Naive Bayes Classifier - Discriminant Function2019 Community Moderator ElectionSPARK, ML: Naive Bayes classifier often assigns 1 as probability predictionHandling underflow in a Gaussian Naive Bayes classifierName Entity Linking with Naive Bayes ClassifierOverfitting Naive BayesBias in Naive Bayes classifierHow do i use the Gaussian function with a Naive Bayes Classifier?Very low probability in naive Bayes classifierBinary classification, precision-recall curve and thresholdsOne class naive bayesNaive Bayes Classifier
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Naive Bayes Classifier - Discriminant Function
2019 Community Moderator ElectionSPARK, ML: Naive Bayes classifier often assigns 1 as probability predictionHandling underflow in a Gaussian Naive Bayes classifierName Entity Linking with Naive Bayes ClassifierOverfitting Naive BayesBias in Naive Bayes classifierHow do i use the Gaussian function with a Naive Bayes Classifier?Very low probability in naive Bayes classifierBinary classification, precision-recall curve and thresholdsOne class naive bayesNaive Bayes Classifier
$begingroup$
To classify my samples, I decided to use Naive Bayes classifier, but I coded it, not used built-in library functions.
If I use this equality, I obtain nice classification accuracy: p1(x) > p2(x) => x belongs to C1
However, I could not understand why discriminant functions produce negative values. If they are probability functions, I think they must generate a value between 0 and 1.
Is there anyone who can explain the reason ?
classification naive-bayes-classifier discriminant-analysis
asked Mar 28 at 18:35
GoktugGoktug
1083
$endgroup$
add a comment |
$begingroup$
To classify my samples, I decided to use Naive Bayes classifier, but I coded it, not used built-in library functions.
If I use this equality, I obtain nice classification accuracy: p1(x) > p2(x) => x belongs to C1
However, I could not understand why discriminant functions produce negative values. If they are probability functions, I think they must generate a value between 0 and 1.
Is there anyone who can explain the reason ?
classification naive-bayes-classifier discriminant-analysis
asked Mar 28 at 18:35
GoktugGoktug
1083
$endgroup$
add a comment |
$begingroup$
To classify my samples, I decided to use Naive Bayes classifier, but I coded it, not used built-in library functions.
If I use this equality, I obtain nice classification accuracy: p1(x) > p2(x) => x belongs to C1
However, I could not understand why discriminant functions produce negative values. If they are probability functions, I think they must generate a value between 0 and 1.
Is there anyone who can explain the reason ?
classification naive-bayes-classifier discriminant-analysis
asked Mar 28 at 18:35
GoktugGoktug
1083
$endgroup$
To classify my samples, I decided to use Naive Bayes classifier, but I coded it, not used built-in library functions.
If I use this equality, I obtain nice classification accuracy: p1(x) > p2(x) => x belongs to C1
However, I could not understand why discriminant functions produce negative values. If they are probability functions, I think they must generate a value between 0 and 1.
Is there anyone who can explain the reason ?
classification naive-bayes-classifier discriminant-analysis
classification naive-bayes-classifier discriminant-analysis
asked Mar 28 at 18:35
GoktugGoktug
1083
asked Mar 28 at 18:35
GoktugGoktug
1083
asked Mar 28 at 18:35
GoktugGoktug
1083
asked Mar 28 at 18:35
GoktugGoktug
1083
asked Mar 28 at 18:35
GoktugGoktug
1083
1083
add a comment |
add a comment |
1 Answer
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In Naive Bayes, for the case of two classes, a discriminant function could be $$D(boldsymbolx) = fracP(boldsymbolx, c=1)P(boldsymbolx, c=0)$$ which can be anywhere in $[0, +infty)$, and decides $c=1$ if $D(boldsymbolx)>1$, $c=0$ otherwise, or it could be the logarithm of that value
$$d(boldsymbolx) = textlogfracP(boldsymbolx, c=1)P(boldsymbolx, c=0)=textlogP(boldsymbolx, c=1)-textlogP(boldsymbolx, c=0)$$
which can be anywhere in $(-infty, +infty)$ (handling zero probability as a special case), and decides $c=1$ if $d(boldsymbolx)>0$, $c=0$ otherwise.
As a side note, $P(boldsymbolx, c=k)$ in Naive Bayes is calculated as
$$P(boldsymbolx, c=k)=P(c=k)prod_i=1^dP(x_i|c=k)$$
or equivalently for log probabilities as
$$textlogP(boldsymbolx, c=k)=textlogP(c=k) + sum_i=1^dtextlogP(x_i|c=k)$$
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
$endgroup$
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
add a comment |
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$begingroup$
In Naive Bayes, for the case of two classes, a discriminant function could be $$D(boldsymbolx) = fracP(boldsymbolx, c=1)P(boldsymbolx, c=0)$$ which can be anywhere in $[0, +infty)$, and decides $c=1$ if $D(boldsymbolx)>1$, $c=0$ otherwise, or it could be the logarithm of that value
$$d(boldsymbolx) = textlogfracP(boldsymbolx, c=1)P(boldsymbolx, c=0)=textlogP(boldsymbolx, c=1)-textlogP(boldsymbolx, c=0)$$
which can be anywhere in $(-infty, +infty)$ (handling zero probability as a special case), and decides $c=1$ if $d(boldsymbolx)>0$, $c=0$ otherwise.
