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I have 38 variables, like oxygen, temperature, pressure, etc and have a task to determine the total yield produced every day from these variables. When I calculate the regression coefficients and intercept value, they seem to be abnormal and very high (Impractical). For example, if 'temperature' coefficient was found to be +375.456, I could not give a meaning to them saying an increase in one unit in temperature would increase yield by 375.456g. That's impractical in my scenario. However, the prediction accuracy seems right. I would like to know, how to interpret these huge intercept( -5341.27355) and huge beta values shown below. One other important point is that I removed multicolinear columns and also, I am not scaling the variables/normalizing them because I need beta coefficients to have meaning such that I could say, increase in temperature by one unit increases yield by 10g or so. Your inputs are highly appreciated!
modl.intercept_
Out[375]: -5341.27354961415
modl.coef_
Out[376]:
array([ 1.38096017e+00, -7.62388829e+00, 5.64611255e+00, 2.26124164e-01,
4.21908571e-01, 4.50695302e-01, -8.15167717e-01, 1.82390184e+00,
-3.32849969e+02, 3.31942553e+02, 3.58830763e+02, -2.05076898e-01,
-3.06404757e+02, 7.86012402e+00, 3.21339318e+02, -7.00817205e-01,
-1.09676321e+04, 1.91481734e+00, 6.02929848e+01, 8.33731416e+00,
-6.23433431e+01, -1.88442804e+00, 6.86526274e+00, -6.76103795e+01,
-1.11406021e+02, 2.48270706e+02, 2.94836048e+01, 1.00279016e+02,
1.42906659e-02, -2.13019683e-03, -6.71427100e+02, -2.03158515e+02,
9.32094007e-03, 5.56457014e+01, -2.91724945e+00, 4.78691176e-01,
8.78121854e+00, -4.93696073e+00])
It's very unlikely that all of these variables are linearly correlated, so I would suggest that you have a look at simple non-linear regression techniques, such as Decision Trees or Kernel Ridge Regression. These are however more difficult to interpret.
Going back to your issue, these high weights might well be due to there being some high amount of correlation between the variables, or that you simply don't have very much training data.
If you instead of linear regression use Lasso Regression, the solution is biased away from high regression coefficients, and the fit will likely improve as well.
A small example on how to do this in scikit-learn, including cross validation of the regularization hyper-parameter:
from sklearn.linear_model LassoCV
# Make up some data
n_samples = 100
n_features = 5
X = np.random.random((n_samples, n_features))
# Make y linear dependent on the features
y = np.sum(np.random.random((1,n_features)) * X, axis=1)
model = LassoCV(cv=5, n_alphas=100, fit_intercept=True)
model.fit(X,y)
print(model.intercept_)
If you have a linear regression, the formula looks like this (y= target, x= features inputs):
y= x1*b1 +x2*b2 + x3*b3 + x4*b4...+ c
where b1,b2,b3,b4... are your modl.coef_. AS you already realized one of your bigges number is 3.319+02 = 331 and the intercept is also quite big with -5431.
As you already mentioned the coeffiecient variables means how much the target variable changes, if the coeffiecient feature changes with 1 unit and all others features are constant.
so for your interpretation, the higher the absoult coeffienct, the higher the influence of your analysis. But it is important to note that the model is using a lot of high coefficient, that means your model is not depending only of one variable
I have recently started experimenting with OneClassSVM ( using Sklearn ) for unsupervised learning and I followed
this example .
I apologize for the silly questions But I’m a bit confused about two things :
Should I train my svm on both regular example case as well as the outliers , or the training is on regular examples only ?
Which of labels predicted by the OSVM and represent outliers is it 1 or -1
Once again i apologize for those questions but for some reason i cannot find this documented anyware
As this example you reference is about novelty-detection, the docs say:
novelty detection:
The training data is not polluted by outliers, and we are interested in detecting anomalies in new observations.
Meaning: you should train on regular examples only.
The approach is based on:
Schölkopf, Bernhard, et al. "Estimating the support of a high-dimensional distribution." Neural computation 13.7 (2001): 1443-1471.
Extract:
Suppose you are given some data set drawn from an underlying probability distribution P and you want to estimate a “simple” subset S of input space such that the probability that a test point drawn from P lies outside of S equals some a priori specied value between 0 and 1.
We propose a method to approach this problem by trying to estimate a function f that is positive on S and negative on the complement.
The above docs also say:
Inliers are labeled 1, while outliers are labeled -1.
