Representing a martingale as the conditional expectation w.r.t filtration?
$begingroup$
Let $M_t$ be an $mathcal{F}_t$ martingale such that $sup_t E|M_t|^p < infty$ for some $p>1$. Show there exists a $Yin L^1(P)$ such that $M_t = E[Y|mathcal{F}_t]$.
I am not sure how to answer this question.
The problem suggests using the result that: the condition in the problem implies the existence of a limiting random variable $M in L^1(P)$ such that $M_t to M$ almost surely and
$$int |M_t - M|dP to 0$$
as $t to infty$.
But I'm not sure how to represent $M_t = E[g(M)|mathcal{F}_t]$ given the corollary doesn't tell us the relationship between $M_t$ and $M$ other than that they converge almost everywhere.
probability self-learning
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $M_t$ be an $mathcal{F}_t$ martingale such that $sup_t E|M_t|^p < infty$ for some $p>1$. Show there exists a $Yin L^1(P)$ such that $M_t = E[Y|mathcal{F}_t]$.
I am not sure how to answer this question.
The problem suggests using the result that: the condition in the problem implies the existence of a limiting random variable $M in L^1(P)$ such that $M_t to M$ almost surely and
$$int |M_t - M|dP to 0$$
as $t to infty$.
But I'm not sure how to represent $M_t = E[g(M)|mathcal{F}_t]$ given the corollary doesn't tell us the relationship between $M_t$ and $M$ other than that they converge almost everywhere.
probability self-learning
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
$endgroup$
add a comment |
$begingroup$
Let $M_t$ be an $mathcal{F}_t$ martingale such that $sup_t E|M_t|^p < infty$ for some $p>1$. Show there exists a $Yin L^1(P)$ such that $M_t = E[Y|mathcal{F}_t]$.
I am not sure how to answer this question.
The problem suggests using the result that: the condition in the problem implies the existence of a limiting random variable $M in L^1(P)$ such that $M_t to M$ almost surely and
$$int |M_t - M|dP to 0$$
as $t to infty$.
But I'm not sure how to represent $M_t = E[g(M)|mathcal{F}_t]$ given the corollary doesn't tell us the relationship between $M_t$ and $M$ other than that they converge almost everywhere.
probability self-learning
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
$endgroup$
Let $M_t$ be an $mathcal{F}_t$ martingale such that $sup_t E|M_t|^p < infty$ for some $p>1$. Show there exists a $Yin L^1(P)$ such that $M_t = E[Y|mathcal{F}_t]$.
I am not sure how to answer this question.
The problem suggests using the result that: the condition in the problem implies the existence of a limiting random variable $M in L^1(P)$ such that $M_t to M$ almost surely and
$$int |M_t - M|dP to 0$$
as $t to infty$.
But I'm not sure how to represent $M_t = E[g(M)|mathcal{F}_t]$ given the corollary doesn't tell us the relationship between $M_t$ and $M$ other than that they converge almost everywhere.
probability self-learning
probability self-learning
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
asked Dec 10 '18 at 4:57
XiaomiXiaomi
1,057115
1,057115
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since $(M_t)$ is a uniformly integrable martingale, it converges a.s. and in $L^1$ to some limit $M$. Then for any $Fin mathcal{F}_t$ and $s>t$,
$$
mathsf{E}[M_t1_F]=mathsf{E}[M_s1_F]tomathsf{E}[M1_F].
$$
Therefore, $M_t=mathsf{E}[Mmid mathcal{F}_t]$ a.s.
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
$endgroup$
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
$begingroup$
Since $(M_t)$ is a uniformly integrable martingale, it converges a.s. and in $L^1$ to some limit $M$. Then for any $Fin mathcal{F}_t$ and $s>t$,
$$
mathsf{E}[M_t1_F]=mathsf{E}[M_s1_F]tomathsf{E}[M1_F].
$$
Therefore, $M_t=mathsf{E}[Mmid mathcal{F}_t]$ a.s.
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
$endgroup$
add a comment |
$begingroup$
Since $(M_t)$ is a uniformly integrable martingale, it converges a.s. and in $L^1$ to some limit $M$. Then for any $Fin mathcal{F}_t$ and $s>t$,
$$
mathsf{E}[M_t1_F]=mathsf{E}[M_s1_F]tomathsf{E}[M1_F].
$$
Therefore, $M_t=mathsf{E}[Mmid mathcal{F}_t]$ a.s.
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
$endgroup$
add a comment |
$begingroup$
Since $(M_t)$ is a uniformly integrable martingale, it converges a.s. and in $L^1$ to some limit $M$. Then for any $Fin mathcal{F}_t$ and $s>t$,
$$
mathsf{E}[M_t1_F]=mathsf{E}[M_s1_F]tomathsf{E}[M1_F].
$$
Therefore, $M_t=mathsf{E}[Mmid mathcal{F}_t]$ a.s.
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
$endgroup$
Since $(M_t)$ is a uniformly integrable martingale, it converges a.s. and in $L^1$ to some limit $M$. Then for any $Fin mathcal{F}_t$ and $s>t$,
$$
mathsf{E}[M_t1_F]=mathsf{E}[M_s1_F]tomathsf{E}[M1_F].
$$
Therefore, $M_t=mathsf{E}[Mmid mathcal{F}_t]$ a.s.
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
answered Dec 10 '18 at 5:20
d.k.o.d.k.o.
8,970628
8,970628
add a comment |
add a comment |
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