It's crazy to me how much more of RL I've understood in these 2 beginning lectures vs multiple days of trying to understand what my professors have cooked up in their lecture. And David actually goes into even more depth.
@VinBhaskara_6 жыл бұрын
Fantastic. The lectures are so consistent, formal, and organized - mathematically as well as pedagogically. Optimal policy for teaching such a course!
@huaweijian22104 жыл бұрын
One of the best lectures here at UCL CS :)
@astaconteazamaiputin3 жыл бұрын
Indeed. One of the best lectures I have seen on a difficult topic. And his attitude is just...rare and priceless actually. I feel outmost respect for his pedagogical methods.
@mathavraj93783 жыл бұрын
I see what you did there. nice
@matthiasho32162 жыл бұрын
I totally agree. Mr David Silver is a great lecturer
@mathavraj93783 жыл бұрын
can we appreciate the audience too? so interactive and they asked all questions that made our foundation even better. Kudos to David for patiently answering everyone with equal dedication
@stevetang51313 жыл бұрын
Best RL course 5 years ago and now still is to me. Thank you.
@BRUNNSPARK3 ай бұрын
If you are reading this in 2024 and thinking what value can a 10 years old course may offer, you are completely wrong. I have looked at many courses and lectures to find out the best startup RL course and found this as the best. Force yourself to start with this.
@happydays3343 ай бұрын
Can we build the machine learning program/code with this course I am interested but I don't know if I would learn how to implement it in the 'coding' part of it?
@muzammilnxs7 жыл бұрын
This lecture series is a gem in reinforcement learning. Best material for RL as yet!!
@shivannawal76876 жыл бұрын
I've heard of this course as well. Do you have a resource that provides the correct way to download the packages required for starting the course ? Thanks
@eyeofhorus13015 жыл бұрын
I would expect nothing less from a guy at the forefront of AlphaZero
@dudley54248 жыл бұрын
Thanks, Petyr Baelish. Great lecture series.
@Chr0nalis8 жыл бұрын
He lacks the moustache.
@rickyhan70236 жыл бұрын
I'm impressed by the size of his head compared to his body.
@falcon202435 жыл бұрын
RL is a ladder.
@NilavraPathak5 жыл бұрын
Nah looks more like Alex Honnold
@user-or7ji5hv8y4 жыл бұрын
He has amazing ability to explain complexity in such clear way.
@opalpearl3051 Жыл бұрын
I have watched lectures from MIT and Stanford and David blows them away.
@whaloob3 жыл бұрын
This guy is an impressively good lecturer. Easily comparable to, if not better than, the best profs I’ve had so far in 7 semesters of university.
@Jacob-wp4km6 жыл бұрын
This intro to MDP is seriously high-quality. Informative yet understandable.
@tylersnard4 жыл бұрын
I looked everywhere for a comprehensible explanation of RL. This is easy to understand and awesome. Thank you!
@adbocode7 жыл бұрын
WARNING: take headphones off 42:25 -> 42:35 and 55:55 -> 56:05
@Stu495836 жыл бұрын
I wish I read this comment before that
@shrishabharadwaj48116 жыл бұрын
Me too.
@huahua716 жыл бұрын
I guess many find this comment after hearing the noise
@chaitragn33795 жыл бұрын
Hahahah I read this comment and searched what is it. True take off headphone or be prepared for it.
@prateekagarawal53115 жыл бұрын
Thanks
@angelachikaebirim10435 жыл бұрын
Prof. Silver is a brilliant teacher. Really enjoying this
@jamesr1412 жыл бұрын
David is an amazing teacher. Because he loves the subject. Now I love it, too. Thanks David. Thanks Deepmind.
@TahlilAhmedchowdhury4 жыл бұрын
I must say this is by far the best course whether you want to learn Deep or Shallow RL. Its so intuitive and easy to understand the basic of RL concepts. The current CS285 Course has a great material but for me its hard to understand. I repeatedly come back here to clear concepts. David Silver is a great Teacher.
@user-gl6os8eh9g Жыл бұрын
Despite its low resolution and not clear sound, this is the Best reinforcement learning lecture. Thanks for super easy, detailed explanation!
@jordan48903 жыл бұрын
"The future is independent of the past given the future" is crazy deep when you think about it. How many people lament over past transgressions and stress over "how it should have gone down" when in reality obtaining your goals is much more along the lines of "what can I do right now in this moment"
@rameshmaddali62085 жыл бұрын
I can not resist from commenting. These presentations are more learnable I am scratching my head since a week. Fortunately I got into your video presentations. Awesome Dave Gold*.
