ML Binary Classifier - Probability of getting result by chance - machine-learning

If I have a trained binary classifier, what is the probability of making a correct prediction by chance?
For example, lets say that I want to make 5 predictions. What is the probability of getting all 5 predictions correct by chance?
Is it: 0.5 * 0.5 * 0.5 * 0.5 * 0.5 = 0.0313 ?

You are correct, however, under the assumption that classes are equally probable.
As a similar thought experiment, if you have a model with 99% accuracy (meaning that for any, randomly chosen sample, it will provide correct label 99% of the time), it also does not have high probability of having all samples correctly. For 100 samples it is just about 36%, and for 300 it is less than 5%... for 1000 it is 0.004%.
In general probability of many event happening one by one will fall down very quickly (exponentially) if the probability of each success is constant.

Related

High AUC and 100% recall, but precision and F1 are low

I have an imbalanced dataset which has 43323 rows and 9 of them belong to 'failure' class, other rows belong to 'normal' class. I trained a classifier with 100% recall and 94.89% AUC for test data (0.75/0.25 split with stratify = y). However, the classifier has 0.18% precision & 0.37% F1 score. I assumed I can find better F1 score by changing the threshold but I failed (I checked the threshold between 0 to 1 with step = 0.01). Also, it seems weired to me that usually when dealing with imbalanced dataset, it is hard to get a high recall. The goal is to get a better F1 score. What can I do for the next step? Thanks!
(To be clear, I used SMOTE to upsample the failure samples in training dataset)
Getting 100% recall is trivial in fact: just classify everything as 1.
Is the precision/recall curve any good? Perhaps a more thorough scan could yield a better result:
probabilities = model.predict_proba(X_test)
precision, recall, thresholds = sklearn.metrics.precision_recall_curve(y_test, probabilities)
f1_scores = 2 * recall * precision / (recall + precision)
best_f1 = np.max(f1_scores)
best_thresh = thresholds[np.argmax(f1_scores)]

How to calculate accuracy score of a random classifier?

Say for example, a dataset contains 60% instances for "Yes" class and 30% instances for "NO" class.
In this scenario, Precision, Recall for the random classifier are
Precision =60%
Recall =50%
Then, what will be the accuracy for random classifier in this scenario?
Some caution is required here, since the very definition of a random classifier is somewhat ambiguous; this is best illustrated in cases of imbalanced data.
By definition, the accuracy of a binary classifier is
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
where P stands for probability.
Indeed, if we stick to the intuitive definition of a random binary classifier as giving
P(prediction=0) = P(prediction=1) = 0.5
then the accuracy computed by the above formula is always 0.5, irrespectively of the class distribution (i.e. the values of P(class=0) and P(class=1)).
However, in this definition, there is an implicit assumption, i.e. that our classes are balanced, each one consisting of 50% of our dataset.
This assumption (and the corresponding intuition) breaks down in cases of class imbalance: if we have a dataset where, say, 90% of samples are of class 0 (i.e. P(class=0)=0.9), then it doesn't make much sense to use the above definition of a random binary classifier; instead, we should use the percentages of the class distributions themselves as the probabilities of our random classifier, i.e.:
P(prediction=0) = P(class=0) = 0.9
P(prediction=1) = P(class=1) = 0.1
Now, plugging these values to the formula defining the accuracy, we get:
acc = P(class=0) * P(prediction=0) + P(class=1) * P(prediction=1)
= (0.9 * 0.9) + (0.1 * 0.1)
= 0.82
which is nowhere close to the naive value of 0.5...
As I already said, AFAIK there are no clear-cut definitions of a random classifier in the literature. Sometimes the "naive" random classifier (always flip a fair coin) is referred to as a "random guess" classifier, while what I have described is referred to as a "weighted guess" one, but still this is far from being accepted as a standard...
The bottom line here is the following: since the main reason for using a random classifier is as a baseline, it makes sense to do so only in relatively balanced datasets. In your case of a 60-40 balance, the result turns out to be 0.52, which is admittedly not far from the naive one of 0.5; but for highly imbalanced datasets (e.g. 90-10), the usefulness itself of the random classifier as a baseline ceases to exist, since the correct baseline has become "always predict the majority class", which here would give an accuracy of 90%, in contrast to the random classifier accuracy of just 82% (let alone the 50% accuracy of the naive approach)...
As #desertnaut mentioned, if you're after a naïve benchmark for your model you're always better using "always predict the majority class" as your benchmark, achieving accuracy of %of_samples_in_majority_class (which is always better than either a random guess or a weighted guess).
In Deepchecks (a package I maintain) we have a check that automatically compares the performance of your model to a simple model (either weighted random, majority class or simple decision tree).
from deepchecks.checks import SimpleModelComparison
from deepchecks import Dataset
SimpleModelComparison().run(Dataset(train_df, label='target'), Dataset(test_df, label='target'), model)