This can also be seen in your example code, extracted:
# Generate some regular novel observations
X = 0.3 * np.random.randn(20, 2)
X_test = np.r_[X + 2, X - 2]
...
# all regular = inliers (defined above)
y_pred_test = clf.predict(X_test)
...
# -1 = outlier <-> error as assumed to be inlier
n_error_test = y_pred_test[y_pred_test == -1].size
I'm a newbie to machine learning and this is one of the first real-world ML tasks challenged.
Some experimental data contains 512 independent boolean features and a boolean result.
There are about 1e6 real experiment records in the provided data set.
In a classic XOR example all 4 out of 4 possible states are required to train NN. In my case its only 2^(10-512) = 2^-505 which is close to zero.
I have no more information about the data nature, just these (512 + 1) * 1e6 bits.
Tried NN with 1 hidden layer on available data. Output of the trained NN on the samples even from the training set are always close to 0, not a single close to "1". Played with weights initialization, gradient descent learning rate.
My code utilizing TensorFlow 1.3, Python 3. Model excerpt:
with tf.name_scope("Layer1"):
#W1 = tf.Variable(tf.random_uniform([512, innerN], minval=-2/512, maxval=2/512), name="Weights_1")
W1 = tf.Variable(tf.zeros([512, innerN]), name="Weights_1")
b1 = tf.Variable(tf.zeros([1]), name="Bias_1")
Out1 = tf.sigmoid( tf.matmul(x, W1) + b1)
with tf.name_scope("Layer2"):
W2 = tf.Variable(tf.random_uniform([innerN, 1], minval=-2/512, maxval=2/512), name="Weights_2")
#W2 = tf.Variable(tf.zeros([innerN, 1]), name="Weights_2")
b2 = tf.Variable(tf.zeros([1]), name="Bias_2")
y = tf.nn.sigmoid( tf.matmul(Out1, W2) + b2)
with tf.name_scope("Training"):
y_ = tf.placeholder(tf.float32, [None,1])
cross_entropy = tf.reduce_mean(
tf.nn.softmax_cross_entropy_with_logits(
labels = y_, logits = y)
)
train_step = tf.train.GradientDescentOptimizer(0.005).minimize(cross_entropy)
with tf.name_scope("Testing"):
# Test trained model
correct_prediction = tf.equal( tf.round(y), tf.round(y_))
# ...
# Train
for step in range(500):
batch_xs, batch_ys = Datasets.train.next_batch(300, shuffle=False)
_, my_y, summary = sess.run([train_step, y, merged_summaries],
feed_dict={x: batch_xs, y_: batch_ys})
I suspect two cases:
my fault – bad NN implementation, wrong architecture;
bad data. Compared to XOR example, incomplete training data would result in a failing NN. However, the training examples fed to the trained NN are supposed to give right predictions, aren't they?
How to evaluate if it is possible at all to train a neural network (a 2-layer perceptron) on the provided data to forecast the result? A case of aceptable set would be the XOR example. Opposed to some random noise.
There are only ad hoc ways to know if it is possible to learn a function with a differentiable network from a dataset. That said, these ad hoc ways do usually work. For example, the network should be able to overfit the training set without any regularisation.
A common technique to gauge this is to only fit the network on a subset of the full dataset. Check that the network can overfit to that, then increase the size of the subset, and increase the size of the network as well. Unfortunately, deciding whether to add extra layers or add more units in a hidden layer is an arbitrary decision you'll have to make.
However, looking at your code, there are a few things that could be going wrong here:
Are your outputs balanced? By that I mean, do you have the same number of 1s as 0s in the dataset targets?
Your initialisation in the first layer is all zeros, the gradient to this will be zero, so it can't learn anything (although, you have a real initialisation above it commented out).
Sigmoid nonlinearities are more difficult to optimise than simpler nonlinearities, such as ReLUs.
I'd recommend using the built-in definitions for layers in Tensorflow to not worry about initialisation, and switching to ReLUs in any hidden layers (you need sigmoid at the output for your boolean target).
Finally, deep learning isn't actually very good at most "bag of features" machine learning problems because they lack structure. For example, the order of the features doesn't matter. Other methods often work better, but if you really want to use deep learning then you could look at this recent paper, showing improved performance by just using a very specific nonlinearity and weight initialisation (change 4 lines in your code above).
I'm trying to train a CNN to categorize text by topic. When I use binary cross-entropy I get ~80% accuracy, with categorical cross-entropy I get ~50% accuracy.
