Pour la première inégalité , on pose
Z
=
X
−
p
{\displaystyle Z=X-p}
Z
¯
n
=
1
n
∑
i
=
1
n
Z
i
{\displaystyle {\overline {Z}}_{n}={\frac {1}{n}}\sum _{i=1}^{n}Z_{i}}
Z
¯
n
{\displaystyle {\overline {Z}}_{n}}
P
(
1
n
∑
i
=
1
n
X
i
≥
p
+
ϵ
)
=
P
(
Z
¯
n
≥
ϵ
)
≤
e
−
h
Z
¯
n
(
ϵ
)
.
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&=P({\overline {Z}}_{n}\geq \epsilon )\\&\leq \mathrm {e} ^{-h_{{\overline {Z}}_{n}}(\epsilon )}.\end{aligned}}}
Or
h
Z
¯
n
(
ϵ
)
=
sup
t
≥
0
{
ϵ
t
−
ln
(
E
[
e
t
Z
¯
n
]
)
}
=
n
h
Z
(
ϵ
)
{\displaystyle h_{{\overline {Z}}_{n}}(\epsilon )=\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{t{\overline {Z}}_{n}}])\}=nh_{Z}(\epsilon )}
{
X
i
}
i
∈
[
1
,
n
]
{\displaystyle \{X_{i}\}_{i\in [\!1,n\!]}}
{
Z
i
}
i
∈
[
1
,
n
]
{\displaystyle \{Z_{i}\}_{i\in [\!1,n\!]}}
E
[
e
t
Z
¯
n
]
=
∏
i
=
1
n
E
[
e
t
n
Z
i
]
=
E
[
e
t
n
Z
]
n
.
{\displaystyle {\begin{aligned}E[\mathrm {e} ^{t{\overline {Z}}_{n}}]&=\prod _{i=1}^{n}E[\mathrm {e} ^{{\frac {t}{n}}Z_{i}}]\\&=E[\mathrm {e} ^{{\frac {t}{n}}Z}]^{n}.\end{aligned}}}
D'où,
h
Z
¯
n
(
ϵ
)
=
sup
t
≥
0
{
ϵ
t
−
ln
(
E
[
e
t
Z
¯
n
]
)
}
=
sup
t
≥
0
{
ϵ
t
−
n
ln
(
E
[
e
t
n
Z
]
)
}
=
n
sup
t
≥
0
{
ϵ
t
n
−
ln
(
E
[
e
t
n
Z
]
)
}
=
n
h
Z
(
ϵ
)
.
{\displaystyle {\begin{aligned}h_{{\overline {Z}}_{n}}(\epsilon )&=\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{t{\overline {Z}}_{n}}])\}\\&=\sup _{t\geq 0}\{\epsilon t-n\ln(E[\mathrm {e} ^{{\frac {t}{n}}Z}])\}\\&=n\sup _{t\geq 0}\{\epsilon {\frac {t}{n}}-\ln(E[\mathrm {e} ^{{\frac {t}{n}}Z}])\}\\&=nh_{Z}(\epsilon ).\end{aligned}}}
Donc,
P
(
1
n
∑
i
=
1
n
X
i
≥
p
+
ϵ
)
≤
e
−
n
sup
t
≥
0
{
ϵ
t
−
ln
(
E
[
e
t
Z
]
)
}
≤
e
n
inf
t
≥
0
{
ln
(
E
[
e
t
Z
]
)
−
ϵ
t
}
≤
e
n
(
ln
(
E
[
e
t
Z
]
)
−
ϵ
t
)
(
pour
t
≥
0
)
.
