[ modifica ] Una funció
g
:
R
→
R
{\displaystyle g:\mathbb {R} \to \mathbb {R} }
absolutament contínua si per a qualsevol
ε
>
0
existeix
δ
>
0
{\displaystyle \varepsilon >0\quad {\text{existeix}}\quad \delta >0}
a
1
<
b
1
<
a
2
<
b
2
<
⋯
<
a
n
<
b
n
{\displaystyle a_{1}<b_{1}<a_{2}<b_{2}<\cdots <a_{n}<b_{n}}
∑
i
=
1
n
(
b
i
−
a
i
)
<
δ
{\displaystyle \sum _{i=1}^{n}(b_{i}-a_{i})<\delta }
∑
i
=
1
n
|
g
(
b
i
)
−
g
(
a
i
)
|
<
ε
.
{\displaystyle \sum _{i=1}^{n}\vert g(b_{i})-g(a_{i})\vert <\varepsilon .}
g
:
[
a
,
b
]
→
R
{\displaystyle g:[a,b]\to \mathbb {R} }
a
<
a
1
<
b
1
<
a
2
<
b
2
<
⋯
<
a
n
<
b
n
<
b
{\displaystyle a<a_{1}<b_{1}<a_{2}<b_{2}<\cdots <a_{n}<b_{n}<b}
Evidentment, una funció absolutament contínua és contínua.
Pel seu especial interès en Teoria de probabilitat ens limitarem a les funcions de distribució absolutament contínues.
Per demostrar aquesta propietat necessitem recordar les notacions de la pàgina Funció de Cantor:
Per a cada nivell
n
≥
1
{\displaystyle n\geq 1}
E
1
n
,
E
2
n
…
,
E
2
n
−
1
n
{\displaystyle E_{1}^{n},E_{2}^{n}\dots ,E_{2^{n}-1}^{n}}
C
1
,
…
,
C
n
{\displaystyle C_{1},\dots ,C_{n}}
E
1
1
=
(
1
3
,
2
3
)
.
{\displaystyle E_{1}^{1}=({\tfrac {1}{3}},{\tfrac {2}{3}}).}
E
1
1
=
(
1
9
,
2
9
)
,
E
2
1
=
(
1
3
,
2
3
)
,
E
3
1
=
(
7
9
,
8
9
)
.
{\displaystyle E_{1}^{1}=({\tfrac {1}{9}},{\tfrac {2}{9}}),\,E_{2}^{1}=({\tfrac {1}{3}},{\tfrac {2}{3}}),\,E_{3}^{1}=({\tfrac {7}{9}},{\tfrac {8}{9}}).}
Vegeu la Figura 1.
Figura 1. Intervals
I
k
n
{\displaystyle I_{k}^{n}}
E
k
n
{\displaystyle E_{k}^{n}}
n
=
1
,
2
,
3.
{\displaystyle n=1,2,3.}
Volem veure que
P
(
X
∈
C
)
=
1
{\displaystyle P(X\in C)=1}
P
(
R
∖
C
)
=
0
{\displaystyle P(\mathbb {R} \backslash C)=0}
R
∖
C
=
(
∞
,
0
)
∪
(
1
,
∞
)
∪
⋃
n
=
1
∞
⋃
k
=
1
2
n
−
1
E
k
n
.
{\displaystyle \mathbb {R} \setminus C=(\infty ,0)\cup (1,\infty )\cup \bigcup _{n=1}^{\infty }\bigcup _{k=1}^{2^{n}-1}E_{k}^{n}.}
F
{\displaystyle F}
(
a
,
b
)
{\displaystyle (a,b)}
P
(
X
∈
(
a
,
b
)
)
=
F
(
b
)
−
F
(
a
)
=
0.