I don't understand why this is. It's a multiclass problem, doesn't that mean that I have to use categorical cross-entropy and that the results with binary cross-entropy are meaningless?
model.add(embedding_layer)
model.add(Dropout(0.25))
# convolution layers
model.add(Conv1D(nb_filter=32,
filter_length=4,
border_mode='valid',
activation='relu'))
model.add(MaxPooling1D(pool_length=2))
# dense layers
model.add(Flatten())
model.add(Dense(256))
model.add(Dropout(0.25))
model.add(Activation('relu'))
# output layer
model.add(Dense(len(class_id_index)))
model.add(Activation('softmax'))
Then I compile it either it like this using categorical_crossentropy as the loss function:
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
or
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
Intuitively it makes sense why I'd want to use categorical cross-entropy, I don't understand why I get good results with binary, and poor results with categorical.
The reason for this apparent performance discrepancy between categorical & binary cross entropy is what user xtof54 has already reported in his answer below, i.e.:
the accuracy computed with the Keras method evaluate is just plain
wrong when using binary_crossentropy with more than 2 labels
I would like to elaborate more on this, demonstrate the actual underlying issue, explain it, and offer a remedy.
This behavior is not a bug; the underlying reason is a rather subtle & undocumented issue at how Keras actually guesses which accuracy to use, depending on the loss function you have selected, when you include simply metrics=['accuracy'] in your model compilation. In other words, while your first compilation option
model.compile(loss='categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
is valid, your second one:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy'])
will not produce what you expect, but the reason is not the use of binary cross entropy (which, at least in principle, is an absolutely valid loss function).
Why is that? If you check the metrics source code, Keras does not define a single accuracy metric, but several different ones, among them binary_accuracy and categorical_accuracy. What happens under the hood is that, since you have selected binary cross entropy as your loss function and have not specified a particular accuracy metric, Keras (wrongly...) infers that you are interested in the binary_accuracy, and this is what it returns - while in fact you are interested in the categorical_accuracy.
Let's verify that this is the case, using the MNIST CNN example in Keras, with the following modification:
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=['accuracy']) # WRONG way
model.fit(x_train, y_train,
batch_size=batch_size,
epochs=2, # only 2 epochs, for demonstration purposes
verbose=1,
validation_data=(x_test, y_test))
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.9975801164627075
# Actual accuracy calculated manually:
import numpy as np
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98780000000000001
score[1]==acc
# False
To remedy this, i.e. to use indeed binary cross entropy as your loss function (as I said, nothing wrong with this, at least in principle) while still getting the categorical accuracy required by the problem at hand, you should ask explicitly for categorical_accuracy in the model compilation as follows:
from keras.metrics import categorical_accuracy
model.compile(loss='binary_crossentropy', optimizer='adam', metrics=[categorical_accuracy])
In the MNIST example, after training, scoring, and predicting the test set as I show above, the two metrics now are the same, as they should be:
# Keras reported accuracy:
score = model.evaluate(x_test, y_test, verbose=0)
score[1]
# 0.98580000000000001
# Actual accuracy calculated manually:
y_pred = model.predict(x_test)
acc = sum([np.argmax(y_test[i])==np.argmax(y_pred[i]) for i in range(10000)])/10000
acc
# 0.98580000000000001
score[1]==acc
# True
System setup:
Python version 3.5.3
Tensorflow version 1.2.1
Keras version 2.0.4
UPDATE: After my post, I discovered that this issue had already been identified in this answer.
It all depends on the type of classification problem you are dealing with. There are three main categories
binary classification (two target classes),
multi-class classification (more than two exclusive targets),
multi-label classification (more than two non exclusive targets), in which multiple target classes can be on at the same time.
In the first case, binary cross-entropy should be used and targets should be encoded as one-hot vectors.
In the second case, categorical cross-entropy should be used and targets should be encoded as one-hot vectors.
In the last case, binary cross-entropy should be used and targets should be encoded as one-hot vectors. Each output neuron (or unit) is considered as a separate random binary variable, and the loss for the entire vector of outputs is the product of the loss of single binary variables. Therefore it is the product of binary cross-entropy for each single output unit.
The binary cross-entropy is defined as
and categorical cross-entropy is defined as
where c is the index running over the number of classes C.