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&\leq \mathrm {e} ^{-n\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{tZ}])\}}\\&\leq \mathrm {e} ^{n\inf _{t\geq 0}\{\ln(E[\mathrm {e} ^{tZ}])-\epsilon t\}}\\&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{tZ}])-\epsilon t)}({\text{pour }}t\geq 0).\end{aligned}}}
On remarque que
E
[
e
t
Z
]
=
e
−
p
t
E
[
e
t
X
]
=
e
−
p
t
(
1
−
p
+
p
e
t
)
{\displaystyle E[\mathrm {e} ^{tZ}]=\mathrm {e} ^{-pt}E[\mathrm {e} ^{tX}]=\mathrm {e} ^{-pt}(1-p+p\mathrm {e} ^{t})}
∀
t
≥
0
,
{\displaystyle \forall t\geq 0,}
ln
(
E
[
e
t
Z
]
)
−
ϵ
t
=
ln
(
1
−
p
+
p
e
t
)
−
(
ϵ
+
p
)
t
=
Ψ
(
t
)
−
ϵ
t
,
{\displaystyle {\begin{aligned}\ln(E[\mathrm {e} ^{tZ}])-\epsilon t&=\ln(1-p+p\mathrm {e} ^{t})-(\epsilon +p)t\\&=\Psi (t)-\epsilon t,\end{aligned}}}
avec
∀
t
∈
R
,
Ψ
(
t
)
=
−
p
t
+
ln
(
1
−
p
+
p
e
t
)
{\displaystyle \forall t\in \mathbb {R} ,~\Psi (t)=-pt+\ln(1-p+p\mathrm {e} ^{t})}
la formule de Taylor Lagrange à l'ordre 2, on calcule les dérivées premières et secondes
Ψ
{\displaystyle \Psi }
∀
t
∈
R
,
Ψ
′
(
t
)
=
−
p
+
p
e
t
1
−
p
+
p
e
t
Ψ
″
(
t
)
=
(
1
−
p
)
p
e
t
(
1
−
p
+
p
e
t
)
2
=
α
β
(
α
+
β
)
2
≤
1
4
,
{\displaystyle {\begin{aligned}\forall t\in \mathbb {R} ,~\Psi ^{'}(t)&=-p+{\frac {p\mathrm {e} ^{t}}{1-p+p\mathrm {e} ^{t}}}\\\Psi ^{''}(t)&={\frac {(1-p)p\mathrm {e} ^{t}}{(1-p+p\mathrm {e} ^{t})^{2}}}\\&={\frac {\alpha \beta }{(\alpha +\beta )^{2}}}\\&\leq {\frac {1}{4}},\end{aligned}}}
avec
α
=
1
−
p
,
β
=
p
e
t
{\displaystyle \alpha =1-p,~\beta =p\mathrm {e} ^{t}}
Ψ
″
(
t
)
{\displaystyle \Psi ^{''}(t)}
1
4
{\displaystyle {\frac {1}{4}}}
(
α
+
β
)
2
=
α
2
+
β
2
+
2
α
β
et
(
α
−
β
)
2
=
α
2
+
β
2
−
2
α
β
≥
0
⇒
2
α
β
≤
α
2
+
β
2
⇒
(
α
+
β
)
2
≥
4
α
β
{\displaystyle (\alpha +\beta )^{2}=\alpha ^{2}+\beta ^{2}+2\alpha \beta {\text{ et }}(\alpha -\beta )^{2}=\alpha ^{2}+\beta ^{2}-2\alpha \beta \geq 0\Rightarrow 2\alpha \beta \leq \alpha ^{2}+\beta ^{2}\Rightarrow (\alpha +\beta )^{2}\geq 4\alpha \beta }
Donc, comme
Ψ
(
0
)
=
Ψ
′
(
0
)
=
0
{\displaystyle \Psi (0)=\Psi ^{'}(0)=0}
la formule de Taylor Lagrange ,
∀
t
∈
R
{\displaystyle \forall t\in \mathbb {R} }
Ψ
(
t
)
=
Ψ
(
0
)
+
t
Ψ
′
(
0
)
+
t
2
2
Ψ
″
(
θ
t
)
≤
t
2
8
,
{\displaystyle {\begin{aligned}\Psi (t)&=\Psi (0)+t\Psi ^{'}(0)+{\frac {t^{2}}{2}}\Psi ^{''}(\theta t)\\&\leq {\frac {t^{2}}{8}},\end{aligned}}}
avec
θ
∈
[
0
,
1
]
{\displaystyle \theta \in [0,1]}
∀
t
≥
0
{\displaystyle \forall t\geq 0}
P
(
1
n
∑
i
=
1
n
X
i
≥
p
+
ϵ
)
≤
e
n
(
ln
(
E
[
e
t
Z
]
)
−
ϵ
t
)
≤
e
n
(
t
2
8
−
ϵ
t
)
.
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{tZ}])-\epsilon t)}\\&\leq \mathrm {e} ^{n({\frac {t^{2}}{8}}-\epsilon t)}.\end{aligned}}}
Soit
∀
t
≥
0
,
g
(
t
)
=
t
2
8
−
ϵ
t
{\displaystyle \forall t\geq 0,~g(t)={\frac {t^{2}}{8}}-\epsilon t}
∀
t
≥
0
,
g
′
(
t
)
=
t
4
−
ϵ
{\displaystyle \forall t\geq 0,~g^{'}(t)={\frac {t}{4}}-\epsilon }
t
=
4
ϵ
{\displaystyle t=4\epsilon }
∀
ϵ
>
0
{\displaystyle \forall \epsilon >0}
P
(
1
n
∑
i
=
1
n
X
i
≥
p
+
ϵ
)
≤
e
n
(
16
ϵ
2
8
−
4
ϵ
2
)
≤
e
−
2
n
ϵ
2
.
