Let U and V be random variables. show that (E (
UV
))
2
≤
E (
U
2
) E (
V
2
)
(This is the Cauchy-Schwarz inequal-
ity).
Hint: Show that if
U
+
tV
= 0
for some constant
t
, then
(E (
UV
))
2
= E (
U
2
) E (
V
2
)
.
Otherwise, note that
0
<
E
(
(
U
+
tV
)
2
)
=
q
(
t
)
,
for all
t
,
where
q
is a quadratic polynomial in
t
. Since
q
has no zeros, its discriminant must be
negative.
(b) Suppose that
U
=
tV
+
b
, with
t
6
= 0
. Show that
ρ
U,V
= 1
if
t >
0
and
ρ
U,V
=
−
1
if
t <
0
.
(c) Show that if
X,Y
are random variables, then
|
Cov (
X,Y
)
|≤
√
Var (
X
)
√
Var (
Y
)
,
and conclude that
|
ρ
X,Y
|≤
1
. In fact,
|
ρ
X,Y
|
= 1
only when
Y
=
tX
+
b
with
t
6
= 0
.
Hint: Apply (a), with
U
=
X
−
E (
X
)
and
V
=
Y
−
E (
Y
)
.
2
Annotations
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Let U And V Be Random Variables Show That E Uv 2 E U 2 E V 2 This Is The Cauchy
/in Uncategorized /by developerLet U and V be random variables. show that (E (
UV
))
2
≤
E (
U
2
) E (
V
2
)
(This is the Cauchy-Schwarz inequal-
ity).
Hint: Show that if
U
+
tV
= 0
for some constant
t
, then
(E (
UV
))
2
= E (
U
2
) E (
V
2
)
.
Otherwise, note that
0
<
E
(
(
U
+
tV
)
2
)
=
q
(
t
)
,
for all
t
,
where
q
is a quadratic polynomial in
t
. Since
q
has no zeros, its discriminant must be
negative.
(b) Suppose that
U
=
tV
+
b
, with
t
6
= 0
. Show that
ρ
U,V
= 1
if
t >
0
and
ρ
U,V
=
−
1
if
t <
0
.
(c) Show that if
X,Y
are random variables, then
|
Cov (
X,Y
)
|≤
√
Var (
X
)
√
Var (
Y
)
,
and conclude that
|
ρ
X,Y
|≤
1
. In fact,
|
ρ
X,Y
|
= 1
only when
Y
=
tX
+
b
with
t
6
= 0
.
Hint: Apply (a), with
U
=
X
−
E (
X
)
and
V
=
Y
−
E (
Y
)
.
2
Annotations
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