I came across an "inverted" issue — I was getting good results with categorical_crossentropy (with 2 classes) and poor with binary_crossentropy. It seems that problem was with wrong activation function. The correct settings were:
for binary_crossentropy: sigmoid activation, scalar target
for categorical_crossentropy: softmax activation, one-hot encoded target
It's really interesting case. Actually in your setup the following statement is true:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
This means that up to a constant multiplication factor your losses are equivalent. The weird behaviour that you are observing during a training phase might be an example of a following phenomenon:
At the beginning the most frequent class is dominating the loss - so network is learning to predict mostly this class for every example.
After it learnt the most frequent pattern it starts discriminating among less frequent classes. But when you are using adam - the learning rate has a much smaller value than it had at the beginning of training (it's because of the nature of this optimizer). It makes training slower and prevents your network from e.g. leaving a poor local minimum less possible.
That's why this constant factor might help in case of binary_crossentropy. After many epochs - the learning rate value is greater than in categorical_crossentropy case. I usually restart training (and learning phase) a few times when I notice such behaviour or/and adjusting a class weights using the following pattern:
class_weight = 1 / class_frequency
This makes loss from a less frequent classes balancing the influence of a dominant class loss at the beginning of a training and in a further part of an optimization process.
EDIT:
Actually - I checked that even though in case of maths:
binary_crossentropy = len(class_id_index) * categorical_crossentropy
should hold - in case of keras it's not true, because keras is automatically normalizing all outputs to sum up to 1. This is the actual reason behind this weird behaviour as in case of multiclassification such normalization harms a training.
After commenting #Marcin answer, I have more carefully checked one of my students code where I found the same weird behavior, even after only 2 epochs ! (So #Marcin's explanation was not very likely in my case).
And I found that the answer is actually very simple: the accuracy computed with the Keras method evaluate is just plain wrong when using binary_crossentropy with more than 2 labels. You can check that by recomputing the accuracy yourself (first call the Keras method "predict" and then compute the number of correct answers returned by predict): you get the true accuracy, which is much lower than the Keras "evaluate" one.
a simple example under a multi-class setting to illustrate
suppose you have 4 classes (onehot encoded) and below is just one prediction
true_label = [0,1,0,0]
predicted_label = [0,0,1,0]
when using categorical_crossentropy, the accuracy is just 0 , it only cares about if you get the concerned class right.
however when using binary_crossentropy, the accuracy is calculated for all classes, it would be 50% for this prediction. and the final result will be the mean of the individual accuracies for both cases.
it is recommended to use categorical_crossentropy for multi-class(classes are mutually exclusive) problem but binary_crossentropy for multi-label problem.
As it is a multi-class problem, you have to use the categorical_crossentropy, the binary cross entropy will produce bogus results, most likely will only evaluate the first two classes only.
50% for a multi-class problem can be quite good, depending on the number of classes. If you have n classes, then 100/n is the minimum performance you can get by outputting a random class.
You are passing a target array of shape (x-dim, y-dim) while using as loss categorical_crossentropy. categorical_crossentropy expects targets to be binary matrices (1s and 0s) of shape (samples, classes). If your targets are integer classes, you can convert them to the expected format via:
from keras.utils import to_categorical
y_binary = to_categorical(y_int)
Alternatively, you can use the loss function sparse_categorical_crossentropy instead, which does expect integer targets.
model.compile(loss='sparse_categorical_crossentropy', optimizer='adam', metrics=['accuracy'])
when using the categorical_crossentropy loss, your targets should be in categorical format (e.g. if you have 10 classes, the target for each sample should be a 10-dimensional vector that is all-zeros except for a 1 at the index corresponding to the class of the sample).
Take a look at the equation you can find that binary cross entropy not only punish those label = 1, predicted =0, but also label = 0, predicted = 1.
However categorical cross entropy only punish those label = 1 but predicted = 1.That's why we make assumption that there is only ONE label positive.
The main point is answered satisfactorily with the brilliant piece of sleuthing by desernaut. However there are occasions when BCE (binary cross entropy) could throw different results than CCE (categorical cross entropy) and may be the preferred choice. While the thumb rules shared above (which loss to select) work fine for 99% of the cases, I would like to add a few new dimensions to this discussion.
The OP had a softmax activation and this throws a probability distribution as the predicted value. It is a multi-class problem. The preferred loss is categorical CE. Essentially this boils down to -ln(p) where 'p' is the predicted probability of the lone positive class in the sample. This means that the negative predictions dont have a role to play in calculating CE. This is by intention.
