Mixing
Convergence of stochastic integral to Brownian motion
$begingroup$
Let $a in mathbb R$, $W(t)$ a standard Brownian motion, and
$$ V(t) = a int_0^{t} e^{-a(t-s)} d W_s. $$
Is it true that
$$ int_0^t V(u) , du = W(t) - W(0) quad text{as} quad a to infty $$
in a certain sense?
convergence brownian-motion stochastic-integrals
edited Feb 3 at 9:02
AddSup
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
add a comment |
$begingroup$
Let $a in mathbb R$, $W(t)$ a standard Brownian motion, and
$$ V(t) = a int_0^{t} e^{-a(t-s)} d W_s. $$
Is it true that
$$ int_0^t V(u) , du = W(t) - W(0) quad text{as} quad a to infty $$
in a certain sense?
convergence brownian-motion stochastic-integrals
edited Feb 3 at 9:02
AddSup
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05
add a comment |
$begingroup$
Let $a in mathbb R$, $W(t)$ a standard Brownian motion, and
$$ V(t) = a int_0^{t} e^{-a(t-s)} d W_s. $$
Is it true that
$$ int_0^t V(u) , du = W(t) - W(0) quad text{as} quad a to infty $$
in a certain sense?
convergence brownian-motion stochastic-integrals
edited Feb 3 at 9:02
AddSup
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
$endgroup$
Let $a in mathbb R$, $W(t)$ a standard Brownian motion, and
$$ V(t) = a int_0^{t} e^{-a(t-s)} d W_s. $$
Is it true that
$$ int_0^t V(u) , du = W(t) - W(0) quad text{as} quad a to infty $$
in a certain sense?
convergence brownian-motion stochastic-integrals
convergence brownian-motion stochastic-integrals
edited Feb 3 at 9:02
AddSup
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
edited Feb 3 at 9:02
AddSup
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
edited Feb 3 at 9:02
AddSup
5001318
edited Feb 3 at 9:02
AddSup
5001318
edited Feb 3 at 9:02
AddSup
5001318
5001318
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
asked Jan 28 at 16:40
Roberto RastapopoulosRoberto Rastapopoulos
950425
950425
$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05
add a comment |
$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05
$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05
add a comment |
1 Answer
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$begingroup$
First,
begin{align}
int_0^t V_u du
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dW_sdu\
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dudW_s,quadtext{stochastic Fubini}.
end{align}
Thus (assuming $W_0=0$ for simplicity),
begin{align}
Eleft[left(int_0^t V_u du-W_tright)^2right]
& = int_0^tleft(
int_0^t 1_{sle u}ae^{-a(u-s)}du-1
right)^2 ds,quadtext{Ito isometry},\
& = frac{1}{2a}(1-e^{-2at})rightarrow 0text{ as }arightarrowinfty.
end{align}
That is,
$$int_0^tV_udurightarrow W_ttext{ in }L^2text{ for each }t>0.$$
In fact, ${X^n:n=1,2,ldots}$, $X^n_t:=int_0^t nint_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.
answered Feb 3 at 8:34
AddSupAddSup
5001318
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
First,
begin{align}
int_0^t V_u du
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dW_sdu\
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dudW_s,quadtext{stochastic Fubini}.
end{align}
Thus (assuming $W_0=0$ for simplicity),
begin{align}
Eleft[left(int_0^t V_u du-W_tright)^2right]
& = int_0^tleft(
int_0^t 1_{sle u}ae^{-a(u-s)}du-1
right)^2 ds,quadtext{Ito isometry},\
& = frac{1}{2a}(1-e^{-2at})rightarrow 0text{ as }arightarrowinfty.
end{align}
That is,
$$int_0^tV_udurightarrow W_ttext{ in }L^2text{ for each }t>0.$$
In fact, ${X^n:n=1,2,ldots}$, $X^n_t:=int_0^t nint_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.
