Differential of Stochastic process
$begingroup$
Given a stochastic process
$$Z_t = e^{4t} int_0^{t} e^{-2s} , ,dB_s$$
where $B$ denotes the standard Brownian Motion.
Determine $dZ_t$.
I tried to make use of Ito's rule, seeing that $Z$ is a function of $t$ and $B$ but I got stuck trying to find the derivative of an integral transform.
I thought maybe applying the product rule of Ito, finding
$$dZ_t = int_0^{t} e^{-2s} , , dB_s cdot 4e^{4t} , dt + e^{4t} cdot e^{-2t} dB$$
But I feel as if this can never be correct...
calculus stochastic-processes stochastic-calculus brownian-motion stochastic-integrals
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
$endgroup$
add a comment |
$begingroup$
Given a stochastic process
$$Z_t = e^{4t} int_0^{t} e^{-2s} , ,dB_s$$
where $B$ denotes the standard Brownian Motion.
Determine $dZ_t$.
I tried to make use of Ito's rule, seeing that $Z$ is a function of $t$ and $B$ but I got stuck trying to find the derivative of an integral transform.
I thought maybe applying the product rule of Ito, finding
$$dZ_t = int_0^{t} e^{-2s} , , dB_s cdot 4e^{4t} , dt + e^{4t} cdot e^{-2t} dB$$
But I feel as if this can never be correct...
calculus stochastic-processes stochastic-calculus brownian-motion stochastic-integrals
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
$endgroup$
$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57
add a comment |
$begingroup$
Given a stochastic process
$$Z_t = e^{4t} int_0^{t} e^{-2s} , ,dB_s$$
where $B$ denotes the standard Brownian Motion.
Determine $dZ_t$.
I tried to make use of Ito's rule, seeing that $Z$ is a function of $t$ and $B$ but I got stuck trying to find the derivative of an integral transform.
I thought maybe applying the product rule of Ito, finding
$$dZ_t = int_0^{t} e^{-2s} , , dB_s cdot 4e^{4t} , dt + e^{4t} cdot e^{-2t} dB$$
But I feel as if this can never be correct...
calculus stochastic-processes stochastic-calculus brownian-motion stochastic-integrals
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
$endgroup$
Given a stochastic process
$$Z_t = e^{4t} int_0^{t} e^{-2s} , ,dB_s$$
where $B$ denotes the standard Brownian Motion.
Determine $dZ_t$.
I tried to make use of Ito's rule, seeing that $Z$ is a function of $t$ and $B$ but I got stuck trying to find the derivative of an integral transform.
I thought maybe applying the product rule of Ito, finding
$$dZ_t = int_0^{t} e^{-2s} , , dB_s cdot 4e^{4t} , dt + e^{4t} cdot e^{-2t} dB$$
But I feel as if this can never be correct...
calculus stochastic-processes stochastic-calculus brownian-motion stochastic-integrals
calculus stochastic-processes stochastic-calculus brownian-motion stochastic-integrals
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
asked Dec 13 '18 at 23:47
L.M. S.L.M. S.
61
61
$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57
add a comment |
$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57
$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57
add a comment |
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$begingroup$
@saz, so could you show me the answer?
$endgroup$
– L.M. S.
Dec 15 '18 at 18:13
$begingroup$
Okay, so $dX = 0 dt + e^{-2t} dB_t$. Also, $frac{partial^2 f}{partial x^2} = 0$. So Ito's formula comes down to: $dZ = 4e^{4t}x dt + e^{2t} dB_t$, which is exactly what I already had....
$endgroup$
– L.M. S.
Dec 15 '18 at 18:19
$begingroup$
I fail to see why the integral cannot be there since $frac{partial f}{partial t} = 4e^{4t}x$, now here the $x$ pops up and $x$ also has the integral.
$endgroup$
– L.M. S.
Dec 15 '18 at 18:25
$begingroup$
Okay, it seems that I made a mess out of this, sorry about that. In your earlier (edited) comment I confused $X$ with $Z$ and therefore I thought that you went wrong. Regarding the answer at the end of your question (which you believed to be wrong): Actually it is correct (I got once more confused with your notation; sorry about that again).
$endgroup$
– saz
Dec 15 '18 at 18:57