{\displaystyle P{\big (}X\in (a,b){\big )}=F(b)-F(a)=0.}
D'on resulta
P
(
R
∖
C
)
=
0
{\displaystyle P(\mathbb {R} \backslash C)=0}
Per veure que
C
{\displaystyle C}
suport tancat de la distribució de Cantor s'ha de provar que per qualsevol entorn
U
{\displaystyle U}
C
{\displaystyle C}
P
(
X
∈
U
)
>
0
{\displaystyle P(X\in U)>0}
Propietats topològiques del conjunt de Cantor) si
x
∈
C
{\displaystyle x\in C}
x
≠
0
,
1
{\displaystyle x\neq 0,1}
U
{\displaystyle U}
x
{\displaystyle x}
x
′
,
x
″
∈
C
{\displaystyle x',x''\in C}
x
′
<
x
<
x
″
{\displaystyle x'<x<x''}
(
x
′
,
x
″
)
⊂
C
{\displaystyle (x',x'')\subset C}
F
{\displaystyle F}
F
(
x
′
)
<
F
(
x
″
)
{\displaystyle F(x')<F(x'')}
Els casos
x
=
0
,
1
{\displaystyle x=0,1}
Volem veure que
P
(
X
∈
I
k
n
)
=
1
2
n
,
k
=
1
,
…
,
2
n
.
{\displaystyle P(X\in I_{k}^{n})={\frac {1}{2^{n}}},\ k=1,\dots ,2^{n}.}
I
k
n
{\displaystyle I_{k}^{n}}
1
/
3
n
{\displaystyle 1/3^{n}}
1
3
n
=
(
0.
0
…
0
⏟
n
−
1
1
)
3
=
(
0.
0
…
0
⏟
n
2
¯
)
3
.
{\displaystyle {\frac {1}{3^{n}}}=(0.\underbrace {0\dots 0} _{n-1}1)_{3}=(0.\underbrace {0\dots 0} _{n}{\overline {2}})_{3}.}
Examinant com són aquest intervals es veu que tenen la forma
I
k
n
=
[
(
0.
b
1
…
b
n
)
3
,
0.
b
1
…
b
n
2
¯
3
]
,
amb
b
1
,
…
,
b
n
∈
{
0
,
2
}
.
{\displaystyle I_{k}^{n}={\big [}(0.b_{1}\dots b_{n})_{3},\,0.b_{1}\dots b_{n}{\overline {2}}_{3}{\big ]},\,{\text{amb}}\ b_{1},\dots ,b_{n}\in \{0,2\}.}
Llavors (vegeu la definició de la funció de Cantor) ,
P
(
X
∈
I
k
n
)
=
F
(
0.
b
1
⋯
b
n
2
¯
3
)
−
F
(
(
0.
b
1
⋯
b
n
)
3
)
=
∑
j
=
n
+
1
∞
2
2
j
+
1
=
1
2
n
.
{\displaystyle P{\big (}X\in I_{k}^{n}{\big )}=F(0.b_{1}\cdots b_{n}{\overline {2}}_{3})-F((0.b_{1}\cdots b_{n})_{3})=\sum _{j=n+1}^{\infty }{\frac {2}{2^{j+1}}}={\frac {1}{2^{n}}}.}
Volem demostrar que la variable aleatòria
S
=
1
−
X
{\displaystyle S=1-X}
H
{\displaystyle H}
S
{\displaystyle S}
H
=
F
{\displaystyle H=F}
F
{\displaystyle F}
G
(
x
)
=
P
(
Y
≤
x
)
=
P
(
1
−
X
≤
x
)
=
P
(
1
−
x
≤
X
)
=
1
−
P
(
X
<
1
−
x
)
=
1
−
F
(
1
−
x
)
=
F
(
x
)
,
{\displaystyle G(x)=P(Y\leq x)=P(1-X\leq x)=P(1-x\leq X)=1-P(X<1-x)=1-F(1-x)=F(x),}
funció de Cantor .
