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In linear regression overfitting occurs when the model is "too complex". This usually happens when there are a large number of parameters compared to the number of observations. Such a model will not generalise well to new data. That is, it will perform well on training data, but poorly on test data.
A simple simulation can show this. Here I use R:
...
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Overfitting happens when the model performs well on the train data but doesn't do well on the test data. This is because the best fit line by your linear regression model is not a generalized one. This might be due to various factors.
Some of the common factors are
Outliers in the train data.
Train and Test data are from different distributions.
So before ...
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Large number of parameters compared to data points
In general, one aspect of overfitting is trying to "invent information out of knowthing" when you want to determine a comparably large number of parameters from a limited amount of actual evidence data points.
For a simple linear regression y = ax + b there are two parameters, so for most sets of ...
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I'll try and provide some intuition for you here, instead of focusing on the mechanics of the math behind the methods.
Imagine you are evaluating whether a coin is fair or not, so you collect a sequence of heads and tails as your data set. In MLE, we simply look at the data we collected and find the maximum likelihood... this works well when we have no prior ...
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To start with, you could use a simple thresholding.
If you have the dataset $S$ where an element has the form $(x,y,c) \in S$, $x$ denotes the year, $y$ is a binary value (exam passed or not), and $c$ is the student id.
you can obtain a classifier by using
$\{(x,y,c) \in S \mid x \leq \theta\}$ and $\{(x,y,c) \in S \mid x > \theta\}$.
Now you can check ...
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You can use a Sigmoid function on the Force values(Scaled to [0-10] based on a max value)
The Threshold should become 5 after scaling.
def predict_proba(y_pred):
y_pred = y_pred*10/100000 # scaled to [0-10]
thresold = 5
proba = np.exp(y_pred - threshold)/(1 + np.exp(y_pred - threshold))
return proba
predict_proba(100000), predict_proba(...
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If you are using pandas, all you need to do is:
import pandas as pd
import seaborn as sns
import matplotlib.pyplot as plt
corrMatrix = df.corr()
Then you can print the correlation matrix and also plot it using seaborn or any other plotting method.
sns.heatmap(corrMatrix, annot=True)
plt.show()
Hope this helps.
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Under/overfitting depends on two things: the amount of data in your dataset and the complexity of your model.
To identify when each of these is happening, you will have to split the data you have into two parts: training data and test data. You then train your model only on the training data, and then evaluate its performance (e.g. calculate its accuracy or ...
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It's a good idea to remember what you're trying to predict and that is: $\mathbb{E}[Y | X=x]$. The simplest estimate is a single tree:
$$
\hat{\mu}(x) = \sum_{i \leq n} w_i(x, \theta) Y_i
$$
with:
$w_i(x, \theta)$ the single tree node weights (equation 4 in the paper)
$Y_i$ your observations
As we all know this is not a great estimate (high variance among ...
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This would be done with a controlled experiment, similar to clinical tests:
Randomly assign customers into to groups 1 and 2. Group 1 is given the attractive offer, group 2 is not (control group).
It's important to avoid any external bias, so for example ideally nobody in the company should know whether a customer is in group 1 or group 2. The goal is to ...
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You could create a second label for your usernames according to whether they contain london or not (pseudocode below):
for idx, username in df['Usernames']:
if 'London' in username:
df['London'].iloc[idx] = 1
else:
df['London'].iloc[idx] = 0
Consequently given you want to go with correlation and that you are comparing binary ...
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Rationale
Some of the terms are a little vague, particularly what you refer to as eligible students and returned students. I'll set some variables for clarity, but tell me if I defined them incorrectly.
I assume them to mean:
eligible students $ = A $ being the set of all students in the after-school program 2019-2020
returning students $ = A\cap S $ where ...
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No, that isn't what it means.
For one thing it is not clear what parameter the confidence interval that you calculated is for.
In any case, some care is needed in the interpretation of (frequentist) confidence intervals.
In frequentist statistics, a confidence interval is random, and the parameter that the interval is for is fixed. In the case of a 99% ...
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