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Is the cross variation (of stochastic processes) bilinear?
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I have not been able to find whether for three stochastic processes (adapted with respect to one constant filtration): $(X_t)_tgeq 0$, $(Y_t)_tgeq 0$ and $(Z_t)_tgeq 0$ we have the following equality:
$langle X+Y, Zrangle_t= langle X,Zrangle_t+langle Y,Zrangle_t$
I ask this because I saw a statement along the lines:
"If $langle L,Nrangle =langle L',Nrangle$ for all $N$ (all three processes are continuous second order martingale), then $langle L-L', L-L'rangle=0$."
This lead me to the more general question, but I assume that if it is true then it's a famous result, and if not there are several classical counterexamples. I would appreciate any hints or answer.
stochastic-processes stochastic-calculus
asked Aug 6 at 13:35
Keen-ameteur
644213
add a comment |Â
up vote
-1
down vote
favorite
I have not been able to find whether for three stochastic processes (adapted with respect to one constant filtration): $(X_t)_tgeq 0$, $(Y_t)_tgeq 0$ and $(Z_t)_tgeq 0$ we have the following equality:
$langle X+Y, Zrangle_t= langle X,Zrangle_t+langle Y,Zrangle_t$
I ask this because I saw a statement along the lines:
"If $langle L,Nrangle =langle L',Nrangle$ for all $N$ (all three processes are continuous second order martingale), then $langle L-L', L-L'rangle=0$."
This lead me to the more general question, but I assume that if it is true then it's a famous result, and if not there are several classical counterexamples. I would appreciate any hints or answer.
stochastic-processes stochastic-calculus
asked Aug 6 at 13:35
Keen-ameteur
644213
3
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
1
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I have not been able to find whether for three stochastic processes (adapted with respect to one constant filtration): $(X_t)_tgeq 0$, $(Y_t)_tgeq 0$ and $(Z_t)_tgeq 0$ we have the following equality:
$langle X+Y, Zrangle_t= langle X,Zrangle_t+langle Y,Zrangle_t$
I ask this because I saw a statement along the lines:
"If $langle L,Nrangle =langle L',Nrangle$ for all $N$ (all three processes are continuous second order martingale), then $langle L-L', L-L'rangle=0$."
This lead me to the more general question, but I assume that if it is true then it's a famous result, and if not there are several classical counterexamples. I would appreciate any hints or answer.
stochastic-processes stochastic-calculus
asked Aug 6 at 13:35
Keen-ameteur
644213
I have not been able to find whether for three stochastic processes (adapted with respect to one constant filtration): $(X_t)_tgeq 0$, $(Y_t)_tgeq 0$ and $(Z_t)_tgeq 0$ we have the following equality:
$langle X+Y, Zrangle_t= langle X,Zrangle_t+langle Y,Zrangle_t$
I ask this because I saw a statement along the lines:
"If $langle L,Nrangle =langle L',Nrangle$ for all $N$ (all three processes are continuous second order martingale), then $langle L-L', L-L'rangle=0$."
This lead me to the more general question, but I assume that if it is true then it's a famous result, and if not there are several classical counterexamples. I would appreciate any hints or answer.
stochastic-processes stochastic-calculus
asked Aug 6 at 13:35
Keen-ameteur
644213
asked Aug 6 at 13:35
Keen-ameteur
644213
asked Aug 6 at 13:35
Keen-ameteur
644213
asked Aug 6 at 13:35
Keen-ameteur
644213
644213
3
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
1
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43
add a comment |Â
3
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
1
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43
3
3
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
1
1
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
answered Aug 6 at 13:44
James Yang
4349
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
answered Aug 6 at 13:44
James Yang
4349
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
add a comment |Â
up vote
3
down vote
accepted
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
answered Aug 6 at 13:44
James Yang
4349
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
answered Aug 6 at 13:44
James Yang
4349
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
answered Aug 6 at 13:44
James Yang
4349
answered Aug 6 at 13:44
James Yang
4349
answered Aug 6 at 13:44
James Yang
4349
answered Aug 6 at 13:44
James Yang
4349
4349
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
add a comment |Â
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
The definition I know for cross variation is of $langle X,Yrangle= frac14 Big( langle X+Y rangle - langle X+Y rangle Big)$ where $langle U rangle$ is the quadratic variation process given by the Doob-Meyer decomposition theorem. Does it follow from the decomposition theorem?
– Keen-ameteur
Aug 6 at 13:46
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
This is a consequence of what I give. The point is that $XY = frac14 ((X+Y)^2 - (X-Y)^2)$.
– James Yang
Aug 6 at 13:51
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
Thanks, I didn't know about this equivalent definition.
– Keen-ameteur
Aug 6 at 14:07
add a comment |Â
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3
This is true. By definition, $langle X+Y, Zrangle$ is the unique process such that $(X+Y)Z - langle X+Y, Zrangle$ is a martingale. It is easy to see that $(X+Y)Z - (langle X,Zrangle+ langle Y,Zrangle)$ is a martingale because $XZ-langle X,Zrangle$ and $YZ-langle Y,Zrangle$ are martingales and sum of martingales is still a martingale.
– James Yang
Aug 6 at 13:42
1
@JamesYang You should probably post that comment as an answer.
– Rhys Steele
Aug 6 at 13:43