Założenie, że
X
{\displaystyle X}
a
{\displaystyle a}
b
{\displaystyle b}
X
{\displaystyle X}
P
(
X
=
0
)
=
1
,
{\displaystyle \textstyle {\mathsf {P}}\left(X=0\right)=1,}
a w tym wypadku dowodzona nierówność jest prawdziwa. Bez straty ogólności można więc założyć, że liczba
a
{\displaystyle a}
b
{\displaystyle b}
Funkcja
s
↦
e
s
x
{\displaystyle s\mapsto e^{sx}}
wypukła , tj.
e
s
x
⩽
b
−
x
b
−
a
e
s
a
+
x
−
a
b
−
a
e
s
b
(
x
∈
[
a
,
b
]
)
.
{\displaystyle e^{sx}\leqslant {\frac {b-x}{b-a}}e^{sa}+{\frac {x-a}{b-a}}e^{sb}\quad (x\in [a,b]).}
Obliczając wartość oczekiwaną obu stron powyższej nierówności, otrzymujemy
E
[
e
s
X
]
⩽
b
−
E
(
X
)
b
−
a
e
s
a
+
E
(
X
)
−
a
b
−
a
e
s
b
=
b
b
−
a
e
s
a
+
−
a
b
−
a
e
s
b
E
(
X
)
=
0
=
(
−
a
b
−
a
)
e
s
a
(
−
b
a
+
e
s
b
−
s
a
)
=
(
−
a
b
−
a
)
e
s
a
(
−
b
−
a
+
a
a
+
e
s
(
b
−
a
)
)
=
(
−
a
b
−
a
)
e
s
a
(
−
b
−
a
a
−
1
+
e
s
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b
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a
)
)
=
(
1
−
θ
+
θ
e
s
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b
−
a
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)
e
−
s
θ
(
b
−
a
)
θ
=
−
a
b
−
a
>
0
{\displaystyle {\begin{aligned}{\mathsf {E}}\left[e^{sX}\right]&\leqslant {\frac {b-{\mathsf {E}}(X)}{b-a}}e^{sa}+{\frac {{\mathsf {E}}(X)-a}{b-a}}e^{sb}\\&={\frac {b}{b-a}}e^{sa}+{\frac {-a}{b-a}}e^{sb}&&{\mathsf {E}}(X)=0\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b}{a}}+e^{sb-sa}\right)\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b-a+a}{a}}+e^{s(b-a)}\right)\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b-a}{a}}-1+e^{s(b-a)}\right)\\&=\left(1-\theta +\theta e^{s(b-a)}\right)e^{-s\theta (b-a)}&&\theta =-{\frac {a}{b-a}}>0\end{aligned}}}
Niech
u
=
s
(
b
−
a
)
.
{\displaystyle u=s(b-a).}
φ
:
R
→
R
{\displaystyle \varphi \colon \mathbb {R} \to \mathbb {R} }
φ
(
u
)
=
−
θ
u
+
log
(
1
−
θ
+
θ
e
u
)
(
u
∈
R
)
.
{\displaystyle \varphi (u)=-\theta u+\log \left(1-\theta +\theta e^{u}\right)\quad (u\in \mathbb {R} ).}
Definicja ta jest poprawna. Istotnie,
1
−
θ
+
θ
e
u
=
θ
(
1
θ
−
1
+
e
u
)
=
θ
(
−
b
a
+
e
u
)
>
0
θ
>
0
,
b
a
<
0
{\displaystyle {\begin{aligned}1-\theta +\theta e^{u}&=\theta \left({\frac {1}{\theta }}-1+e^{u}\right)\\&=\theta \left(-{\frac {b}{a}}+e^{u}\right)\\&>0&&\theta >0,\quad {\frac {b}{a}}<0\end{aligned}}}
W konsekwencji,
E
[
e
s
X
]
⩽
e
φ
(
u
)
.
{\displaystyle {\mathsf {E}}\left[e^{sX}\right]\leqslant e^{\varphi (u)}.}
Ze wzoru Taylora , dla każdej liczby rzeczywistej
u
{\displaystyle u}
v
{\displaystyle v}
[
0
,
u
]
,
{\displaystyle [0,u],}
φ
(
u
)
=
φ
(
0
)
+
u
φ
′
(
0
)
+
1
2
u
2
φ
″
(
v
)
.
