Download MTH/STA 562 Exercise 6.40. Since .# and .$ are independent

MTH/STA 562
Exercise 6.40. Since Y1 and Y2 are independent standard normal random variables, it follows from
Example 6.11 that Y12 and Y22 are independent chi-square random variables with 1 degree of freedom. Then
the moment-generating function of Y1 and Y2 are
mY12 (t) = (1
2t)
1=2
mY22 (t) = (1
and
1=2
2t)
;
respectively, so that the moment-generating function of U = Y12 + Y22 is
mU (t) = mY12 (t) mY22 (t) = (1
2t)
1=2
(1
1=2
2t)
= (1
2t)
1
which is the moment-generating function for a chi-square random variable with 2 degrees of freedom. Hence,
U has a chi-square distribution with 2 degrees of freedom.
Exercise 6.41. Since Y1 , Y2 ,
, Yn are independent normal random variables, each with mean
variance 2 , the moment-generating function of Yi , i = 1; 2;
; n, is given by
1
t + t2
2
mYi (t) = exp
Then
2
:
h
i
h
i
1
mai Yi (t) = E et(ai Yi ) = E e(ai t)Yi = mYi (ai t) = exp ai t + a2i t2
2
By the independence of Y1 , Y2 ,
+ an Yn is
mU (t)
exp
n
X
1
t
ai + t2
2
i=1
man Yn (t) =
2
n
X
i=1
a2i
!
n
Y
1
exp ai t + a2i t2
2
i=1
n
X
a2i .
:
Hence, U is normally distributed with mean
i=1
n
X
2
;
which is the moment-generating function for a normal random variable with mean
2
2
, Yn , it follows that the moment-generating function of U = a1 Y1 + a2 Y2 +
= ma1 Y1 (t) ma2 Y2 (t)
=
and
ai and variance
i=1
2
n
X
n
X
ai and variance
i=1
a2i .
i=1
Exercise 6.45. Since Y1 and Y2 are independent normal random variables with E (Y1 ) = 10, E (Y2 ) = 4,
V ar (Y1 ) = 0:5, and V ar (Y2 ) = 0:2, it follows from Theorem 6.3 that U = 100 + 7Y1 + 3Y2 is a normal
random variable with
E (U ) = 100 + 7E (Y1 ) + 3E (Y2 ) = 100 + 7 (10) + 3 (4) = 182
and
V ar (U ) = 72 V ar (Y1 ) + 32 V ar (Y2 ) = 49 (0:5) + 9 (0:2) = 12:61:
It is necessary to …nd a value c such that P fU > cg = 0:91. Now
c 182
Z> p
12:61
P fU > cg = P
:
The value of the standard normal random variable that satis…es the requirement is z = 2:33. Hence,
c 182
p
= 2:33
12:61
and so c = $190:27.
1
Exercise 6.49. The moment-generating functions of Y1 and Y2 are
mY1 (t) = pet + q
n1
mY2 (t) = pet + q
and
n2
;
respectively. Since Y1 and Y2 are independent, we have
n1
mY1 +Y2 (t) = mY1 (t) mY2 (t) = pet + q
pet + q
n2
= pet + q
n1 +n2
which is the moment-generating function of the binomial random variable with parameters n1 + n2 and p.
Hence, Y1 + Y2 is a binomial random variable with parameters n1 + n2 and p.
Exercise 6.52.
(a) The moment-generating functions of Y1 and Y2 are
t
mY1 (t) = e 1 (e
1)
t
mY2 (t) = e 2 (e
and
1)
;
respectively. Since Y1 and Y2 are independent, we have
t
mY1 +Y2 (t) = mY1 (t) mY2 (t) = e 1 (e
1)
t
e 2 (e
1)
= e(
1+ 2)
(et
which is the moment-generating function of the Poisson random variable with mean
is a Poisson random variable with mean 1 + 2 .
1)
1
+
2.
Hence, Y1 + Y2
(b) By de…nition,
P fY1 = kjY1 + Y2 = mg =
P fY1 = k; Y2 = m kg
P fY1 = k; Y1 + Y2 = mg
=
P fY1 + Y2 = mg
P fY1 + Y2 = mg
k
1e
=
=
for k = 0; 1; 2;
m
2
k
e 2
(m k)!
1
k!
(
1+ 2)
m
e
m!
( 1+ 2)
=
m!
k! (m k)! (
k
m
k
m k
2
1
1
+
k m k
1 2
m
1 + 2)
2
1
+
2
; m, which is a binomial distribution with parameters m and
1= ( 1
+
2 ).
