How to evaluate the total probability using conditional probabilities?
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Suppose we have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ 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,
begin{equation*}
P(X)=P(X|C_1,C_2)P(C_1,C_2)+P(X|C_1',C_2)P(C_1',C_2)+P(X|C_1,C_2')P(C_1',C_2)+P(X|C_1',C_2')P(C_1',C_2')
end{equation*}
Using conditional independence, we evaluate this as follows,
begin{equation*}
P(X) =P(C_1|X)P(C_2|X)P(X)+P(C_1'|X)P(C_2|X)P(X)+P(C_1|X)P(C_2'|X)P(X)+P(C_1'|X)P(C_2'|X)P(X)
end{equation*}
But then I have $P(X)$ on both side of the equality in the second equation. Is there a difference between these two (are the ones on the RHS the prior probability and the $P(X)$ on the LHS the posterior)? Any clarification would be appreciated.
In my case $P(X)$ is unknown, and I'm trying to solve for it. Would I go about this by guessing values of $P(X)$ for those on the RHS and evaluate this for the value of $P(X)$ on the LHS?
probability probability-theory bayesian
asked Nov 20 at 15:01
user617643
226
|
show 2 more comments
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0
down vote
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Suppose we have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ 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,
begin{equation*}
P(X)=P(X|C_1,C_2)P(C_1,C_2)+P(X|C_1',C_2)P(C_1',C_2)+P(X|C_1,C_2')P(C_1',C_2)+P(X|C_1',C_2')P(C_1',C_2')
end{equation*}
Using conditional independence, we evaluate this as follows,
begin{equation*}
P(X) =P(C_1|X)P(C_2|X)P(X)+P(C_1'|X)P(C_2|X)P(X)+P(C_1|X)P(C_2'|X)P(X)+P(C_1'|X)P(C_2'|X)P(X)
end{equation*}
But then I have $P(X)$ on both side of the equality in the second equation. Is there a difference between these two (are the ones on the RHS the prior probability and the $P(X)$ on the LHS the posterior)? Any clarification would be appreciated.
In my case $P(X)$ is unknown, and I'm trying to solve for it. Would I go about this by guessing values of $P(X)$ for those on the RHS and evaluate this for the value of $P(X)$ on the LHS?
probability probability-theory bayesian
asked Nov 20 at 15:01
user617643
226
What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17
|
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose we have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ 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,
begin{equation*}
P(X)=P(X|C_1,C_2)P(C_1,C_2)+P(X|C_1',C_2)P(C_1',C_2)+P(X|C_1,C_2')P(C_1',C_2)+P(X|C_1',C_2')P(C_1',C_2')
end{equation*}
Using conditional independence, we evaluate this as follows,
begin{equation*}
P(X) =P(C_1|X)P(C_2|X)P(X)+P(C_1'|X)P(C_2|X)P(X)+P(C_1|X)P(C_2'|X)P(X)+P(C_1'|X)P(C_2'|X)P(X)
end{equation*}
But then I have $P(X)$ on both side of the equality in the second equation. Is there a difference between these two (are the ones on the RHS the prior probability and the $P(X)$ on the LHS the posterior)? Any clarification would be appreciated.
In my case $P(X)$ is unknown, and I'm trying to solve for it. Would I go about this by guessing values of $P(X)$ for those on the RHS and evaluate this for the value of $P(X)$ on the LHS?
probability probability-theory bayesian
asked Nov 20 at 15:01
user617643
226
Suppose we have 2 events $C_1$ and $C_2$ so that $C_1$, $C_2$, $C_1'$ 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,
begin{equation*}
P(X)=P(X|C_1,C_2)P(C_1,C_2)+P(X|C_1',C_2)P(C_1',C_2)+P(X|C_1,C_2')P(C_1',C_2)+P(X|C_1',C_2')P(C_1',C_2')
end{equation*}
Using conditional independence, we evaluate this as follows,
begin{equation*}
P(X) =P(C_1|X)P(C_2|X)P(X)+P(C_1'|X)P(C_2|X)P(X)+P(C_1|X)P(C_2'|X)P(X)+P(C_1'|X)P(C_2'|X)P(X)
end{equation*}
But then I have $P(X)$ on both side of the equality in the second equation. Is there a difference between these two (are the ones on the RHS the prior probability and the $P(X)$ on the LHS the posterior)? Any clarification would be appreciated.
In my case $P(X)$ is unknown, and I'm trying to solve for it. Would I go about this by guessing values of $P(X)$ for those on the RHS and evaluate this for the value of $P(X)$ on the LHS?
probability probability-theory bayesian
probability probability-theory bayesian
asked Nov 20 at 15:01
user617643
226
asked Nov 20 at 15:01
user617643
226
edited Nov 20 at 15:15
asked Nov 20 at 15:01
user617643
226
asked Nov 20 at 15:01
user617643
226
asked Nov 20 at 15:01
user617643
226
226
What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17
|
show 2 more comments
What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17
What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17
|
show 2 more comments
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What if you divide both sides by $P(X)?$
– saulspatz
Nov 20 at 15:07
Then I get $1$ on the LHS.
– user617643
Nov 20 at 15:08
Yes, but look at the right-hand side.
– saulspatz
Nov 20 at 15:15
That is the sum of the probabilities of the possible outcomes.
– user617643
Nov 20 at 15:16
Doesn't that answer you question?
– saulspatz
Nov 20 at 15:17