Let
X
n
,
n
≥
0
{\displaystyle X_{n},n\geq 0}
X
=
{
1
,
2
}
{\displaystyle \mathrm {X} =\{1,2\}}
P
{\displaystyle P}
P
11
=
1
/
3
,
P
12
=
2
/
3
,
P
21
=
P
22
=
1
/
2
{\displaystyle P_{11}=1/3,P_{12}=2/3,P_{21}=P_{22}=1/2}
f
:
X
→
R
{\displaystyle f:\mathrm {X} \to \mathbb {R} }
f
(
1
)
=
1
{\displaystyle f(1)=1}
f
(
2
)
=
4
{\displaystyle f(2)=4}
g
:
X
→
R
{\displaystyle g:\mathrm {X} \to \mathbb {R} }
Y
n
=
f
(
X
n
)
−
f
(
X
0
)
−
∑
i
=
0
n
−
1
g
(
X
i
)
,
n
≥
1
,
{\displaystyle Y_{n}=f(X_{n})-f(X_{0})-\sum _{i=0}^{n-1}g(X_{i}),n\geq 1,}
is a martingale relative to the filtration
F
n
X
{\displaystyle {\mathcal {F}}_{n}^{X}}
X
n
{\displaystyle X_{n}}
Notice that since
f
,
g
{\displaystyle f,g}
Y
n
{\displaystyle Y_{n}}
F
n
{\displaystyle {\mathcal {F}}_{n}}
Y
n
{\displaystyle Y_{n}}
F
n
{\displaystyle {\mathcal {F}}_{n}}
n
{\displaystyle n}
Y
n
{\displaystyle Y_{n}}
L
1
{\displaystyle L^{1}}
Therefore, we only need to check the conditional martingale property, i.e. we want to show
Y
n
=
E
(
Y
n
+
1
|
F
n
{\displaystyle Y_{n}=E(Y_{n+1}|{\mathcal {F}}_{n}}
That is, we want
f
(
X
n
)
−
f
(
X
0
)
−
∑
i
=
1
n
−
1
g
(
X
i
)
=
E
[
f
(
X
n
+
1
)
−
f
(
X
0
)
−
∑
i
=
1
n
g
(
X
i
)
|
F
n
]
f
(
X
n
)
=
E
[
f
(
X
n
+
1
)
|
F
n
]
−
g
(
X
n
)
{\displaystyle {\begin{aligned}f(X_{n})-f(X_{0})-\sum _{i=1}^{n-1}g(X_{i})&=E[f(X_{n+1})-f(X_{0})-\sum _{i=1}^{n}g(X_{i})|{\mathcal {F}}_{n}]\\f(X_{n})&=E[f(X_{n+1})|{\mathcal {F}}_{n}]-g(X_{n})\end{aligned}}}
Therefore, if
Y
n
{\displaystyle Y_{n}}
g
(
X
n
)
=
E
[
f
(
X
n
+
1
)
|
F
n
]
−
f
(
X
n
)
{\displaystyle g(X_{n})=E[f(X_{n+1})|{\mathcal {F}}_{n}]-f(X_{n})}
Since
X
=
{
1
,
2
}
{\displaystyle \mathrm {X} =\{1,2\}}
g
(
1
)
=
E
[
f
(
X
n
+
1
)
|
X
n
=
1
]
−
f
(
1
)
=
(
1
⋅
1
/
3
+
4
⋅
2
/
3
)
−
1
=
2
{\displaystyle g(1)=E[f(X_{n+1})|X_{n}=1]-f(1)=(1\cdot 1/3+4\cdot 2/3)-1=2}
g
(
2
)
=
E
[
f
(
X
n
+
1
)
|
X
n
=
2
]
−
f
(
2
)
=
(
1
⋅
1
/
2
+
4
⋅
1
/
2
)
=
−
3
2
{\displaystyle g(2)=E[f(X_{n+1})|X_{n}=2]-f(2)=(1\cdot 1/2+4\cdot 1/2)=-{\frac {3}{2}}}
This explicitly defines the function
g
{\displaystyle g}
Y
n
{\displaystyle Y_{n}}
Let
ξ
n
{\displaystyle \xi _{n}}
α
>
0
{\displaystyle \alpha >0}
∑
n
=
1
∞
(
ξ
n
+
n
−
α
)
(
n
α
+
1
)
{\displaystyle \sum _{n=1}^{\infty }(\xi _{n}+n^{-\alpha })^{(n^{\alpha +1})}}
converge almost surely?
![{\displaystyle {\begin{aligned}f(X_{n})-f(X_{0})-\sum _{i=1}^{n-1}g(X_{i})&=E[f(X_{n+1})-f(X_{0})-\sum _{i=1}^{n}g(X_{i})|{\mathcal {F}}_{n}]\\f(X_{n})&=E[f(X_{n+1})|{\mathcal {F}}_{n}]-g(X_{n})\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/adb5970b78e698d7eb80b4081a2d2783ca687050)
![{\displaystyle g(X_{n})=E[f(X_{n+1})|{\mathcal {F}}_{n}]-f(X_{n})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/65c9854d1043a0bcd6be2ad358af779ee639998c)
![{\displaystyle g(1)=E[f(X_{n+1})|X_{n}=1]-f(1)=(1\cdot 1/3+4\cdot 2/3)-1=2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6be323eb6cec306fb6f0d0c8e494cabd365a4ed3)
![{\displaystyle g(2)=E[f(X_{n+1})|X_{n}=2]-f(2)=(1\cdot 1/2+4\cdot 1/2)=-{\frac {3}{2}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/558c5f4139b637b6a4b476c6ba09ac903dba5430)
