(ML 14.7) Forward algorithm (part 1)

[@zhuokaizhao5926]

Thank you for your video. I have read the paper by Lawrence R. Rabiner explaining HMM and Forward/Backward algorithms, but it was not as clear as you explained! Very helpful!

[@behshadmohebali6234]

Nothing new. Just wanted to say you deserve every bit of praise you are getting here and more. Cheers.

[@CherryPauper]

I'm failing my test today lol.

[@nijiasheng711]

For those who may be confused about the underlying logic flow, the prerequisites are bayes theorem, conditional probability, conditional independence, D-separation algorithm, bayes networks

[@lancelofjohn6995]

It is the fifth time for me to hear the lecture,finally I understand the equation after 1 year!

[@hanchen2355]

Can't imagine how some comments claim this crystal clear tutorial "confusing"..

[@nikos_kafritsas]

Great video and explanation, unfortunately the variables m and n are not well defined and that's why people get confused

[@Cookies4125]

n is the number of z variables (or x variables) i.e. the number of time steps in the HMM m is the number of states each z variable (the hidden variable) can take

[@amirkhalili82]

Your good teaching addicted me to go forward seeing more videos Good that you described everything slowly in details, Bad that you described everything slowly in details ;)

[@ilyasaroui7745]

watch the previous videos on HMM to understand m and n and all the variables explained. great explanation!

[@보쿠링]

very clear and intuitive explanation. Thanks a lot!!

[@p.z.8355]

The paper by Rabiner explains it quite clear

[@yunshanchen9252]

Great explanation ! you make it so easy to understand!

[@wenkunwu3644]

For the complexity calculation. I understand where the first m comes from. Then it looks like there are only k z_{k}'s instead of m z_{k}'s. How do you get \Theta(m^2) for each k? After checking out the backward algorithm I understand that the complexity for the whole forward-backward algorithm is \Theta(nm^2), this is because in backward algorithm, we have (m-k) z_{k}'s. So the sum of both algorithm should give m z_{k}'s, the complexity for each k is \Theta(m^2) and the final complexity should be \Theta(nm^2). Could you explain a little bit more why the complexity for each individual algorithm is also \Theta(nm^2)?

[@saparagus]

Yes, it was not very clear, but see Jason Wu's note below: when we write \alpha(z_k), we really mean the joint probabilities p(z_k,x_k) for all possible outcomes of z_k -- of which there are m. So there are a total of m values \alpha(z_k), one for each outcome of z_k. Next, in the computation of EACH \alpha(z_k), there is the summation over all possible outcomes of z_{k-1} , of which there are also m. That's how we get a total of \Theta(m^2) for computing all (m) values of \alpha(z_k). And then since there is a total of n values that k takes, the (recursive!) computation of all alphas is m*m*n.

[@subhabratabanerjee8600]

Looks nice, I am looking for a worked example, preferably from Natural Language Processing. Any idea?

[@nielsnielsen5905]

You're the best 🥳

[@barabum2]

Can you explain why you are summing (at 2:15)? It seems to me that it should be a product there, not a sum. How did you come out with that formula for p(z_k, x_1:k)? Thanks.

[@melainineelbou4869]

barabum2 its the relation between joint probability and marginale probability

[@thomaspeterson2568]

www.quora.com/What-is-marginalization-in-probability

[@Raven-bi3xn]

At minute 10:55, shouldn't P(z[1],x[1]) be equal to P(z[1] | x[1]). P(x[1])? The emission matrix gives us the likelihood of an observation (Z) given a hidden state (X), not the other way around.

[@TheMeltone1]

Do you have any videos that work through example problems?

[@medic0re]

You are a great man :D

[@ZbiggySmall]

A lot in this and previous videos, you refer to "separation rule" and that you "condition on something". I don't understand what those mean? Can you give a link to any video where you have already explained them in details? i cannot followed this videos without proper understanding. Thanks.

[@houdayaqine1166]

Actually he was talking about D-separation, you can get better understanding of this concept in that link www.andrew.cmu.edu/user/scheines/tutor/d-sep.html :)

[@artnovikovdotru]

Hello. Could tell me what to do when emission probability equals to zero at some step? Then all further alphas equals to zero. Re-estimation formulas (Baum-Welch) don't make any sense then. I'm trying to implement HMM with guassian mixtures. So I can't use smoothing techniques since those are only for discrete distributions. How to deal with such a problem?

[@lancelofjohn6995]

I hope you were my professor.

[@chriswalsh5925]

a bit lost at this point after watching the prev 6 vids... suddenly the rate of new stuff ramped up!

[@keweichen3638]

very well explained

[@PranavKhade]

I am writing a code for this and I can't understand the 'm' variable.

[@artomeri7266]

18 people do not know how to prove independence, and do not know probability chain rule :'(

[@theWujiechen]

I believe \Theta(m): For all the situations of z_{k-1}. \Theta(m^2):For all the situations of (z_{k-1}, z_k) .

[@MaxDiscere]

11:26 wort written "known" ever

[@Aaron041288]

how can I get p(z1) ? any help?

[@MrSengkruy]

hard to follow . a lot of new things which are not clearly explain.

[@waqaramjad1638]

very complicated to understand

[@alizamani5223]

i wish a forward code in the matlab program.

[@thomaspeterson2568]

I wish you code it

[@souslicer]

i dont understand the summation

[@remusomega]

m=n ffs

[@gattra]

no, m = # of training examples. n = number of timesteps

[@piggvar123]

define things more proper please...

[@abdulrahmanahmed7405]

try to relate your explanations to real life applications

[@lancelofjohn6995]

It is the fifth time for me to hear the lecture,finally I understand the equation after 1 year!