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## Logistic Regression

TOTAL POINTS 5

EXPLANATION:

Our estimate for P(y=1|x;θ) is 0.7  =>because =>hθ(x) = 0.7

Our estimate for P(y=0|x;θ) is 0.3 =>because => P(y=0|x;θ) = 1 - P(y = 1| x; θ); the former is 1 - 0.7= 0.3

EXPLANATION:

J(θ) will be a convex function, so gradient descent should converge to the global minimum.(true)
=>
fact

Adding polynomial features (e.g., instead using hθ(x) = g(θ0 + θ1x1 + θ2x2 + θ3x2 + θ4x1x2 + θ5x2 )) could increase how well we can fit the training data (true)
=>Adding new features can only improve the fit on the training set: since setting θ3 = θ4 = θ5 = 0 makes the hypothesis the same as the original one, gradient descent will use those features (by making the corresponding non-zero) only if doing so improves the training set fit

other statements that can occur in this question:

At the optimal value of θ (e.g., found by fminunc), we will have J(θ) ≥ 0. (true)

variation to 3rd  question is provided at the end.

EXPLANATION:

The cost function J(θ) for logistic regression trained with examples is always greater than or equal to zero.(true)
=>The cost for any example x(i) is always ≥ 0 since it is the negative log of a quantity less than one. The cost function J(θ) is a summation over the cost for each eample, so the cost function itself must be greater than or equal to zero.

The sigmoid function is never greater than one(true)
=>fact

other statements that can occur in this question:

The one-vs-all technique allows you to use logistic regression for problems in which each y(i)comes from a fixed, discrete set of values. (true)
=>If each y(i) is one of k different values, we can give a label to each y(i)belongs{1,2,....,k} and use one-vs-all as described in the lecture.

EXPLANATION:

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.

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variations in 5 th question:

variations in 3 th question:

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reference : coursera

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