@adi3312 жыл бұрын
The part from 54:30 to 1:01:00 was a lot to take into. But it's actually not that much once you get a grip on it. First there is the *Bellman Equation* and the *Bellman Expectation Equation* The difference is that the *Bellman Expectation Equation* also incorporates actions. *Bellman Expectation Equation* = *Bellman Equation* + actions. After that we basically just have 4 variations of this: 1. *Vπ* in terms of *Qπ* 2. *Qπ* in terms of *Vπ* 3. *Vπ* in terms of *itself* 4. *Qπ* in terms of *itself*
@fahmikhoirurrijal94279 ай бұрын
i still don't understand how the calculations that we got the values for student MDP in 53:30, it was said prior that in MDP there's not just one expectation value anymore but here every nodes has one value? e.g how do we get the +2.7 value in the second class?
@shresthapatir64953 ай бұрын
@@fahmikhoirurrijal9427if you understood it now can you please explain?
@theahmedmustafa2 ай бұрын
@shresthapatir6495 Idk if this a little too late but here is how you get it: The value of the state = expected value of all actions that can be taken from it The value of an action = reward + expected value of all states it leads to There are two actions from that state: Sleep or Study. Lets talk about Sleep. This action always leads to terminal state so the expected value of states it leads to is 0. The reward is also 0 as stated in slides. Value of Sleep = reward + expected value of all states it leads to = 0 + 0 = 0 Same can be done for Study. This action always leads to next state which has a value of 7.4 so the expected value of states it leads to is 7.4. The reward stated is -2. Value of Study = reward + expected value of all states it leads to = -2 + 7.4 = 5.4 Now we have values of all actions for this state. The value of the state is the expected value of all its actions. In this case, the policy is to have a 50/50 choice if actions so each action has a 0.5 probability. Value of state = expected value of all actions = (0.5 * value of Sleep) + (0.5 * value of Study) = (0.5 * 0) + (0.5 * 5.4) = 2.7 So the value of this state is 2.7
@elio31016 жыл бұрын
Great lecture. I really liked the approach, building from MP to MRP and finally MDP. It was well connected, like listening to a story. Thank you very much!
@user-qw4jf1th7c6 ай бұрын
Absolutely fantastic !! Thank you Dr. Silver !
@khaled46352 жыл бұрын
Way better than the 2021 lecture with a similar title. Never found such a detailed explanation of MDPs Thanks
@astaconteazamaiputin3 жыл бұрын
Thank you for the question at 1:04:30 about states turning into actions / decisions, I appreciate the clarification !
@yidingji75524 жыл бұрын
Best introductory RL course on KZbin!
@jacksong87484 жыл бұрын
1:41:24 did anyone else feel like an agent in a partially observable environment when the camera shut off while he was talking about partially observable environments?
@consecuencias.imprevistas8 жыл бұрын
If you're confused like I was at slide 32 (53:27) when he goes over the state-value function for the student MDP, the calculation for each node, using the look-ahead step from Bellman Equation is: v_pi(s) = E_pi[R_t+1 + gamma*v(S_t+1)|S_t = s]. For instance, for the case of the -1.3 node, we calculate, given that the policy assigns 0.5 probabilities for all actions: 0.5*(-2 + 2.7) + 0.5*(-1 + -2.3) ~= -1.3 Where the first term of the sum is the contribution from the Study node, and the second term is the contribution from the Facebook node
@wonchulkim30087 жыл бұрын
Could you be more specific??? I don't know how to get -1.3
@MinhVu-fo6hd7 жыл бұрын
It is 0.5*[-2+2.7*(1)] + 0.5*[-1-2.3*(1)] = -1.3
@seongmukgang98437 жыл бұрын
Thanks very much.
@wahabfiles62604 жыл бұрын
I dont understand difference between state value function and action value function. In state value, the value of being in some state depends on the immediate reward in the next state. But we need an "action" to got to the next state so isnt it essentially action value function?
@ludmiladeute63533 жыл бұрын
Thank you very much, professor David, for these great lectures!
@Bromine973 жыл бұрын
Thank you for such a great set of lectures on Sutton and Barto's book. The lectures are so apt, thank you David for all your lectures
@mangaart3366 Жыл бұрын
What an amazing lecture! I had so much fun watching it and it helped clarify some things in my head about MDP.