Designem la funció de distribució de
Y
{\displaystyle Y}
G
{\displaystyle G}
G
=
F
{\displaystyle G=F}
t
∈
[
0
,
1
]
{\displaystyle t\in [0,1]}
teorema de les probabilitats totals ,
G
(
t
)
=
P
(
Y
≤
t
)
=
P
(
Z
=
1
)
P
(
Y
≤
t
|
Z
=
1
)
+
P
(
Z
=
0
)
P
(
Y
≤
t
|
Z
=
0
)
=
1
2
P
(
X
1
3
≤
t
)
+
1
2
P
(
X
2
3
+
2
3
≤
t
)
=
1
2
P
(
X
1
≤
3
t
)
+
1
2
P
(
X
2
≤
3
t
−
2
)
=
1
2
F
(
3
t
)
+
1
2
F
(
3
t
−
2
)
=
F
(
t
)
,
{\displaystyle {\begin{aligned}G(t)=P(Y\leq t)&=P(Z=1)\,P(Y\leq t\,|\,Z=1)+P(Z=0)\,P(Y\leq t\,|\,Z=0)\\&={\frac {1}{2}}\,P{\Big (}{\frac {X_{1}}{3}}\leq t{\Big )}+{\frac {1}{2}}\,P{\Big (}{\frac {X_{2}}{3}}+{\frac {2}{3}}\leq t{\Big )}={\frac {1}{2}}\,P(X_{1}\leq 3t)+{\frac {1}{2}}\,P(X_{2}\leq 3t-2)\\&={\frac {1}{2}}\,F(3t)+{\frac {1}{2}}\,F(3t-2)=F(t),\end{aligned}}}
on hem aplicat la propietat 7 de la funció de Cantor .
Del fet que
X
{\displaystyle X}
1
−
X
{\displaystyle 1-X}
E
[
X
]
=
E
[
1
−
X
]
,
{\displaystyle E[X]=E[1-X],}
E
[
X
]
=
1
2
.
{\displaystyle E[X]={\frac {1}{2}}.}
[ modifica ] De la propietat d'autosemblança tenim
E
[
X
2
]
=
E
[
(
Z
X
1
3
+
(
1
−
Z
)
(
X
2
3
+
2
3
)
)
2
]
=
1
9
E
[
Z
2
]
E
[
X
2
]
+
1
9
E
[
Z
2
]
E
[
(
X
+
2
)
2
]
=
1
18
E
[
X
2
]
+
1
18
(
E
[
X
2
]
+
4
E
[
X
]
+
4
)
=
1
18
E
[
X
2
]
+
1
18
E
[
X
2
]
+
1
3
,
{\displaystyle {\begin{array}{rl}E[X^{2}]&=E{\bigg [}{\bigg (}Z\,{\dfrac {X_{1}}{3}}+(1-Z)\,{\Big (}{\dfrac {X_{2}}{3}}+{\dfrac {2}{3}}{\Big )}{\bigg )}^{2}{\bigg ]}={\dfrac {1}{9}}\,E[Z^{2}]E[X^{2}]+{\dfrac {1}{9}}\,E[Z^{2}]\,E[(X+2)^{2}]\\&={\dfrac {1}{18}}\,E[X^{2}]+{\dfrac {1}{18}}\,{\Big (}E[X^{2}]+4\,E[X]+4{\Big )}={\dfrac {1}{18}}\,E[X^{2}]+{\frac {1}{18}}\ E[X^{2}]+{\dfrac {1}{3}},\end{array}}}
E
[
X
2
]
=
3
8
.
{\displaystyle E[X^{2}]={\frac {3}{8}}.}
var
(
X
)
=
E
[
X
2
]
−
(
E
[
X
]
)
2
=
1
8
.
{\displaystyle \operatorname {var} (X)=E[X^{2}]-{\big (}E[X]{\big )}^{2}={\frac {1}{8}}.}
[ modifica ] Escrivim
m
n
=
E
[
X
n
]
.
