Deriving individual probability from conditional probability











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Very simple question - say you have discrete variables $a in S$ and $x in Q$, where $S$ and $Q$ are finite sets. Say also you know the value of $P(a|x)$ for all $S$ and $Q$. How do you derive $P(a)$ for all $a in S$? I'm thinking the formula is something like the following:



$P(a) = frac{sum_{x in Q}P(a|x)}{|Q|}$



Is this correct? What would the formula be if $Q$ were continuous?










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  • What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
    – Kavi Rama Murthy
    Nov 17 at 0:39






  • 1




    I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
    – Michael
    Nov 17 at 1:19

















up vote
1
down vote

favorite












Very simple question - say you have discrete variables $a in S$ and $x in Q$, where $S$ and $Q$ are finite sets. Say also you know the value of $P(a|x)$ for all $S$ and $Q$. How do you derive $P(a)$ for all $a in S$? I'm thinking the formula is something like the following:



$P(a) = frac{sum_{x in Q}P(a|x)}{|Q|}$



Is this correct? What would the formula be if $Q$ were continuous?










share|cite|improve this question







New contributor




ahelwer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
    – Kavi Rama Murthy
    Nov 17 at 0:39






  • 1




    I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
    – Michael
    Nov 17 at 1:19















up vote
1
down vote

favorite









up vote
1
down vote

favorite











Very simple question - say you have discrete variables $a in S$ and $x in Q$, where $S$ and $Q$ are finite sets. Say also you know the value of $P(a|x)$ for all $S$ and $Q$. How do you derive $P(a)$ for all $a in S$? I'm thinking the formula is something like the following:



$P(a) = frac{sum_{x in Q}P(a|x)}{|Q|}$



Is this correct? What would the formula be if $Q$ were continuous?










share|cite|improve this question







New contributor




ahelwer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Very simple question - say you have discrete variables $a in S$ and $x in Q$, where $S$ and $Q$ are finite sets. Say also you know the value of $P(a|x)$ for all $S$ and $Q$. How do you derive $P(a)$ for all $a in S$? I'm thinking the formula is something like the following:



$P(a) = frac{sum_{x in Q}P(a|x)}{|Q|}$



Is this correct? What would the formula be if $Q$ were continuous?







probability conditional-probability






share|cite|improve this question







New contributor




ahelwer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




ahelwer is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 17 at 0:10









ahelwer

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Check out our Code of Conduct.












  • What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
    – Kavi Rama Murthy
    Nov 17 at 0:39






  • 1




    I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
    – Michael
    Nov 17 at 1:19




















  • What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
    – Kavi Rama Murthy
    Nov 17 at 0:39






  • 1




    I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
    – Michael
    Nov 17 at 1:19


















What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
– Kavi Rama Murthy
Nov 17 at 0:39




What is $|Q|$? If it is the cardinality of $Q$ then the equation is surely false.
– Kavi Rama Murthy
Nov 17 at 0:39




1




1




I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
– Michael
Nov 17 at 1:19






I assume by $P[a]$ and $P[a|x]$ you really mean $P[A=a]$ and $P[A=a|X=x]$ for some random variables $A, X$ and some particular parameters $a,x$.
– Michael
Nov 17 at 1:19












1 Answer
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up vote
1
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If $A$ and $X$ are random variables and we know $X$ takes values in a discrete set $mathcal{Q}$ then for any real number $a$ we have by the law of total probability:
$$ boxed{P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]P[X=x]} quad (1) $$
In the special case when $mathcal{Q}$ is a finite set with $|mathcal{Q}|$ elements, and when $X$ takes values equally likely over all elements of $|mathcal{Q}|$, then $P[X=x] = frac{1}{|mathcal{Q}|}$ for all $x in mathcal{Q}$ and the above formula (1) reduces to:
$$ P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]frac{1}{|mathcal{Q}|}$$
which is similar in form to your conjectured formula.