@volkerleukhardt4830 Жыл бұрын
Very cool, having this available free of charge. Thank you so much 😊🙌
@zhongchuxiong Жыл бұрын
0:45 Markov Processes {S, P}. used to describe reinforcement learning environment. 3:26 Markov Property: the current state contains all the information of history. 4:35 State transition Matix. transition probability from s to s`. 7:27 Example: student decision chain. 13:00 Markov Reward Processes {S, P, R, γ} 29:10 Bellman Equation for MRPs: Vs = Rt + γ * Σ(Ps+1 * Vs+1) 39:01 Bellman Equation in Matrix Form. 43:14 Markov Decision Processes {S, A, P, R, γ} 46:15 Policies: for taking actions. π(a|s) 50:51 Value Function. state-value funciton action-value function 53:37 Bellman Expectation Equation, following a policy. Bellman expectation equation for Vπ(s) Bellman expectation equation for Qπ(s,a) 1:02:52 Bellman Expectation Equation in matrix form 1:04:01 Q&A 1:08:06 Optimal Value Function. q* 1:15:41 Optimal policy. π ≥ π` if v(s) ≥ v`(s), ∀s (for all states) Theorem: for any MDP, there's an optimal policy that's ≥ all other policies. Theorem: all optimal policies leads to the same optimal value funciton. 1:19:54 find the optimal policy 1:19:55 Bellman optimality equation. v*(s) = max(q*(a,s)) q*(a,s) = R + γ * ΣP * v*(s`) 1:29:39 how to solve bellman optimality equation? it's non-linear. use iterative solution. value iteration (next lecture) policy iteration (next lecture) q-learning (future lecture) sarsa 1:31:49 Q&A 1:40:18 Extensions to MDPs continuous & infinite MDP partially observable MDP undiscounted, average reward MDP
@user-gl6os8eh9g Жыл бұрын
Thanks. very helpful
@olenapanasovska99003 жыл бұрын
The transition probability of this course exponentially lows down from 1 mln ( the first lesson) to 50 thousand in 10 lessons. Only 5% have come out from Class 3)))
@Chanakya063 жыл бұрын
Best course material to start Reinforcement learning. However the math seems to be difficult at least for me.
@0mane04 жыл бұрын
just to make it clear about the question asked at about 23:00: I actually think that the sequence is finite not by an arbitrary definition but due to the fact that every loop in that scenario will eventually terminate with proabability of 1. For instance the "stay in facebook" loop has a probability of 0.9^n of happening during n stages, wich tends to zero as n tends to infinity, thus the probability of (eventually) leaving the loop tends to 1.
@andriesrosseau50484 жыл бұрын
Yes, he kinda dismisses the question out of hand, but the answer is more mathematically subtle than he makes it out to be.
@DonMalchimo2 жыл бұрын
I searched for a commentary like that, because I got confused from the answer of this question in the video. Thanks. :) I think this can be seen by a short analysis of the transition matrix, e.g., the eigenvalues which are all between zero and one for each state except for the state "Sleep", which is one. This means that P^n for n to infinity converges to a matrix only with zeroes except for the column "to Sleep" which are all 1 then.
@edmonddantes4705 Жыл бұрын
Yes, episodic Markov processes terminate almost surely :). It is subtle. Also MDPs with a.s. absolutely bounded immediate rewards have finite discounted return (gamma < 1). But he is not very rigorous, he just wants to convey the main ideas.
@FedericoBaldassarre6 жыл бұрын
My question is related to when the reward for a state-action pair is computed in a Markov Decision Process. When showing the Student MDP on page 25 (min 45:00), the reward for (study, pub) is given immediately after the action is taken and exiting the state. This is compatible with the way we indicate the reward as R^a_s, meaning that it is defined by the action you've taken and the state you're exiting. However, later on in the video (min 56:30) and on the slides (page 32) it looks like the reward is given when the environment pushes you to the next state, once the action has been chosen. In formulas, it seems like r depends on the action and the next state, so R^a_{s'} To me, it makes more sense that the agent receives the reward as a function of the action and the state it'll end up in, rather than the state it is leaving. Consider a game where one can decide to attack or run: if I decide to attack and the environment decides that I should be defeated, I would like to get a negative reward immediately and not later. So, which one is the right way of thinking? Do the equations and the plot need fixing or am I misunderstanding something?
@MsCalcioo5 жыл бұрын
Hi, The initial example is an example of Markov Reward Process. So, the reward function only depends on the state (you can check the slide for the definition of MRP). Therefore, you get a reward at time t+1 when you exit the state at time t no matter what the action is. The later example is Markov Decision Process. In MDP, your reward function definition also considers which action you take. So you take some action and exit the state at time t ("study" or "pub") and get a reward at time t+1 based on your actions and where you'll end up with.
@ercanatam63442 жыл бұрын
@@MsCalcioo You seem an expert on MDP. I have a question, "Pub" was a state in MRP example but in MDP it is removed being a state. Why? This confuses me. Moreover, rewards in MDPs are shown on the arc from (s,a) to s', but in the MDP example this convention is not followed.
