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Question 1

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Question 2

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Answer

$F(x_i) = sign(w_1 f_1(x_i) + w_2 f_2(x_i) + w_3 f_3(x_i) + w_4 f_4(x_i)$

$ = sign(0.61\times(+1) + 0.53\times(-1) + 0.88\times(-1) + 0.34\times(+1))$

$ = sign(-0.46)$

$ => \hat y_i = -1 $

Question 3

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Question 4

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Question 5

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Answer

$$\frac{1}{2} ln\big(\frac{1 - \text{weighted_error}(f_t)}{\text{weighted_error}(f_t)} \big)$$$$= \frac{1}{2} ln\big(\frac{1 - 0.25}{0.25} \big)$$$$= 0.55$$

Question 6

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Question 7

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Answer

• Question 7 needs a calculation of the weighted error. From the lectures weighted error is given by weight of mistakes / total weight.
• Since three of the four data points have the same label I assume that there is just one mistake and since all the weights sum to 1 that weight is then equal to the weighted error, which is 0.35

Question 8

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Question 9

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Question 10

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Question 11

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