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Clarifications on the definition of $P(cdot|X=x, Yneq y)$?
$begingroup$
I have a question on the following: take two random variables $Y,X$ with support $mathcal{Y}, mathcal{X}$ respectively.
Let $yin mathcal{Y}$, $xin mathcal{X}$.
Let $P(cdot | X=x)$ be the probability distribution of $Y$ conditional on $X=x$.
I have some doubts on the definition of the probability distribution of $Y$ conditional on $X=x$ and $Yneq y$, denoted by $P(cdot| X=x, Yneq y)$.
Does this probability have subsets of $mathcal{Y}$ containing $y$ in its domain? In other words, imaging $mathcal{Y}$ finite for example, is $P(y|X=x, Yneq y)$ well defined?
What is the relation between $P(cdot| X=x, Yneq y)$ and $P(cdot| X=x)$? In other words, if we fix $P(cdot| X=x)$ does this mean that $P(cdot| X=x, Yneq y)$ is automatically determined? And viceversa?
probability-theory conditional-probability
edited Jan 19 at 19:57
Did
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
$endgroup$
add a comment |
$begingroup$
I have a question on the following: take two random variables $Y,X$ with support $mathcal{Y}, mathcal{X}$ respectively.
Let $yin mathcal{Y}$, $xin mathcal{X}$.
Let $P(cdot | X=x)$ be the probability distribution of $Y$ conditional on $X=x$.
I have some doubts on the definition of the probability distribution of $Y$ conditional on $X=x$ and $Yneq y$, denoted by $P(cdot| X=x, Yneq y)$.
Does this probability have subsets of $mathcal{Y}$ containing $y$ in its domain? In other words, imaging $mathcal{Y}$ finite for example, is $P(y|X=x, Yneq y)$ well defined?
What is the relation between $P(cdot| X=x, Yneq y)$ and $P(cdot| X=x)$? In other words, if we fix $P(cdot| X=x)$ does this mean that $P(cdot| X=x, Yneq y)$ is automatically determined? And viceversa?
probability-theory conditional-probability
edited Jan 19 at 19:57
Did
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
$endgroup$
$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56
add a comment |
$begingroup$
I have a question on the following: take two random variables $Y,X$ with support $mathcal{Y}, mathcal{X}$ respectively.
Let $yin mathcal{Y}$, $xin mathcal{X}$.
Let $P(cdot | X=x)$ be the probability distribution of $Y$ conditional on $X=x$.
I have some doubts on the definition of the probability distribution of $Y$ conditional on $X=x$ and $Yneq y$, denoted by $P(cdot| X=x, Yneq y)$.
Does this probability have subsets of $mathcal{Y}$ containing $y$ in its domain? In other words, imaging $mathcal{Y}$ finite for example, is $P(y|X=x, Yneq y)$ well defined?
What is the relation between $P(cdot| X=x, Yneq y)$ and $P(cdot| X=x)$? In other words, if we fix $P(cdot| X=x)$ does this mean that $P(cdot| X=x, Yneq y)$ is automatically determined? And viceversa?
probability-theory conditional-probability
edited Jan 19 at 19:57
Did
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
$endgroup$
I have a question on the following: take two random variables $Y,X$ with support $mathcal{Y}, mathcal{X}$ respectively.
Let $yin mathcal{Y}$, $xin mathcal{X}$.
Let $P(cdot | X=x)$ be the probability distribution of $Y$ conditional on $X=x$.
I have some doubts on the definition of the probability distribution of $Y$ conditional on $X=x$ and $Yneq y$, denoted by $P(cdot| X=x, Yneq y)$.
Does this probability have subsets of $mathcal{Y}$ containing $y$ in its domain? In other words, imaging $mathcal{Y}$ finite for example, is $P(y|X=x, Yneq y)$ well defined?
What is the relation between $P(cdot| X=x, Yneq y)$ and $P(cdot| X=x)$? In other words, if we fix $P(cdot| X=x)$ does this mean that $P(cdot| X=x, Yneq y)$ is automatically determined? And viceversa?
probability-theory conditional-probability
probability-theory conditional-probability
edited Jan 19 at 19:57
Did
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
edited Jan 19 at 19:57
Did
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
edited Jan 19 at 19:57
Did
248k23224463
edited Jan 19 at 19:57
Did
248k23224463
edited Jan 19 at 19:57
Did
248k23224463
248k23224463
asked Jan 18 at 15:32
STFSTF
431422
asked Jan 18 at 15:32
STFSTF
431422
asked Jan 18 at 15:32
STFSTF
431422
431422
$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56
add a comment |
$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56
$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56
$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56
add a comment |
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$begingroup$
First, $P(cdot | X=x)$ is not the probability distribution of $Y$ conditional on $X=x$ but the probability $P$ conditional on $X=x$. Second, you seem to be unaware that, in general, conditional distributions $P( mid X=x)$ have a rather precise definition. You might want to start with this...
$endgroup$
– Did
Jan 19 at 19:56