@edmonddantes4705 Жыл бұрын
@@ercanatam6344 The states in the student MDP are literally the same as the states in the MRP, David just did not write the names in the circles/squares to avoid saturating the notation. What is confusing you is that he is using the words "pub" or "study" in order to describe actions as well. He means "go to pub" or "go to study". If you carry out the action "go to study" or "go to pub", you end up in the states "study"/"pub". Sometimes "go" could also mean "keep in", as in "keep in the pub", "keep studying", but I hope you get what I mean.
@fahmikhoirurrijal94279 ай бұрын
@@edmonddantes4705 i still don't understand how the calculations that we got the values for student MDP in 53:30, it was said prior that in MDP there's not just one expectation value anymore but here every nodes has one value? e.g how do we get the +2.7 value in the second class?
@edmonddantes47059 ай бұрын
@@fahmikhoirurrijal9427 For any MDP, given a fixed policy pi, every state has a value given by the state-value function. The state-value function is literally defined in the previous slide and the chosen policy pi is uniform, so I am not sure what you are missing. Anyway, he applies the formula in the previous slide (you can do Monte Carlo sampling for an approximation or apply the Bellman expectation equation). PS: By the way, any MDP with a fixed policy is an MRP.
@morganjones83684 жыл бұрын
Exceptional lecture series so far. Very clear useful examples and explainations. Thank you very much Dr. Silver.
@JohnHAdams-vo2pk4 жыл бұрын
Fantastic Lectures. I use them along with the Sutton/Barto text book. Excellent stuff
@ogsconnect13126 жыл бұрын
Well done! At 6:05, I guess the lower matrix value should be Pn1
@luisjalabert83665 жыл бұрын
Same error at 39:30
@astaconteazamaiputin3 жыл бұрын
Yes, it should. And he says this during the talk, so no doubt about it.
@adi3313 жыл бұрын
@@astaconteazamaiputin It's corrected in the slides at: www.davidsilver.uk/wp-content/uploads/2020/03/MDP.pdf
@mohamedmusa71492 жыл бұрын
@@luisjalabert8366 Yes, that is an error.
@Anonymous-vh9tc4 жыл бұрын
best teacher on RL, there is no contest
@FrostW1nd2 жыл бұрын
19:07 The guy taking a call walking across the lecture room missing out the great content, -10 immediate reward there.
@peterkimemiah96696 ай бұрын
I have lost many months of my thesis, wish I had these lectures earlier
@slack1284 жыл бұрын
The explanation of the Bellman Optimality Equation for V* was really enjoyable to watch. Thank you.
@kiuhnmmnhuik26277 жыл бұрын
@1:37:00 Here's an easier example to understand the idea behind the optimality conditions. Let's say you want to find the shortest path from A to C and someone gives you a supposedly optimal solution which passes through B, i.e. the (supposedly) optimal path is A->B->C. If the path A->B is not the shortest path from A to B, then the total solution can't be optimal because we can find a better path A~>B and then form A~>B->C which is shorter than A->B->C. Now let's say someone gives you a supposedly optimal policy pi. This is more tricky than the shortest path example because of recursion. You notice that pi doesn't always choose the optimal action at every given time. In other words, you compute q_pi for the policy pi and you notice that q_pi(s,a) = R(a|s) + gamma sum_{s'} P(s'|s,a) sum_{a'} pi(a'|s') q_pi(s',a') < R(a|s) + gamma sum_{s'} P(s'|s,a) max_{a'} q_pi(s',a') for some pair (s,a). Let pi' be the improved policy which always chooses the best action according to q_pi. We get q_pi(s,a) = R(a|s) + gamma sum_{s'} P(s'|s,a) sum_{a'} pi(a'|s') q_pi(s',a') < R(a|s) + gamma sum_{s'} P(s'|s,a) max_{a'} q_pi(s',a') = R(a|s) + gamma sum_{s'} P(s'|s,a) sum_{a'} pi'(a'|s') q_pi(s',a') Note that we have pi' and q_pi (not q_pi'!) in the same expression. We would like to have q_pi' instead of q_pi. We can update q_pi with the new estimations and then repeat the maximization process. Note that at each maximization step we increase our estimation of q_pi until we converge to q_pi'. Since we're always increasing q_pi, we must have q_pi < q_pi' which means that pi' is strictly better than pi. I hope this makes sense. I've never read the official proof and this is just the sketch of one direction of the proof (i.e. optimality => optimal condition). In conclusion, whenever q_pi doesn't satisfy the bellman optimality equation (BOE), then we can increase q_pi until, estimation after estimation, we converge to a q_pi' which satisfy the BOE and such that pi' > pi.
@jieshi1695 жыл бұрын
You save my day! Thank you so much for sharing this video on youtube!