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&\leq \mathrm {e} ^{n({\frac {16\epsilon ^{2}}{8}}-4\epsilon ^{2})}\\&\leq \mathrm {e} ^{-2n\epsilon ^{2}}.\end{aligned}}}
Pour la deuxième inégalité ,
∀
ϵ
>
0
{\displaystyle \forall \epsilon >0}
P
(
1
n
∑
i
=
1
n
X
i
≤
p
−
ϵ
)
=
P
(
Z
¯
n
≤
−
ϵ
)
=
P
(
−
Z
¯
n
≥
ϵ
)
≤
e
−
h
−
Z
¯
n
(
t
)
d'après l'inégalité de Chernoff
≤
e
−
n
h
−
Z
(
t
)
≤
e
n
inf
t
≥
0
{
ln
(
E
[
e
−
t
Z
]
)
−
ϵ
t
}
≤
e
n
(
ln
(
E
[
e
−
t
Z
]
)
−
ϵ
t
)
(
pour
t
≥
0
)
.
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\leq p-\epsilon )&=P({\overline {Z}}_{n}\leq -\epsilon )\\&=P(-{\overline {Z}}_{n}\geq \epsilon )\\&\leq \mathrm {e} ^{-h_{-{\overline {Z}}_{n}}(t)}{\text{ d'après l'inégalité de Chernoff}}\\&\leq \mathrm {e} ^{-nh_{-Z}(t)}\\&\leq \mathrm {e} ^{n\inf _{t\geq 0}\{\ln(E[\mathrm {e} ^{-tZ}])-\epsilon t\}}\\&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{-tZ}])-\epsilon t)}({\text{pour }}t\geq 0).\end{aligned}}}
On remarque que :
∀
t
≥
0
{\displaystyle \forall t\geq 0}
E
[
e
−
t
Z
]
=
e
p
t
E
[
e
−
t
X
]
=
e
p
t
(
1
−
p
+
p
e
−
t
)
⇒
ln
(
E
[
e
−
t
Z
]
)
=
p
t
+
ln
(
1
−
p
+
p
e
−
t
)
=
Ψ
(
−
t
)
≤
t
2
8
.
{\displaystyle {\begin{aligned}E[\mathrm {e} ^{-tZ}]&=\mathrm {e} ^{pt}E[\mathrm {e} ^{-tX}]\\&=\mathrm {e} ^{pt}(1-p+p\mathrm {e} ^{-t})\\\Rightarrow \ln(E[\mathrm {e} ^{-tZ}])&=pt+\ln(1-p+p\mathrm {e} ^{-t})\\&=\Psi (-t)\\&\leq {\frac {t^{2}}{8}}.\end{aligned}}}
Donc,
∀
ϵ
>
0
,
∀
t
≥
0
{\displaystyle \forall \epsilon >0,~\forall t\geq 0}
P
(
1
n
∑
i
=
1
n
X
i
≤
p
−
ϵ
)
≤
e
n
(
t
2
8
−
ϵ
t
)
≤
e
−
2
n
ϵ
2
,
{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\leq p-\epsilon )&\leq \mathrm {e} ^{n({\frac {t^{2}}{8}}-\epsilon t)}\\&\leq \mathrm {e} ^{-2n\epsilon ^{2}},\end{aligned}}}
par un argument similaire qui a servi à démontrer la première inégalité.
![{\displaystyle \phi (t)=\mathbb {E} [e^{tX}]<+\infty ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef15eb49257030fe7df225bcf2acc825e85ac80b)
![{\displaystyle {\text{si }}t>0,\,\mathbb {P} \left(X\geq a\right)\leq e^{-ta}\mathbb {E} [e^{tX}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83b7851d234360f5461d49e53508c9160bcc046a)
![{\displaystyle {\text{si }}t<0,\,\mathbb {P} \left(X\leq -a\right)\leq e^{ta}\mathbb {E} [e^{tX}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5587cff653928dcb35cdfb271d154550bb99526b)
![{\displaystyle \mathbb {E} [X_{i}]=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb2d224569b261f80aefcf837e798e35bd5d2346)
![{\displaystyle {\begin{aligned}\mathrm {e} ^{t(X-a)}&\geq {1}_{\{X\geq a\}}\\\Rightarrow E\left[\mathrm {e} ^{t(X-a)}\right]&\geq P(X\geq a)\\\Rightarrow E\left[\mathrm {e} ^{tX}\right]\mathrm {e} ^{-ta}&\geq P(X\geq a).\\\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d5afb3348ab91f699d55094609bd29ed0b1d3659)
![{\displaystyle {\begin{aligned}\mathrm {e} ^{t(X+a)}\geq {1}_{\{X\leq -a\}}\\\Rightarrow P(X\leq -a)&\leq E\left[\mathrm {e} ^{t(X+a)}\right]\\&\leq \mathrm {e} ^{ta}\mathrm {e} ^{\ln(\phi (t))}\\&\leq \mathrm {e} ^{-(-ta-\ln(\phi (t)))},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8d7a505c8e0cba499f24d9e26932e39e696646f)
