Mixing
Uniform integrability of the following martingale
$begingroup$
Let $A_{n,k}in F$ where $1leq kleq 2^n$ such that
$A_{n,k}$ are disjoint for fixed $n$ and different $k$,
$A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$.
Let $F_n=sigma({A_{n,k}:1leq kleq 2^n})$ and let $mu$ a probability measure on $F$ such that $mu<<mathbb{P}$.
Define $$X_n(omega) = left{
begin{array}{lr}
frac{mu(A_{n,k})}{mathbb{P}(A_{n,k})} & : omega in (A_{n,k}),mathbb{P}(A_{n,k})neq 0\
0 & : text{else}.
end{array}
right.$$
Why is $X_n$ uniformly integrable and why does this imply that there exists an integrable $X_infty$ such that $mu(A)=int_A X_infty dmathbb{P}$?
I have shown that $X_n$ is a martingale and that for all $Ain F_n$ we have
$$mu(A)=int_A X_n dmathbb{P}.$$
By absolute continuity I know that for all $epsilon$ there is a $delta$ such that $mathbb{P}(A)<deltaimpliesmu(A)<epsilon$.
Then we want to show that
$$limlimits_{Ktoinfty}sup_nint_{{X_n>K}}X_ndmathbb{P}=0.$$
We know that ${X_n>K}=cup_{1leq kleq 2^n}{mu(A_{n,k})>Kmathbb{P}(A_{n,k})}$, how do I continue from here?
convergence martingales uniform-integrability
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $A_{n,k}in F$ where $1leq kleq 2^n$ such that
$A_{n,k}$ are disjoint for fixed $n$ and different $k$,
$A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$.
Let $F_n=sigma({A_{n,k}:1leq kleq 2^n})$ and let $mu$ a probability measure on $F$ such that $mu<<mathbb{P}$.
Define $$X_n(omega) = left{
begin{array}{lr}
frac{mu(A_{n,k})}{mathbb{P}(A_{n,k})} & : omega in (A_{n,k}),mathbb{P}(A_{n,k})neq 0\
0 & : text{else}.
end{array}
right.$$
Why is $X_n$ uniformly integrable and why does this imply that there exists an integrable $X_infty$ such that $mu(A)=int_A X_infty dmathbb{P}$?
I have shown that $X_n$ is a martingale and that for all $Ain F_n$ we have
$$mu(A)=int_A X_n dmathbb{P}.$$
By absolute continuity I know that for all $epsilon$ there is a $delta$ such that $mathbb{P}(A)<deltaimpliesmu(A)<epsilon$.
Then we want to show that
$$limlimits_{Ktoinfty}sup_nint_{{X_n>K}}X_ndmathbb{P}=0.$$
We know that ${X_n>K}=cup_{1leq kleq 2^n}{mu(A_{n,k})>Kmathbb{P}(A_{n,k})}$, how do I continue from here?
convergence martingales uniform-integrability
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
$endgroup$
add a comment |
$begingroup$
Let $A_{n,k}in F$ where $1leq kleq 2^n$ such that
$A_{n,k}$ are disjoint for fixed $n$ and different $k$,
$A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$.
Let $F_n=sigma({A_{n,k}:1leq kleq 2^n})$ and let $mu$ a probability measure on $F$ such that $mu<<mathbb{P}$.
Define $$X_n(omega) = left{
begin{array}{lr}
frac{mu(A_{n,k})}{mathbb{P}(A_{n,k})} & : omega in (A_{n,k}),mathbb{P}(A_{n,k})neq 0\
0 & : text{else}.
end{array}
right.$$
Why is $X_n$ uniformly integrable and why does this imply that there exists an integrable $X_infty$ such that $mu(A)=int_A X_infty dmathbb{P}$?
I have shown that $X_n$ is a martingale and that for all $Ain F_n$ we have
$$mu(A)=int_A X_n dmathbb{P}.$$
By absolute continuity I know that for all $epsilon$ there is a $delta$ such that $mathbb{P}(A)<deltaimpliesmu(A)<epsilon$.
Then we want to show that
$$limlimits_{Ktoinfty}sup_nint_{{X_n>K}}X_ndmathbb{P}=0.$$
We know that ${X_n>K}=cup_{1leq kleq 2^n}{mu(A_{n,k})>Kmathbb{P}(A_{n,k})}$, how do I continue from here?
convergence martingales uniform-integrability
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
$endgroup$
Let $A_{n,k}in F$ where $1leq kleq 2^n$ such that
$A_{n,k}$ are disjoint for fixed $n$ and different $k$,
$A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$.
Let $F_n=sigma({A_{n,k}:1leq kleq 2^n})$ and let $mu$ a probability measure on $F$ such that $mu<<mathbb{P}$.
