Mixing
conditional distribution of $X$ given $min(X,alpha)$
1
$begingroup$
let $alpha in mathbb{R}.$ So I want to find a conditional distribution for $X$ given $Y=min(X,alpha).$
I know that for all $B in mathcal{B}(mathbb{R})$
$$P_Y(B)=int_B 1_{]-infty, alpha]}(x)dP_X+P(X>alpha)delta_{alpha}(B).$$
I don't know this will help us to find out the conditional distribution.
Thank you for any idea.
probability-theory measure-theory conditional-expectation conditional-probability
asked Jan 31 at 15:25
mathexmathex
1088
$endgroup$
add a comment |
1
$begingroup$
let $alpha in mathbb{R}.$ So I want to find a conditional distribution for $X$ given $Y=min(X,alpha).$
I know that for all $B in mathcal{B}(mathbb{R})$
$$P_Y(B)=int_B 1_{]-infty, alpha]}(x)dP_X+P(X>alpha)delta_{alpha}(B).$$
I don't know this will help us to find out the conditional distribution.
Thank you for any idea.
probability-theory measure-theory conditional-expectation conditional-probability
asked Jan 31 at 15:25
mathexmathex
1088
$endgroup$
$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07
add a comment |
1
1
1
0
$begingroup$
let $alpha in mathbb{R}.$ So I want to find a conditional distribution for $X$ given $Y=min(X,alpha).$
I know that for all $B in mathcal{B}(mathbb{R})$
$$P_Y(B)=int_B 1_{]-infty, alpha]}(x)dP_X+P(X>alpha)delta_{alpha}(B).$$
I don't know this will help us to find out the conditional distribution.
Thank you for any idea.
probability-theory measure-theory conditional-expectation conditional-probability
asked Jan 31 at 15:25
mathexmathex
1088
$endgroup$
let $alpha in mathbb{R}.$ So I want to find a conditional distribution for $X$ given $Y=min(X,alpha).$
I know that for all $B in mathcal{B}(mathbb{R})$
$$P_Y(B)=int_B 1_{]-infty, alpha]}(x)dP_X+P(X>alpha)delta_{alpha}(B).$$
I don't know this will help us to find out the conditional distribution.
Thank you for any idea.
probability-theory measure-theory conditional-expectation conditional-probability
probability-theory measure-theory conditional-expectation conditional-probability
asked Jan 31 at 15:25
mathexmathex
1088
asked Jan 31 at 15:25
mathexmathex
1088
asked Jan 31 at 15:25
mathexmathex
1088
asked Jan 31 at 15:25
mathexmathex
1088
asked Jan 31 at 15:25
mathexmathex
1088
1088
$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07
add a comment |
$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07
$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07
add a comment |
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$begingroup$
It seems that any $$(A,y)mapsto 1_{yneq alpha} delta_{y}(A) + 1_{y=alpha}P_{X|Y=y}(A)$$ where $P_{X|Y=alpha}$ is an arbitrary probability measure is a regular conditional probability of $X|min(X,alpha)$.
$endgroup$
– Gabriel Romon
Feb 2 at 11:55
$begingroup$
if we take a non negative function $h$ from $mathbb{R^2}$ to $mathbb{R}$ a way to find the conditional distribution is to write :$$int_{mathbb{R^2}}h(x,y)dP_{(X,Y)}(x,y)=int_{mathbb{R}}(int_{mathbb{R}}h(x,y)dP_{X|Y=y}(x))dP_Y(y)$$
$endgroup$
– mathex
Feb 2 at 20:44
$begingroup$
And we have $$int h(X,min(alpha,X))dP=int_{]-infty,alpha]}h(x,x)dP_X(x)+int_{]alpha,+infty[}h(x,alpha)dP_X(x)$$ I wonder how to find $P_{X|Y=y}$ from this equality
$endgroup$
– mathex
Feb 2 at 20:54
$begingroup$
So that $$forall y in mathbb{R}, forall B in mathcal{B}(mathbb{R}), P_{X|Y=y}(B)=1_{]-infty,alpha]}(y)delta_y(B)+1_{]alpha,infty[}(y)frac{P_X(B cap ]alpha,+infty[)}{P_X(]alpha,+infty[)}$$ supposing that $P_X(]alpha,+infty[)>0$
$endgroup$
– mathex
Feb 2 at 21:07