马尔科夫不等式(Markov’s inequality)
对于随机变量
X
X
X,有
P
(
∣
X
∣
⩾
ε
)
⩽
E
∣
X
∣
k
ε
k
,
ε
>
0
,
k
<
∞
P\left( \left| X \right|\geqslant \varepsilon \right) \leqslant \frac{E\left| X \right|^k}{\varepsilon ^k},\varepsilon >0,k<\infty
P(∣X∣⩾ε)⩽εkE∣X∣k,ε>0,k<∞
证明：
P
(
∣
X
∣
⩾
ε
)
=
∫
∣
x
∣
⩾
ε
f
(
x
)
d
x
⩽
∫
∣
x
∣
⩾
ε
∣
x
∣
k
ε
k
f
(
x
)
d
x
⩽
1
ε
k
∫
−
∞
+
∞
∣
x
∣
k
f
(
x
)
d
x
=
E
∣
X
∣
k
ε
k
P\left( \left| X \right|\geqslant \varepsilon \right) =\int_{\left| x \right|\geqslant \varepsilon}{f\left( x \right) dx}\leqslant \int_{\left| x \right|\geqslant \varepsilon}{\frac{\left| x \right|^k}{\varepsilon ^k}f\left( x \right) dx} \\ \leqslant \frac{1}{\varepsilon ^k}\int_{-\infty}^{+\infty}{\left| x \right|^kf\left( x \right) dx}=\frac{E\left| X \right|^k}{\varepsilon ^k}
P(∣X∣⩾ε)=∫∣x∣⩾εf(x)dx⩽∫∣x∣⩾εεk∣x∣kf(x)dx⩽εk1∫−∞+∞∣x∣kf(x)dx=εkE∣X∣k
坎泰利不等式(Cantelli’s inequality)
对于随机变量
X
X
X(
X
>
μ
X>\mu
X>μ)，存在有限均值和方差，则有
P
(
X
−
μ
⩾
ε
)
⩽
σ
2
σ
2
+
ε
2
P\left( X-\mu \geqslant \varepsilon \right) \leqslant \frac{\sigma ^2}{\sigma ^2+\varepsilon ^2}
P(X−μ⩾ε)⩽σ2+ε2σ2文章来源：https://www.toymoban.com/news/detail-718337.html
证明：
利用马尔科夫不等式令
Y
Y
Y等于
X
−
μ
X-\mu
X−μ，则有
P
(
Y
⩾
ε
)
⩽
E
(
Y
2
)
ε
2
P\left( Y\geqslant \varepsilon \right) \leqslant \frac{E\left( Y^2 \right)}{\varepsilon ^2}
P(Y⩾ε)⩽ε2E(Y2)取
β
>
0
\beta>0
β>0,有
P
(
Y
+
β
⩾
ε
+
β
)
⩽
E
(
Y
+
β
)
2
(
ε
+
β
)
2
P\left( Y+\beta \geqslant \varepsilon +\beta \right) \leqslant \frac{E\left( Y+\beta \right) ^2}{\left( \varepsilon +\beta \right) ^2}
P(Y+β⩾ε+β)⩽(ε+β)2E(Y+β)2
E
(
Y
+
β
)
2
=
E
(
Y
2
+
2
Y
β
+
β
2
)
=
σ
2
+
β
2
(
E
Y
=
0
)
E\left( Y+\beta \right) ^2=E\left( Y^2+2Y\beta +\beta ^2 \right) =\sigma ^2+\beta ^2\left( EY=0 \right)
E(Y+β)2=E(Y2+2Yβ+β2)=σ2+β2(EY=0),所以
P
(
Y
+
β
⩾
ε
+
β
)
⩽
σ
2
+
β
2
(
ε
+
β
)
2
P\left( Y+\beta \geqslant \varepsilon +\beta \right) \leqslant \frac{\sigma ^2+\beta ^2}{\left( \varepsilon +\beta \right) ^2}
P(Y+β⩾ε+β)⩽(ε+β)2σ2+β2令
g
(
β
)
=
σ
2
+
β
2
(
ε
+
β
)
2
g\left( \beta \right) =\frac{\sigma ^2+\beta ^2}{\left( \varepsilon +\beta \right) ^2}
g(β)=(ε+β)2σ2+β2
则
g
′
(
β
)
=
2
(
ε
β
−
σ
2
)
(
ε
+
β
)
3
g\prime\left( \beta \right) =\frac{2\left( \varepsilon \beta -\sigma ^2 \right)}{\left( \varepsilon +\beta \right) ^3}
g′(β)=(ε+β)32(εβ−σ2)
当
β
=
σ
2
ε
\beta =\frac{\sigma ^2}{\varepsilon}
β=εσ2时
g
(
β
)
g\left(\beta\right)
g(β)有最小值，
g
m
i
n
(
β
)
=
g
(
σ
2
ε
)
=
σ
2
+
σ
4
ε
2
(
ε
+
σ
2
ε
)
2
=
σ
2
ε
2
+
σ
2
g_{min}\left( \beta \right) =g\left( \frac{\sigma ^2}{\varepsilon} \right) =\frac{\sigma ^2+\frac{\sigma ^4}{\varepsilon ^2}}{\left( \varepsilon +\frac{\sigma ^2}{\varepsilon} \right) ^2}=\frac{\sigma ^2}{\varepsilon ^2+\sigma ^2}
gmin(β)=g(εσ2)=(ε+εσ2)2σ2+ε2σ4=ε2+σ2σ2
因此，
P
(
Y
⩾
ε
)
⩽
σ
2
σ
2
+
ε
2
P\left( Y\geqslant \varepsilon \right) \leqslant \frac{\sigma ^2}{\sigma ^2+\varepsilon ^2}
P(Y⩾ε)⩽σ2+ε2σ2文章来源地址https://www.toymoban.com/news/detail-718337.html
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