{\displaystyle m_{n}=E[X^{n}].}
m
n
+
1
=
E
[
(
Z
X
1
3
+
(
1
−
Z
)
(
X
2
3
+
2
3
)
)
n
+
1
]
=
1
3
n
+
1
E
[
Z
n
+
1
X
1
n
+
1
+
(
1
−
Z
)
n
+
1
(
X
2
+
2
)
n
+
1
]
=
1
2
3
n
+
1
(
m
n
+
1
+
∑
j
=
0
n
+
1
(
n
+
1
j
)
m
j
2
n
+
1
−
j
)
{\displaystyle {\begin{aligned}m_{n+1}&=E{\Big [}{\Big (}Z\,{\frac {X_{1}}{3}}+(1-Z)\,{\Big (}{\frac {X_{2}}{3}}+{\frac {2}{3}}{\Big )}{\Big )}^{n+1}{\Big ]}\\&={\frac {1}{3^{n+1}}}E{\big [}Z^{n+1}\,X_{1}^{n+1}+(1-Z)^{n+1}{\big (}X_{2}+2{\big )}^{n+1}{\big ]}\\&={\frac {1}{2\,3^{n+1}}}{\bigg (}m_{n+1}+\sum _{j=0}^{n+1}{\binom {n+1}{j}}m_{j}\,2^{n+1-j}{\bigg )}\end{aligned}}}
Per la versió per esperances de la fórmula de les probabilitats totals,
φ
(
t
)
=
E
[
e
i
t
X
]
=
P
(
Z
=
1
)
E
[
exp
{
i
t
(
Z
X
1
/
3
+
(
1
−
Z
)
(
X
2
/
3
+
2
/
3
)
)
}
|
Z
=
1
]
+
P
(
Z
=
0
)
E
[
exp
{
i
t
(
Z
X
1
/
3
+
(
1
−
Z
)
(
X
2
/
3
+
2
/
3
)
)
}
|
Z
=
0
]
=
1
2
E
[
e
i
t
X
1
/
3
]
+
1
2
E
[
e
i
t
(
X
2
/
3
+
2
/
3
)
]
=
1
2
φ
(
t
/
3
)
+
1
2
e
2
i
t
/
3
φ
(
t
/
3
)
=
1
2
φ
(
t
/
3
)
(
1
+
e
2
i
t
/
3
)
.
{\displaystyle {\begin{aligned}\varphi (t)&=E{\big [}e^{itX}{\big ]}\\&=P(Z=1)\,E{\bigg [}\exp {\Big \{}it{\Big (}Z\,X_{1}/3+(1-Z)\,{\big (}X_{2}/3+2/3{\big )}{\Big )}{\Big \}}\,\vert \,Z=1{\bigg ]}\\&+P(Z=0)\,E{\bigg [}\exp {\Big \{}it{\Big (}Z\,X_{1}/3+(1-Z)\,{\big (}X_{2}/3+2/3{\big )}{\Big )}{\Big \}}\,\vert \,Z=0{\bigg ]}\\&={\frac {1}{2}}\,E{\big [}e^{itX_{1}/3}{\big ]}+{\frac {1}{2}}\,E{\big [}e^{it(X_{2}/3+2/3)}{\big ]}={\frac {1}{2}}\varphi (t/3)+{\frac {1}{2}}e^{2it/3}\varphi (t/3)\\&={\frac {1}{2}}\varphi (t/3){\big (}1+e^{2it/3}{\big )}.\end{aligned}}}
e
i
a
=
cos
a
+
i
sin
a
,
{\displaystyle e^{ia}=\cos a+i\,\sin a,}
cos
a
=
1
2
(
e
−
i
a
+
e
i
a
)
=
1
2
e
−
i
a
(
1
+
e
2
i
a
)
.
{\displaystyle \cos a={\frac {1}{2}}{\big (}e^{-ia}+e^{ia}{\big )}={\frac {1}{2}}e^{-ia}{\big (}1+e^{2ia}{\big )}.}
a
=
t
/
3
{\displaystyle a=t/3}
φ
(
t
)
=
φ
(
t
/
3
)
e
i
t
/
3
cos
(
t
/
3
)
.
{\displaystyle \varphi (t)=\varphi (t/3)\,e^{it/3}\cos(t/3).}
φ
(
t
)
=
φ
(
t
/
3
n
)
exp
{
i
t
∑
j
=
1
n
1
3
n
}
∏
j
=
1
n
cos
(
t
/
3
n
)
.
{\displaystyle \varphi (t)=\varphi (t/3^{n})\,\exp {\Big \{}it\sum _{j=1}^{n}{\frac {1}{3^{n}}}{\Big \}}\,\prod _{j=1}^{n}\cos(t/3^{n}).}
n
→
∞
{\displaystyle n\to \infty }
φ
(
t
)
=
e
i
t
/
2
∏
n
=
1
∞
cos
(
t
/
3
n
)
.
