A friendly introduction to Bayes Theorem and Hidden Markov Models 

Happy I found this video.. even though there's a Corona lockdown :D

It's coincidentally rainy outside 😂

Based on previous experiance, because it is rainy at your side, i predict that you were probably not happy 😔

@@a7md944 Bob was more likely not happy, we are the hidden state - whats the probability that the lockdown was not justified and that people were dying because of lack of medical help instead of the illness.

Eigenvectors and SVD for sure.

@@jacobmoore8734 check out 3blue1brown's channel for the Essence of Linear Algebra. He explains that matrices are linear functions like y=f(x) or like a line 'y=mx', with y-intercept b=0. Eigenvectors are special inputs 'x' such that f(x) = kx, where k is some scalar coefficient (k is the eigenvalue associated with the special input x). For certain types of NxN matrices, (the covariance matrix used in PCA for example) are super interesting because any point in N-dimensional coordinates can be represented as a linear combination (ax1 + bx2+...) of the eigenvectors. The eigenvectors form a 'basis' for that space. This is where SVD (singular value decomposition) comes in. SVD essentially asks "instead of just multiplying x by your matrix, why don't you decompose this task into 3 easier tasks?" Let's say your matrix is C for covariance. Then SVD says that C = ULU' where U is made up of the eigenvectors for columns, U' is the transpose of U, and L is a diagonal matrix with the eigenvalues. Pretend we're doing y = C*x. Then first we do w = U'*x. This essentially represents x as a linear combination of eigenvectors. Said another way, you've changed the representation of point x from the original coordinate system to the eigenvector coordinate system. Next we do z = L*w, which scales every value of vector w by an eigenvalue. Some of these eigenvalues are very small and the result in z is perhaps closer to 0. Some of these eigenvalues are relatively large and upscale the result in z. Finally, when you do y = U*z, all you're doing it translating your scaled z vector back into the original coordinate system. So SVD basically splits a matrix into 3 different operations: 1. represents an input vector in terms of eigenvector coordinates 2. scales each coordinate by an eigenvalue 3. represents the scaled result back in terms of the original coordinates When you look at PCA (principal components analysis), you take your covariance matrix and decompose it to look at how much your eigenvalues scale the eigenvector coordinates. The largest eigenvalues correspond to the direction (eigenvector) of largest variation

Definitely eigenvectors! Please!

Yes, please, do that.

You may have already found some, this is an attempt by University of Calcutta, not so coolly done, but please see if it makes sense ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-C6fH5Nfoj40.html

Isn't this opposite of calculus? Discrete vs continuous functions.

Similar situation here, I have a master thesis in anomaly detection, and using HMM is a candidate. I'm afraid it's much more complicated than this, but it sure made it look less scary

Thank you Changyu!

Thank you for your note, I hope the midterm went great! :)

You are right

I agree. seems there is a minor mistake there. choosing the "best" probability in each day doesnt ensure the optimal path we are looking for. If I understand correctly, you should start from the end, looking for the best final probability. then go "backwards", looking for the specific path which led to this final probability.

I agree with you, Junghoon. I reach the same conclusion. and I think the best way is to actually register the path we came for calculating each maximum value. and at the end, we can start with maximum and print the road that we had registered to print result. or instead of using memory, as you said we can calculate to see which one is matching along the path from the maximum value on the last day to reach to the start day.

Yes, I noticed that. so the results in this demonstration are wrong.

Thank you so much Brandon! Always happy to hear from you.

Luis , your way of teaching is so good , even a 10 year old will be able to understand such a complex topic. Will definitely check you book as well once my current stack is finished.

I second that!

Did you mean in such a simple way?

That is correct in principle, but if two paths visit the same node, then the max/best path up-to-that-node (so far) will certainly be a part of the best path visiting that node. And in this example each node has only two previous nodes. As mentioned above, same principle as with Dijkstra's algorithm.