@willlogs4 жыл бұрын
Why would 59 people dislike this free gold?!
@minhdang17037 жыл бұрын
I can't thank you enough for sharing such valuable lectures.
@kiuhnmmnhuik26277 жыл бұрын
Nah... you can always read the papers. The information is already out there.
@memechannel33453 жыл бұрын
Yea
@sagarhukkire7426 жыл бұрын
Thanks ..the way he explains math is simply amazing...
@kevinmcinerney95527 жыл бұрын
At approx 22mins somebody asks about an infinite sequence. David says this is impossible, but even it was possible it's interesting to observe that the sequence of rewards would sum a finite value because it's actually just geometric series. This was discussed in the Udacity course a bit. Loving these lectures btw. David is magnificent.
@JaysonSunshine7 жыл бұрын
Also, since there is a sink, the probability goes to zero of generating a sequence of length n as n goes to infinity.
@monktastic16 жыл бұрын
I think he was talking about the case where there's no reduction factor across time.
@edmonddantes4705 Жыл бұрын
To be more precise, the discounted return converges almost surely as long as the rewards are uniformly bounded (almost surely) and gamma is strictly less than one.
@edmonddantes4705 Жыл бұрын
@@monktastic1 That statement requires some mathematical formalisation, since one could have two different states pointing at each other with probability one, and hence arriving to any of those states would give rise to an infinite alternating sequence. However, those states could be formalised as one state and then you could consider the class of those two states to be a fixed point. There is some truth in your statement, but it requires more formalisation, it is hand-wavy.
@kiuhnmmnhuik26277 жыл бұрын
I just want to point out that the recursive formulas are very easy to understand and derive if one disentangle them. For instance, let's say *s* is the current state, *a* an action and *s'* the new state after the action, i.e. s->a->s'. Then v_pi(s) = sum_{a,s'} P(a,s'|s) [R(a|s) + gamma v_pi(s')] where P(a,s'|s) is the probability of taking action *a* and ending up in state *s'* given that we are in state *s*. Now we note that P(a,s'|s) = P(s'|s,a)P(a|s) = P(s'|s,a)pi(a|s). Also, sum_{a,s'} (...) is sum_a sum_{s'} (...), so, we get v_pi(s) = sum_{a,s'} P(a,s'|s) [R(a|s) + gamma v_pi(s')] = sum_a sum_{s'} P(s'|s,a)pi(a|s) [R(a|s) + gamma v_pi(s')] = sum_a pi(a|s) sum_{s'} P(s'|s,a) [R(a|s) + gamma v_pi(s')] = sum_a pi(a|s) [ sum_{s'} P(s'|s,a) R(a|s) + sum_{s'} P(s'|s,a) gamma v_pi(s') ] = sum_a pi(a|s) [ R(a|s) + gamma sum_{s'} P(s'|s,a) v_pi(s') ] which is the formula @57:38. Same thing with q_pi(s,a). We start from state *s* and do action *a*. What happens is that we end up in state *s'* and then do another action *a'*. In other words, the sequence is s->a->s'->a'. Since we know *s* and *a*, but *s'* and *a'* are random, we compute the mean of the total reward over *s'* and *a'*: q_pi(s,a) = sum_{s',a'} P(s',a'|s,a) [R(a|s) + gamma q_pi(s',a')] That's it! Now note that P(s',a'|s,a) = P(a'|s,a,s')P(s'|s,a) = P(a'|s')P(s'|s,a) = pi(a'|s')P(s'|s,a) where we used the fact that *a'*, given *s'*, is independent of *s* and *a*. Therefore, similarly to before, we get: q_pi(s,a) = sum_{s',a'} P(s',a'|s,a) [R(a|s) + gamma q_pi(s',a')] = [sum_{s',a'} P(s',a'|s,a) R(a|s) ] + [sum_{s',a'} P(s',a'|s,a) gamma q_pi(s',a')] = R(a|s) + gamma sum_{s',a'} P(s',a'|s,a) q_pi(s',a') = R(a|s) + gamma sum_{s'} sum_{a'} pi(a'|s')P(s'|s,a) q_pi(s',a') = R(a|s) + gamma sum_{s'} P(s'|s,a) sum_{a'} pi(a'|s') q_pi(s',a') which is the formula @59:00. This is little more than basic probability.
@SuperShank766 жыл бұрын
fantastic lecture. Prof. David Silver is amazing. The questions are a bit silly. The lectures are amazing.
@xxgimpl0rdxx224 жыл бұрын
Those poor students who had to take all of this in in one sitting.
