Uniformly integrable local martingale

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Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

probability-theory martingales uniform-integrability local-martingales

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asked Jul 23 at 12:16

Protawn

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Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

probability-theory martingales uniform-integrability local-martingales

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asked Jul 23 at 12:16

Protawn

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up vote
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down vote

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Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

probability-theory martingales uniform-integrability local-martingales

share|cite|improve this question

asked Jul 23 at 12:16

Protawn

32129

Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

probability-theory martingales uniform-integrability local-martingales

share|cite|improve this question

asked Jul 23 at 12:16

Protawn

32129

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 12:16

Protawn

32129

asked Jul 23 at 12:16

Protawn

32129

asked Jul 23 at 12:16

Protawn

32129

32129

add a comment | 

add a comment | 

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It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $mathbbR^3$ and define $X_t = |B_t|^-1$ for $t in (varepsilon, infty)$ where $0<varepsilon<1$. Let $mathcalF_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_1+t$ and the filtration $mathcalG_t = mathcalF_1+t$. We have that $Y_t$ is adapted to $mathcalG_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x mapsto |x|^-1$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $mathbbE[X_t^2] = t^-1$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t to 0$ almost surely as $t to infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = mathbbE[Y_infty mid mathcalG_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.

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edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $mathbbR^3$ and define $X_t = |B_t|^-1$ for $t in (varepsilon, infty)$ where $0<varepsilon<1$. Let $mathcalF_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_1+t$ and the filtration $mathcalG_t = mathcalF_1+t$. We have that $Y_t$ is adapted to $mathcalG_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x mapsto |x|^-1$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $mathbbE[X_t^2] = t^-1$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t to 0$ almost surely as $t to infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = mathbbE[Y_infty mid mathcalG_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.

share|cite|improve this answer

edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

5,5751828

add a comment | 

up vote
3
down vote



accepted

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $mathbbR^3$ and define $X_t = |B_t|^-1$ for $t in (varepsilon, infty)$ where $0<varepsilon<1$. Let $mathcalF_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_1+t$ and the filtration $mathcalG_t = mathcalF_1+t$. We have that $Y_t$ is adapted to $mathcalG_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x mapsto |x|^-1$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $mathbbE[X_t^2] = t^-1$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t to 0$ almost surely as $t to infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = mathbbE[Y_infty mid mathcalG_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.

share|cite|improve this answer

edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

5,5751828

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $mathbbR^3$ and define $X_t = |B_t|^-1$ for $t in (varepsilon, infty)$ where $0<varepsilon<1$. Let $mathcalF_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_1+t$ and the filtration $mathcalG_t = mathcalF_1+t$. We have that $Y_t$ is adapted to $mathcalG_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x mapsto |x|^-1$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $mathbbE[X_t^2] = t^-1$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t to 0$ almost surely as $t to infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = mathbbE[Y_infty mid mathcalG_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.

share|cite|improve this answer

edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

5,5751828

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $mathbbR^3$ and define $X_t = |B_t|^-1$ for $t in (varepsilon, infty)$ where $0<varepsilon<1$. Let $mathcalF_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_1+t$ and the filtration $mathcalG_t = mathcalF_1+t$. We have that $Y_t$ is adapted to $mathcalG_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x mapsto |x|^-1$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $mathbbE[X_t^2] = t^-1$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t to 0$ almost surely as $t to infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = mathbbE[Y_infty mid mathcalG_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.

share|cite|improve this answer

edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

5,5751828

share|cite|improve this answer

share|cite|improve this answer

edited Jul 23 at 12:55

edited Jul 23 at 12:55

edited Jul 23 at 12:55

answered Jul 23 at 12:46

Rhys Steele

5,5751828

answered Jul 23 at 12:46

Rhys Steele

5,5751828

answered Jul 23 at 12:46

Rhys Steele

5,5751828

5,5751828

add a comment | 

add a comment | 

 

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