Exercise 6.53. Let Y1 , Y2 ,
, Yn be independent binomial random variables with ni trials and
probability of success given by pi , i = 1; 2;
; n. Then The moment-generating function of Yi is
mYi (t) = pi et + (1
for i = 1; 2;
pi )
ni
; n.
(a) If n1 = n2 =
n
X
U=
Yi is
= nn = m and p1 = p2 =
= pn = p, then the moment-generating function of
i=1
mU (t)
= mY1 (t) mY2 (t) mYn (t)
m
= pet + (1 p)
pet + (1
=
pet + (1
p)
p)
m
pet + (1
p)
m
mn
which is the moment-generating function of the binomial random variable with mn trials and probability of
n
X
success given by p. Hence,
Yi is a binomial random variable with mn trials and probability of success
i=1
given by p.
2
(b) If p1 = p2 =
= pn = p, then the moment-generating function of U =
n
X
Yi is
i=1
mU (t)
= mY1 (t) mY2 (t) mYn (t)
n1
= pet + (1 p)
pet + (1
pet + (1
=
p)
n1 +n2 +
n2
p)
pet + (1
p)
nn
+nn
which is the moment-generating function of the binomial random variable with n1 + n2 +
+ nn trials and
n
X
Yi is a binomial random variable with n1 + n2 +
+ nn trials
probability of success given by p. Hence,
i=1
and probability of success given by p.
(c) By de…nition,
P
(
Y1 = k
n
X
Yi = m
i=1
)
P
=
(
n
X
Y1 = k;
P
( n
X
)
P fY1 = kg P
( n
X
( n
X
P
P
=
(
Y1 = k;
Yi = m
i=1
=
Yi = m
i=1
)
Yi = m
i=2
)
k
)
Yi = m
( ni=2
X
)
k
)
Yi = m
i=1
:
Yi = m
i=1
By a similar argument to part (b), it can be shown that
P
n
X
n
X
Yi is a binomial random variable with n2 + n3 +
i=2
+ nn trials and probability of success given by p. Thus,
)
(
n
X
P Y1 = k
Yi = m
i=1
n1 k
p (1
k
n2 + n3 +
+ nn m k
n +n +
p
(1 p) 2 3
m k
n1 + n 2 +
+ nn m
n +n + +nn m
p (1 p) 1 2
m
n1
n2 + n3 +
+ nn
k
m k
n1 + n2 +
+ nn
m
=
=
for k = 0; 1; 2;
n1 k
p)
+nn (m k)
; min fn1 ; mg, which is a hypergeometric distribution with parameters N =
r = n1 .
(d) By de…nition,
P
P
=
(
Y1 + Y2 = k
(
Y1 + Y2 = k;
P
( n
X
i=1
n
X
i=1
n
X
i=1
Yi = m
)
)
P
Yi = m
)
=
Yi = m
(
Y1 + Y2 = k;
P
( n
X
i=1
3
n
X
i=3
Yi = m
)
Yi = m
k
)
:
n
X
i=1
ni and
P fY1 + Y2 = kg P
=
P
( n
X
( n
X
Yi = m
i=3
Yi = m
i=1
)
k
)
:
By a similar argument to part (b), it can be shown that Y1 + Y2 is a binomial random variable with n1 + n2
n
X
trials and probability of success given by p and
Yi is a binomial random variable with n3 + n4 +
+ nn
i=3
trials and probability of success given by p. Thus,
(
)
n
X
P Y1 + Y2 = k
Yi = m
i=1
n1 + n2 k
p (1
k
n3 + n 4 +
+ nn m k
n +n +
p
(1 p) 3 4
m k
n1 + n 2 +
+ nn m
n +n + +nn m
p (1 p) 1 2
m
n1 + n2
n3 + n4 +
+ nn
k
m k
n1 + n 2 +
+ nn
m
=
=
for k = 0; 1; 2;
n1 +n2 k
p)
+nn (m k)
; min fn1 + n2 ; mg, which is a hypergeometric distribution with parameters N =
and r = n1 + n2 .
(e) No, the moment-generating function of
n
X
n
X
ni
i=1
Yi cannot be easily simpli…ed to the desired form.
i=1
Exercise 6.54.
(a) The moment-generating function of Yi is
mYi (t) = exp
for i = 1; 2;
; n. Since Y1 , Y2 ,
i
et
1
, Yn are independent, the moment-generating functions of
n
X
Yi =
i=1
Y1 + Y2 +
+ Yn is
n
mX
(t)
= mY1 (t) mY2 (t)
mYn (t)
Yi
i=1
=
=
exp
exp
"
1
et
n
X
i=1
i
1 exp
!
et
2
1
#
et
1
exp
n
et
which is the moment-generating function of the Poisson random variable with mean
Hence,
n
X
Yi is a Poisson random variable with mean
i=1
n
X
i=1
(b) By de…nition,
4
n
X
i=1
i.
