Cluster probabilites: Bayesian network (sprinkler example, Russel/ Norvig) as a clustered network
$begingroup$
like others here I am also learning with Russel's and Norvig's book about artificial intelligence. My question is about the conditional probability tables of a clustered multiply connected network with a meganode.
We have the following Bayesian multiply connected network.
In order to reduce computation (according to the book). We combine Sprinkler and Rain and get a meganode.
I have my difficulties with understanding the conditional probability table for the new meganode.
I assume that the values
- $P(S+R|C) = .08$ can be calculated with $.08 = P(S|C) cdot P(R|C) = .10 cdot .80$.
- Same goes for $.02$ with $.02 = P(S|C) cdot P(R|neg C) = .10 cdot .20$.
However I do not understand why
- $P(S+R|neg C) = .40$ could be computed with $.40 = P(S|neg C) cdot P(R|C) = .40 cdot .80$: Since according to the $TT$ column Sprinkler should be true and Cloudy should be true, same with Rain. However we have to use Sprinkler true but Cloudy false.
- or $0.72$ could be computed with $.72 = P(R|C) - ( P(S|C) cdot P(R|C)) = .80 - .10 cdot .80 = .80 - .08$
Hence my overall question is how $P(S+R=x)$ can be computed.
Thank you!
probability machine-learning bayesian naive-bayes
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
$endgroup$
add a comment |
$begingroup$
like others here I am also learning with Russel's and Norvig's book about artificial intelligence. My question is about the conditional probability tables of a clustered multiply connected network with a meganode.
We have the following Bayesian multiply connected network.
In order to reduce computation (according to the book). We combine Sprinkler and Rain and get a meganode.
I have my difficulties with understanding the conditional probability table for the new meganode.
I assume that the values
- $P(S+R|C) = .08$ can be calculated with $.08 = P(S|C) cdot P(R|C) = .10 cdot .80$.
- Same goes for $.02$ with $.02 = P(S|C) cdot P(R|neg C) = .10 cdot .20$.
However I do not understand why
- $P(S+R|neg C) = .40$ could be computed with $.40 = P(S|neg C) cdot P(R|C) = .40 cdot .80$: Since according to the $TT$ column Sprinkler should be true and Cloudy should be true, same with Rain. However we have to use Sprinkler true but Cloudy false.
- or $0.72$ could be computed with $.72 = P(R|C) - ( P(S|C) cdot P(R|C)) = .80 - .10 cdot .80 = .80 - .08$
Hence my overall question is how $P(S+R=x)$ can be computed.
Thank you!
probability machine-learning bayesian naive-bayes
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
$endgroup$
add a comment |
$begingroup$
like others here I am also learning with Russel's and Norvig's book about artificial intelligence. My question is about the conditional probability tables of a clustered multiply connected network with a meganode.
We have the following Bayesian multiply connected network.
In order to reduce computation (according to the book). We combine Sprinkler and Rain and get a meganode.
I have my difficulties with understanding the conditional probability table for the new meganode.
I assume that the values
- $P(S+R|C) = .08$ can be calculated with $.08 = P(S|C) cdot P(R|C) = .10 cdot .80$.
- Same goes for $.02$ with $.02 = P(S|C) cdot P(R|neg C) = .10 cdot .20$.
However I do not understand why
- $P(S+R|neg C) = .40$ could be computed with $.40 = P(S|neg C) cdot P(R|C) = .40 cdot .80$: Since according to the $TT$ column Sprinkler should be true and Cloudy should be true, same with Rain. However we have to use Sprinkler true but Cloudy false.
- or $0.72$ could be computed with $.72 = P(R|C) - ( P(S|C) cdot P(R|C)) = .80 - .10 cdot .80 = .80 - .08$
Hence my overall question is how $P(S+R=x)$ can be computed.
Thank you!
probability machine-learning bayesian naive-bayes
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
$endgroup$
like others here I am also learning with Russel's and Norvig's book about artificial intelligence. My question is about the conditional probability tables of a clustered multiply connected network with a meganode.
We have the following Bayesian multiply connected network.
In order to reduce computation (according to the book). We combine Sprinkler and Rain and get a meganode.
I have my difficulties with understanding the conditional probability table for the new meganode.
I assume that the values
- $P(S+R|C) = .08$ can be calculated with $.08 = P(S|C) cdot P(R|C) = .10 cdot .80$.
- Same goes for $.02$ with $.02 = P(S|C) cdot P(R|neg C) = .10 cdot .20$.
However I do not understand why
- $P(S+R|neg C) = .40$ could be computed with $.40 = P(S|neg C) cdot P(R|C) = .40 cdot .80$: Since according to the $TT$ column Sprinkler should be true and Cloudy should be true, same with Rain. However we have to use Sprinkler true but Cloudy false.
- or $0.72$ could be computed with $.72 = P(R|C) - ( P(S|C) cdot P(R|C)) = .80 - .10 cdot .80 = .80 - .08$
Hence my overall question is how $P(S+R=x)$ can be computed.
Thank you!
probability machine-learning bayesian naive-bayes
probability machine-learning bayesian naive-bayes
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
asked Apr 3 '15 at 18:55
Peter F.wellPeter F.well
1
1
add a comment |
add a comment |
1 Answer
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$begingroup$
I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,
$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4,$$
$$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4.$$
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
$endgroup$
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,
$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4,$$
$$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4.$$
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
$endgroup$
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
add a comment |
$begingroup$
I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,
$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4,$$
$$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4.$$
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
$endgroup$
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
add a comment |
$begingroup$
I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,
$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4,$$
$$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4.$$
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
$endgroup$
I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,
$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5cdot 0.2 =0.1,$$
$$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4,$$
$$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5cdot 0.8 =0.4.$$
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
answered Apr 3 '15 at 19:48
Alex R.Alex R.
24.9k12452
24.9k12452
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
add a comment |
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
$begingroup$
Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1cdot 0.8 =0.08 checkmark \ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1cdot 0.2 =0.02 checkmark \ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5cdot 0.8 =0.4 bigstar\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5cdot 0.2 =0.1 bigstar\ $$ What is my calculation error in the last two lines?
$endgroup$
– Peter F.well
Apr 3 '15 at 20:10
add a comment |
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