Total probability as the sum of conditional probabilities
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Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, and their complements ($C_1'$ and $C_2'$) partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
Using the product rule, this can be expressed as follow,
$$P(X|C_1)P(C_1)P(C_2|X,C_1)+P(X|C_1')P(C_1')P(C_2|X,C_1')+P(X|C_1)P(C_1)P(C_2'|X,C_1)+P(X|C_1)P(C_1')P(C_2'|X,C_1').$$
Secondly, using the conditional independence of $C_1$ and $C_2$ (and their complements) given $X$, this is,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked or failed to invoke.
probability probability-theory
asked Nov 21 at 17:31
user617643
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up vote
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Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, and their complements ($C_1'$ and $C_2'$) partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
Using the product rule, this can be expressed as follow,
$$P(X|C_1)P(C_1)P(C_2|X,C_1)+P(X|C_1')P(C_1')P(C_2|X,C_1')+P(X|C_1)P(C_1)P(C_2'|X,C_1)+P(X|C_1)P(C_1')P(C_2'|X,C_1').$$
Secondly, using the conditional independence of $C_1$ and $C_2$ (and their complements) given $X$, this is,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked or failed to invoke.
probability probability-theory
asked Nov 21 at 17:31
user617643
226
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user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
Yes, they are complements.
– user617643
Nov 21 at 18:37
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, and their complements ($C_1'$ and $C_2'$) partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
Using the product rule, this can be expressed as follow,
$$P(X|C_1)P(C_1)P(C_2|X,C_1)+P(X|C_1')P(C_1')P(C_2|X,C_1')+P(X|C_1)P(C_1)P(C_2'|X,C_1)+P(X|C_1)P(C_1')P(C_2'|X,C_1').$$
Secondly, using the conditional independence of $C_1$ and $C_2$ (and their complements) given $X$, this is,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked or failed to invoke.
probability probability-theory
asked Nov 21 at 17:31
user617643
226
New contributor
user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Suppose I have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, and their complements ($C_1'$ and $C_2'$) partition the probability space. Let the event I'm interested in be denoted by $X$. $C_1$ and $C_2$ and their complements are conditionally independent given $X$. The probability of $X$ is given by,
$$P(X)=P(X,C_1,C_2)+P(X,C_1',C_2)+P(X,C_1,C_2')+P(X,C_1',C_2')$$
Using the product rule, this can be expressed as follow,
$$P(X|C_1)P(C_1)P(C_2|X,C_1)+P(X|C_1')P(C_1')P(C_2|X,C_1')+P(X|C_1)P(C_1)P(C_2'|X,C_1)+P(X|C_1)P(C_1')P(C_2'|X,C_1').$$
Secondly, using the conditional independence of $C_1$ and $C_2$ (and their complements) given $X$, this is,
$$P(X|C_1)P(C_1)P(C_2|X)+P(X|C_1')P(C_1')P(C_2|X)+P(X|C_1)P(C_1)P(C_2'|X)+P(X|C_1')P(C_1')P(C_2'|X)$$
Have I done this correctly? I'll accept your answer if you can say yes or no, and refer to the relevant rules I've invoked or failed to invoke.
probability probability-theory
probability probability-theory
asked Nov 21 at 17:31
user617643
226
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user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 21 at 17:31
user617643
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edited Nov 22 at 0:18
asked Nov 21 at 17:31
user617643
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asked Nov 21 at 17:31
user617643
226
asked Nov 21 at 17:31
user617643
226
226
New contributor
user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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user617643 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
Yes, they are complements.
– user617643
Nov 21 at 18:37
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58
add a comment |
"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
Yes, they are complements.
– user617643
Nov 21 at 18:37
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58
"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
Yes, they are complements.
– user617643
Nov 21 at 18:37
Yes, they are complements.
– user617643
Nov 21 at 18:37
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Indeed, if you do have that conditional independence, then that will be okay.
PS: It is the non-empty pairwise intersections of the sets and their complements that partition the space.
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Indeed, if you do have that conditional independence, then that will be okay.
PS: It is the non-empty pairwise intersections of the sets and their complements that partition the space.
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
add a comment |
up vote
1
down vote
accepted
Indeed, if you do have that conditional independence, then that will be okay.
PS: It is the non-empty pairwise intersections of the sets and their complements that partition the space.
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Indeed, if you do have that conditional independence, then that will be okay.
PS: It is the non-empty pairwise intersections of the sets and their complements that partition the space.
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
Indeed, if you do have that conditional independence, then that will be okay.
PS: It is the non-empty pairwise intersections of the sets and their complements that partition the space.
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
answered Nov 22 at 0:18
Graham Kemp
84.2k43378
84.2k43378
add a comment |
add a comment |
user617643 is a new contributor. Be nice, and check out our Code of Conduct.
user617643 is a new contributor. Be nice, and check out our Code of Conduct.
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"$C_1,C_2,C_1', text{ and }C_2'$ partition the probability space..." Are thry disjoint? Or $C_j'=C_j^c$?
– d.k.o.
Nov 21 at 17:50
Yes, they are complements.
– user617643
Nov 21 at 18:37
@d.k.o. Any idea?
– user617643
Nov 21 at 22:58