Mixing
Levy upward in $L^2$
$begingroup$
Let $(F_n)_n$ be a filtration and $F=sigma(cup_n F_n)$.
Let $Z$ be a random variable such that $mathbb{E}Z^2<infty$.
Let $X_n=mathbb{E}[Z|F_n]$.
Given:
If $X_nto Z$ a.s. then $||X_n-Z||_2to0$ iff $||X_n||_2to ||Z||_2$.
For a $pi$-system $Pi$ such that $sigma(Pi)subset F$, if we have $int_A Ydmathbb{P}=int_A Xdmathbb{P}$ for all $Ain Pi$ then
$$Y=mathbb{E}[X|sigma(Pi)].$$
How do we show that $||X_n-Z||_2to0$?
I know that
$$||X_n||_2=left(mathbb{E}X_n^2right)^{1/2}=left(mathbb{E}[mathbb{E}[Z|F_n]^2]right)^{1/2}.$$
By conditional Jensen I thought that then $||X_n||_2leqleft(mathbb{E}[mathbb{E}[Z^2|F_n]]right)^{1/2}=left(mathbb{E}[Z^2]right)^{1/2}=||Z||_2$.
Is this correct? How do I continue from here?
convergence lp-spaces martingales
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
$endgroup$
|
show 2 more comments
$begingroup$
Let $(F_n)_n$ be a filtration and $F=sigma(cup_n F_n)$.
Let $Z$ be a random variable such that $mathbb{E}Z^2<infty$.
Let $X_n=mathbb{E}[Z|F_n]$.
Given:
If $X_nto Z$ a.s. then $||X_n-Z||_2to0$ iff $||X_n||_2to ||Z||_2$.
For a $pi$-system $Pi$ such that $sigma(Pi)subset F$, if we have $int_A Ydmathbb{P}=int_A Xdmathbb{P}$ for all $Ain Pi$ then
$$Y=mathbb{E}[X|sigma(Pi)].$$
How do we show that $||X_n-Z||_2to0$?
I know that
$$||X_n||_2=left(mathbb{E}X_n^2right)^{1/2}=left(mathbb{E}[mathbb{E}[Z|F_n]^2]right)^{1/2}.$$
By conditional Jensen I thought that then $||X_n||_2leqleft(mathbb{E}[mathbb{E}[Z^2|F_n]]right)^{1/2}=left(mathbb{E}[Z^2]right)^{1/2}=||Z||_2$.
Is this correct? How do I continue from here?
convergence lp-spaces martingales
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
$endgroup$
1
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
1
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
1
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50
|
show 2 more comments
$begingroup$
Let $(F_n)_n$ be a filtration and $F=sigma(cup_n F_n)$.
Let $Z$ be a random variable such that $mathbb{E}Z^2<infty$.
Let $X_n=mathbb{E}[Z|F_n]$.
Given:
If $X_nto Z$ a.s. then $||X_n-Z||_2to0$ iff $||X_n||_2to ||Z||_2$.
For a $pi$-system $Pi$ such that $sigma(Pi)subset F$, if we have $int_A Ydmathbb{P}=int_A Xdmathbb{P}$ for all $Ain Pi$ then
$$Y=mathbb{E}[X|sigma(Pi)].$$
How do we show that $||X_n-Z||_2to0$?
I know that
$$||X_n||_2=left(mathbb{E}X_n^2right)^{1/2}=left(mathbb{E}[mathbb{E}[Z|F_n]^2]right)^{1/2}.$$
By conditional Jensen I thought that then $||X_n||_2leqleft(mathbb{E}[mathbb{E}[Z^2|F_n]]right)^{1/2}=left(mathbb{E}[Z^2]right)^{1/2}=||Z||_2$.
Is this correct? How do I continue from here?
convergence lp-spaces martingales
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
$endgroup$
Let $(F_n)_n$ be a filtration and $F=sigma(cup_n F_n)$.
Let $Z$ be a random variable such that $mathbb{E}Z^2<infty$.
Let $X_n=mathbb{E}[Z|F_n]$.
Given:
If $X_nto Z$ a.s. then $||X_n-Z||_2to0$ iff $||X_n||_2to ||Z||_2$.
For a $pi$-system $Pi$ such that $sigma(Pi)subset F$, if we have $int_A Ydmathbb{P}=int_A Xdmathbb{P}$ for all $Ain Pi$ then
$$Y=mathbb{E}[X|sigma(Pi)].$$
How do we show that $||X_n-Z||_2to0$?
I know that
$$||X_n||_2=left(mathbb{E}X_n^2right)^{1/2}=left(mathbb{E}[mathbb{E}[Z|F_n]^2]right)^{1/2}.$$
By conditional Jensen I thought that then $||X_n||_2leqleft(mathbb{E}[mathbb{E}[Z^2|F_n]]right)^{1/2}=left(mathbb{E}[Z^2]right)^{1/2}=||Z||_2$.
Is this correct? How do I continue from here?
convergence lp-spaces martingales
convergence lp-spaces martingales
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
asked Jan 28 at 20:59
Joachim DoyleJoachim Doyle
818
818
1
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
1
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
1
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50
|
show 2 more comments
1
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
1
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
1
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50
1
1
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
1
1
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
1
1
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50
|
show 2 more comments
0
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1
$begingroup$
What do you know about limit theorems for martingales? Note that $(X_n)_n$ is an $L^2$-bounded martingale, hence .....
$endgroup$
– saz
Jan 29 at 8:36
$begingroup$
@saz Then it converges in $L^2$?
$endgroup$
– Joachim Doyle
Jan 29 at 8:40
1
$begingroup$
Yes, it does.. so all what remains to do is to identity the limit, let's call it $X_{infty}$. To this end you have to use that $$ int_F Z , dmathbb{P} = int_F X_n , dmathbb{P}$$ for any $F in F_n$, $n in mathbb{N}$. You can let $n to infty$ on the right-hand side (why?) to obtain $$int_F Z , dmathbb{P} = int_F X_{infty} , dmathbb{P}$$ for all $F in F_n$, $n in mathbb{N}$. Conclude that $X_{infty}=Z$ a.s.
$endgroup$
– saz
Jan 29 at 8:46
$begingroup$
@saz Why can we let $ntoinfty$ in $int_F X_n dmathbb{P}$? Is it because of monotone convergence and the martingale property?
$endgroup$
– Joachim Doyle
Jan 29 at 9:26
1
$begingroup$
Note that $$int_F (X_n-X_{infty})^2 , dmathbb{P} leq int (X_n-X_{infty})^2 , dmathbb{P}.$$ Now use the fact that $Y_n to Y$ in $L^2$ implies $int Y_n , dmathbb{P} to int Y , dmathbb{P}$.
$endgroup$
– saz
Jan 29 at 9:50