Mixing
A confusion in the proof of-“A convex function over a Martingale is a Submartingale.”
$begingroup$
Suppose $left{ Z_{n},ngeq1right} $ is a (discrete) martingale, $f$ is a convex function. Now, we want to prove that ${f(Z_n),ngeq1}$ is a submartingale.
As it is stated in A convex function over a Martingale is a Submartingale --- Proof, the proof can be written as begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq & fleft(Eleft[left.Z_{n+1}right|Z_{1},ldots,Z_{n}right]right)\
& = & fleft(Z_{n}right).
end{eqnarray*}
My question is: why can we check
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq
& = & fleft(Z_{n}right)
end{eqnarray*}
instead of
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|f(Z_{1}),ldots,f(Z_{n})right] & geq
& = & fleft(Z_{n}right)?
end{eqnarray*}
Since now, ${f(Z_n),ngeq1}$ should be seen as a stochastic process.
probability-theory stochastic-processes martingales
asked Jan 20 at 1:56
whereamIwhereamI
327115
$endgroup$
|
show 1 more comment
$begingroup$
Suppose $left{ Z_{n},ngeq1right} $ is a (discrete) martingale, $f$ is a convex function. Now, we want to prove that ${f(Z_n),ngeq1}$ is a submartingale.
As it is stated in A convex function over a Martingale is a Submartingale --- Proof, the proof can be written as begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq & fleft(Eleft[left.Z_{n+1}right|Z_{1},ldots,Z_{n}right]right)\
& = & fleft(Z_{n}right).
end{eqnarray*}
My question is: why can we check
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq
& = & fleft(Z_{n}right)
end{eqnarray*}
instead of
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|f(Z_{1}),ldots,f(Z_{n})right] & geq
& = & fleft(Z_{n}right)?
end{eqnarray*}
Since now, ${f(Z_n),ngeq1}$ should be seen as a stochastic process.
probability-theory stochastic-processes martingales
asked Jan 20 at 1:56
whereamIwhereamI
327115
$endgroup$
$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45
|
show 1 more comment
$begingroup$
Suppose $left{ Z_{n},ngeq1right} $ is a (discrete) martingale, $f$ is a convex function. Now, we want to prove that ${f(Z_n),ngeq1}$ is a submartingale.
As it is stated in A convex function over a Martingale is a Submartingale --- Proof, the proof can be written as begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq & fleft(Eleft[left.Z_{n+1}right|Z_{1},ldots,Z_{n}right]right)\
& = & fleft(Z_{n}right).
end{eqnarray*}
My question is: why can we check
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq
& = & fleft(Z_{n}right)
end{eqnarray*}
instead of
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|f(Z_{1}),ldots,f(Z_{n})right] & geq
& = & fleft(Z_{n}right)?
end{eqnarray*}
Since now, ${f(Z_n),ngeq1}$ should be seen as a stochastic process.
probability-theory stochastic-processes martingales
asked Jan 20 at 1:56
whereamIwhereamI
327115
$endgroup$
Suppose $left{ Z_{n},ngeq1right} $ is a (discrete) martingale, $f$ is a convex function. Now, we want to prove that ${f(Z_n),ngeq1}$ is a submartingale.
As it is stated in A convex function over a Martingale is a Submartingale --- Proof, the proof can be written as begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq & fleft(Eleft[left.Z_{n+1}right|Z_{1},ldots,Z_{n}right]right)\
& = & fleft(Z_{n}right).
end{eqnarray*}
My question is: why can we check
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|Z_{1},ldots,Z_{n}right] & geq
& = & fleft(Z_{n}right)
end{eqnarray*}
instead of
begin{eqnarray*}
Eleft[left.fleft(Z_{n+1}right)right|f(Z_{1}),ldots,f(Z_{n})right] & geq
& = & fleft(Z_{n}right)?
end{eqnarray*}
Since now, ${f(Z_n),ngeq1}$ should be seen as a stochastic process.
probability-theory stochastic-processes martingales
probability-theory stochastic-processes martingales
asked Jan 20 at 1:56
whereamIwhereamI
327115
asked Jan 20 at 1:56
whereamIwhereamI
327115
asked Jan 20 at 1:56
whereamIwhereamI
327115
asked Jan 20 at 1:56
whereamIwhereamI
327115
asked Jan 20 at 1:56
whereamIwhereamI
327115
327115
$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45
|
show 1 more comment
$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45
$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The result should be stated as follows:
Suppose that ${Z_n}$ is a martingale w.r.t. ${mathcal{F}_n:=sigma(Z_1,ldots,Z_n)}$ and $f$ is a convex function. Then ${f(Z_n)}$ is a submartingale w.r.t. ${mathcal{F}_n}$.
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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$begingroup$
The result should be stated as follows:
Suppose that ${Z_n}$ is a martingale w.r.t. ${mathcal{F}_n:=sigma(Z_1,ldots,Z_n)}$ and $f$ is a convex function. Then ${f(Z_n)}$ is a submartingale w.r.t. ${mathcal{F}_n}$.
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
add a comment |
$begingroup$
The result should be stated as follows:
Suppose that ${Z_n}$ is a martingale w.r.t. ${mathcal{F}_n:=sigma(Z_1,ldots,Z_n)}$ and $f$ is a convex function. Then ${f(Z_n)}$ is a submartingale w.r.t. ${mathcal{F}_n}$.
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
$endgroup$
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
add a comment |
$begingroup$
The result should be stated as follows:
Suppose that ${Z_n}$ is a martingale w.r.t. ${mathcal{F}_n:=sigma(Z_1,ldots,Z_n)}$ and $f$ is a convex function. Then ${f(Z_n)}$ is a submartingale w.r.t. ${mathcal{F}_n}$.
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
$endgroup$
The result should be stated as follows:
Suppose that ${Z_n}$ is a martingale w.r.t. ${mathcal{F}_n:=sigma(Z_1,ldots,Z_n)}$ and $f$ is a convex function. Then ${f(Z_n)}$ is a submartingale w.r.t. ${mathcal{F}_n}$.
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
answered Jan 20 at 3:45
d.k.o.d.k.o.
10k629
10k629
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
add a comment |
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
$begingroup$
I agree with this statement. I was not sure about this since the book does not mention "w.r.t" a fixed sigma algebra. Thank you very much.
$endgroup$
– whereamI
Jan 20 at 3:54
add a comment |
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$begingroup$
There must be a filtration associated with the sequence ${Z_n}$. Is it the natural one?
$endgroup$
– d.k.o.
Jan 20 at 3:06
$begingroup$
@d.k.o. I don't think so. In the book "An introduction to Stochastic Differential Equations" by Evans, the martingale is defined by: Let $X_1,...,X_n$... be a sequence of real-valued random variables. If $X_k=E(X_j|X_1,...,X_k)$ for all $jgeq k$, we call ${X_i}_{i=1}^infty$ a discrete martingale.
$endgroup$
– whereamI
Jan 20 at 3:12
$begingroup$
Check this.
$endgroup$
– d.k.o.
Jan 20 at 3:35
$begingroup$
@d.k.o Do you mean the general definition? But why can't we use the basic definition here?
$endgroup$
– whereamI
Jan 20 at 3:41
$begingroup$
How do you show that $$ mathsf{E}[Z_{n+1}mid f(Z_1),ldots,f(Z_n)]=Z_n quadtext{a.s.?} $$
$endgroup$
– d.k.o.
Jan 20 at 3:45