Classification with imbalanced dataset using Multi Layer Perceptrons

I am having a trouble in classification problem.
I have almost 400k number of vectors in training data with two labels, and I'd like to train MLP which classifies data into two classes.
However, the dataset is so imbalanced. 95% of them have label 1, and others have label 0. The accuracy grows as training progresses, and stops after reaching 95%. I guess this is because the network predict the label as 1 for all vectors.
So far, I tried dropping out layers with 0.5 probabilities. But, the result is the same. Is there any ways to improve the accuracy?
I think the best way to deal with unbalanced data is to use weights for your class. For example, you can weight your classes such that sum of weights for each class will be equal.
import pandas as pd
df = pd.DataFrame({'x': range(7),
'y': [0] * 2 + [1] * 5})
df['weight'] = df['y'].map(len(df)/2/df['y'].value_counts())
print(df)
print(df.groupby('y')['weight'].agg({'samples': len, 'weight': sum}))
output:
x y weight
0 0 0 1.75
1 1 0 1.75
2 2 1 0.70
3 3 1 0.70
4 4 1 0.70
5 5 1 0.70
6 6 1 0.70
samples weight
y
0 2.0 3.5
1 5.0 3.5
You could try another classifier on subset of examples. SVMs, may work good with small data, so you can take let's say 10k examples only, with 5/1 proportion in classes.
You could also oversample small class somehow and under-sample the another.
You can also simply weight your classes.
Think also about proper metric. It's good that you noticed that the output you have predicts only one label. It is, however, not easily seen using accuracy.
Some nice ideas about unbalanced dataset here:
https://machinelearningmastery.com/tactics-to-combat-imbalanced-classes-in-your-machine-learning-dataset/
Remember not to change your test set.
That's a common situation: the network learns a constant and can't get out of this local minimum.
When the data is very unbalanced, like in your case, one possible solution is a weighted cross entropy loss function. For instance, in tensorflow, apply a built-in tf.nn.weighted_cross_entropy_with_logits function. There is also a good discussion of this idea in this post.
But I should say that getting more data to balance both classes (if that's possible) will always help.

Machine learning using ReLu return NaN

I am trying to figure out this whole machine learning thing, so I was making some testing. I wanted to make it learn the sinus function (with a radian angle). The neural network is:
1 Input (radian angle) / 2 hidden layer / 1 output (prediction of the sinus)
For the squash activation I am using: RELU and it's important to note that when I was using the Logistic function instead of RELU the script was working.
So to do that, I've made a loop that start at 0 and finish at 180, and it will translate the number in radian (radian = loop_index*Math.PI/180) then it'll basically do the sinus of this radian angle and store the radian and the sinus result.
So my table look like this for an entry: {input:[RADIAN ANGLE], output:[sin(radian)]}
for(var i = 0; i <= 180; i++) {
radian = (i*(Math.PI / 180));
train_table.push({input:[radian],output:[Math.sin(radian)]})
}
I use this table to train my Neural Network using Cross Entropy and a learning rate of 0.3 with 20000 iterations.
The problem is that it fail, when I try to predict anything it returns "NaN"
I am using the framework Synaptic (https://github.com/cazala/synaptic) and here is a JSfiddle of my code: https://jsfiddle.net/my7xe9ks/2/
A learning rate must be carefully tuned, this parameter matters a lot, specially when the gradients explode and you get a nan. When this happens, you have to reduce the learning rate, usually by a factor of 10.
In your specific case, the learning rate is too high, if you use 0.05 or 0.01 the network now trains and works properly.
Also another important detail is that you are using cross-entropy as a loss, this loss is used for classification, and you have a regression problem. You should prefer a mean squared error loss instead.

Gradient descent stochastic update - Stopping criterion and update rule - Machine Learning

My dataset has m features and n data points. Let w be a vector (to be estimated). I'm trying to implement gradient descent with stochastic update method. My minimizing function is least mean square.
The update algorithm is shown below:
for i = 1 ... n data:
for t = 1 ... m features:
w_t = w_t - alpha * (<w>.<x_i> - <y_i>) * x_t
where <x> is a raw vector of m features, <y> is a column vector of true labels, and alpha is a constant.
My questions:
Now according to wiki, I don't need to go through all data points and I can stop when error is small enough. Is it true?
I don't understand what should be the stopping criterion here. If anyone can help with this that would be great.
With this formula - which I used in for loop - is it correct? I believe (<w>.<x_i> - <y_i>) * x_t is my ∆Q(w).
Now according to wiki, I don't need to go through all data points and I can stop when error is small enough. Is it true?
This is especially true when you have a really huge training set and going through all the data points is so expensive. Then, you would check the convergence criterion after K stochastic updates (i.e. after processing K training examples). While it's possible, it doesn't make much sense to do this with a small training set. Another thing people do is randomizing the order in which training examples are processed to avoid having too many correlated examples in a raw which may result in "fake" convergence.
I don't understand what should be the stopping criterion here. If anyone can help with this that would be great.
There are a few options. I recommend trying as many of them and deciding based on empirical results.
difference in the objective function for the training data is smaller than a threshold.
difference in the objective function for held-out data (aka. development data, validation data) is smaller than a threshold. The held-out examples should NOT include any of the examples used for training (i.e. for stochastic updates) nor include any of the examples in the test set used for evaluation.
the total absolute difference in parameters w is smaller than a threshold.
in 1, 2, and 3 above, instead of specifying a threshold, you could specify a percentage. For example, a reasonable stopping criterion is to stop training when |squared_error(w) - squared_error(previous_w)| < 0.01 * squared_error(previous_w) $$.
sometimes, we don't care if we have the optimal parameters. We just want to improve the parameters we originally had. In such case, it's reasonable to preset a number of iterations over the training data and stop after that regardless of whether the objective function actually converged.
With this formula - which I used in for loop - is it correct? I believe (w.x_i - y_i) * x_t is my ∆Q(w).
It should be 2 * (w.x_i - y_i) * x_t but it's not a big deal given that you're multiplying by the learning rate alpha anyway.

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