C1_W3 Non-linear decision boundaries

Course Q&A Machine Learning Specialization Supervised ML: Regression and Classification
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KakaoTalk_Photo_2024-02-23-21-10-45
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In the picture above, I don’t understand why y_hat = 1 inside the ellipse or more complex shaped decision boundary and y_hat = 0 outside both decision boundaries, instead of the other way around.

Actually I saw the same question before but I still have a question.
In the case of z = w1x1+w2x2+b and considering x1=0, x2=0, z is determined by b. And becasue the shape is ellipse b should be negative. If b is positive every z will be positive. Why y_hat = 1 is inside the ellipse?

Can someone pls help out and explain, thanks?

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Which region is 1 vs 0 depends on the labels in the training set.

You also cannot assume that w or b is always positive. They are real numbers, they can be positive or negative.

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Thanks

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Oh, I got it. Thanks for answering!!

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The y_hat = 1 and y_hat = 0 are just examples here, and you switch them around depending on how you choose to label y_hat.

For example, say you are trying set up a decision boundary for a circle with z = w1 * x1^2 + w2 * x2^2 with logistic regression. So you’ll need to apply the sigmoid activation function:

y_predict = sigmoid(w1 * x1^2 + w2 * x2^2)

What you’ll need next is to determine the threshold, and to make things easier in the lectures, we usually choose threshold to be 0.5.

if y_predict > 0.5, then y_hat = 0
if y_predict <= 0.5, then y_hat = 1

In that case, the model is basically predicting y_hat = 0 if:

y_predict > 0.5
sigmoid(w1 * x1^2 + w2 * x2^2) > 0.5
w1 * x1^2 + w2 * x2^2 > 0.5

And that’s the example you see on the slide above. Above, we chose y_hat = 0 if y_predict > 0.5. However, you can choose the opposite:

if y_predict > 0.5, then y_hat = 1
if y_predict <= 0.5, then y_hat = 0

It’s your decision as the person designing the model.

In order for the decision boundary to look like an ellipsis, you need the squared terms, similar to the example I gave above.

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You have made a very good point :wink: You are the first one here!

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Oh I meant z=w1x1^2+w2x2^2+b. I wrote it wrong.

So you mean that the important point of this lecture is the decision boundary act as the baseline of prediction?

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Oh, it was just a minor mistake! Thanks for answering me!

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