If $M_t$ and $M^2_t-t$ are martingales, do we have $langle Mrangle_t=t$? The 2019 Stack Overflow Developer Survey Results Are InDetermining a square integrable martingaleWhy is it true that the continuous local martingale with quadratic variation “t” is a square integrable continuous martingale?About exponential martingalesAn application of the Dambis-Dubins-Schwarz theorem. Is my argument correct?Sufficient condition for time-changed quadratic covariation to vanish in probabilityQuadratic variation of $X_t = tB_t$?Marginally Gaussian not Bivariate Gaussian - Ito IntegralCan Local Martingales be characterized only using their FV process and BM?Show that this processes are martingalesContinuous local martingale $M$ with $langle Mrangle_t=t$ is a martingale
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If $M_t$ and $M^2_t-t$ are martingales, do we have $langle Mrangle_t=t$?
The 2019 Stack Overflow Developer Survey Results Are InDetermining a square integrable martingaleWhy is it true that the continuous local martingale with quadratic variation “t” is a square integrable continuous martingale?About exponential martingalesAn application of the Dambis-Dubins-Schwarz theorem. Is my argument correct?Sufficient condition for time-changed quadratic covariation to vanish in probabilityQuadratic variation of $X_t = tB_t$?Marginally Gaussian not Bivariate Gaussian - Ito IntegralCan Local Martingales be characterized only using their FV process and BM?Show that this processes are martingalesContinuous local martingale $M$ with $langle Mrangle_t=t$ is a martingale
$begingroup$
If $M_t$ and $M^2_t-t$ are martingales, can we have $langle Mrangle_t=t$ ? I meet this in the proof of Levy' theorem.
The levy's characterization of brownian motion can be stated as follows:
Suppose that both $M_t$ and $M^2_t-t$ are martingales. Then $M$ is a Brownian notion.
In the proof, we can frist do the Ito formular to $fin C^2(mathbbR)$ which gives us
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d langle Mrangle_t
$$
where $langle Mrangle_t$ is the quadratic variation. However, in the lecture notes, the proof says we can obtain that
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d s.
$$
So, why we have $langle Mrangle_t=t$ here?
stochastic-processes martingales
asked Mar 30 at 22:01
whereamIwhereamI
368115
$endgroup$
add a comment |
$begingroup$
If $M_t$ and $M^2_t-t$ are martingales, can we have $langle Mrangle_t=t$ ? I meet this in the proof of Levy' theorem.
The levy's characterization of brownian motion can be stated as follows:
Suppose that both $M_t$ and $M^2_t-t$ are martingales. Then $M$ is a Brownian notion.
In the proof, we can frist do the Ito formular to $fin C^2(mathbbR)$ which gives us
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d langle Mrangle_t
$$
where $langle Mrangle_t$ is the quadratic variation. However, in the lecture notes, the proof says we can obtain that
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d s.
$$
So, why we have $langle Mrangle_t=t$ here?
stochastic-processes martingales
asked Mar 30 at 22:01
whereamIwhereamI
368115
$endgroup$
2
$begingroup$
Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01
add a comment |
$begingroup$
If $M_t$ and $M^2_t-t$ are martingales, can we have $langle Mrangle_t=t$ ? I meet this in the proof of Levy' theorem.
The levy's characterization of brownian motion can be stated as follows:
Suppose that both $M_t$ and $M^2_t-t$ are martingales. Then $M$ is a Brownian notion.
In the proof, we can frist do the Ito formular to $fin C^2(mathbbR)$ which gives us
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d langle Mrangle_t
$$
where $langle Mrangle_t$ is the quadratic variation. However, in the lecture notes, the proof says we can obtain that
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d s.
$$
So, why we have $langle Mrangle_t=t$ here?
stochastic-processes martingales
asked Mar 30 at 22:01
whereamIwhereamI
368115
$endgroup$
If $M_t$ and $M^2_t-t$ are martingales, can we have $langle Mrangle_t=t$ ? I meet this in the proof of Levy' theorem.
The levy's characterization of brownian motion can be stated as follows:
Suppose that both $M_t$ and $M^2_t-t$ are martingales. Then $M$ is a Brownian notion.
In the proof, we can frist do the Ito formular to $fin C^2(mathbbR)$ which gives us
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d langle Mrangle_t
$$
where $langle Mrangle_t$ is the quadratic variation. However, in the lecture notes, the proof says we can obtain that
$$
fleft(M_tright)=f(0)+int_0^t f^primeleft(M_sright) d M_s+frac12 int_0^t f^prime primeleft(M_sright) d s.
$$
So, why we have $langle Mrangle_t=t$ here?
stochastic-processes martingales
stochastic-processes martingales
asked Mar 30 at 22:01
whereamIwhereamI
368115
asked Mar 30 at 22:01
whereamIwhereamI
368115
asked Mar 30 at 22:01
whereamIwhereamI
368115
asked Mar 30 at 22:01
whereamIwhereamI
368115
asked Mar 30 at 22:01
whereamIwhereamI
368115
368115
2
$begingroup$
Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01
add a comment |
2
$begingroup$
Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01
2
2
$begingroup$
Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01
$begingroup$
Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01
add a comment |
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Yes, by the uniqueness of the Doob-Meyer decomposition.
$endgroup$
– Yu Ding
Mar 30 at 23:01