Is it possible to give an intuitive idea of the notation at the base of the transition Kernel for a Markov...
Given $A in sigma(mathcal{S})$, with $mathcal{S}$ is the state space, the transition kernel is a function $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$
$forall x in mathcal{S}, K(x,cdot)$ is a probability measure;
$forall A in mathcal{B}(mathcal{S}), K(cdot,A)$ is measurable.
If the process at the initial time (t = 0) starts from a fixed state $x_0$:
$$Pr{(X_1,X_2) in A_1 times A_2}=int_{A_1 } K(y,A_2)K(x_0,dy)$$
is it possible to give an intuitive idea of the concepts: "probability measure" and "measurable" and then give an intuitive idea of the meaning of the following notation?
$$int_{A_1 } K(y,A_2)K(x_0,dy)$$
in particular of
$$K(x_0,dy)$$
?
measure-theory markov-chains intuition
asked Nov 30 at 9:37
Mike9
113
|
show 1 more comment
Given $A in sigma(mathcal{S})$, with $mathcal{S}$ is the state space, the transition kernel is a function $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$
$forall x in mathcal{S}, K(x,cdot)$ is a probability measure;
$forall A in mathcal{B}(mathcal{S}), K(cdot,A)$ is measurable.
If the process at the initial time (t = 0) starts from a fixed state $x_0$:
$$Pr{(X_1,X_2) in A_1 times A_2}=int_{A_1 } K(y,A_2)K(x_0,dy)$$
is it possible to give an intuitive idea of the concepts: "probability measure" and "measurable" and then give an intuitive idea of the meaning of the following notation?
$$int_{A_1 } K(y,A_2)K(x_0,dy)$$
in particular of
$$K(x_0,dy)$$
?
measure-theory markov-chains intuition
asked Nov 30 at 9:37
Mike9
113
Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27
|
show 1 more comment
Given $A in sigma(mathcal{S})$, with $mathcal{S}$ is the state space, the transition kernel is a function $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$
$forall x in mathcal{S}, K(x,cdot)$ is a probability measure;
$forall A in mathcal{B}(mathcal{S}), K(cdot,A)$ is measurable.
If the process at the initial time (t = 0) starts from a fixed state $x_0$:
$$Pr{(X_1,X_2) in A_1 times A_2}=int_{A_1 } K(y,A_2)K(x_0,dy)$$
is it possible to give an intuitive idea of the concepts: "probability measure" and "measurable" and then give an intuitive idea of the meaning of the following notation?
$$int_{A_1 } K(y,A_2)K(x_0,dy)$$
in particular of
$$K(x_0,dy)$$
?
measure-theory markov-chains intuition
asked Nov 30 at 9:37
Mike9
113
Given $A in sigma(mathcal{S})$, with $mathcal{S}$ is the state space, the transition kernel is a function $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$
$forall x in mathcal{S}, K(x,cdot)$ is a probability measure;
$forall A in mathcal{B}(mathcal{S}), K(cdot,A)$ is measurable.
If the process at the initial time (t = 0) starts from a fixed state $x_0$:
$$Pr{(X_1,X_2) in A_1 times A_2}=int_{A_1 } K(y,A_2)K(x_0,dy)$$
is it possible to give an intuitive idea of the concepts: "probability measure" and "measurable" and then give an intuitive idea of the meaning of the following notation?
$$int_{A_1 } K(y,A_2)K(x_0,dy)$$
in particular of
$$K(x_0,dy)$$
?
measure-theory markov-chains intuition
measure-theory markov-chains intuition
asked Nov 30 at 9:37
Mike9
113
asked Nov 30 at 9:37
Mike9
113
edited Dec 2 at 10:13
asked Nov 30 at 9:37
Mike9
113
asked Nov 30 at 9:37
Mike9
113
asked Nov 30 at 9:37
Mike9
113
113
Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27
|
show 1 more comment
Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27
Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27
|
show 1 more comment
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Probability is already a word very intuitive. I recommend you Ross' introductory probability book for intuitive explanations if you don't mind his approach of sacrificing mathematical rigors for probabilistic interpretations.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:44
I've got to study on Ross and the concepts of Markov chains and probabilities of transitions I am quite clear. But I can not interpret the following general notation: $K(x_0,dy)$ and I asked if it was possible to have an intuitive idea
– Mike9
Nov 30 at 9:50
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
Nov 30 at 9:56
And regard this integral as a convolution.
– Jean Marie
Nov 30 at 10:04
So being $K(cdot , cdot): mathcal{S} times mathcal{B}(mathcal{S}) to [0,1]$, in this case is it same things to write $K(x_0,z)$ with $z=y+dy$? With $z in mathcal{B}(mathcal{S})$?
– Mike9
Nov 30 at 17:27