answered Feb 3 at 8:34
AddSupAddSup
5001318
$endgroup$
add a comment |
$begingroup$
First,
begin{align}
int_0^t V_u du
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dW_sdu\
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dudW_s,quadtext{stochastic Fubini}.
end{align}
Thus (assuming $W_0=0$ for simplicity),
begin{align}
Eleft[left(int_0^t V_u du-W_tright)^2right]
& = int_0^tleft(
int_0^t 1_{sle u}ae^{-a(u-s)}du-1
right)^2 ds,quadtext{Ito isometry},\
& = frac{1}{2a}(1-e^{-2at})rightarrow 0text{ as }arightarrowinfty.
end{align}
That is,
$$int_0^tV_udurightarrow W_ttext{ in }L^2text{ for each }t>0.$$
In fact, ${X^n:n=1,2,ldots}$, $X^n_t:=int_0^t nint_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.
answered Feb 3 at 8:34
AddSupAddSup
5001318
$endgroup$
add a comment |
$begingroup$
First,
begin{align}
int_0^t V_u du
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dW_sdu\
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dudW_s,quadtext{stochastic Fubini}.
end{align}
Thus (assuming $W_0=0$ for simplicity),
begin{align}
Eleft[left(int_0^t V_u du-W_tright)^2right]
& = int_0^tleft(
int_0^t 1_{sle u}ae^{-a(u-s)}du-1
right)^2 ds,quadtext{Ito isometry},\
& = frac{1}{2a}(1-e^{-2at})rightarrow 0text{ as }arightarrowinfty.
end{align}
That is,
$$int_0^tV_udurightarrow W_ttext{ in }L^2text{ for each }t>0.$$
In fact, ${X^n:n=1,2,ldots}$, $X^n_t:=int_0^t nint_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.
answered Feb 3 at 8:34
AddSupAddSup
5001318
$endgroup$
First,
begin{align}
int_0^t V_u du
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dW_sdu\
& = int_0^tint_0^t 1_{sle u}ae^{-a(u-s)}dudW_s,quadtext{stochastic Fubini}.
end{align}
Thus (assuming $W_0=0$ for simplicity),
begin{align}
Eleft[left(int_0^t V_u du-W_tright)^2right]
& = int_0^tleft(
int_0^t 1_{sle u}ae^{-a(u-s)}du-1
right)^2 ds,quadtext{Ito isometry},\
& = frac{1}{2a}(1-e^{-2at})rightarrow 0text{ as }arightarrowinfty.
end{align}
That is,
$$int_0^tV_udurightarrow W_ttext{ in }L^2text{ for each }t>0.$$
In fact, ${X^n:n=1,2,ldots}$, $X^n_t:=int_0^t nint_0^u e^{-n(u-s)}dW_sdu$, converges uniformly to $W$ on compact intervals.
answered Feb 3 at 8:34
AddSupAddSup
5001318
answered Feb 3 at 8:34
AddSupAddSup
5001318
answered Feb 3 at 8:34
AddSupAddSup
5001318
answered Feb 3 at 8:34
AddSupAddSup
5001318
5001318
add a comment |
add a comment |
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$begingroup$
Why the vote to close?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:47
$begingroup$
Hint: for all Itô-integrable functions $f(t, omega) = g(t)$ (that is, for non-random integrands), $int_0^t g(s)mathrm{d}B_s$ follows the normal distribution with zero mean and variance which can be computed using the Itô isometry.
$endgroup$
– Pantelis Sopasakis
Jan 28 at 16:51
$begingroup$
Ah yeah, that is reminiscent of a course I took some years ago. Thank you very much!
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 16:53
$begingroup$
@PantelisSopasakis The convergence which is asked for, is stronger than convergence in distribution.
$endgroup$
– Did
Jan 28 at 17:04
$begingroup$
@PantelisSopasakis Would you know whether the resulting Brownian motion is the same as the Brownian motion we started with?
$endgroup$
– Roberto Rastapopoulos
Jan 28 at 17:05