mock exam 2020

mock exam 2020

by Peter El Hachem -
Number of replies: 2

Hello, 

For Question 3, I don't see how logistic regression is robust ot outliers. Isn't this one of the main reasons we use SVMs ? So that we can be robust to outliers by allowing missclassifications ?


And for the third one, with the aerial images. How is it appropriate to use logistic regression immediately on the images. Shouldn't we use CNN first then apply a logistic regression ?


In reply to Peter El Hachem

Re: mock exam 2020

by Tsz Kin Brian Tsang -
For first question, the logistic regression is compared to the liner regression. While logistic regression can be treated as the "improved" version of the linear regression to deal with the samples that is too close to the boundary. So, we can see logistic regression more robust.

For the second question, the image has been annotated with 0 / 1 already.
In reply to Tsz Kin Brian Tsang

Re: mock exam 2020

by Sena Kiciroglu -
Thanks for answering the question, I'll add some more details :)

1) I'll give an example to illustrate the point further. Let's say that we are trying to classify the following data points with the labels:

X = np.array([[0,0.5,1,6,8,10]]).T
y = np.array([[-1,-1,-1,1,1,1]]).T

It's a dataset of how many hours per week students spend studying for Intro to ML. The labels are -1 if they don't study enough, and +1 if they study enough. I'll first try to predict the labels using linear regression. 

w = np.linalg.inv(X.T@X)@X.T@y
y_pred = w@X.T
print("before thresh", y_pred)
>> before thresh [[0. 0.05590062 0.11180124 0.67080745 0.89440994 1.11801242]]

Since linear regression predicts continous values, we'll have to threshold this. I set a threshold at 0.5 (eyeballed), anything less than this will be labelled -1, more than this will be labelled +1.

After thresholding, my prediction is: 
after thresh [[-1. -1. -1.  1.  1.  1.]]
As for logistic regression on the same data:
clf = LogisticRegression().fit(X, y[:,0])
y_pred = clf.predict(X)
print(y_pred)

>> [-1 -1 -1 1 1 1]

So both algorithms work (for now). Now let's add a new student to this dataset, who studies 40 hours per week (wow!)
X = np.array([[0,0.5,1,6,8,10,40]]).T
y = np.array([[-1,-1,-1,1,1,1,1]]).T

Our linear regression results are: 
[[0.         0.01734906 0.03469813 0.20818876 0.27758501 0.34698126  1.38792505]]
If I keep thresholding at 0.5:[[-1. -1. -1. -1. -1. -1. 1.]]

So we actually have to move our threshold, otherwise the poor students who study 6,7,10 hours would be classified as "not enough". I move the threshold to 0.2. 
[[-1. -1. -1.  1.  1.  1.  1.]]
So it still works, but we had to adjust our threshold due to the outlier. This in itself is not nice. If I have a test data now of a student who studies 5 hours, linear regression gives result 0.17 which is below the threshold of 0.2, and would get classified as "not enough" (seems a bit unfair).

Whereas this outlier did nothing to logistic regression's performance, and we didn't have to tune any threshold. In fact, the classification results are:
[-1 -1 -1  1  1  1  1]
So what happened? With linear regression, we tried to fit a line to the data, which was greatly influenced by the outlier. With logistic regression, the sigmoid function comes to our rescue. Here's a figure the illustrate this:
Linear vs Logistic Regression | Linear and Logistic Regression

2) Each pixel is annotated with 0 and 1, as the previous reply pointed out. This means that each data point can be treated as a sample (we can engineer some features for it, such as the pixel's RGB values, maybe the pixel's neighbour's RGB values, etc.) with a label of 0 or 1. 

What you suggested is also correct by the way, you can feed the images to a CNN to extract per pixel features instead of engineering them by hand, then feed these enhanced data samples to a logistic regression model. Or even better, you can use the CNN for the end task as well: directly get the classification results. You might remember that this is what happens with UNet.

Best,
Sena