@17teacmrocks3 жыл бұрын
not that much at the university level actually. very standard and presented in an easy to follow manner. also I assume they had to have some prereqs in math and probability graph theory to get to this class. MDP and bellman equation are needed for solving optimization equations of stochastic systems
@SergioGonzalez-gu1bk3 жыл бұрын
@@17teacmrocks I love how at the end he starts questions and states that he is expecting that not everyone understood. I'm following because I read the topics before watching, that isn't a light read at all.
@Samuel-wl4fw3 жыл бұрын
Just watched the video over 5-6 hours lol, took some notes meanwhile and talked about the difficult parts of the subject with my mate. Definitely a lot of stuff covered, and happy to be able to watch it in my own tempo, would be a little overwhelming to watch and understand in 2 hours.
@AakarshNair10 ай бұрын
shockingly clear explanations!
@MrBeatronic8 жыл бұрын
At 1:11:47, isn't optimal action-value function q*("going to pub") =1 + (0.2*6 + 0.4*8 + 0.4*10) = 9.4 instead of 8.4?
@consecuencias.imprevistas8 жыл бұрын
I got the same result
@shaileshjain27457 жыл бұрын
Yep, I got 9.4 too.
@lush93yt7 жыл бұрын
same result here
@muhammadusama60407 жыл бұрын
exactly
@jingyaohuang85607 жыл бұрын
me too
@akarshrastogi36825 жыл бұрын
David pointlessly pointing at himself from 23:08 - lol
@abhi2204 жыл бұрын
So funny you pointed it out!
@mihailtal75504 жыл бұрын
I keep missing what's really important in those lectures. Thank God we have people like you to to guide us.
@channagirijagadish12015 жыл бұрын
Excellent lecture series. Thanks, David.
@poojanpatel24376 жыл бұрын
at 5:09 last element of matrix is Pn1 not P11. refer to latest slides for correction. www0.cs.ucl.ac.uk/staff/d.silver/web/Teaching_files/MDP.pdf
@SimonLimygoogleplus8 жыл бұрын
Why there is a period of time muted? Is this copyright concern?
@alexnichol31387 жыл бұрын
that's what i was wondering (around 35:00).
@paulgarcia28874 жыл бұрын
cannot reveal the top secret information
@valken6668 жыл бұрын
Great lecture! but volume is really low
@halibite7 жыл бұрын
increasing volume with headphone
@timothdev4 жыл бұрын
Yep, I'd to download it and use VLC to boost volume.
@abhijeetsharma57152 жыл бұрын
Yeah, you can use a chrome extension to boost audio if you're on a PC. Suggestion: Volume Master Chrome Extension.
@yoloswaggins21617 жыл бұрын
Got my crew of Markov reward values, G unit.
@machine-wisdom6 жыл бұрын
One thing I'd like to discuss is whether the discount factor should be the same for both positive and negative rewards. Losses loom larger than gains, and it is fair to imagine that a risk aversive agent would not discount a negative reward.
@edmonddantes4705 Жыл бұрын
Note that, in general, for the fixed point convergence theorems (see lectures 4 and 5) to converge mathematically (and hence to be usable), negative rewards need to be discounted. How much discount you add is up to you, but I am guessing you can run simulations with different discount factors and see what works better, use something like TD-lambda where there are combined returns with different discounts, etc. On a different note, I have the impression that from a mathematical point of view, using negative or positive rewards is kind of equivalent. Indeed, let the immediate returns be uniformly bounded by below by a negative constant that we will call -C. Then by summing C to all the rewards, you construct a MDP that only has positive rewards and the optimal policies of the original process (the one with negative rewards) and the modified one and clearly the same, they are shared. Now you could ask: what is better for convergence? And that seems like a nontrivial question that depends on the problem, algorithm, spaces, etc.
@43SunSon4 жыл бұрын
There is a typo @1:22:08 ?? q*=8.4 ? I think should be 9.4 = 1+1(0.4*10+0.4*8+0.2*6) base on the equation q*(s,a) @1:26:55 . Someone verify this with me?
@DeathStocker4 жыл бұрын
That's exactly what I found weird. I think it has to be 9.4.
@43SunSon4 жыл бұрын
@@DeathStocker it is. I am right, i checked other comments, they said the same thing.
@DeathStocker4 жыл бұрын
@@43SunSon thanks for confirming!
@kaspar82923 жыл бұрын
Yup, I just got stuck there too. Should be 9,4
@tljstewart3 жыл бұрын
@57:17 "[State-Value Function] V is telling how good it is to be in a particular state, [Action-Value Function] Q is telling how good it is to take a particular actions from a given state"
@alfcnz6 жыл бұрын
What's the convention he's mentioning at 31:48, of the reward indexing?
