Mixing
Smoothing out a Stochastic Process
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For a stochastic process $G(t)$, one technique of making the process continuous is to define a process $G^m(t) := int_0^t m e^m (s-t) G(s) ds$. I would like to understand the intuition behind such a definition and verify that it is in fact continuous almost surely. Furthermore, it is claimed that $G^m$ converges to $G$ in the $L_2$ sense as $m rightarrow infty$.
Are there more natural methods to smooth out stochastic processes?
calculus stochastic-processes
asked Jul 24 at 3:33
JohnKnoxV
828113
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up vote
2
down vote
favorite
For a stochastic process $G(t)$, one technique of making the process continuous is to define a process $G^m(t) := int_0^t m e^m (s-t) G(s) ds$. I would like to understand the intuition behind such a definition and verify that it is in fact continuous almost surely. Furthermore, it is claimed that $G^m$ converges to $G$ in the $L_2$ sense as $m rightarrow infty$.
Are there more natural methods to smooth out stochastic processes?
calculus stochastic-processes
asked Jul 24 at 3:33
JohnKnoxV
828113
2
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
For a stochastic process $G(t)$, one technique of making the process continuous is to define a process $G^m(t) := int_0^t m e^m (s-t) G(s) ds$. I would like to understand the intuition behind such a definition and verify that it is in fact continuous almost surely. Furthermore, it is claimed that $G^m$ converges to $G$ in the $L_2$ sense as $m rightarrow infty$.
Are there more natural methods to smooth out stochastic processes?
calculus stochastic-processes
asked Jul 24 at 3:33
JohnKnoxV
828113
For a stochastic process $G(t)$, one technique of making the process continuous is to define a process $G^m(t) := int_0^t m e^m (s-t) G(s) ds$. I would like to understand the intuition behind such a definition and verify that it is in fact continuous almost surely. Furthermore, it is claimed that $G^m$ converges to $G$ in the $L_2$ sense as $m rightarrow infty$.
Are there more natural methods to smooth out stochastic processes?
calculus stochastic-processes
asked Jul 24 at 3:33
JohnKnoxV
828113
asked Jul 24 at 3:33
JohnKnoxV
828113
asked Jul 24 at 3:33
JohnKnoxV
828113
asked Jul 24 at 3:33
JohnKnoxV
828113
828113
2
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48
add a comment |Â
2
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48
2
2
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48
add a comment |Â
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2
Well, $G^m$ is the convolution of $G$ with a smooth function... and therefore it has very nice regularity properties.
– saz
Jul 24 at 5:35
I suppose you are assuming that $G$ is square-integrable?
– Math1000
Jul 24 at 5:37
Yes, it's square integrable. thanks!
– JohnKnoxV
Jul 24 at 12:48