Mixing
How to represent conditional distribution of a given Markov Chain?
$begingroup$
Let $mathbf{X} = (X_1,dots, X_n) sim P_{mathbf{X}}$ be i.i.d samples and
$mathbf{x}:=(x_1,dots, x_n)$ be its realizations, respectively.
Let $T:= t(mathbf{X})$ be a statistic, where
$t(cdot)$ is a function such as $t(mathbf{x}):=frac{1}{n}sum_{i=1}x_i$ (sample mean).
Then, I want to find a probability distribution of $Y$ given $mathbf{X}=mathbf{x}$ i.e.,
$P_{Ymid mathbf{X}=mathbf{x}}$ given a following Markov chain:
$$
mathbf{X} to Tto Y.
$$
In other words, I want to represent $P_{Ymid mathbf{X}=mathbf{x}}$ by
$P_mathbf{X}, t(cdot)$ and $P_{Ymid T}$.
My attempt:
$P_{Tmid mathbf{X}=mathbf{x}}$ (, which is measure-theoretically defined via some probability space $(Omega, mathcal{F}, mathbb{P})$ as a special case of a conditional expectation of some random variable given $X$), is represented as follows, which is owing to an @Song's thorough answer in my previous question:
$$
P_{Tmid mathbf{X}=mathbf{x}}(B) = 1_{{t(mathbf{x})in B}} :=
begin{cases}
1, & t(mathbf{x})in B, \
0, & text{otherwise},
end{cases}
$$
where $B$ is some Borel set.
From here, I have no clue to proceed.
When $Y$ is defined as $Y := t(mathbf{x}) + Z$ and $Zsim N(0, sigma^2)$, $Y$ given $T=t(mathbf{x})$ is distributed to $N(t(mathbf{x}), sigma^2)$.
Therefore I suppose the distribution $P_{Ymid mathbf{X}=mathbf{x}}$ is equal to
$P_{Ymid T=t(mathbf{x})}$.
Is it (always) true?
probability markov-chains conditional-probability
asked Jan 12 at 16:02
ykkxyzykkxyz
83
$endgroup$
add a comment |
$begingroup$
Let $mathbf{X} = (X_1,dots, X_n) sim P_{mathbf{X}}$ be i.i.d samples and
$mathbf{x}:=(x_1,dots, x_n)$ be its realizations, respectively.
Let $T:= t(mathbf{X})$ be a statistic, where
$t(cdot)$ is a function such as $t(mathbf{x}):=frac{1}{n}sum_{i=1}x_i$ (sample mean).
Then, I want to find a probability distribution of $Y$ given $mathbf{X}=mathbf{x}$ i.e.,
$P_{Ymid mathbf{X}=mathbf{x}}$ given a following Markov chain:
$$
mathbf{X} to Tto Y.
$$
In other words, I want to represent $P_{Ymid mathbf{X}=mathbf{x}}$ by
$P_mathbf{X}, t(cdot)$ and $P_{Ymid T}$.
My attempt:
$P_{Tmid mathbf{X}=mathbf{x}}$ (, which is measure-theoretically defined via some probability space $(Omega, mathcal{F}, mathbb{P})$ as a special case of a conditional expectation of some random variable given $X$), is represented as follows, which is owing to an @Song's thorough answer in my previous question:
$$
P_{Tmid mathbf{X}=mathbf{x}}(B) = 1_{{t(mathbf{x})in B}} :=
begin{cases}
1, & t(mathbf{x})in B, \
0, & text{otherwise},
end{cases}
$$
where $B$ is some Borel set.
From here, I have no clue to proceed.
When $Y$ is defined as $Y := t(mathbf{x}) + Z$ and $Zsim N(0, sigma^2)$, $Y$ given $T=t(mathbf{x})$ is distributed to $N(t(mathbf{x}), sigma^2)$.
Therefore I suppose the distribution $P_{Ymid mathbf{X}=mathbf{x}}$ is equal to
$P_{Ymid T=t(mathbf{x})}$.
Is it (always) true?
probability markov-chains conditional-probability
asked Jan 12 at 16:02
ykkxyzykkxyz
83
$endgroup$
$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34
add a comment |
$begingroup$
Let $mathbf{X} = (X_1,dots, X_n) sim P_{mathbf{X}}$ be i.i.d samples and
$mathbf{x}:=(x_1,dots, x_n)$ be its realizations, respectively.