@alfcnz6 жыл бұрын
"Sutton & Barto convention" and he's referring to this textbook incompleteideas.net/book/the-book-2nd.html
@preethiabi2008 Жыл бұрын
Thank you professor
@hanwang25614 жыл бұрын
At 6:05, the lower matrix value should be Pn1
@lawlietlight25922 жыл бұрын
There is is typo in transition matrix suffix at 40:12 , it should have been Pn1
@Moustafa8657 жыл бұрын
I think there is a problem with the State transition matrix. First entry in the last row has a wrong subscript index. It should be {n,1} instead of {1,1}.
@Alley00Cat6 жыл бұрын
A student in the class mentions this
@Avivalious6 жыл бұрын
I agree
@nittur10086 жыл бұрын
Being {n,1}, it would not make sense as it would be , calculating probability of going to state 1 knowing state n. right?
@Morpho326 жыл бұрын
It does make sense. If we are in state n, we can go to state 1. Unless state n is "absorbing", in this case we cannot go anywhere after reaching state n but it would still appear in the transition matrix with values 0.
@HellYeah2236 жыл бұрын
I think you might be confusing states and steps. Assume you can go from state n to state 1, but not from step n back (in time) to step 1.
@udayadampage731611 ай бұрын
5:23: State Transition Probability Matrix: Last entry of Row n, Column 1: to be corrected as Pn1
@juleswombat53097 жыл бұрын
Yeah well David really now needs to write a book on this subject material. Enjoying the videos but really need a captured analysis. There are few books on RL, Sutton's book is nearly twenty years old (draft updated for next year) But this lecture series is easier to comprehend than Standards.Any book needs to have the algorithms described, to show how the theory can be applied.Well done.
@edmonddantes4705 Жыл бұрын
Most of the material is taken from Sutton & Barto, even most of the pictures, examples, etc, so that book you are talking about is already written.
@jont72074 жыл бұрын
37:18 How do we calculate the value of states that have circular transitions? i.e. the value of class 3 (v(s)=4.3) depends on the value of pub (v(s)=0.8). How do we arrive at 0.8 for the pub then if the value of the pub depends on the value of class 3?
@jont72074 жыл бұрын
I guess this is done in the Dynamic Programming for the next lecture. It is solved recursively by dynamic programming, since there is no closed form for bellman equation.
@donamincorleone8 жыл бұрын
Great lecture. Thanks!
@udayadampage731611 ай бұрын
39:34: State Transition Probability Matrix: Last entry of Row n, Column 1: to be corrected as Pn1
@JD0ng3 жыл бұрын
props to the person who asked the question at 31:27. Thank you!
@nimishbansal4234 Жыл бұрын
The difference between the state value function and the action value function is that the state value function means the expected total return starting from a state s towards the end of the sequence by following a policy pi that means sampling some actions according to that policy pi and choosing a particular action in every state whereas according to action value function we sample some actions according to a policy pi and then by moving from the present state to the successor state we compute the expected total return for all the sampled actions. Is the understanding correct??
@zerox25365 күн бұрын
Is discout factor can be equal to 0 or 1??? I think brackets are incorrect
@TheTaekwondo19962 жыл бұрын
Thanks a lot for the great lecture!
@Hasan-sn2gb3 ай бұрын
Very good course! I feel the audience are not ready for this course by asking low quality questions.
@eladnachmias11972 жыл бұрын
Around minute 51:00, the bellman equation for the action-value function shouldn’t have the policy subscript for the expectation, as the current action is already given, so it is an averaging goes only over the “environment dice”.
@edmonddantes4705 Жыл бұрын
Your comment indicates lack of understanding. Of course it should have the subscript pi. G_t is the discounted return, so you need to take many more actions in the future states in order to calculate that expectation, even if your immediate action a is fixed.
@sreenivasanac6 жыл бұрын
WARNING: TAKE HEADPHONES OFF 55:50 - 56:05 AND ALSO FROM 42:25 -> 42:35 VERY LOUD NOISE
@andyyuan978 жыл бұрын
Can someone enable the automatic CC?
@abhi2206 жыл бұрын
by someone you mean, the guy who uploaded the video?
@wahabfiles62604 жыл бұрын
@@abhi220 lol
@TXxJiggle Жыл бұрын
A little difficult to keep up with but very rewarding.
@kaspar82924 жыл бұрын
Am I wrong or should q* of a=Pub at 1:13:10 be 9.4 ?? In the example of (s=class2, a=study), the optimal state-value function is calculated as q*=R+sum(P_state(i)*v(state(i)))=-2+1,0*10=-8
@xiaoshisong98363 жыл бұрын
should be 9.4
@achyuthvishwamithra2 жыл бұрын
In reality, must we program an agent to store the optimal action calculated for each state, in the data structure of the state thus adhering to markov property or will the agent calculate optimal value at every step every time?
