Is it possible to give an intuitive idea of the notation at the base of the transition Kernel for a Markov...
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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
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked 18 hours ago
Saverio Mazza
63
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Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
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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
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked 18 hours ago
Saverio Mazza
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
18 hours ago
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
– Saverio Mazza
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago
|
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked 18 hours ago
Saverio Mazza
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked 18 hours ago
Saverio Mazza
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
asked 18 hours ago
Saverio Mazza
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
edited 18 hours ago
GNUSupporter 8964民主女神 地下教會
12.4k72344
12.4k72344
asked 18 hours ago
Saverio Mazza
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 18 hours ago
Saverio Mazza
63
asked 18 hours ago
Saverio Mazza
63
63
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Saverio Mazza is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
18 hours ago
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
– Saverio Mazza
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago
|
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
18 hours ago
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
– Saverio Mazza
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago
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
18 hours ago
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
18 hours ago
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
– Saverio Mazza
18 hours ago
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
– Saverio Mazza
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago
|
show 1 more comment
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Saverio Mazza is a new contributor. Be nice, and check out our Code of Conduct.
Saverio Mazza is a new contributor. Be nice, and check out our Code of Conduct.
Saverio Mazza is a new contributor. Be nice, and check out our Code of Conduct.
Saverio Mazza is a new contributor. Be nice, and check out our Code of Conduct.
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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
18 hours ago
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
– Saverio Mazza
18 hours ago
Regard dy as an infinitesimal "neighborhood" at y.
– GNUSupporter 8964民主女神 地下教會
18 hours ago
And regard this integral as a convolution.
– Jean Marie
18 hours ago
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 $z in mathcal{B}(mathcal{S})$?
– Saverio Mazza
10 hours ago