Derivation of Ito's Lemma (Strong)
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Regarding the proof in Thm 1.1, I am confused as to why we can assume that $X_t$ can be substituted into the formula at the top of the image since that formula was derived using a Brownian motion. Is it because the transformation $mu_tdt +sigma_tdB_t$ is a Brownian motion too? and if so, how can one show this?
stochastic-processes stochastic-calculus
asked Dec 16 '18 at 13:34
DLBDLB
548
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add a comment |
$begingroup$
Regarding the proof in Thm 1.1, I am confused as to why we can assume that $X_t$ can be substituted into the formula at the top of the image since that formula was derived using a Brownian motion. Is it because the transformation $mu_tdt +sigma_tdB_t$ is a Brownian motion too? and if so, how can one show this?
stochastic-processes stochastic-calculus
asked Dec 16 '18 at 13:34
DLBDLB
548
$endgroup$
$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
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– Song
Dec 16 '18 at 13:38
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So do all general processes have $(dX_t)^2=dt$
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– DLB
Dec 16 '18 at 14:08
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The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21
add a comment |
$begingroup$
Regarding the proof in Thm 1.1, I am confused as to why we can assume that $X_t$ can be substituted into the formula at the top of the image since that formula was derived using a Brownian motion. Is it because the transformation $mu_tdt +sigma_tdB_t$ is a Brownian motion too? and if so, how can one show this?
stochastic-processes stochastic-calculus
asked Dec 16 '18 at 13:34
DLBDLB
548
$endgroup$
Regarding the proof in Thm 1.1, I am confused as to why we can assume that $X_t$ can be substituted into the formula at the top of the image since that formula was derived using a Brownian motion. Is it because the transformation $mu_tdt +sigma_tdB_t$ is a Brownian motion too? and if so, how can one show this?
stochastic-processes stochastic-calculus
stochastic-processes stochastic-calculus
asked Dec 16 '18 at 13:34
DLBDLB
548
asked Dec 16 '18 at 13:34
DLBDLB
548
asked Dec 16 '18 at 13:34
DLBDLB
548
asked Dec 16 '18 at 13:34
DLBDLB
548
asked Dec 16 '18 at 13:34
DLBDLB
548
548
$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
$endgroup$
– Song
Dec 16 '18 at 13:38
$begingroup$
So do all general processes have $(dX_t)^2=dt$
$endgroup$
– DLB
Dec 16 '18 at 14:08
$begingroup$
The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21
add a comment |
$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
$endgroup$
– Song
Dec 16 '18 at 13:38
$begingroup$
So do all general processes have $(dX_t)^2=dt$
$endgroup$
– DLB
Dec 16 '18 at 14:08
$begingroup$
The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21
$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
$endgroup$
– Song
Dec 16 '18 at 13:38
$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
$endgroup$
– Song
Dec 16 '18 at 13:38
$begingroup$
So do all general processes have $(dX_t)^2=dt$
$endgroup$
– DLB
Dec 16 '18 at 14:08
$begingroup$
So do all general processes have $(dX_t)^2=dt$
$endgroup$
– DLB
Dec 16 '18 at 14:08
$begingroup$
The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21
add a comment |
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$begingroup$
That is because Ito calculus applies not only to Brownian motion, but also to general (continuous semimartingale) processes.
$endgroup$
– Song
Dec 16 '18 at 13:38
$begingroup$
So do all general processes have $(dX_t)^2=dt$
$endgroup$
– DLB
Dec 16 '18 at 14:08
$begingroup$
The equation in the last line is true. But $(dX_t)^2=dt$ is not a general property.
$endgroup$
– Song
Dec 16 '18 at 14:11
$begingroup$
But doesn't this have to be a property of all S.P. satisfying the theorem?
$endgroup$
– DLB
Dec 16 '18 at 15:21