If $X$ is a continuous random variable with PDF $f_X(x)$ then the law of total probability formula (1) is changed to:
$$ boxed{P[A=a] = int_{-infty}^{infty} P[A=a|X=x]f_X(x)dx} $$





Note also that we can only take probabilities of events. So if $Y$ is a random variable, example events are ${Yleq 12}$ or ${Y=8}$ and we can speak of $P[Yleq 12]$ (the probability that $Y$ is less than or equal to 12) and $P[Y=8]$ (the probability that $Y$ is 8) but it makes no sense to speak of $P[Y]$ (the probability that $Y$ is...what???).



So I do not like your notation $P[a]$ and $P[a|x]$ since (i) I do not know if $a$ is supposed to be a random variable or a parameter; (ii) $a$ certainly is not an event so $P[a]$ makes no sense (the probability that $a$ is...what???)






share|cite|improve this answer























  • Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
    – ahelwer
    Nov 17 at 1:29













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










If $A$ and $X$ are random variables and we know $X$ takes values in a discrete set $mathcal{Q}$ then for any real number $a$ we have by the law of total probability:
$$ boxed{P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]P[X=x]} quad (1) $$
In the special case when $mathcal{Q}$ is a finite set with $|mathcal{Q}|$ elements, and when $X$ takes values equally likely over all elements of $|mathcal{Q}|$, then $P[X=x] = frac{1}{|mathcal{Q}|}$ for all $x in mathcal{Q}$ and the above formula (1) reduces to:
$$ P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]frac{1}{|mathcal{Q}|}$$
which is similar in form to your conjectured formula.



If $X$ is a continuous random variable with PDF $f_X(x)$ then the law of total probability formula (1) is changed to:
$$ boxed{P[A=a] = int_{-infty}^{infty} P[A=a|X=x]f_X(x)dx} $$





Note also that we can only take probabilities of events. So if $Y$ is a random variable, example events are ${Yleq 12}$ or ${Y=8}$ and we can speak of $P[Yleq 12]$ (the probability that $Y$ is less than or equal to 12) and $P[Y=8]$ (the probability that $Y$ is 8) but it makes no sense to speak of $P[Y]$ (the probability that $Y$ is...what???).



So I do not like your notation $P[a]$ and $P[a|x]$ since (i) I do not know if $a$ is supposed to be a random variable or a parameter; (ii) $a$ certainly is not an event so $P[a]$ makes no sense (the probability that $a$ is...what???)






share|cite|improve this answer























  • Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
    – ahelwer
    Nov 17 at 1:29

















up vote
1
down vote



accepted










If $A$ and $X$ are random variables and we know $X$ takes values in a discrete set $mathcal{Q}$ then for any real number $a$ we have by the law of total probability:
$$ boxed{P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]P[X=x]} quad (1) $$
In the special case when $mathcal{Q}$ is a finite set with $|mathcal{Q}|$ elements, and when $X$ takes values equally likely over all elements of $|mathcal{Q}|$, then $P[X=x] = frac{1}{|mathcal{Q}|}$ for all $x in mathcal{Q}$ and the above formula (1) reduces to:
$$ P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]frac{1}{|mathcal{Q}|}$$
which is similar in form to your conjectured formula.



If $X$ is a continuous random variable with PDF $f_X(x)$ then the law of total probability formula (1) is changed to:
$$ boxed{P[A=a] = int_{-infty}^{infty} P[A=a|X=x]f_X(x)dx} $$





Note also that we can only take probabilities of events. So if $Y$ is a random variable, example events are ${Yleq 12}$ or ${Y=8}$ and we can speak of $P[Yleq 12]$ (the probability that $Y$ is less than or equal to 12) and $P[Y=8]$ (the probability that $Y$ is 8) but it makes no sense to speak of $P[Y]$ (the probability that $Y$ is...what???).