@MrPneunomia5 жыл бұрын
Best RL course series I've tackled
@akarshrastogi36825 жыл бұрын
If you've seen other courses / university lectures like Stanford's CS234 or UCB CS285, could you please recommend some playlist that's very useful like this one, probably with a little focus on python implementation?
@piotr780 Жыл бұрын
4:05 there is one problem with markov property, if S is (S1,S2) which are two correlated random variables, then we can't simply estimate correlation between S1 and S2 using last realization of those variables - we need full history (the longer the more accurate is our estimate), so we simply lose some information in this formulation until we extend S with additional informations (with whole history or values allowing for...
@ilfat_khairullin3 жыл бұрын
Thank you! Amazing lecture!
@natureoflife54904 жыл бұрын
Great lecture, however, if there is an autogenerated subtitle then it is easy to realize.
@TechnologyBudda7 жыл бұрын
31:05 - That's the linearity of expectations. Iterated expectations is nesting by conditioning.
@estherderman84726 жыл бұрын
According to what sigma-algebra is G_t+1 conditioned? I am missing something in the last equality
@anselmoufc6 жыл бұрын
I think he was a bit confusing here. Isn't he using the fact the expectation is a linear operator (E[x+y] = E[x]+E[y]) and then using the fact that E[E[x]] = E[x], since E[x] is a constant and expectation of a constant is the constant itself?
@o200211065 жыл бұрын
@@anselmoufc my understandin also
@edmonddantes4705 Жыл бұрын
@@estherderman8472 He is just using E[X+Y] = E[X + E[Y]]. In this case, X := R_{t+1} | S_t=s, Y := G_{t+1} | S_t=s, which are random variables. You are confusing it with E[X|Y], which is a random variable and not a number since it is defined as E[X|sigma(Y)], where sigma(Y) is the sigma-algebra generated by Y. Anyway, what he is doing is trivial.
@BruinChang2 жыл бұрын
I have a question around 57:00. After an agent takes an action, should the next state not be specified? If the next state is specified, how can I take the expectation over different states?
@kaierliang2 жыл бұрын
The environment decides which state you end up in, sometimes there is only one option, sometimes there are multiable options like the pub action
@Enderman-en3dv4 жыл бұрын
1:12:20 Shouldn't the q* value for the Pub be 9.4? It's 1 + (.2*6 + .4*8 + .4*10) because of the R+1 of going to the pub, right?
@Zantorc4 жыл бұрын
The general consensus is that you're right, it should be 9.4 which is what I got.
@43SunSon4 жыл бұрын
@@Zantorc same here! I think should be 9.4, not 8.4
@ichbins1613 жыл бұрын
in 37:18 - Why is there no reward for going in the pub, so why is it 4.3 and not 5.3
@subhambhowmick80445 жыл бұрын
On slide : State Transition Matrix 6:19 Should the matrix be ? P11.....P1n .................. Pn1......Pnn
@leandronw14 жыл бұрын
It should. A student in that class asked about that right after he was done explaining it.
@SangrezKhan Жыл бұрын
David is love
@krishcoolguy16 жыл бұрын
why the audio was muted at 35:00 - 35:40. It seems to be interesting part. Can any one please explain what Mr.David has explained during these 40secs?
@kevingatimu83195 жыл бұрын
It's simply a look-ahead diagram to evaluate the value of the current state, v(s). To calculate v(s), we add 2 values: (1) the reward for being in that current state, i.e., R_s, and (2) the discounted probability-weighted sum of the values of all possible future states one step later from s.
@mahmoudfathy20746 жыл бұрын
@1:02:37 Lecturer: "Is that Clear?" Me: "No it's not, there's a problem and YOU know what it is and you know that I know that you wished nobody notices it 😂" No matter what, these are the best lectures in RL , I really enjoy these underlying details. Thanks David Silver
@mahmoudfathy20746 жыл бұрын
Hahaha, it's clear right on the next slide , my bad !
@BobTheZealot7 жыл бұрын
Those mic knocks hurt cuz they are so much louder than his voice
@10603204 жыл бұрын
at 1:07:40, when he says stochastic transitions, the transitions are by two separate agencies. So the agent takes an action from a state. The environment depending on the action puts you in a new state. Correct?
@indianbeatboxersunited43114 жыл бұрын
Yup somewhat like that. So depending on your action, there is an independent randomness operating which helps in deciding the state you end up in. Deterministic transition would be where your action solely decides which state you end up in. Stochastic transitions adds randomness to the transition from one state to the next state