# Coursera: Machine Learning (Week 3) Quiz - Logistic Regression | Andrew NG

# ▸ Logistic Regression :

- Suppose that you have trained a logistic regression classifier, and it outputs on a new example a prediction = 0.2. This means (check all that apply):
- Our estimate for P(y = 1|x; θ) is 0.8.
h(x) gives P(y=1|x; θ), not 1 - P(y=1|x; θ)

- Our estimate for P(y = 0|x; θ) is 0.8.
Since we must have P(y=0|x;θ) = 1 - P(y=1|x; θ), the former is

1 - 0.2 = 0.8. - Our estimate for P(y = 1|x; θ) is 0.2.
h(x) is precisely P(y=1|x; θ), so each is 0.2.

- Our estimate for P(y = 0|x; θ) is 0.2.
h(x) is P(y=1|x; θ), not P(y=0|x; θ)

- Our estimate for P(y = 1|x; θ) is 0.8.

- Suppose you have the following training set, and fit a logistic regression classifier .

**Which of the following are true? Check all that apply.**- Adding polynomial features (e.g., instead using ) could increase how well we can fit the training data.
- At the optimal value of θ (e.g., found by fminunc), we will have J(θ) ≥ 0.
- Adding polynomial features (e.g., instead using ) would increase J(θ) because we are now summing over more terms.
- If we train gradient descent for enough iterations, for some examples in the training set it is possible to obtain .

- For logistic regression, the gradient is given by . Which of these is a correct gradient descent update for logistic regression with a learning rate of ? Check all that apply.

- Which of the following statements are true? Check all that apply.
- The one-vs-all technique allows you to use logistic regression for problems in which each comes from a fixed, discrete set of values.
If each is one of k different values, we can give a label to each and use one-vs-all as described in the lecture.

- For logistic regression, sometimes gradient descent will converge to a local minimum (and fail to find the global minimum). This is the reason we prefer more advanced optimization algorithms such as fminunc (conjugate gradient/BFGS/L-BFGS/etc).
The cost function for logistic regression is convex, so gradient descent will always converge to the global minimum. We still might use a more advanced optimisation algorithm since they can be faster and don’t require you to select a learning rate.

- The cost function for logistic regression trained with examples is always greater than or equal to zero.
The cost for any example is always since it is the negative log of a quantity less than one. The cost function is a summation over the cost for each sample, so the cost function itself must be greater than or equal to zero.

- Since we train one classifier when there are two classes, we train two classifiers when there are three classes (and we do one-vs-all classification).
We will need 3 classfiers. One-for-each class.

- The one-vs-all technique allows you to use logistic regression for problems in which each comes from a fixed, discrete set of values.

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- Suppose you train a logistic classifier . Suppose , , . Which of the following figures represents the decision boundary found by your classifier?
- Figure:

In this figure, we transition from negative to positive when x1 goes from left of 6 to right of 6 which is true for the given values of θ.

- Figure:

- Figure:

- Figure:

- Figure:

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Fifth question is wrong

ReplyDeleteOk. What do you think the correct answer is??

Deletecorrect answer is D.

DeleteThanks for the response.But simply Not possible.

DeleteIf you substitute to value of theta1 & theta2 in the equation given in the question, you will get x1=6. If you plot the graph x1=6 line, that line will always be parallel to x2 axis.

Correct answer is "A" only.

Please check your question. Only possible reason is your question is slightly different than mine but you didn't read it carefully and copied the answer directly.

Please check once again.

no ur answer is wrong ..it will be D..THANK ME LATER

Deletecan you please provide an explanation supporting to your answer??

DeleteThanks in advance.

2nd question has stupid options which dont make much sense without knowing formula

ReplyDeletetheta =

ReplyDelete6

-1

0

y = 1 if 6 + (-1(x1) + (0*x2) => GE(greaterThanEqualTo) 0

6 - x1 => = 0

-x1 => -6

x1 <= (lessThanEqualto) <= 6

therefore the decision boundary is a vertical line where x1=6 and everything to the left

of that denotes y = 1 , while everything to the right denotes y = 0

you don't have to copy somebody's answer it is in the course notes.

Thanks for the detailed explanation.

DeleteSo For other viewers, to avoid the confusion, The correct answer for Que 5 is "A" only.