Mean square continuity and differentiable sample paths

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There are a number of questions involved in this post so even if you do not feel able to answer all please answer any that you can. Consider a stochastic process $Z(x)$ where $x$ is the index parameter. This process is said to be mean-square continuous if $mathbbEleft[Z(x)^2right] < +infty$ and:

beginequation
limlimits_x rightarrow s mathbbEleft[left(Z(x) - Z(s)right)^2right] = 0
endequation

Question 1): Are these two conditions necessary and sufficient or are other conditions required for mean square continuity of $Z(x)$?

Question 2): Does the differentiability of realisation $z(x)$, also known as a sample path, of $Z(x)$ imply the existence of a derivative process $dotZ(x)$ with finite variance?

Question 3): Is saying that $Z(x)$ is differentiable nowhere equivalent to saying that its derivative process $dotZ(x)$ has infinite variance? For instance the Wiener process is mean-square continuous yet nowhere differentiable and is often used to represent the integral of white noise which has infinite variance.

Question 4) Is it correct to say that mean-square differentiability of $Z(x)$ is a necessary but not sufficient condition for the differentiability of any realisation $z(x)$? But rather that if a process is both mean-square differentiable and separable then its samples paths are differentiable.

derivatives stochastic-processes

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edited Jul 24 at 11:09

asked Jul 24 at 10:48

7Jack

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There are a number of questions involved in this post so even if you do not feel able to answer all please answer any that you can. Consider a stochastic process $Z(x)$ where $x$ is the index parameter. This process is said to be mean-square continuous if $mathbbEleft[Z(x)^2right] < +infty$ and:

beginequation
limlimits_x rightarrow s mathbbEleft[left(Z(x) - Z(s)right)^2right] = 0
endequation

Question 1): Are these two conditions necessary and sufficient or are other conditions required for mean square continuity of $Z(x)$?

Question 2): Does the differentiability of realisation $z(x)$, also known as a sample path, of $Z(x)$ imply the existence of a derivative process $dotZ(x)$ with finite variance?

Question 3): Is saying that $Z(x)$ is differentiable nowhere equivalent to saying that its derivative process $dotZ(x)$ has infinite variance? For instance the Wiener process is mean-square continuous yet nowhere differentiable and is often used to represent the integral of white noise which has infinite variance.

Question 4) Is it correct to say that mean-square differentiability of $Z(x)$ is a necessary but not sufficient condition for the differentiability of any realisation $z(x)$? But rather that if a process is both mean-square differentiable and separable then its samples paths are differentiable.

derivatives stochastic-processes

share|cite|improve this question

edited Jul 24 at 11:09

asked Jul 24 at 10:48

7Jack

212113

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

There are a number of questions involved in this post so even if you do not feel able to answer all please answer any that you can. Consider a stochastic process $Z(x)$ where $x$ is the index parameter. This process is said to be mean-square continuous if $mathbbEleft[Z(x)^2right] < +infty$ and:

beginequation
limlimits_x rightarrow s mathbbEleft[left(Z(x) - Z(s)right)^2right] = 0
endequation

Question 1): Are these two conditions necessary and sufficient or are other conditions required for mean square continuity of $Z(x)$?

Question 2): Does the differentiability of realisation $z(x)$, also known as a sample path, of $Z(x)$ imply the existence of a derivative process $dotZ(x)$ with finite variance?

Question 3): Is saying that $Z(x)$ is differentiable nowhere equivalent to saying that its derivative process $dotZ(x)$ has infinite variance? For instance the Wiener process is mean-square continuous yet nowhere differentiable and is often used to represent the integral of white noise which has infinite variance.

Question 4) Is it correct to say that mean-square differentiability of $Z(x)$ is a necessary but not sufficient condition for the differentiability of any realisation $z(x)$? But rather that if a process is both mean-square differentiable and separable then its samples paths are differentiable.

derivatives stochastic-processes

share|cite|improve this question

edited Jul 24 at 11:09

asked Jul 24 at 10:48

7Jack

212113

There are a number of questions involved in this post so even if you do not feel able to answer all please answer any that you can. Consider a stochastic process $Z(x)$ where $x$ is the index parameter. This process is said to be mean-square continuous if $mathbbEleft[Z(x)^2right] < +infty$ and:

beginequation
limlimits_x rightarrow s mathbbEleft[left(Z(x) - Z(s)right)^2right] = 0
endequation

Question 1): Are these two conditions necessary and sufficient or are other conditions required for mean square continuity of $Z(x)$?

Question 2): Does the differentiability of realisation $z(x)$, also known as a sample path, of $Z(x)$ imply the existence of a derivative process $dotZ(x)$ with finite variance?

Question 3): Is saying that $Z(x)$ is differentiable nowhere equivalent to saying that its derivative process $dotZ(x)$ has infinite variance? For instance the Wiener process is mean-square continuous yet nowhere differentiable and is often used to represent the integral of white noise which has infinite variance.

Question 4) Is it correct to say that mean-square differentiability of $Z(x)$ is a necessary but not sufficient condition for the differentiability of any realisation $z(x)$? But rather that if a process is both mean-square differentiable and separable then its samples paths are differentiable.

derivatives stochastic-processes

share|cite|improve this question

edited Jul 24 at 11:09

asked Jul 24 at 10:48

7Jack

212113

share|cite|improve this question

share|cite|improve this question

edited Jul 24 at 11:09

edited Jul 24 at 11:09

edited Jul 24 at 11:09

asked Jul 24 at 10:48

7Jack

212113

asked Jul 24 at 10:48

7Jack

212113

asked Jul 24 at 10:48

7Jack

212113

212113

add a comment | 

add a comment | 

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