{\displaystyle \varphi (u)=\varphi (0)+u\varphi '(0)+{\tfrac {1}{2}}u^{2}\varphi ''(v).}
Wynika stąd, że
φ
(
0
)
=
0
φ
′
(
0
)
=
−
θ
+
θ
e
u
1
−
θ
+
θ
e
u
|
u
=
0
=
0
φ
″
(
v
)
=
θ
e
v
(
1
−
θ
+
θ
e
v
)
−
θ
2
e
2
v
(
1
−
θ
+
θ
e
v
)
2
=
θ
e
v
1
−
θ
+
θ
e
v
(
1
−
θ
e
v
1
−
θ
+
θ
e
v
)
=
t
(
1
−
t
)
t
=
θ
e
v
1
−
θ
+
θ
e
v
⩽
1
4
t
>
0
{\displaystyle {\begin{aligned}\varphi (0)&=0\\\varphi '(0)&=-\theta +\left.{\frac {\theta e^{u}}{1-\theta +\theta e^{u}}}\right|_{u=0}\\&=0\\[6pt]\varphi ''(v)&={\frac {\theta e^{v}\left(1-\theta +\theta e^{v}\right)-\theta ^{2}e^{2v}}{\left(1-\theta +\theta e^{v}\right)^{2}}}\\[6pt]&={\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\left(1-{\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\right)\\[6pt]&=t(1-t)&&t={\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\\&\leqslant {\tfrac {1}{4}}&&t>0\end{aligned}}}
Oznacza to, że
φ
(
u
)
⩽
0
+
u
⋅
0
+
1
2
u
2
⋅
1
4
=
1
8
u
2
=
1
8
s
2
(
b
−
a
)
2
.
{\displaystyle \varphi (u)\leqslant 0+u\cdot 0+{\tfrac {1}{2}}u^{2}\cdot {\tfrac {1}{4}}={\tfrac {1}{8}}u^{2}={\tfrac {1}{8}}s^{2}(b-a)^{2}.}
Ostatecznie
E
[
e
s
X
]
⩽
exp
(
1
8
s
2
(
b
−
a
)
2
)
.
{\displaystyle {\mathsf {E}}\left[e^{sX}\right]\leqslant \exp \left({\tfrac {1}{8}}s^{2}(b-a)^{2}\right).}
Bibliografia
edytuj
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle {\mathsf {E}}\left[e^{\lambda X}\right]\leqslant \exp \left({\frac {\lambda ^{2}(b-a)^{2}}{8}}\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/12652c36e05ea0993d99b010654a936bd4c155b6)
![{\displaystyle e^{sx}\leqslant {\frac {b-x}{b-a}}e^{sa}+{\frac {x-a}{b-a}}e^{sb}\quad (x\in [a,b]).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb215ef3afeeba67710b907c9f1664ae3fa400cc)
![{\displaystyle {\begin{aligned}{\mathsf {E}}\left[e^{sX}\right]&\leqslant {\frac {b-{\mathsf {E}}(X)}{b-a}}e^{sa}+{\frac {{\mathsf {E}}(X)-a}{b-a}}e^{sb}\\&={\frac {b}{b-a}}e^{sa}+{\frac {-a}{b-a}}e^{sb}&&{\mathsf {E}}(X)=0\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b}{a}}+e^{sb-sa}\right)\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b-a+a}{a}}+e^{s(b-a)}\right)\\&=\left(-{\frac {a}{b-a}}\right)e^{sa}\left(-{\frac {b-a}{a}}-1+e^{s(b-a)}\right)\\&=\left(1-\theta +\theta e^{s(b-a)}\right)e^{-s\theta (b-a)}&&\theta =-{\frac {a}{b-a}}>0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39b7cc92edba972bb4e3e55e73cc7810205c10ec)
![{\displaystyle {\mathsf {E}}\left[e^{sX}\right]\leqslant e^{\varphi (u)}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28f6a825d023b30bb7cb77a094bfc7ac5f205933)
![{\displaystyle [0,u],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae6ddeb3f4c1a1eede273b630e70e462264bddc0)
![{\displaystyle {\begin{aligned}\varphi (0)&=0\\\varphi '(0)&=-\theta +\left.{\frac {\theta e^{u}}{1-\theta +\theta e^{u}}}\right|_{u=0}\\&=0\\[6pt]\varphi ''(v)&={\frac {\theta e^{v}\left(1-\theta +\theta e^{v}\right)-\theta ^{2}e^{2v}}{\left(1-\theta +\theta e^{v}\right)^{2}}}\\[6pt]&={\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\left(1-{\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\right)\\[6pt]&=t(1-t)&&t={\frac {\theta e^{v}}{1-\theta +\theta e^{v}}}\\&\leqslant {\tfrac {1}{4}}&&t>0\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5476674c98f0499ccdbe2d6b7d433c5b7458329)
![{\displaystyle {\mathsf {E}}\left[e^{sX}\right]\leqslant \exp \left({\tfrac {1}{8}}s^{2}(b-a)^{2}\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/921c89dd0e9fa5555dc1d1af6a493004c3c108f2)