Let $T:= t(mathbf{X})$ be a statistic, where
$t(cdot)$ is a function such as $t(mathbf{x}):=frac{1}{n}sum_{i=1}x_i$ (sample mean).
Then, I want to find a probability distribution of $Y$ given $mathbf{X}=mathbf{x}$ i.e.,
$P_{Ymid mathbf{X}=mathbf{x}}$ given a following Markov chain:
$$
mathbf{X} to Tto Y.
$$
In other words, I want to represent $P_{Ymid mathbf{X}=mathbf{x}}$ by
$P_mathbf{X}, t(cdot)$ and $P_{Ymid T}$.
My attempt:
$P_{Tmid mathbf{X}=mathbf{x}}$ (, which is measure-theoretically defined via some probability space $(Omega, mathcal{F}, mathbb{P})$ as a special case of a conditional expectation of some random variable given $X$), is represented as follows, which is owing to an @Song's thorough answer in my previous question:
$$
P_{Tmid mathbf{X}=mathbf{x}}(B) = 1_{{t(mathbf{x})in B}} :=
begin{cases}
1, & t(mathbf{x})in B, \
0, & text{otherwise},
end{cases}
$$
where $B$ is some Borel set.
From here, I have no clue to proceed.
When $Y$ is defined as $Y := t(mathbf{x}) + Z$ and $Zsim N(0, sigma^2)$, $Y$ given $T=t(mathbf{x})$ is distributed to $N(t(mathbf{x}), sigma^2)$.
Therefore I suppose the distribution $P_{Ymid mathbf{X}=mathbf{x}}$ is equal to
$P_{Ymid T=t(mathbf{x})}$.
Is it (always) true?
probability markov-chains conditional-probability
asked Jan 12 at 16:02
ykkxyzykkxyz
83
$endgroup$
Let $mathbf{X} = (X_1,dots, X_n) sim P_{mathbf{X}}$ be i.i.d samples and
$mathbf{x}:=(x_1,dots, x_n)$ be its realizations, respectively.
Let $T:= t(mathbf{X})$ be a statistic, where
$t(cdot)$ is a function such as $t(mathbf{x}):=frac{1}{n}sum_{i=1}x_i$ (sample mean).
Then, I want to find a probability distribution of $Y$ given $mathbf{X}=mathbf{x}$ i.e.,
$P_{Ymid mathbf{X}=mathbf{x}}$ given a following Markov chain:
$$
mathbf{X} to Tto Y.
$$
In other words, I want to represent $P_{Ymid mathbf{X}=mathbf{x}}$ by
$P_mathbf{X}, t(cdot)$ and $P_{Ymid T}$.
My attempt:
$P_{Tmid mathbf{X}=mathbf{x}}$ (, which is measure-theoretically defined via some probability space $(Omega, mathcal{F}, mathbb{P})$ as a special case of a conditional expectation of some random variable given $X$), is represented as follows, which is owing to an @Song's thorough answer in my previous question:
$$
P_{Tmid mathbf{X}=mathbf{x}}(B) = 1_{{t(mathbf{x})in B}} :=
begin{cases}
1, & t(mathbf{x})in B, \
0, & text{otherwise},
end{cases}
$$
where $B$ is some Borel set.
From here, I have no clue to proceed.
When $Y$ is defined as $Y := t(mathbf{x}) + Z$ and $Zsim N(0, sigma^2)$, $Y$ given $T=t(mathbf{x})$ is distributed to $N(t(mathbf{x}), sigma^2)$.
Therefore I suppose the distribution $P_{Ymid mathbf{X}=mathbf{x}}$ is equal to
$P_{Ymid T=t(mathbf{x})}$.
Is it (always) true?
probability markov-chains conditional-probability
probability markov-chains conditional-probability
asked Jan 12 at 16:02
ykkxyzykkxyz
83
asked Jan 12 at 16:02
ykkxyzykkxyz
83
asked Jan 12 at 16:02
ykkxyzykkxyz
83
asked Jan 12 at 16:02
ykkxyzykkxyz
83
asked Jan 12 at 16:02
ykkxyzykkxyz
83
83
$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34
add a comment |
$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34
$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34
$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34
add a comment |
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$begingroup$
Yes it is always true. $P_{Y mid X = x} = P_{Y mid T = t(x)}$ if and only if $Y$ is conditionally independent of $X$ given $T$. And in the case that $Y = T + Z$, where $Z, X$ are independent, $Y$ is conditionally independent of $X$ given $T$.
$endgroup$
– Alex
Jan 12 at 21:34