So I do not like your notation $P[a]$ and $P[a|x]$ since (i) I do not know if $a$ is supposed to be a random variable or a parameter; (ii) $a$ certainly is not an event so $P[a]$ makes no sense (the probability that $a$ is...what???)






share|cite|improve this answer























  • Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
    – ahelwer
    Nov 17 at 1:29















up vote
1
down vote



accepted







up vote
1
down vote



accepted






If $A$ and $X$ are random variables and we know $X$ takes values in a discrete set $mathcal{Q}$ then for any real number $a$ we have by the law of total probability:
$$ boxed{P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]P[X=x]} quad (1) $$
In the special case when $mathcal{Q}$ is a finite set with $|mathcal{Q}|$ elements, and when $X$ takes values equally likely over all elements of $|mathcal{Q}|$, then $P[X=x] = frac{1}{|mathcal{Q}|}$ for all $x in mathcal{Q}$ and the above formula (1) reduces to:
$$ P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]frac{1}{|mathcal{Q}|}$$
which is similar in form to your conjectured formula.



If $X$ is a continuous random variable with PDF $f_X(x)$ then the law of total probability formula (1) is changed to:
$$ boxed{P[A=a] = int_{-infty}^{infty} P[A=a|X=x]f_X(x)dx} $$





Note also that we can only take probabilities of events. So if $Y$ is a random variable, example events are ${Yleq 12}$ or ${Y=8}$ and we can speak of $P[Yleq 12]$ (the probability that $Y$ is less than or equal to 12) and $P[Y=8]$ (the probability that $Y$ is 8) but it makes no sense to speak of $P[Y]$ (the probability that $Y$ is...what???).



So I do not like your notation $P[a]$ and $P[a|x]$ since (i) I do not know if $a$ is supposed to be a random variable or a parameter; (ii) $a$ certainly is not an event so $P[a]$ makes no sense (the probability that $a$ is...what???)






share|cite|improve this answer














If $A$ and $X$ are random variables and we know $X$ takes values in a discrete set $mathcal{Q}$ then for any real number $a$ we have by the law of total probability:
$$ boxed{P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]P[X=x]} quad (1) $$
In the special case when $mathcal{Q}$ is a finite set with $|mathcal{Q}|$ elements, and when $X$ takes values equally likely over all elements of $|mathcal{Q}|$, then $P[X=x] = frac{1}{|mathcal{Q}|}$ for all $x in mathcal{Q}$ and the above formula (1) reduces to:
$$ P[A=a] = sum_{x in mathcal{Q}} P[A=a|X=x]frac{1}{|mathcal{Q}|}$$
which is similar in form to your conjectured formula.



If $X$ is a continuous random variable with PDF $f_X(x)$ then the law of total probability formula (1) is changed to:
$$ boxed{P[A=a] = int_{-infty}^{infty} P[A=a|X=x]f_X(x)dx} $$





Note also that we can only take probabilities of events. So if $Y$ is a random variable, example events are ${Yleq 12}$ or ${Y=8}$ and we can speak of $P[Yleq 12]$ (the probability that $Y$ is less than or equal to 12) and $P[Y=8]$ (the probability that $Y$ is 8) but it makes no sense to speak of $P[Y]$ (the probability that $Y$ is...what???).



So I do not like your notation $P[a]$ and $P[a|x]$ since (i) I do not know if $a$ is supposed to be a random variable or a parameter; (ii) $a$ certainly is not an event so $P[a]$ makes no sense (the probability that $a$ is...what???)







share|cite|improve this answer














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share|cite|improve this answer








edited Nov 17 at 1:16

























answered Nov 17 at 1:10









Michael

13k11325




13k11325












  • Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
    – ahelwer
    Nov 17 at 1:29




















  • Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
    – ahelwer
    Nov 17 at 1:29


















Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
– ahelwer
Nov 17 at 1:29






Brilliant! Thank you for taking the time to decode my question, especially relating it to the special case where $X$ takes values equally likely over all elements of $Q$. To clarify, $a in S$ were possible outcomes given some precondition $x in Q$ which changed the probabilities of outcomes $a in S$. Thank you for introducing this improved notation.
– ahelwer
Nov 17 at 1:29












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