You need to understand the cross-entropy for binary and multi-class problems.

**Multi-class cross-entropy**

Your formula is correct and it directly corresponds to tf.nn.softmax_cross_entropy_with_logits.

**For example:**

-tf.reduce_sum(p * tf.log(q), axis=1)

p and q are expected for probability distributions over N classes. In particular, N can be 2, as in the following example:

p = tf.placeholder(tf.float32, shape=[None, 2])

logit_q = tf.placeholder(tf.float32, shape=[None, 2])

q = tf.nn.softmax(logit_q)

feed_dict = {

p: [[0, 1],

[1, 0],

[1, 0]],

logit_q: [[0.2, 0.8],

[0.7, 0.3],

[0.5, 0.5]]

}

prob1 = -tf.reduce_sum(p * tf.log(q), axis=1)

prob2 = tf.nn.softmax_cross_entropy_with_logits(labels=p, logits=logit_q)

print(prob1.eval(feed_dict)) # [ 0.43748799 0.51301527 0.69314718]

print(prob2.eval(feed_dict)) # [ 0.43748799 0.51301527 0.69314718]

Note that q is computing tf.nn.softmax, i.e. outputs a probability distribution. So it's still multi-class cross-entropy formula, only for N = 2.

**Binary cross-entropy**

This correct formula for this problem is

p * -tf.log(q) + (1 - p) * -tf.log(1 - q)

But it's a case of the multi-class case, the meaning of p and q is different here. Each p and q is a number, corresponding to a probability of class A.

The common part is the p * -tf.log(q) part and the sum. p was a one-hot vector, it can a number, zero or one. Same for q - it was a probability distribution, now's it's a number (probability).

If p is a vector, then each individual component is an independent binary classification. This answer outlines the difference between softmax and sigmoid functions in tensorflow. So the definition p = [0, 0, 0, 1, 0] doesn't mean a one-hot vector, but 5 different features, 4 of which are off and 1 is on. The definition q = [0.2, 0.2, 0.2, 0.2, 0.2] means that each of 5 features is on with 20% probability.

The goal of the sigmoid function is to squash the logit to [0, 1] interval.

This formula can also be used for multiple independent features, and that's what tf.nn.sigmoid_cross_entropy_with_logits computes:

p = tf.placeholder(tf.float32, shape=[None, 5])

logit_q = tf.placeholder(tf.float32, shape=[None, 5])

q = tf.nn.sigmoid(logit_q)

feed_dict = {

p: [[0, 0, 0, 1, 0],

[1, 0, 0, 0, 0]],

logit_q: [[0.2, 0.2, 0.2, 0.2, 0.2],

[0.3, 0.3, 0.2, 0.1, 0.1]]

}

prob1 = -p * tf.log(q)

prob2 = p * -tf.log(q) + (1 - p) * -tf.log(1 - q)

prob3 = p * -tf.log(tf.sigmoid(logit_q)) + (1-p) * -tf.log(1-tf.sigmoid(logit_q))

prob4 = tf.nn.sigmoid_cross_entropy_with_logits(labels=p, logits=logit_q)

print(prob1.eval(feed_dict))

print(prob2.eval(feed_dict))

print(prob3.eval(feed_dict))

print(prob4.eval(feed_dict))

You can notice that the last three tensors are equal, while the prob1 is only a part of cross-entropy, so it contains the correct value only when p is 1:

[[ 0. 0. 0. 0.59813893 0. ]

[ 0.55435514 0. 0. 0. 0. ]]

[[ 0.79813886 0.79813886 0.79813886 0.59813887 0.79813886]

[ 0.5543552 0.85435522 0.79813886 0.74439669 0.74439669]]

[[ 0.7981388 0.7981388 0.7981388 0.59813893 0.7981388 ]

[ 0.55435514 0.85435534 0.7981388 0.74439663 0.74439663]]

[[ 0.7981388 0.7981388 0.7981388 0.59813893 0.7981388 ]

[ 0.55435514 0.85435534 0.7981388 0.74439663 0.74439663]]

If you take the sum of -p * tf.log(q) along axis=1, that doesn't make sense in this syntax, so it'd be a valid formula in multi-class case.

Hope this answer helps

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