Define $$X_n(omega) = left{
begin{array}{lr}
frac{mu(A_{n,k})}{mathbb{P}(A_{n,k})} & : omega in (A_{n,k}),mathbb{P}(A_{n,k})neq 0\
0 & : text{else}.
end{array}
right.$$
Why is $X_n$ uniformly integrable and why does this imply that there exists an integrable $X_infty$ such that $mu(A)=int_A X_infty dmathbb{P}$?
I have shown that $X_n$ is a martingale and that for all $Ain F_n$ we have
$$mu(A)=int_A X_n dmathbb{P}.$$
By absolute continuity I know that for all $epsilon$ there is a $delta$ such that $mathbb{P}(A)<deltaimpliesmu(A)<epsilon$.
Then we want to show that
$$limlimits_{Ktoinfty}sup_nint_{{X_n>K}}X_ndmathbb{P}=0.$$
We know that ${X_n>K}=cup_{1leq kleq 2^n}{mu(A_{n,k})>Kmathbb{P}(A_{n,k})}$, how do I continue from here?
convergence martingales uniform-integrability
convergence martingales uniform-integrability
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
asked Jan 29 at 15:30
Joachim DoyleJoachim Doyle
818
818
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $Xge 0$ such that
$$
mu(A) =int_A X mathrm{d}Bbb P,quad forall Ain F.
$$ We can find that for each $kle 2^n$
$$
X_n(omega) =frac{mu(A_{n,k})}{Bbb P(A_{n,k})}=frac{1}{Bbb P(A_{n,k})}int_{A_{n,k}} Xmathrm{d}Bbb Pquadtext{ on } A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{kle 2^n}$, this is equivalent to
$$
X_n =Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_nsubset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $Bbb E[|X|]=Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
int_{{X_n>M}}X_nmathrm{d}Bbb P = int_{{X_n>M}}Xmathrm{d}Bbb P.
$$ For given $epsilon>0$ there exists $delta>0$ such that $int_A Xmathrm{d}Bbb P<epsilon$ for all $Bbb P(A)<delta$. Since for all $M>Bbb E[X]/delta$,$$
Bbb P(X_n>M)le frac{Bbb EX_n}{M}le frac{Bbb EX}{M}<delta
$$ it follows $int_{{X_n>M}}X_nmathrm{d}Bbb P=int_{{X_n>M}}Xmathrm{d}Bbb P <epsilon$ for all $nge 1$.
answered Jan 29 at 15:50
SongSong
18.5k21651
$endgroup$
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $Xge 0$ such that
$$
mu(A) =int_A X mathrm{d}Bbb P,quad forall Ain F.
$$ We can find that for each $kle 2^n$
$$
X_n(omega) =frac{mu(A_{n,k})}{Bbb P(A_{n,k})}=frac{1}{Bbb P(A_{n,k})}int_{A_{n,k}} Xmathrm{d}Bbb Pquadtext{ on } A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{kle 2^n}$, this is equivalent to
$$
X_n =Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_nsubset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $Bbb E[|X|]=Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
int_{{X_n>M}}X_nmathrm{d}Bbb P = int_{{X_n>M}}Xmathrm{d}Bbb P.
$$ For given $epsilon>0$ there exists $delta>0$ such that $int_A Xmathrm{d}Bbb P<epsilon$ for all $Bbb P(A)<delta$. Since for all $M>Bbb E[X]/delta$,$$
Bbb P(X_n>M)le frac{Bbb EX_n}{M}le frac{Bbb EX}{M}<delta
$$ it follows $int_{{X_n>M}}X_nmathrm{d}Bbb P=int_{{X_n>M}}Xmathrm{d}Bbb P <epsilon$ for all $nge 1$.
answered Jan 29 at 15:50
SongSong
18.5k21651
$endgroup$
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
add a comment |
$begingroup$
I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $Xge 0$ such that
$$
mu(A) =int_A X mathrm{d}Bbb P,quad forall Ain F.
$$ We can find that for each $kle 2^n$
$$
X_n(omega) =frac{mu(A_{n,k})}{Bbb P(A_{n,k})}=frac{1}{Bbb P(A_{n,k})}int_{A_{n,k}} Xmathrm{d}Bbb Pquadtext{ on } A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{kle 2^n}$, this is equivalent to
$$
X_n =Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_nsubset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $Bbb E[|X|]=Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
int_{{X_n>M}}X_nmathrm{d}Bbb P = int_{{X_n>M}}Xmathrm{d}Bbb P.
$$ For given $epsilon>0$ there exists $delta>0$ such that $int_A Xmathrm{d}Bbb P<epsilon$ for all $Bbb P(A)<delta$. Since for all $M>Bbb E[X]/delta$,$$
Bbb P(X_n>M)le frac{Bbb EX_n}{M}le frac{Bbb EX}{M}<delta
$$ it follows $int_{{X_n>M}}X_nmathrm{d}Bbb P=int_{{X_n>M}}Xmathrm{d}Bbb P <epsilon$ for all $nge 1$.
