definition of a sample path of a stochastic process
1
Why do we fix $omega$ in the definition of a sample path of a stochastic process? If $f:T to Omega$ is an arbitrary function, can't we define sample path to be equal to ${X(t, f(t)): t in T}$?
stochastic-processes
asked Nov 27 at 13:22
Denisof
856
|
show 7 more comments
1
Why do we fix $omega$ in the definition of a sample path of a stochastic process? If $f:T to Omega$ is an arbitrary function, can't we define sample path to be equal to ${X(t, f(t)): t in T}$?
stochastic-processes
asked Nov 27 at 13:22
Denisof
856
1
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
1
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
1
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
1
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
1
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45
|
show 7 more comments
1
1
1
Why do we fix $omega$ in the definition of a sample path of a stochastic process? If $f:T to Omega$ is an arbitrary function, can't we define sample path to be equal to ${X(t, f(t)): t in T}$?
stochastic-processes
asked Nov 27 at 13:22
Denisof
856
Why do we fix $omega$ in the definition of a sample path of a stochastic process? If $f:T to Omega$ is an arbitrary function, can't we define sample path to be equal to ${X(t, f(t)): t in T}$?
stochastic-processes
stochastic-processes
asked Nov 27 at 13:22
Denisof
856
asked Nov 27 at 13:22
Denisof
856
asked Nov 27 at 13:22
Denisof
856
asked Nov 27 at 13:22
Denisof
856
asked Nov 27 at 13:22
Denisof
856
856
1
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
1
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
1
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
1
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
1
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45
|
show 7 more comments
1
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
1
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
1
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
1
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
1
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45
1
1
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
1
1
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
1
1
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
1
1
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
1
1
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45
|
show 7 more comments
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1
A sample path is usually function $Ttomathbb R$ prescribed by $tmapsto X_t(omega)$. Every $omegainOmega$ induces such a function. In that sense the label "sample path" is reserved. What is the use placing the label also on sets (so not functions) as described in your question?
– drhab
Nov 27 at 13:35
1
They are not exactly the same. You can at most say that the function $tmapsto X_t(omega)$ is a special case of the (more general) $tmapsto X_t(f(t))$. It arises if $f$ is a constant function. If you are interested in this more general case then it might be handy to find a suitable label for it. Labels that allready exist must not be used for that.
– drhab
Nov 27 at 14:19
1
I don't understand what you are trying to say. What is the role of the concept "iid" in this? If $f$ is not a constant function then it is most unlikely that you can find an $omegainOmega$ such that the functions $tmapsto X_t(omega)$ and $tmapsto X_t(f(t))$ are the same.
– drhab
Nov 27 at 14:35
1
If we identify heads with $0$ and tails with $1$ then in that situation $Omega={0,1}^T$ for $T=mathbb N$. Then a stochastic process can be looked at as a function $X:mathbb NtimesOmegatomathbb R$. Setting $X_n:=X(n,-):Omegatomathbb R$ it can be looked at as collection of random variables: ${X_nmid ninmathbb N}$. A sample path is then a function $mathbb Ntomathbb R$ prescribed by $nmapsto X_n(omega)$ or (if you like that more) a sequence $(X_1(omega),X_2(omega),cdots)$
– drhab
Nov 28 at 13:43
1
You call $omega$ an element of $Omega^T_{X_t}$ in that context. I don't know what you mean by that. $omega$ is an element of $Omega$. In this case that means that any $omega$ can be looked at as a sequence like $(0,0,1,0,1,1,0,dots)$.
– drhab
Nov 28 at 13:45