![{\displaystyle h_{{\overline {Z}}_{n}}(\epsilon )=\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{t{\overline {Z}}_{n}}])\}=nh_{Z}(\epsilon )}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9c26fcd8cf07de5080991c021afe6dc28e05c50)
![{\displaystyle \{X_{i}\}_{i\in [\!1,n\!]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e80444a9b19b1566f5374fbdc08e679dd72a75e1)
![{\displaystyle \{Z_{i}\}_{i\in [\!1,n\!]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3960c6c696e7785a45e5a5fba08440c6fc00719b)
![{\displaystyle {\begin{aligned}E[\mathrm {e} ^{t{\overline {Z}}_{n}}]&=\prod _{i=1}^{n}E[\mathrm {e} ^{{\frac {t}{n}}Z_{i}}]\\&=E[\mathrm {e} ^{{\frac {t}{n}}Z}]^{n}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27e32f8d6867080a086f3da03b7a377b40828056)
![{\displaystyle {\begin{aligned}h_{{\overline {Z}}_{n}}(\epsilon )&=\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{t{\overline {Z}}_{n}}])\}\\&=\sup _{t\geq 0}\{\epsilon t-n\ln(E[\mathrm {e} ^{{\frac {t}{n}}Z}])\}\\&=n\sup _{t\geq 0}\{\epsilon {\frac {t}{n}}-\ln(E[\mathrm {e} ^{{\frac {t}{n}}Z}])\}\\&=nh_{Z}(\epsilon ).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0836287a88d695915fa92f0a1220798ea5039298)
![{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&\leq \mathrm {e} ^{-n\sup _{t\geq 0}\{\epsilon t-\ln(E[\mathrm {e} ^{tZ}])\}}\\&\leq \mathrm {e} ^{n\inf _{t\geq 0}\{\ln(E[\mathrm {e} ^{tZ}])-\epsilon t\}}\\&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{tZ}])-\epsilon t)}({\text{pour }}t\geq 0).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fb617683833d36f2b62da74fb1aad47fa6e7aa12)
![{\displaystyle E[\mathrm {e} ^{tZ}]=\mathrm {e} ^{-pt}E[\mathrm {e} ^{tX}]=\mathrm {e} ^{-pt}(1-p+p\mathrm {e} ^{t})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2fe17e18abdf7b568a98a70d08efe745c919f4c3)
![{\displaystyle {\begin{aligned}\ln(E[\mathrm {e} ^{tZ}])-\epsilon t&=\ln(1-p+p\mathrm {e} ^{t})-(\epsilon +p)t\\&=\Psi (t)-\epsilon t,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4e9c74d69d21d904eefb82f213ace5377af2193)
![{\displaystyle \theta \in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fead1e7dceab4be5ab2e91f5108144722daa8c36)
![{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\geq p+\epsilon )&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{tZ}])-\epsilon t)}\\&\leq \mathrm {e} ^{n({\frac {t^{2}}{8}}-\epsilon t)}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9bf3c5438cae35a6e9937887e8ac5c93f3147b3)
![{\displaystyle {\begin{aligned}P({\frac {1}{n}}\sum _{i=1}^{n}X_{i}\leq p-\epsilon )&=P({\overline {Z}}_{n}\leq -\epsilon )\\&=P(-{\overline {Z}}_{n}\geq \epsilon )\\&\leq \mathrm {e} ^{-h_{-{\overline {Z}}_{n}}(t)}{\text{ d'après l'inégalité de Chernoff}}\\&\leq \mathrm {e} ^{-nh_{-Z}(t)}\\&\leq \mathrm {e} ^{n\inf _{t\geq 0}\{\ln(E[\mathrm {e} ^{-tZ}])-\epsilon t\}}\\&\leq \mathrm {e} ^{n(\ln(E[\mathrm {e} ^{-tZ}])-\epsilon t)}({\text{pour }}t\geq 0).\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/26aa4854dae1730ef9b67076f177d2b35e4f73bd)
![{\displaystyle {\begin{aligned}E[\mathrm {e} ^{-tZ}]&=\mathrm {e} ^{pt}E[\mathrm {e} ^{-tX}]\\&=\mathrm {e} ^{pt}(1-p+p\mathrm {e} ^{-t})\\\Rightarrow \ln(E[\mathrm {e} ^{-tZ}])&=pt+\ln(1-p+p\mathrm {e} ^{-t})\\&=\Psi (-t)\\&\leq {\frac {t^{2}}{8}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf07fee8b313cb37ea8758ab60f6c7500fa98cf2)
