How do you multiply two conditional probabilities?
I have to multiply this:
$P(a|b,c)·P(b|c)$
How do you multiply those two expressions?
It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.
probability
asked 5 hours ago
VansFannelVansFannel
13018
add a comment |
I have to multiply this:
$P(a|b,c)·P(b|c)$
How do you multiply those two expressions?
It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.
probability
asked 5 hours ago
VansFannelVansFannel
13018
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago
add a comment |
I have to multiply this:
$P(a|b,c)·P(b|c)$
How do you multiply those two expressions?
It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.
probability
asked 5 hours ago
VansFannelVansFannel
13018
I have to multiply this:
$P(a|b,c)·P(b|c)$
How do you multiply those two expressions?
It seems that $P(a|b,c)·P(b|c) = P(a,b|c)$ but I don't know how to obtain that expression multiplying the two previous conditional probabilities.
probability
probability
asked 5 hours ago
VansFannelVansFannel
13018
asked 5 hours ago
VansFannelVansFannel
13018
edited 4 hours ago
VansFannel
asked 5 hours ago
VansFannelVansFannel
13018
asked 5 hours ago
VansFannelVansFannel
13018
asked 5 hours ago
VansFannelVansFannel
13018
13018
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago
add a comment |
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}
So, in all, we have:
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
add a comment |
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}
So, in all, we have:
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
add a comment |
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}
So, in all, we have:
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
add a comment |
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}
So, in all, we have:
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
Those two expressions simply mean:
(the Probability of event $a$ given the events $b$ and $c$) $times$ (the Probability of event $b$ given event $c$).
The two terms are probabilities which are scalar values between the 0 and 1 (including 0 and 1). So you simply multiply the two values together as you would any two numbers. So for example, if the probability of $a$, given $b$ and $c$ were 0.40 and the probability of $b$ given $c$ where .70, then:
$P(a|b,c)·P(b|c)=0.40times0.70=0.28$
UPDATED BASED ON YOUR EDITED QUESTION:
The proof is straight-forward and is based on three applications of the the definition of conditional probability, which states that if $D$ and $E$ are two events in space $S$, and $P(E)>0$, then the conditional probability of $D$ given $E$, written by $P(D|E)$ is $frac{P(DE)}{P(E)}$
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(a|bc)frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(bc)}frac{P(bc)}{P(c)}\
& = & frac{P(abc)}{P(c)}\
& = & frac{P[(ab)c]}{P(c)}\
& = & P(ab|c)
end{eqnarray*}
So, in all, we have:
begin{eqnarray*}
P(a|bc)P(b|c) & = & P(ab|c)
end{eqnarray*}
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
edited 1 hour ago
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
answered 4 hours ago
StatsStudentStatsStudent
4,89032042
4,89032042
add a comment |
add a comment |
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
add a comment |
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
add a comment |
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
If you don't know the probabilities $P(a|b,c)$ or $P(b|c)$ themselves, you can try to reformulate them in terms of probabilities that you do know. The chain rule, or Baye's theorem would be useful for doing so. For example, by the chain rule:
$$
P(a,b,c) = P(a|b,c) P(b|c) P(c)
$$
which would imply:
$$
P(a|b,c) P(b|c) = frac{P(a,b,c)}{P(c)}
$$
so if you had the probabilities $P(a,b,c)$ and $P(c)$, you would be able to calculate the product of $P(a|b,c)$ and $P(b|c)$.
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
answered 4 hours ago
Stat_ProgrammerStat_Programmer
3011
3011
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
add a comment |
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
I have updated my question with more details. Sorry to do it so late. I want to know the expression, not the value, resulting of multiplying the two conditional probabilities.
– VansFannel
4 hours ago
add a comment |
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Intuitively you could compare this with P(a|b)P(b) = P(a,b)
– Martijn Weterings
4 hours ago
Are you asking about the arithmetic as in @StatsStudent 's answer? Or are you asking "what the nomenclature of the product of these two probabilities?"
– Alexis
4 hours ago
The nomenglature.
– VansFannel
4 hours ago
Note, that this is only true so long as $P(c)>0$ and $P(bc)>0$.
– StatsStudent
3 hours ago
@VansFannel Since you are a new user, please be sure to accept the answer that helped you most by clicking the left check mark. Accepts help encourage users to provide answers.
– StatsStudent
2 hours ago