{\displaystyle \varphi (t)=e^{it/2}\prod _{n=1}^{\infty }\cos(t/3^{n}).}
![{\displaystyle g:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b1d5103d0e36767a27715bcfc57137119294aad)
![{\displaystyle I_{k}^{n}={\big [}(0.b_{1}\dots b_{n})_{3},\,0.b_{1}\dots b_{n}{\overline {2}}_{3}{\big ]},\,{\text{amb}}\ b_{1},\dots ,b_{n}\in \{0,2\}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/034ff9beb7a5c8312ee51aef161b7bafdb82653f)
![{\displaystyle t\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/31a5c18739ff04858eecc8fec2f53912c348e0e5)
![{\displaystyle E[X]=E[1-X],}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c12af1fc795d9c7aea8d2d1d7bf3c648a372fce7)
![{\displaystyle E[X]={\frac {1}{2}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f07d34ee0189bfd85f49e784543c99df37566bad)
![{\displaystyle {\begin{array}{rl}E[X^{2}]&=E{\bigg [}{\bigg (}Z\,{\dfrac {X_{1}}{3}}+(1-Z)\,{\Big (}{\dfrac {X_{2}}{3}}+{\dfrac {2}{3}}{\Big )}{\bigg )}^{2}{\bigg ]}={\dfrac {1}{9}}\,E[Z^{2}]E[X^{2}]+{\dfrac {1}{9}}\,E[Z^{2}]\,E[(X+2)^{2}]\\&={\dfrac {1}{18}}\,E[X^{2}]+{\dfrac {1}{18}}\,{\Big (}E[X^{2}]+4\,E[X]+4{\Big )}={\dfrac {1}{18}}\,E[X^{2}]+{\frac {1}{18}}\ E[X^{2}]+{\dfrac {1}{3}},\end{array}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c57db9d4a00625da3dfd9591256f789d3b81629e)
![{\displaystyle E[X^{2}]={\frac {3}{8}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/393f0f45458b2abb3722d7f3784cdc1b556316ea)
![{\displaystyle \operatorname {var} (X)=E[X^{2}]-{\big (}E[X]{\big )}^{2}={\frac {1}{8}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bfbb60e5c39ac0136c3528114146c4599d2dce65)
![{\displaystyle m_{n}=E[X^{n}].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0cefb032ac0ce18d81d13eee4171b9df8b27fa88)
![{\displaystyle {\begin{aligned}m_{n+1}&=E{\Big [}{\Big (}Z\,{\frac {X_{1}}{3}}+(1-Z)\,{\Big (}{\frac {X_{2}}{3}}+{\frac {2}{3}}{\Big )}{\Big )}^{n+1}{\Big ]}\\&={\frac {1}{3^{n+1}}}E{\big [}Z^{n+1}\,X_{1}^{n+1}+(1-Z)^{n+1}{\big (}X_{2}+2{\big )}^{n+1}{\big ]}\\&={\frac {1}{2\,3^{n+1}}}{\bigg (}m_{n+1}+\sum _{j=0}^{n+1}{\binom {n+1}{j}}m_{j}\,2^{n+1-j}{\bigg )}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bbb9db2e2df6416c4ab7ea3f40e651646bee8aee)
![{\displaystyle {\begin{aligned}\varphi (t)&=E{\big [}e^{itX}{\big ]}\\&=P(Z=1)\,E{\bigg [}\exp {\Big \{}it{\Big (}Z\,X_{1}/3+(1-Z)\,{\big (}X_{2}/3+2/3{\big )}{\Big )}{\Big \}}\,\vert \,Z=1{\bigg ]}\\&+P(Z=0)\,E{\bigg [}\exp {\Big \{}it{\Big (}Z\,X_{1}/3+(1-Z)\,{\big (}X_{2}/3+2/3{\big )}{\Big )}{\Big \}}\,\vert \,Z=0{\bigg ]}\\&={\frac {1}{2}}\,E{\big [}e^{itX_{1}/3}{\big ]}+{\frac {1}{2}}\,E{\big [}e^{it(X_{2}/3+2/3)}{\big ]}={\frac {1}{2}}\varphi (t/3)+{\frac {1}{2}}e^{2it/3}\varphi (t/3)\\&={\frac {1}{2}}\varphi (t/3){\big (}1+e^{2it/3}{\big )}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7529e09a9b7c6d4f43a313d0085a62f0cd746557)