On a rare occasion, it may be needed to make the -ve voices count. This can be done by treating the above sample as a series of binary predictions. So if expected is [1 0 0 0 0] and predicted is [0.1 0.5 0.1 0.1 0.2], this is further broken down into:
expected = [1,0], [0,1], [0,1], [0,1], [0,1]
predicted = [0.1, 0.9], [.5, .5], [.1, .9], [.1, .9], [.2, .8]
Now we proceed to compute 5 different cross entropies - one for each of the above 5 expected/predicted combo and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.8)]
The CE has a different scale but continues to be a measure of the difference between the expected and predicted values. The only difference is that in this scheme, the -ve values are also penalized/rewarded along with the +ve values. In case your problem is such that you are going to use the output probabilities (both +ve and -ves) instead of using the max() to predict just the 1 +ve label, then you may want to consider this version of CE.
How about a multi-label situation where expected = [1 0 0 0 1]? Conventional approach is to use one sigmoid per output neuron instead of an overall softmax. This ensures that the output probabilities are independent of each other. So we get something like:
expected = [1 0 0 0 1]
predicted is = [0.1 0.5 0.1 0.1 0.9]
By definition, CE measures the difference between 2 probability distributions. But the above two lists are not probability distributions. Probability distributions should always add up to 1. So conventional solution is to use same loss approach as before - break the expected and predicted values into 5 individual probability distributions, proceed to calculate 5 cross entropies and sum them up. Then:
CE = -[ ln(.1) + ln(0.5) + ln(0.9) + ln(0.9) + ln(0.9)] = 3.3
The challenge happens when the number of classes may be very high - say a 1000 and there may be only couple of them present in each sample. So the expected is something like: [1,0,0,0,0,0,1,0,0,0.....990 zeroes]. The predicted could be something like: [.8, .1, .1, .1, .1, .1, .8, .1, .1, .1.....990 0.1's]
In this case the CE =
- [ ln(.8) + ln(.8) for the 2 +ve classes and 998 * ln(0.9) for the 998 -ve classes]
= 0.44 (for the +ve classes) + 105 (for the negative classes)
You can see how the -ve classes are beginning to create a nuisance value when calculating the loss. The voice of the +ve samples (which may be all that we care about) is getting drowned out. What do we do? We can't use categorical CE (the version where only +ve samples are considered in calculation). This is because, we are forced to break up the probability distributions into multiple binary probability distributions because otherwise it would not be a probability distribution in the first place. Once we break it into multiple binary probability distributions, we have no choice but to use binary CE and this of course gives weightage to -ve classes.
One option is to drown the voice of the -ve classes by a multiplier. So we multiply all -ve losses by a value gamma where gamma < 1. Say in above case, gamma can be .0001. Now the loss comes to:
= 0.44 (for the +ve classes) + 0.105 (for the negative classes)
The nuisance value has come down. 2 years back Facebook did that and much more in a paper they came up with where they also multiplied the -ve losses by p to the power of x. 'p' is the probability of the output being a +ve and x is a constant>1. This penalized -ve losses even further especially the ones where the model is pretty confident (where 1-p is close to 1). This combined effect of punishing negative class losses combined with harsher punishment for the easily classified cases (which accounted for majority of the -ve cases) worked beautifully for Facebook and they called it focal loss.
So in response to OP's question of whether binary CE makes any sense at all in his case, the answer is - it depends. In 99% of the cases the conventional thumb rules work but there could be occasions when these rules could be bent or even broken to suit the problem at hand.
For a more in-depth treatment, you can refer to: https://towardsdatascience.com/cross-entropy-classification-losses-no-math-few-stories-lots-of-intuition-d56f8c7f06b0
The binary_crossentropy(y_target, y_predict) doesn't need to apply to binary classification problem.
In the source code of binary_crossentropy(), the nn.sigmoid_cross_entropy_with_logits(labels=target, logits=output) of tensorflow was actually used.
And, in the documentation, it says that:
Measures the probability error in discrete classification tasks in which each class is independent and not mutually exclusive. For instance, one could perform multilabel classification where a picture can contain both an elephant and a dog at the same time.
Is it a good practice to use sigmoid or tanh output layers in Neural networks directly to estimate probabilities?
i.e the probability of given input to occur is the output of sigmoid function in the NN
EDIT
I wanted to use neural network to learn and predict the probability of a given input to occur..
You may consider the input as State1-Action-State2 tuple.
Hence the output of NN is the probability that State2 happens when applying Action on State1..
I Hope that does clear things..
