Q and A Compulsory Exercise part 2

Questions that contribute to the clarity of the tasks are answered here.

Q2: In problem 2a, I need some help to figure out the likelihood  p(yt|xt).

A2: Note that yt is the observation. It has a deterministic link to xt. The only information you get from xt???????  through yt, is which interval it is in, so the Likelihood becomes zero if there is a miss match, and one if there is a match, i.e. L(y=1|x=-1.3)=1, whereas    L(y=1|x=1.3)=0, similarly    L(y=3|x=-1.3)=0 and L(y=3|x=1.3) = 1.

 

Q1: I have problems with 1b. I do not get how we derive g(x). Why are the transitions between the models  -0.5 and 0.5 when we do the gradient at -1 0 and 1?

A1:   if the function is -0.5x^2,  then the derivative is -x.   Thus the linear tangent in the three locations will  be 
-1:  x-0.5
0 : 0
1 : 0.5-x 

All of these three tangents  will always be above the function  -0.5x^2, so to have the one that is closest we use the minimum of the three. The point where the crossings are at -0.5 and 0.5.  so this gives the points where we change the approximation. 

 

Published Mar. 20, 2024 1:27 AM - Last modified Apr. 1, 2024 11:35 AM