... n
n
nn
n
nn
23
2
10
21
⎛
⎝
⎜
⎞
⎠
⎟
+
⎛
⎝
⎜
⎞
⎠
⎟
+⋅⋅⋅+
⎛
⎝
⎜
⎞
⎠
⎟
=−
⎛
⎝
⎜
⎞
⎠
⎟
−
⎛
⎝
⎜
⎞
⎠
⎟
=−−
relationships characterizing the independence of A
j
, j = 1, , n and they are
all necessary. For example, for n = 3 we will have:
PA A A PA PA PA
PA A PA PA
PA A PA PA
PA A PA PA
12 3 1 2 3
12 1 ... following examples:
Let S = {1, 2, 3, 4}, P( {1} ) = ···= P({4}) =
1
4
,...
... (See also Figs. 13 .14 and 13 .15 ; F
n 1, m 1
stands for an r.v. distributed as F
n 1, m 1
.)
The power of the test is easily determined by the fact that
1
1
1
1
11
1
2
2
2
1
1
2
2
1
2
1
2
1
σ
σ
τ
YY ... test.)
13 .6.3 Five resistance measurements are taken on two test pieces and the
observed values (in ohms) are as follows:
xxxxx
yyyy y
12 345
12 345
0 11 8 0 12 5 0...
... justi-
fiable by a purely mathematical reasoning, it does provide a method for
producing estimates in many cases of practical importance. In addition, an
MLE is often shown to have several desirable ... p
r
j
j
r
x
r
x
j
j
r
x
r
x
r
x
r
r
r
θθ
1
1
1
1
11 1 1
1
11
1
, ,
!
!
!
!
,
()
= ⋅⋅⋅
= ⋅⋅⋅ − −⋅⋅⋅−
()
=
=
−−
∏
∏
−
where n =∑
r
j =1
x
j
. Then
log , , log
!
!
log
log log...
... as Negative Exponential
with parameter
λ
= 1. Then:
Exercises 233
SJ SJ
SJ SJ
SJ SJ
12 3 12 3 2 31 2 31
132 13 2 312 312
213 213 3 21 3 21
100
010
0 01
1
0 01
100
010
1
100
0 01
010
1
010
0 01
100
1
010
10 0
0 01
1
0 01
010
10 0
1
::
:;:
::.
== ... consider the following
result. Let Z
1
, , Z
k
be independent N(0, 1) , and set
Y
k
Z
k
Z
k
Z
YZZ
YZZZ
Y
kk
Z
k...
... EY EY
1 2 11 12 21 22
11 21 12 22 11 22 12 21
11 21 12 22 11 22 12 21
11 21 12 22 11
()
=+
()
+
()
[]
=−
()
++
()
=
()
−
()
[]
+
()
+
()
[]
=
()()
−
()()
[]
+
()
2222 12 21
11 12 21 22 1 2
()
−
()()
[]
=+
()
+
()
=
()()
EY ... probability space (S, A, P) and let A
1
, A
2
be events. Set
X
1
= I
A
1
, X
2
= I
A
2
and show that X
1
, X
2
are independ...
... f(x
2
|x
1
) ≥ 0 and
fx x
fx
fx x
fx
fx
xx
21
11
12
11
11
11
1
22
()
=
()
()
=
()
⋅
()
=
∑∑
,,
fx x dx
fx
fx x dx
fx
fx
21 2
11
12 2
11
11
11
1
()
=
()
()
=
()
⋅
()
=
−∞
∞
−∞
∞
∫∫
,.
In a similar fashion, ... Y
XY
EXY
1
1
2
2
12
12 12
12
12
Cov ,
.
From the Cauchy–Schwarz inequality, we have that
ρ
2
≤ 1; that is 1 ≤
ρ
≤ 1,
and
ρ
= 1 if and only if
YX=+ −...
... ∩
()
+⋅⋅⋅
+
PAPABAPABABA
PA B A B A
cc cc cc
cc
n
c
n
c
n
1 11 2 11 223
11 1
=
()
+∩
()
()
+∩
()
∩
()
()
+⋅⋅⋅+ ∩
()
⋅⋅⋅ ∩
()
()
+⋅⋅⋅
()
+
PA PA B PA PA B PA B PA
PA B PA B PA
cc cc cc
cc
n
c
n
c
n
1 11 ... black
at least 2 cards in are red
exactly 1 card in is an ace
the first card in is a diamond,
the second is a heart and the third is a club
card in is a d...