1 ϩ v2 b
2
ϭ v1v1>2v2v2>2u(v1>2)Ϫ1(v2 ϩ v1u)Ϫ(v1ϩv2)>2
v1 v2
⌫ a 2 b ⌫ a 2 b

if u Ͼ 0 and 0 if u Յ 0, which is the required result.

CHAPTER 4 Special Probability Distributions 139

4.46. Prove that the F distribution is unimodal at the value a v1 Ϫ 2 b a v2 2b if v1 Ͼ 2.
v1 ϩ
v2

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)Ϫ1(v2 ϩ v1u)Ϫ(v1ϩv2)>2

If this has a relative maximum, it will occur where the derivative is zero, i.e.,

a v1 Ϫ 1 b u(v1>2)Ϫ2(v2 ϩ v1u)Ϫ(v1ϩv2)>2 Ϫ u(v1>2)Ϫ1v1 a v1 ϩ v2 b(v2 ϩ v1u)Ϫ[(v1ϩv2)>2]Ϫ1 ϭ 0
2 2

Dividing by u(v1>2)Ϫ2(v2 ϩ v1u)Ϫ[(v1ϩv2)>2]Ϫ1, u 2 0, we find

a v1 Ϫ 1 b (v2 ϩ v1u) Ϫ uv1a v1 ϩ v2 b ϭ 0 or u ϭ a v1 Ϫ 2 v2 2b
2 2 v1 b a v2 ϩ

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 ϭ 10, v2 ϭ 15, we find F0.95,10,15 ϭ 2.54.

(b) From Appendix F, where v1 ϭ 15, v2 ϭ 9, we find F0.99,15,9 ϭ 4.96.

(c) By Theorem 4-8, page 117, F0.05,8,30 ϭ 1 ϭ 1 ϭ 0.325.
F0.95,30,8 3.08

(d) By Theorem 4-8, page 117, F0.01,15,9 ϭ 1 ϭ 1 ϭ 0.257.
F0.99,9,15 3.89

Relationships among F, x2, and t distributions

4.48. Verify that (a) F0.95 ϭ t02.975, (b) F0.99 ϭ t02.995.

(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 ϭ (12.71)2, 18.5 ϭ (4.30)2, 10.1 ϭ (3.18)2, 7.71 ϭ (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 ϭ (63.66)2, 98.5 ϭ (9.92)2, 34.1 ϭ (5.84)2, 21.2 ϭ (4.60)2, etc.

4.49. Prove Theorem 4-9, page 117, which can be briefly stated as
F1Ϫp ϭ t21Ϫ(p>2)

and therefore generalize the results of Problem 4.48.

Let v1 ϭ 1, v2 ϭ v in the density function for the F distribution [(45), page 116]. Then

⌫av ϩ 1b
2
f (u) ϭ vv>2uϪ1>2(v ϩ u)Ϫ(vϩ1)>2
1 v
⌫ a 2 b ⌫ a 2 b

⌫av ϩ 1b u Ϫ(vϩ1)>2
2 v
ϭ vv>2uϪ1>2vϪ(vϩ1)>2 a1 ϩ b
v
! p⌫ a 2 b

⌫av ϩ 1b u Ϫ(vϩ1)>2
2 v
ϭ uϪ1>2 a 1 ϩ b
v
! vp⌫ a 2 b

140 CHAPTER 4 Special Probability Distributions

for u Ͼ 0, and f (u) ϭ 0 for u Յ 0. Now, by the definition of a percentile value, F1Ϫp is the number such that
P(U Յ F1Ϫp) ϭ 1 Ϫ p. Therefore,

⌫av ϩ 1b F1 Ϫ p u Ϫ(vϩ1)>2
2 v
3 uϪ1>2a1 ϩ b du ϭ 1 Ϫ p
v
! vp⌫ a 2 b 0

In the integral make the change of variable t ϭ ϩ !u:

⌫av ϩ 1b ϩ1F1Ϫp t2 Ϫ(vϩ1)>2
2 v
2 3 a1 ϩ b dt ϭ 1 Ϫ p
v 0
!vp⌫ a 2 b

Comparing with (42), page 115, we see that the left-hand side of the last equation equals

