Hello,
I have a short question regarding the solution of Question 15. When you derivate the MS Loss you get according to me 2 *Sum(yi -axi-b) = 0, Therefore when then putting in the values one would get value 5 for b instead of 2.5. Or am I wrong?
Thank you in advance
instead of only \(y_n=1\), you should put all values of \(y_n \in {1,3,2,4} and x_n \in {-2,-1,0,3} \). As a result, \(\sum_{n=1 }^{4} x_n = 0\) and you are left with b
Yes of course, that I do understand, however, in the solution it is divided by 4, and in my opinion this is true but it again needs to be multiplied by 2 because of having derivated the square term...
@Anonymous said:
Yes of course, that I do understand, however, in the solution it is divided by 4, and in my opinion this is true but it again needs to be multiplied by 2 because of having derivated the square term...
When you make that equal to 0 you probably can get rid of -1 and 2
Exam 2021 Q 15 (possible mistake)
Hello,
I have a short question regarding the solution of Question 15. When you derivate the MS Loss you get according to me 2 *Sum(yi -axi-b) = 0, Therefore when then putting in the values one would get value 5 for b instead of 2.5. Or am I wrong?
Thank you in advance
instead of only \(y_n=1\), you should put all values of \(y_n \in {1,3,2,4} and x_n \in {-2,-1,0,3} \). As a result, \(\sum_{n=1 }^{4} x_n = 0\) and you are left with b
Yes of course, that I do understand, however, in the solution it is divided by 4, and in my opinion this is true but it again needs to be multiplied by 2 because of having derivated the square term...
When you make that equal to 0 you probably can get rid of -1 and 2
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