Mixing
Power of stochastic kernel (transposition probability function)
$begingroup$
I am reading the following book:
Markov Chains and Invariant Probabilities, by Herndadez-Lerma and Lasserre (p.23)
- Given measure space $(X,mathcal{B})$
For each $xin X$ and $Bin mathcal{B}$, let $$P(x,B)=mathcal{P}(xi_{n+1}in B mid xi_n = x).$$ This defines a stochastic kernel on $X$, which means
$P(x,cdot)$ is a probability measure on $mathcal{B}$ for each fixed $xin X$.
$P(cdot,B)$ is a measurable function on $X$ for each fixed $Bin mathcal{B}$.
The $n-$step stochastic kernel $P^n(x,B)$ can be defined recursively by
$$P^n(x,B) = int_XP(x,dy)P^{n-1}(y,B) = int_XP^{n-1}(x,dy)P(y,B)$$
with $P^0(x,cdot) = delta_x(cdot)$.
Based on this definition,
$$P^1(x,B) = int_X P^0(x,dy)P(y,B) = int_x P(y,B)delta_x(dy) = P(x,B).$$
$$P^2(x,B) = int_X P^1(x,dy)P(y,B) = int_X P(y,B) int_X P(y,dy)delta_x(dy)$$
and then we have
$$P^2(x,B) = = int_X P(y,B) P(y,dy)delta_x(dy) = P(x,B) int_X P(y,dy)delta_x(dy)$$
I have no idea how to say $int_X P(y,dy)delta_x(dy) =P(x,B) $.
Moreover, can anyone let me know the meaning of such definition of $P^n(x,B)$ from the perspective of Markov Chain (this definition seems difficult to understand for me)?
probability-theory measure-theory markov-chains
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
$endgroup$
add a comment |
$begingroup$
I am reading the following book:
Markov Chains and Invariant Probabilities, by Herndadez-Lerma and Lasserre (p.23)
- Given measure space $(X,mathcal{B})$
For each $xin X$ and $Bin mathcal{B}$, let $$P(x,B)=mathcal{P}(xi_{n+1}in B mid xi_n = x).$$ This defines a stochastic kernel on $X$, which means
$P(x,cdot)$ is a probability measure on $mathcal{B}$ for each fixed $xin X$.
$P(cdot,B)$ is a measurable function on $X$ for each fixed $Bin mathcal{B}$.
The $n-$step stochastic kernel $P^n(x,B)$ can be defined recursively by
$$P^n(x,B) = int_XP(x,dy)P^{n-1}(y,B) = int_XP^{n-1}(x,dy)P(y,B)$$
with $P^0(x,cdot) = delta_x(cdot)$.
Based on this definition,
$$P^1(x,B) = int_X P^0(x,dy)P(y,B) = int_x P(y,B)delta_x(dy) = P(x,B).$$
$$P^2(x,B) = int_X P^1(x,dy)P(y,B) = int_X P(y,B) int_X P(y,dy)delta_x(dy)$$
and then we have
$$P^2(x,B) = = int_X P(y,B) P(y,dy)delta_x(dy) = P(x,B) int_X P(y,dy)delta_x(dy)$$
I have no idea how to say $int_X P(y,dy)delta_x(dy) =P(x,B) $.
Moreover, can anyone let me know the meaning of such definition of $P^n(x,B)$ from the perspective of Markov Chain (this definition seems difficult to understand for me)?
probability-theory measure-theory markov-chains
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
$endgroup$
add a comment |
$begingroup$
I am reading the following book:
Markov Chains and Invariant Probabilities, by Herndadez-Lerma and Lasserre (p.23)
- Given measure space $(X,mathcal{B})$
For each $xin X$ and $Bin mathcal{B}$, let $$P(x,B)=mathcal{P}(xi_{n+1}in B mid xi_n = x).$$ This defines a stochastic kernel on $X$, which means
$P(x,cdot)$ is a probability measure on $mathcal{B}$ for each fixed $xin X$.
$P(cdot,B)$ is a measurable function on $X$ for each fixed $Bin mathcal{B}$.
The $n-$step stochastic kernel $P^n(x,B)$ can be defined recursively by
$$P^n(x,B) = int_XP(x,dy)P^{n-1}(y,B) = int_XP^{n-1}(x,dy)P(y,B)$$
with $P^0(x,cdot) = delta_x(cdot)$.
Based on this definition,
$$P^1(x,B) = int_X P^0(x,dy)P(y,B) = int_x P(y,B)delta_x(dy) = P(x,B).$$
$$P^2(x,B) = int_X P^1(x,dy)P(y,B) = int_X P(y,B) int_X P(y,dy)delta_x(dy)$$
and then we have
$$P^2(x,B) = = int_X P(y,B) P(y,dy)delta_x(dy) = P(x,B) int_X P(y,dy)delta_x(dy)$$
I have no idea how to say $int_X P(y,dy)delta_x(dy) =P(x,B) $.