EDIT
When training NN, I do random Action on State1 and observe resultant State2; then teach NN that input State1-Action-State2 should result in output 1.0
First, just a couple of small points on the conventional MLP lexicon (might help for internet searches, etc.): 'sigmoid' and 'tanh' are not 'output layers' but functions, usually referred to as "activation functions". The return value of the activation function is indeed the output from each layer, but they are not the output layer themselves (nor do they calculate probabilities).
Additionally, your question recites a choice between two "alternatives" ("sigmoid and tanh"), but they are not actually alternatives, rather the term 'sigmoidal function' is a generic/informal term for a class of functions, which includes the hyperbolic tangent ('tanh') that you refer to.
The term 'sigmoidal' is probably due to the characteristic shape of the function--the return (y) values are constrained between two asymptotic values regardless of the x value. The function output is usually normalized so that these two values are -1 and 1 (or 0 and 1). (This output behavior, by the way, is obviously inspired by the biological neuron which either fires (+1) or it doesn't (-1)). A look at the key properties of sigmoidal functions and you can see why they are ideally suited as activation functions in feed-forward, backpropagating neural networks: (i) real-valued and differentiable, (ii) having exactly one inflection point, and (iii) having a pair of horizontal asymptotes.
In turn, the sigmoidal function is one category of functions used as the activation function (aka "squashing function") in FF neural networks solved using backprop. During training or prediction, the weighted sum of the inputs (for a given layer, one layer at a time) is passed in as an argument to the activation function which returns the output for that layer. Another group of functions apparently used as the activation function is piecewise linear function. The step function is the binary variant of a PLF:
def step_fn(x) :
if x <= 0 :
y = 0
if x > 0 :
y = 1
(On practical grounds, I doubt the step function is a plausible choice for the activation function, but perhaps it helps understand the purpose of the activation function in NN operation.)
I suppose there an unlimited number of possible activation functions, but in practice, you only see a handful; in fact just two account for the overwhelming majority of cases (both are sigmoidal). Here they are (in python) so you can experiment for yourself, given that the primary selection criterion is a practical one:
# logistic function
def sigmoid2(x) :
return 1 / (1 + e**(-x))
# hyperbolic tangent
def sigmoid1(x) :
return math.tanh(x)
what are the factors to consider in selecting an activation function?
First the function has to give the desired behavior (arising from or as evidenced by sigmoidal shape). Second, the function must be differentiable. This is a requirement for backpropagation, which is the optimization technique used during training to 'fill in' the values of the hidden layers.
For instance, the derivative of the hyperbolic tangent is (in terms of the output, which is how it is usually written) :
def dsigmoid(y) :
return 1.0 - y**2
Beyond those two requriements, what makes one function between than another is how efficiently it trains the network--i.e., which one causes convergence (reaching the local minimum error) in the fewest epochs?
#-------- Edit (see OP's comment below) ---------#
I am not quite sure i understood--sometimes it's difficult to communicate details of a NN, without the code, so i should probably just say that it's fine subject to this proviso: What you want the NN to predict must be the same as the dependent variable used during training. So for instance, if you train your NN using two states (e.g., 0, 1) as the single dependent variable (which is obviously missing from your testing/production data) then that's what your NN will return when run in "prediction mode" (post training, or with a competent weight matrix).
You should choose the right loss function to minimize.
The squared error does not lead to the maximum likelihood hypothesis here.
The squared error is derived from a model with Gaussian noise:
P(y|x,h) = k1 * e**-(k2 * (y - h(x))**2)
You estimate the probabilities directly. Your model is:
P(Y=1|x,h) = h(x)
P(Y=0|x,h) = 1 - h(x)
P(Y=1|x,h) is the probability that event Y=1 will happen after seeing x.
The maximum likelihood hypothesis for your model is:
h_max_likelihood = argmax_h product(
h(x)**y * (1-h(x))**(1-y) for x, y in examples)
This leads to the "cross entropy" loss function.
See chapter 6 in Mitchell's Machine Learning
for the loss function and its derivation.
There is one problem with this approach: if you have vectors from R^n and your network maps those vectors into the interval [0, 1], it will not be guaranteed that the network represents a valid probability density function, since the integral of the network is not guaranteed to equal 1.
E.g., a neural network could map any input form R^n to 1.0. But that is clearly not possible.
So the answer to your question is: no, you can't.
However, you can just say that your network never sees "unrealistic" code samples and thus ignore this fact. For a discussion of this (and also some more cool information on how to model PDFs with neural networks) see contrastive backprop.