Stochastic version of the Radon-Nikodým theorem

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Let

• $(Omega,mathcal A,operatorname P)$ be a probability space


• $T>0$


• $I:=(0,T]$


• $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$


• $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$


• $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$


• $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$

Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.

Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?

probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym

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asked Aug 5 at 22:46

0xbadf00d

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up vote
1
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Let

• $(Omega,mathcal A,operatorname P)$ be a probability space


• $T>0$


• $I:=(0,T]$


• $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$


• $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$


• $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$


• $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$

Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.

Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?

probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym

share|cite|improve this question

asked Aug 5 at 22:46

0xbadf00d

2,04141028

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Let

• $(Omega,mathcal A,operatorname P)$ be a probability space


• $T>0$


• $I:=(0,T]$


• $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$


• $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$


• $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$


• $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$

Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.

Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?

probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym

share|cite|improve this question

asked Aug 5 at 22:46

0xbadf00d

2,04141028

Let

• $(Omega,mathcal A,operatorname P)$ be a probability space


• $T>0$


• $I:=(0,T]$


• $(mathcal F_t)_tinoverline I$ be a filtration on $(Omega,mathcal A)$


• $mathcal P$ denote the predictable $sigma$-algebra on $Omegatimesoverline I$


• $A:Omegatimesoverline Itomathbb R$ be $mathcal Aotimesmathcal B(overline I)$-measurable such that $A_0(omega)=0$ and $A(omega)$ is continuous and of bounded variation for all $omegainOmega$ and $$operatorname Eleft[|A|_Tright]<inftytag1,$$ where $|A(omega)|$ denotes the total variation function of $A(omega)$ for $omegainOmega$


• $B:Omegatimesoverline Itomathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(omega)$ is nondecreasing for all $omegainOmega$

Assume $$int_(0,:t]xi:rm dA=0;;;textalmost surely for all tin Itag1$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$ with $$int_(0,:t]xi:rm dB=0;;;textalmost surely for all tin I.tag2$$

Now, let $$mu(P):=operatorname Eleft[int1_P:rm dAright]$$ and $$nu(P):=operatorname Eleft[int1_P:rm dBright]$$ for $Pinmathcal P$. By assumption, $mu$ is absolutely continuous with respect to $nu$ and hence the Radon-Nikodým derivative $$alpha:=fracrm dmurm dnu$$ is well-defined and $mathcal P$-measurable with $$operatorname Eleft[int|alpha|:rm dBright]<inftytag3$$ and $$operatorname Eleft[intxi:rm dBright]=intxi:rm dmu=intalphaxi:rm dnutag4$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$.

Let $$M:=A-intalpha:rm dB$$ and note that $$operatorname Eleft[intxi:rm dMright]=0tag5$$ for all bounded nonnegative $mathcal P$-measurable $xi:Omegatimesoverline Itomathbb R$. Why does $(5)$ yield that $M$ is an $mathcal F$-martingale?

probability-theory stochastic-processes martingales stochastic-analysis radon-nikodym

share|cite|improve this question

asked Aug 5 at 22:46

0xbadf00d

2,04141028

share|cite|improve this question

share|cite|improve this question

asked Aug 5 at 22:46

0xbadf00d

2,04141028

asked Aug 5 at 22:46

0xbadf00d

2,04141028

asked Aug 5 at 22:46

0xbadf00d

2,04141028

2,04141028

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

Fix $s leq t$. For fixed $F in mathcalF_s$ the process

$$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$

is bounded, non-negative and progressively measurable. By assumption, this implies that

$$mathbbE left( int xi , dM right)=0$$

which means that

$$mathbbE(1_F (M_t-M_s)) =0.$$

Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that

$$mathbbE(M_t-M_s mid mathcalF_s)=0,$$

and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
  – 0xbadf00d
  Aug 8 at 20:27
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Fix $s leq t$. For fixed $F in mathcalF_s$ the process

$$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$

is bounded, non-negative and progressively measurable. By assumption, this implies that

$$mathbbE left( int xi , dM right)=0$$

which means that

$$mathbbE(1_F (M_t-M_s)) =0.$$

Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that

$$mathbbE(M_t-M_s mid mathcalF_s)=0,$$

and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
  – 0xbadf00d
  Aug 8 at 20:27
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Fix $s leq t$. For fixed $F in mathcalF_s$ the process

$$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$

is bounded, non-negative and progressively measurable. By assumption, this implies that

$$mathbbE left( int xi , dM right)=0$$

which means that

$$mathbbE(1_F (M_t-M_s)) =0.$$

Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that

$$mathbbE(M_t-M_s mid mathcalF_s)=0,$$

and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
  – 0xbadf00d
  Aug 8 at 20:27
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Fix $s leq t$. For fixed $F in mathcalF_s$ the process

$$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$

is bounded, non-negative and progressively measurable. By assumption, this implies that

$$mathbbE left( int xi , dM right)=0$$

which means that

$$mathbbE(1_F (M_t-M_s)) =0.$$

Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that

$$mathbbE(M_t-M_s mid mathcalF_s)=0,$$

and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

Fix $s leq t$. For fixed $F in mathcalF_s$ the process

$$xi(u,omega) := 1_F(omega) 1_(s,t](u)$$

is bounded, non-negative and progressively measurable. By assumption, this implies that

$$mathbbE left( int xi , dM right)=0$$

which means that

$$mathbbE(1_F (M_t-M_s)) =0.$$

Since $F in mathcalF_s$ is arbitrary, this is equivalent to saying that

$$mathbbE(M_t-M_s mid mathcalF_s)=0,$$

and so $(M_t,mathcalF_t)_t geq 0$ is a martingale.

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

share|cite|improve this answer

share|cite|improve this answer

answered Aug 7 at 17:56

saz

73k552112

answered Aug 7 at 17:56

saz

73k552112

answered Aug 7 at 17:56

saz

73k552112

73k552112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
  – 0xbadf00d
  Aug 8 at 20:27
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
  – 0xbadf00d
  Aug 8 at 20:27
  

  
  
  
  

Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
– 0xbadf00d
Aug 8 at 20:27

Oops, that was trivial. Thank you for answer, anyway. I've got a related question. Maybe you can help there too.
– 0xbadf00d
Aug 8 at 20:27

add a comment | 

 

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