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What loss function to use when labels are probabilities?

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2

$begingroup$

What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].

It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.

Would something like MSE (after applying softmax) make sense, or is there a better loss function?

neural-networks loss-functions probability-distribution

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asked 7 hours ago

Thomas JohnsonThomas Johnson

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2

$begingroup$

What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].

It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.

Would something like MSE (after applying softmax) make sense, or is there a better loss function?

neural-networks loss-functions probability-distribution

share|improve this question

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

New contributor

Thomas Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].

It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.

Would something like MSE (after applying softmax) make sense, or is there a better loss function?

neural-networks loss-functions probability-distribution

share|improve this question

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

New contributor

Thomas Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

New contributor

Thomas Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

New contributor

Thomas Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

New contributor

Thomas Johnson is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 7 hours ago

Thomas JohnsonThomas Johnson

1133

1 Answer
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3

$begingroup$

Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.

You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,

$$H(p,q)=-sum_xin X p(x) log q(x).$$
$ $

Note that one-hot labels would mean that
$$
p(x) =
begincases
1 & textif x text is the true label\
0 & textotherwise
endcases$$

which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:

$$H(p,q) = -log q(x_label)$$

share|improve this answer

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

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3

$begingroup$

Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.

You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,

$$H(p,q)=-sum_xin X p(x) log q(x).$$
$ $

Note that one-hot labels would mean that
$$
p(x) =
begincases
1 & textif x text is the true label\
0 & textotherwise
endcases$$

which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:

$$H(p,q) = -log q(x_label)$$

share|improve this answer

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

$endgroup$

add a comment | 

3

$begingroup$

Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.

You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,

$$H(p,q)=-sum_xin X p(x) log q(x).$$
$ $

Note that one-hot labels would mean that
$$
p(x) =
begincases
1 & textif x text is the true label\
0 & textotherwise
endcases$$

which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:

$$H(p,q) = -log q(x_label)$$

share|improve this answer

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

$endgroup$

add a comment | 

$begingroup$

Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.

You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,

$$H(p,q)=-sum_xin X p(x) log q(x).$$
$ $

Note that one-hot labels would mean that
$$
p(x) =
begincases
1 & textif x text is the true label\
0 & textotherwise
endcases$$

which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:

$$H(p,q) = -log q(x_label)$$

share|improve this answer

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

$endgroup$

share|improve this answer

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119

answered 7 hours ago

Philip RaeisghasemPhilip Raeisghasem

998119