As a side note, $P(boldsymbolx, c=k)$ in Naive Bayes is calculated as
$$P(boldsymbolx, c=k)=P(c=k)prod_i=1^dP(x_i|c=k)$$
or equivalently for log probabilities as
$$textlogP(boldsymbolx, c=k)=textlogP(c=k) + sum_i=1^dtextlogP(x_i|c=k)$$
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
$endgroup$
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
add a comment |
$begingroup$
In Naive Bayes, for the case of two classes, a discriminant function could be $$D(boldsymbolx) = fracP(boldsymbolx, c=1)P(boldsymbolx, c=0)$$ which can be anywhere in $[0, +infty)$, and decides $c=1$ if $D(boldsymbolx)>1$, $c=0$ otherwise, or it could be the logarithm of that value
$$d(boldsymbolx) = textlogfracP(boldsymbolx, c=1)P(boldsymbolx, c=0)=textlogP(boldsymbolx, c=1)-textlogP(boldsymbolx, c=0)$$
which can be anywhere in $(-infty, +infty)$ (handling zero probability as a special case), and decides $c=1$ if $d(boldsymbolx)>0$, $c=0$ otherwise.
As a side note, $P(boldsymbolx, c=k)$ in Naive Bayes is calculated as
$$P(boldsymbolx, c=k)=P(c=k)prod_i=1^dP(x_i|c=k)$$
or equivalently for log probabilities as
$$textlogP(boldsymbolx, c=k)=textlogP(c=k) + sum_i=1^dtextlogP(x_i|c=k)$$
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
$endgroup$
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
add a comment |
$begingroup$
In Naive Bayes, for the case of two classes, a discriminant function could be $$D(boldsymbolx) = fracP(boldsymbolx, c=1)P(boldsymbolx, c=0)$$ which can be anywhere in $[0, +infty)$, and decides $c=1$ if $D(boldsymbolx)>1$, $c=0$ otherwise, or it could be the logarithm of that value
$$d(boldsymbolx) = textlogfracP(boldsymbolx, c=1)P(boldsymbolx, c=0)=textlogP(boldsymbolx, c=1)-textlogP(boldsymbolx, c=0)$$
which can be anywhere in $(-infty, +infty)$ (handling zero probability as a special case), and decides $c=1$ if $d(boldsymbolx)>0$, $c=0$ otherwise.
As a side note, $P(boldsymbolx, c=k)$ in Naive Bayes is calculated as
$$P(boldsymbolx, c=k)=P(c=k)prod_i=1^dP(x_i|c=k)$$
or equivalently for log probabilities as
$$textlogP(boldsymbolx, c=k)=textlogP(c=k) + sum_i=1^dtextlogP(x_i|c=k)$$
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
$endgroup$
In Naive Bayes, for the case of two classes, a discriminant function could be $$D(boldsymbolx) = fracP(boldsymbolx, c=1)P(boldsymbolx, c=0)$$ which can be anywhere in $[0, +infty)$, and decides $c=1$ if $D(boldsymbolx)>1$, $c=0$ otherwise, or it could be the logarithm of that value
$$d(boldsymbolx) = textlogfracP(boldsymbolx, c=1)P(boldsymbolx, c=0)=textlogP(boldsymbolx, c=1)-textlogP(boldsymbolx, c=0)$$
which can be anywhere in $(-infty, +infty)$ (handling zero probability as a special case), and decides $c=1$ if $d(boldsymbolx)>0$, $c=0$ otherwise.
As a side note, $P(boldsymbolx, c=k)$ in Naive Bayes is calculated as
$$P(boldsymbolx, c=k)=P(c=k)prod_i=1^dP(x_i|c=k)$$
or equivalently for log probabilities as
$$textlogP(boldsymbolx, c=k)=textlogP(c=k) + sum_i=1^dtextlogP(x_i|c=k)$$
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
edited Mar 28 at 19:01
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
answered Mar 28 at 18:51
EsmailianEsmailian
2,805318
2,805318
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
add a comment |
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
1
1
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
$begingroup$
Thank you so much. While I was designing my bayes classifier, I took natural logarithm of posterior probability to make probability formula less complicated. Because I benefited from normal distribution function when I define probability discriminant function.
$endgroup$
– Goktug
Mar 28 at 19:15
add a comment |
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