Geology Reference
In-Depth Information
2
2
2
=
Ex
(
. .
xy
y
)
(
)
x
y
2
2
2
2
=
Ex
() .
Exy
(. )
Ey
()
.
.
x
x
y
y
2
2
2
2
=
(( )
Ex
) (( )
Ey
) 2[(.)
Exy
. ]
x
y
x
y
=
Vx
( )
Vy
( )
2Cov( ,
xy
)
In fact
Cov (
x
,
y
)=
[(
Ex
)(
y
)]
x
y
=
Ex y
(.
.
y
.
x
.
)
x
y
x
y
=
Exy
(.)
. ( )
Ey
. ()
Ex
x
y
x
y
=
Exy
(.)
.
.
.
Exy
(.)
. )
xy
xy
xy
xy
Cov (
x
2
)=
2
2
2
=
(Cov( . ))
xx
E x x
( ,
)
m
Ex
()
m
E
(
x
2
) =
2
Cov( ,
xx
)
m
2. Properties of Variogram and Covariance Models
Question 2.1
Let
Z
(
x
) be a stationary random function. Show that the co-variance that
is equal to
C
(
h
) =
E
{[
Z
(
x
+
h
) -
m
)[
Z
(
x
) -
m
]} has the following properties:
(i)
C
(0)
0
(ii)
C
(
h
) =
C
(-
h
)
(iii) |
C
(
h
)|
$
C
(0)
Answer
(i)
C
(
h
) =
EZx h mZx m
{[
(
)
][
( )
]}
2
C
(0) =
EZx m
{[ ()
]}
Vx
() 0
(ii)
C
(-
h
) =
Put
xh y
EZx h mZx m
{[
(
)
][
( )
]}
=
EZy mZy h m
{[
(
)
][
(
)
]}
Ch
Because the random variable is stationary.
(iii) It is to be shown that
=
()
$ $
CCh
(0)
( )
C
(0)
to show that
Ch
()
$
2
C
(0)
2
0
$
EZxh Zx
{[
(
)
( )] }
E Zxh m
{[(
(
)
)
(
Zx m
( )
)] }
The expectation of a square is always
0
2
2
=
EZx h
[
(
)
m
]
EZx
[
( )
m
]
2[
EZx h
(
.
)
m
]
EZx
[
( )
m
]}
= 2(0) 2() 0
C
C h
;$
C h
()
C
(0)
to show that
Ch
()
C
(0)
again:
2
2
EZx h m
{[
(
)
)
(
Zx m
( )
)] }
;
0
EZx h m
[
(
)
]
