Intuition behind logloss function
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2
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I have a difficulty understanding the intuition behind the logloss function since it seems to totally ignore negative examples where y = 0.
The images below visualize my question to some extend:
Your advice will be appreciated!
log-loss
asked Nov 27 at 8:40
user8270077
1462
add a comment |
up vote
2
down vote
favorite
I have a difficulty understanding the intuition behind the logloss function since it seems to totally ignore negative examples where y = 0.
The images below visualize my question to some extend:
Your advice will be appreciated!
log-loss
asked Nov 27 at 8:40
user8270077
1462
1
I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have a difficulty understanding the intuition behind the logloss function since it seems to totally ignore negative examples where y = 0.
The images below visualize my question to some extend:
Your advice will be appreciated!
log-loss
asked Nov 27 at 8:40
user8270077
1462
I have a difficulty understanding the intuition behind the logloss function since it seems to totally ignore negative examples where y = 0.
The images below visualize my question to some extend:
Your advice will be appreciated!
log-loss
log-loss
asked Nov 27 at 8:40
user8270077
1462
asked Nov 27 at 8:40
user8270077
1462
asked Nov 27 at 8:40
user8270077
1462
asked Nov 27 at 8:40
user8270077
1462
asked Nov 27 at 8:40
user8270077
1462
1462
1
I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48
add a comment |
1
I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48
1
1
I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48
I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48
add a comment |
1 Answer
1
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up vote
7
down vote
The formula you used, seems to be
$$
H(X) = -P(X)log P(X),
$$
the definition of entropy. You seem to be asking about cross-entropy loss, also known as log-loss, which is defined as
$$
L(y, hat y) = underbrace{-y log(hat y)}_{text{when } y=1} ;underbrace{- (1-y) log(1-hat y)}_{text{when } y=0}
$$
where $y in {0, 1}$ is the label and $hat y$ is the predicted probability for the label. So the loss is zero for perfect classifications $y = hat y = 1$, or $y = hat y = 0$, and logarithmically increases otherwise.
answered Nov 27 at 9:18
Tim♦
55.2k9124212
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
7
down vote
The formula you used, seems to be
$$
H(X) = -P(X)log P(X),
$$
the definition of entropy. You seem to be asking about cross-entropy loss, also known as log-loss, which is defined as
$$
L(y, hat y) = underbrace{-y log(hat y)}_{text{when } y=1} ;underbrace{- (1-y) log(1-hat y)}_{text{when } y=0}
$$
where $y in {0, 1}$ is the label and $hat y$ is the predicted probability for the label. So the loss is zero for perfect classifications $y = hat y = 1$, or $y = hat y = 0$, and logarithmically increases otherwise.
answered Nov 27 at 9:18
Tim♦
55.2k9124212
add a comment |
up vote
7
down vote
The formula you used, seems to be
$$
H(X) = -P(X)log P(X),
$$
the definition of entropy. You seem to be asking about cross-entropy loss, also known as log-loss, which is defined as
$$
L(y, hat y) = underbrace{-y log(hat y)}_{text{when } y=1} ;underbrace{- (1-y) log(1-hat y)}_{text{when } y=0}
$$
where $y in {0, 1}$ is the label and $hat y$ is the predicted probability for the label. So the loss is zero for perfect classifications $y = hat y = 1$, or $y = hat y = 0$, and logarithmically increases otherwise.
answered Nov 27 at 9:18
Tim♦
55.2k9124212
add a comment |
up vote
7
down vote
up vote
7
down vote
The formula you used, seems to be
$$
H(X) = -P(X)log P(X),
$$
the definition of entropy. You seem to be asking about cross-entropy loss, also known as log-loss, which is defined as
$$
L(y, hat y) = underbrace{-y log(hat y)}_{text{when } y=1} ;underbrace{- (1-y) log(1-hat y)}_{text{when } y=0}
$$
where $y in {0, 1}$ is the label and $hat y$ is the predicted probability for the label. So the loss is zero for perfect classifications $y = hat y = 1$, or $y = hat y = 0$, and logarithmically increases otherwise.
answered Nov 27 at 9:18
Tim♦
55.2k9124212
The formula you used, seems to be
$$
H(X) = -P(X)log P(X),
$$
the definition of entropy. You seem to be asking about cross-entropy loss, also known as log-loss, which is defined as
$$
L(y, hat y) = underbrace{-y log(hat y)}_{text{when } y=1} ;underbrace{- (1-y) log(1-hat y)}_{text{when } y=0}
$$
where $y in {0, 1}$ is the label and $hat y$ is the predicted probability for the label. So the loss is zero for perfect classifications $y = hat y = 1$, or $y = hat y = 0$, and logarithmically increases otherwise.
answered Nov 27 at 9:18
Tim♦
55.2k9124212
edited Nov 27 at 9:27
answered Nov 27 at 9:18
Tim♦
55.2k9124212
answered Nov 27 at 9:18
Tim♦
55.2k9124212
answered Nov 27 at 9:18
Tim♦
55.2k9124212
55.2k9124212
add a comment |
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I don't know what you are trying to do, however read en.wikipedia.org/wiki/… for the cross entropy for logistic regression. Your loss is not specified correctly if this is what you intended to do.
– Cowboy Trader
Nov 27 at 8:48