2 ? P(0 Ͻ T Յ ϩ !F1Ϫp)

where T is a random variable having Student’s t distribution with v degrees of freedom. Therefore,
1Ϫp
2 ϭ P(0 Ͻ T Յ ϩ !F1Ϫp)

ϭ P(T Յ ϩ !F1Ϫp) Ϫ P(T Յ 0)

ϭ P(T Յ ϩ !F1Ϫp) Ϫ 1
2

where we have used the symmetry of the t distribution. Solving, we have

p
P(T Յ ϩ !F1Ϫp) ϭ 1 Ϫ 2

But, by definition, t1Ϫ(p/2) is the number such that
p

P(T Յ t1Ϫ(p>2)) ϭ 1 Ϫ 2

and this number is uniquely determined, since the density function of the t distribution is strictly positive.
Therefore,

ϩ !F1Ϫp ϭ t1Ϫ(p>2) or F1Ϫp ϭ t12Ϫ(p>2)
which was to be proved.

4.50. Verify Theorem 4-10, page 117, for (a) p ϭ 0.95, (b) p ϭ 0.99.

(a) Compare the entries in the last row of the F0.95 table in Appendix F (corresponding to v2 ϭ `) with the
entries under x20.95 in Appendix E. Then we see that

3.84 ϭ 3.184, 3.00 ϭ 5.299, 2.60 ϭ 7.381, 2.37 ϭ 9.449, 2.21 ϭ 115.1, 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 ϭ `) with the
entries under x02.99 in Appendix E. Then we see that

6.63 ϭ 6.163, 4.61 ϭ 9.221, 3.78 ϭ 113.3, 3.32 ϭ 134.3, 3.02 ϭ 155.1, 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 distribu-
tion. Show that X and Y are independent if and only if their correlation coefficient is zero.

CHAPTER 4 Special Probability Distributions 141

If the correlation coefficient r ϭ 0, then the bivariate normal density function (49), page 117, becomes

f (x, y) ϭ c1 e dϪ(xϪm1)2>2s12 c1 e dϪ(yϪm2)2>2s22
s1 !2p s2 !2p

and since this is a product of a function of x alone and a function of y alone for all values of x and y, it follows
that X and Y are independent.

Conversely, if X and Y are independent, f (x, y) given by (49) must for all values of x and y be the product of
a function of x alone and a function of y alone. This is possible only if r ϭ 0.

Miscellaneous distributions

4.52. Find the probability that in successive tosses of a fair die, a 3 will come up for the first time on the
fifth toss.

Method 1
The probability of not getting a 3 on the first toss is 5>6. Similarly, the probability of not getting a 3 on the second
toss is 5>6, etc. Then the probability of not getting a 3 on the first 4 tosses is (5>6) (5>6) (5>6) (5>6) ϭ (5/6)4.
Therefore, since the probability of getting a 3 on the fifth toss is 1>6, the required probability is

5 4 1 625
6 6 7776
a b a b ϭ

Method 2 (using formula)
Using the geometric distribution, page 117, with p ϭ 1>6, q ϭ 5>6, x ϭ 5, we see that the required proba-
bility is

1 5 4 625
6 6 7776
a b a b ϭ

4.53. Verify the expressions given for (a) the mean, (b) the variance, of the Weibull distribution, page 118.

`

(a) m ϭ E(X) ϭ 3 abxb eϪaxb dx
0

ϭ ab ` a u b eϪu 1 u(1>b)Ϫ1 du
a1>b a b
3
0

`

ϭ aϪ1>b 3 u1>beϪu du
0

ϭ aϪ1>b⌫ a 1 ϩ 1 b
b

where we have used the substitution u ϭ axb to evaluate the integral.

`

(b) E(X2) ϭ 3 abxbϩ1eϪaxbdx
0

ϭ ab ` a u b 1ϩ(1>b) 1 u(1>b)Ϫ1 du
a1>b a b
3 eϪu
0

`

ϭ aϪ2>b 3 u2>b eϪu du
0

ϭ aϪ2>b⌫ a 1 ϩ 2 b
b

Then

s2 ϭ E[(X Ϫ m)2] ϭ E(X2) Ϫ m2

ϭ aϪ2>b c ⌫ a1 ϩ 2 b Ϫ ⌫2 a1 ϩ 1 b d
b b