1
i
=
1+ 2+
+
n.
P
(
Y1 = k
n
X
Yi = m
i=1
)
P
=
(
Y1 = k;
P
( n
X
n
X
P fY1 = kg P
P
)
=
Y1 = k;
( n
X
( n
X
P
Yi = m
i=2
k
)
)
n
X
Yi = m
( ni=2
X
)
n
X
By a similar argument to part (a), it can be shown that
Yi = m
i=1
:
Yi is a Poisson random variable with mean
i=2
i.
Thus,
i=2
(
P
Y1 = kj
n
X
)
Yi = m
i=1
k
1e
0
@
=
0
@
1
k!
=
=
iA
i=2
n
X
i=1
1m
iA
0
k
n
X
i=2
n
X
B
B 1
B
n
BX
@
1k 0
i
C
C
C
C
A
iA
i=1
i
i=1
0
1
X
i
iA
i=2
(m k)!
0
n
m!
1
n
X
exp@
exp@
k
1
i=1
for k = 0; 1; 2;
1m
n
X
m!
k! (m k)!
m
k
!m
k
!m
1m
n
X
B
B i=2
B
n
BX
@
k
iC
i
i=1
C
C
C
A
; m, which is a binomial distribution with parameters m and
1
, n
X
i=1
(c) By de…nition,
P
(
Y1 + Y2 = k
n
X
i=1
Yi = m
)
P
=
(
Y1 + Y2 = k;
P
P
=
(
( n
X
n
X
i=1
Yi = m
)
)
Yi = m
i=1
Y1 + Y2 = k;
P
( n
X
i=1
5
k
)
Yi = m
i=1
n
X
P
(
Yi = m
i=1
=
Yi = m
i=1
)
n
X
i=3
Yi = m
)
Yi = m
k
)
i.
=
P fY1 + Y2 = kg P
P
( n
X
( n
X
Yi = m
i=3
Yi = m
i=1
k
)
)
:
From Exercise 6.52, Y1 + Y2 is a Poisson random variable with mean 1 + 2 . Also, by a similar argument
n
n
X
X
to part (a), it can be shown that
Yi is a Poisson random variable with mean
i . Thus,
i=3
P
(
Y1 + Y2 = k
n
X
i=3
Yi = m
i=1
)
(
1+ 2)
k
=
exp[ (
k!
0
@
1 + 2 )]
=
1+
@
n
X
i=3
0
exp@
2)
iA
n
X
i=1
m!
k
1m
n
X
i
i=3
0
1k 0
B
B 1+
B
n
B X
@
i
n
X
i
i=1
n
X
C B
B i=3
B
n
C BX
A @
2C
C
i=1
for k = 0; 1; 2;
iA
i=1
m!
k! (m k)!
m
k
1m
n
X
(
=
0
!m
0
k
iA
i=3
(m k)!
1
iA
!m
1m
1
n
X
exp@
k
k
iC
i
i=1
C
C
C
A
; m, which is a binomial distribution with parameters m and (
1
+
2)
, n
X
i.
i=1
Exercise 6.57. The moment-generating functions of Yi is
mY1 (t) = (1
for i = 1; 2;
+ Yn is
; n. Since Y1 , Y2 ,
mU (t)
(1
i
, Yn are independent, the moment-generating functions of U = Y1 + Y2 +
= mY1 (t) mY2 (t)
=
t)
t)
(
1+
mYn (t) = (1
2+
+
t)
1
(1
2
t)
(1
n
t)
n)
which is the moment-generating function of the gamma random variable with parameters
and . Hence, U is a gamma random variable with parameters 1 + 2 +
+ n and .
1
+
2
+
+
n
Exercise 6.59. The moment-generating functions of Y1 and Y2 are
mY1 (t) = (1
2t)
1
and
mY2 (t) = (1
2t)
2
;
respectively. Since Y1 and Y2 are independent, the moment-generating functions of U = Y1 + Y2 is
mU (t) = mY1 (t) mY2 (t) = (1
2t)
1
(1
2t)
2
= (1
2t)
which is the moment-generating function of the chi-square random variable with
Hence, U is a chi-square random variable with 1 + 2 degrees of freedom.
6
(
1
1+ 2)
+
2
degrees of freedom.