Moreover, can anyone let me know the meaning of such definition of $P^n(x,B)$ from the perspective of Markov Chain (this definition seems difficult to understand for me)?
probability-theory measure-theory markov-chains
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
$endgroup$
I am reading the following book:
Markov Chains and Invariant Probabilities, by Herndadez-Lerma and Lasserre (p.23)
- Given measure space $(X,mathcal{B})$
For each $xin X$ and $Bin mathcal{B}$, let $$P(x,B)=mathcal{P}(xi_{n+1}in B mid xi_n = x).$$ This defines a stochastic kernel on $X$, which means
$P(x,cdot)$ is a probability measure on $mathcal{B}$ for each fixed $xin X$.
$P(cdot,B)$ is a measurable function on $X$ for each fixed $Bin mathcal{B}$.
The $n-$step stochastic kernel $P^n(x,B)$ can be defined recursively by
$$P^n(x,B) = int_XP(x,dy)P^{n-1}(y,B) = int_XP^{n-1}(x,dy)P(y,B)$$
with $P^0(x,cdot) = delta_x(cdot)$.
Based on this definition,
$$P^1(x,B) = int_X P^0(x,dy)P(y,B) = int_x P(y,B)delta_x(dy) = P(x,B).$$
$$P^2(x,B) = int_X P^1(x,dy)P(y,B) = int_X P(y,B) int_X P(y,dy)delta_x(dy)$$
and then we have
$$P^2(x,B) = = int_X P(y,B) P(y,dy)delta_x(dy) = P(x,B) int_X P(y,dy)delta_x(dy)$$
I have no idea how to say $int_X P(y,dy)delta_x(dy) =P(x,B) $.
Moreover, can anyone let me know the meaning of such definition of $P^n(x,B)$ from the perspective of Markov Chain (this definition seems difficult to understand for me)?
probability-theory measure-theory markov-chains
probability-theory measure-theory markov-chains
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
edited Jan 9 at 21:54
sleeve chen
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
asked Jan 9 at 9:50
sleeve chensleeve chen
3,08641852
3,08641852
add a comment |
add a comment |
1 Answer
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$begingroup$
$int f(y) ddelta_x(y)=f(x)$ for any non-negative measurable function $f$.
$P^{n}(x,B)$ is the probability of starting at $x$ at time $0$ and landing inside $B$ in $n$ steps.
You have made a mistake with $P^{2}(x,B)$. You should write $P^{2}(x,B)=int_X P^{1}(y,B) P^{1}(x,dy)$. It is not possible to pull out $P(x,B)$ from this.
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
add a comment |
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1 Answer
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$begingroup$
$int f(y) ddelta_x(y)=f(x)$ for any non-negative measurable function $f$.
$P^{n}(x,B)$ is the probability of starting at $x$ at time $0$ and landing inside $B$ in $n$ steps.
You have made a mistake with $P^{2}(x,B)$. You should write $P^{2}(x,B)=int_X P^{1}(y,B) P^{1}(x,dy)$. It is not possible to pull out $P(x,B)$ from this.
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
add a comment |
$begingroup$
$int f(y) ddelta_x(y)=f(x)$ for any non-negative measurable function $f$.
$P^{n}(x,B)$ is the probability of starting at $x$ at time $0$ and landing inside $B$ in $n$ steps.
You have made a mistake with $P^{2}(x,B)$. You should write $P^{2}(x,B)=int_X P^{1}(y,B) P^{1}(x,dy)$. It is not possible to pull out $P(x,B)$ from this.
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
add a comment |
$begingroup$
$int f(y) ddelta_x(y)=f(x)$ for any non-negative measurable function $f$.
$P^{n}(x,B)$ is the probability of starting at $x$ at time $0$ and landing inside $B$ in $n$ steps.
You have made a mistake with $P^{2}(x,B)$. You should write $P^{2}(x,B)=int_X P^{1}(y,B) P^{1}(x,dy)$. It is not possible to pull out $P(x,B)$ from this.
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
$endgroup$
$int f(y) ddelta_x(y)=f(x)$ for any non-negative measurable function $f$.
$P^{n}(x,B)$ is the probability of starting at $x$ at time $0$ and landing inside $B$ in $n$ steps.
You have made a mistake with $P^{2}(x,B)$. You should write $P^{2}(x,B)=int_X P^{1}(y,B) P^{1}(x,dy)$. It is not possible to pull out $P(x,B)$ from this.
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
edited Jan 9 at 10:22
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
answered Jan 9 at 9:56
Kavi Rama MurthyKavi Rama Murthy
57.5k42160
57.5k42160
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
add a comment |
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Your first answer is what I used in deriving $P(x,B)$. Would you mind to give me more detailed comment about my derivation of $P^2(x,B)$?
$endgroup$
– sleeve chen
Jan 9 at 10:09
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
$begingroup$
Why can't I put out $P(x,B)$? The reason I do this is because $x$ and $B$ are not function of $y$. Is it not true?
$endgroup$
– sleeve chen
Jan 9 at 10:54
1
1
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
$begingroup$
In your formula for $P^{2}$ there should be a third variable $z$. You cannot have $dy$ appearing twice in that formula. You will not have $P(x,B)$ in that formula.
$endgroup$
– Kavi Rama Murthy
Jan 9 at 11:48
add a comment |
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