Kolmogorov extension theorem for measures on the space of continuous functions
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In a text I am reading occurs the following:
One has a family of probability measures $(P_T)_{Tgeq 0}$ such that $P_T$ is a measure on the measurable space $(C([0,T],Bbb R^d), mathcal B (C([0,T],Bbb R^d)) )$.
It is stated that consistency of the family $(P_T)_{Tgeq 0}$ is yielding a measure $P$ on $C([0,infty),Bbb R^d)$.
I think consistency means if we take $pi^S_T: C([0,S],Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$, then for every $Sgeq T$ we have $P_T = P_S circ (pi_T^S)^{-1}$.
P seems to satisfy $P_T = Pcirc(pi_T)^{-1}$, where $pi_T: C([0,infty),Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$
Is this a trivial fact? My first thought was to apply the Kolmogorov extension theorem but I did not come far enough with it.
probability-theory measure-theory stochastic-processes
asked Jan 4 at 15:11
FalrachFalrach
1,767225
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add a comment |
$begingroup$
In a text I am reading occurs the following:
One has a family of probability measures $(P_T)_{Tgeq 0}$ such that $P_T$ is a measure on the measurable space $(C([0,T],Bbb R^d), mathcal B (C([0,T],Bbb R^d)) )$.
It is stated that consistency of the family $(P_T)_{Tgeq 0}$ is yielding a measure $P$ on $C([0,infty),Bbb R^d)$.
I think consistency means if we take $pi^S_T: C([0,S],Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$, then for every $Sgeq T$ we have $P_T = P_S circ (pi_T^S)^{-1}$.
P seems to satisfy $P_T = Pcirc(pi_T)^{-1}$, where $pi_T: C([0,infty),Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$
Is this a trivial fact? My first thought was to apply the Kolmogorov extension theorem but I did not come far enough with it.
probability-theory measure-theory stochastic-processes
asked Jan 4 at 15:11
FalrachFalrach
1,767225
$endgroup$
add a comment |
$begingroup$
In a text I am reading occurs the following:
One has a family of probability measures $(P_T)_{Tgeq 0}$ such that $P_T$ is a measure on the measurable space $(C([0,T],Bbb R^d), mathcal B (C([0,T],Bbb R^d)) )$.
It is stated that consistency of the family $(P_T)_{Tgeq 0}$ is yielding a measure $P$ on $C([0,infty),Bbb R^d)$.
I think consistency means if we take $pi^S_T: C([0,S],Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$, then for every $Sgeq T$ we have $P_T = P_S circ (pi_T^S)^{-1}$.
P seems to satisfy $P_T = Pcirc(pi_T)^{-1}$, where $pi_T: C([0,infty),Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$
Is this a trivial fact? My first thought was to apply the Kolmogorov extension theorem but I did not come far enough with it.
probability-theory measure-theory stochastic-processes
asked Jan 4 at 15:11
FalrachFalrach
1,767225
$endgroup$
In a text I am reading occurs the following:
One has a family of probability measures $(P_T)_{Tgeq 0}$ such that $P_T$ is a measure on the measurable space $(C([0,T],Bbb R^d), mathcal B (C([0,T],Bbb R^d)) )$.
It is stated that consistency of the family $(P_T)_{Tgeq 0}$ is yielding a measure $P$ on $C([0,infty),Bbb R^d)$.
I think consistency means if we take $pi^S_T: C([0,S],Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$, then for every $Sgeq T$ we have $P_T = P_S circ (pi_T^S)^{-1}$.
P seems to satisfy $P_T = Pcirc(pi_T)^{-1}$, where $pi_T: C([0,infty),Bbb R^d) to C([0,T],Bbb R^d)$,$fmapsto fvert_{[0,T]}$
Is this a trivial fact? My first thought was to apply the Kolmogorov extension theorem but I did not come far enough with it.
probability-theory measure-theory stochastic-processes
probability-theory measure-theory stochastic-processes
asked Jan 4 at 15:11
FalrachFalrach
1,767225
asked Jan 4 at 15:11
FalrachFalrach
1,767225
edited Jan 4 at 16:18
Falrach
asked Jan 4 at 15:11
FalrachFalrach
1,767225
asked Jan 4 at 15:11
FalrachFalrach
1,767225
asked Jan 4 at 15:11
FalrachFalrach
1,767225
1,767225
add a comment |
add a comment |
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