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求证：
R
X
(
τ
)
=
R
X
(
?
τ
)
R_X (\tau)=R_X(-\tau)
RX?(τ)=RX?(?τ)
证明一：
∵ 针对平稳过程，
R
X
(
s
,
t
)
R_X(s,t)
RX?(s,t) 记为
R
X
(
t
?
s
)
R_X(t-s)
RX?(t?s)
∴ 针对
R
X
(
τ
)
R_X(\tau)
RX?(τ)，可令
τ
=
t
?
s
\tau=t-s
τ=t?s
则
R
X
(
τ
)
=
R
X
(
t
?
s
)
=
R
X
(
s
,
t
)
R_X(\tau)=R_X(t-s)=R_X(s,t)
RX?(τ)=RX?(t?s)=RX?(s,t)
R
X
(
?
τ
)
=
R
X
(
s
?
t
)
=
R
X
(
t
,
s
)
R_X(-\tau)=R_X(s-t)=R_X(t,s)
RX?(?τ)=RX?(s?t)=RX?(t,s)
而
R
X
(
s
,
t
)
=
E
[
X
(
s
)
X
(
t
)
]
R_X(s,t)=E[X(s)X(t)]
RX?(s,t)=E[X(s)X(t)]，
R
X
(
t
,
s
)
=
E
[
X
(
t
)
X
(
s
)
]
R_X(t,s)=E[X(t)X(s)]
RX?(t,s)=E[X(t)X(s)]
显然，
R
X
(
s
,
t
)
=
R
X
(
t
,
s
)
R_X(s,t)=R_X(t,s)
RX?(s,t)=RX?(t,s)
所以，
R
X
(
τ
)
=
R
X
(
?
τ
)
R_X(\tau)=R_X(-\tau)
RX?(τ)=RX?(?τ)
证明二：
R
X
(
τ
)
=
E
[
X
(
t
)
X
(
t
+
τ
)
]
R_X(\tau)=E[X(t)X(t+\tau)]
RX?(τ)=E[X(t)X(t+τ)]
=
E
[
X
(
t
+
τ
)
X
(
t
)
]
=E[X(t+\tau)X(t)]
=E[X(t+τ)X(t)]
=
R
X
(
?
τ
)
=R_X(-\tau)
=RX?(?τ)，得证
本文的LaTeX代码如下：
求证：$R_X (\tau)=R_X(-\tau)$
证明一：
∵ 针对平稳过程，$R_X(s,t)$ 记为 $R_X(t-s)$
∴ 针对 $R_X(\tau)$，可令 $\tau=t-s$
则 $R_X(\tau)=R_X(t-s)=R_X(s,t)$
$R_X(-\tau)=R_X(s-t)=R_X(t,s)$
而 $R_X(s,t)=E[X(s)X(t)]$，$R_X(t,s)=E[X(t)X(s)]$
显然，$R_X(s,t)=R_X(t,s)$
所以，$R_X(\tau)=R_X(-\tau)$
\
证明二：
$R_X(\tau)=E[X(t)X(t+\tau)]$
$=E[X(t+\tau)X(t)]$
$=R_X(-\tau)$，得证
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