answered Jan 29 at 15:50
SongSong
18.5k21651
$endgroup$
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
add a comment |
$begingroup$
I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $Xge 0$ such that
$$
mu(A) =int_A X mathrm{d}Bbb P,quad forall Ain F.
$$ We can find that for each $kle 2^n$
$$
X_n(omega) =frac{mu(A_{n,k})}{Bbb P(A_{n,k})}=frac{1}{Bbb P(A_{n,k})}int_{A_{n,k}} Xmathrm{d}Bbb Pquadtext{ on } A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{kle 2^n}$, this is equivalent to
$$
X_n =Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_nsubset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $Bbb E[|X|]=Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
int_{{X_n>M}}X_nmathrm{d}Bbb P = int_{{X_n>M}}Xmathrm{d}Bbb P.
$$ For given $epsilon>0$ there exists $delta>0$ such that $int_A Xmathrm{d}Bbb P<epsilon$ for all $Bbb P(A)<delta$. Since for all $M>Bbb E[X]/delta$,$$
Bbb P(X_n>M)le frac{Bbb EX_n}{M}le frac{Bbb EX}{M}<delta
$$ it follows $int_{{X_n>M}}X_nmathrm{d}Bbb P=int_{{X_n>M}}Xmathrm{d}Bbb P <epsilon$ for all $nge 1$.
answered Jan 29 at 15:50
SongSong
18.5k21651
$endgroup$
I think the formulation is unnecessarily complicated than it actually is. First note that by Radon-Nikodym theorem, there exists $Xge 0$ such that
$$
mu(A) =int_A X mathrm{d}Bbb P,quad forall Ain F.
$$ We can find that for each $kle 2^n$
$$
X_n(omega) =frac{mu(A_{n,k})}{Bbb P(A_{n,k})}=frac{1}{Bbb P(A_{n,k})}int_{A_{n,k}} Xmathrm{d}Bbb Pquadtext{ on } A_{n,k}.
$$ Since $F_n$ is generated by disjoint family $(A_{n,k})_{kle 2^n}$, this is equivalent to
$$
X_n =Bbb E[X|F_n].
$$ (Recall discrete version definition of conditional expectation.) We can also check that $(F_n)_n$ is a filtration, i.e. $F_nsubset F_{n+1}$ by the assumption $A_{n,k}=A_{n+1,2k-1}cup A_{n+1,2k}$. Since $X$ is integrable, i.e. $Bbb E[|X|]=Bbb E[X]=1$, this immediately shows that $(X_n,F_n)$ is a uniformly integrable martingale. To see uniform integrability, observe that
$$
int_{{X_n>M}}X_nmathrm{d}Bbb P = int_{{X_n>M}}Xmathrm{d}Bbb P.
$$ For given $epsilon>0$ there exists $delta>0$ such that $int_A Xmathrm{d}Bbb P<epsilon$ for all $Bbb P(A)<delta$. Since for all $M>Bbb E[X]/delta$,$$
Bbb P(X_n>M)le frac{Bbb EX_n}{M}le frac{Bbb EX}{M}<delta
$$ it follows $int_{{X_n>M}}X_nmathrm{d}Bbb P=int_{{X_n>M}}Xmathrm{d}Bbb P <epsilon$ for all $nge 1$.
answered Jan 29 at 15:50
SongSong
18.5k21651
edited Jan 29 at 15:56
answered Jan 29 at 15:50
SongSong
18.5k21651
answered Jan 29 at 15:50
SongSong
18.5k21651
answered Jan 29 at 15:50
SongSong
18.5k21651
18.5k21651
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
add a comment |
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
$begingroup$
Why does it follow from $X_n=frac{1}{mathbb{P}(A_{n,k})}int_{A_{n,k}}X dmathbb{P}$ on $A_{n,k}$ that $X_n=mathbb{E}[X|F_n]$?
$endgroup$
– Joachim Doyle
Jan 29 at 16:02
1
1
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
$begingroup$
@JoachimDoyle To see this, note that $F =sigma(A_{k})$ (dropped the index $n$) implies any $F$ measurable r.v. is of the form $sum_{kle 2^n} c_k 1_{A_k}$ for some $(c_k)_{kle 2^n}$. Hence $Bbb E[X|F]$ is also of this form, say $sum_k d_k 1_{A_k}$. It suffices to show $d_k = int_{A_k} Xmathrm{d}Bbb P /Bbb P(A_k)$ for each $k$. It follows from $$int_{A_k} Xmathrm{d}Bbb P=int_{A_k} Bbb E[X|F]mathrm{d}Bbb P =d_k Bbb P(A_k).$$
$endgroup$
– Song
Jan 29 at 16:09
add a comment |
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