 So, in the previous lecture we were looking at the linear quadratic Gaussian team and I had mentioned to you that the this particular team has the property that every person by person optimal solution is also team optimal. Now, remember that if the linear quadratic in a linear quadratic team what we have is that the cost is quadratic in the in the actions of the players. So, it is a quadratic in so it is a quadratic function like this u transpose q u where u is the vector formed by stacking up all the vector actions of the n players q here is a matrix which is positive definite. The dependence on psi is linear in this sort of way the information that that player i receives is a is a is a linear function of psi. So, it is some h i times psi where h i is a full row rank matrix and what one has to choose is u i which is a function of y i u i equal to gamma i of y i. Now, because of because of this particular nature of this problem so because u is is because l is strictly convex in because q is positive definite l is strictly convex in u and because it is quadratic it is also continuously differentiable in u and this holds for each fixed psi. Now, if psi itself is Gaussian that means it is a Gaussian vector then in that case we have a linear quadratic Gaussian team. Now, otherwise it is just some linear quadratic team now in in either case because l is l satisfies the hypothesis of this theorem here so if you have a linear if you have a static team in which u 1 to u n l of u 1 to u n comma psi has the property that l is continuously differentiable and strictly convex in u 1 to u n for each fixed psi then every person by person optimal solution is also team optimal. So, then so consequently every person by person optimal solution of your of the of of this particular team problem this is your LQ this is this the LQG team would also be a team optimal. So, it is easy to see that these hypothesis are being satisfied most the this follows from the positive definiteness of Q the quadraticness of this and the fact that h the information structure is that in y i is a function of only psi. So, y so this problem is therefore a static team problem alright. So, now let us come to let us come to the special case of the linear quadratic Gaussian team this the above result has only told us that every person by person optimal solution is team optimal it has not told us what the nature of the of the solution actually is. So, now in order to do that let us let us look at the following make a few observations here. So, we so the cost that we have let us we can we can write out this this cost in the in this following form. So, we can write out so our cost is has the form u transpose u transpose q u plus u transpose u transpose s times psi. Now, we can it turns out that one can write out this cost in the in a in the following sort of form where you have you write u minus let us say q inverse q inverse another vector r which could potentially depend on psi times q and u minus q inverse another vector r that potentially depends on psi plus some other stuff which does not depend on u independent of u all I have done here is really completed the squares. So, I noticing that the expression on the left here is actually a convex quadratic it means it is very much very close to being a perfect square I have sought I all I have done here is basically completed the squares completed the squares. So, this complete completion of squares can all you would have seen it in scalars you can but you can also do it in vectors another way of thinking about this is that notice that q here is some matrix which is potentially a dense matrix but it is positive definite and it is symmetric. So, I forgot to mention q is positive definite and symmetric it is positive definite we do not need a symmetric. So, q notice that q here is a is potentially a dense matrix. So, thanks to this q you can we can we can write q using its eigenvalue decomposition and find that q can in fact be broken down into some matrix like this into something like u minus u sorry into p capital lambda p capital lambda t. So, this where this capital lambda is a diagonal matrix diagonal matrix of eigenvalues eigenvalues of q. So, because of this kind of this kind of transformation or what I just called completing the squares what one can do is basically write out this the expression that you have on the left in this sort of form. Now, once you have written it in this sort of form in fact we you can later see that this R in fact is actually linear the it turns out here in this case because of this because of the dependence or because of this particular because this term is linear in psi it actually turns out that R itself is linear in psi. So, this itself turns out to be linear in psi. So, once this is linear in psi and psi itself is a Gaussian vector then this cost now starts resembling the cost here starts resembling is similar to is akin to a mean square estimation it is akin to a mean square estimation of a Gaussian of a Gaussian given given another vector which is Gaussian. So, it is as if what one is doing is basically doing a mean square estimation of some kind of Gaussian vector given the knowledge of another Gaussian vector. And these Gaussian vectors how are they derived well they are all derived from taking linear combinations of the same Gaussian vector psi. So, consequently the vector that we are estimating is and the vector that who is through which we are getting the information they are all jointly Gaussian. And therefore, the theory of jointly Gaussian random variables applies and consequently what we find is that what each player is doing in the optimal strategy for each player to do in this problem turns out to be to play a linear strategy. So, the so the our main result we find is that u i equal to some lambda i times y i is the optimal strategy. And the way to prove this actually goes through using the earlier result. So, so if you assume that each player is playing a linear strategy then it becomes then all of these then all the use also become all the other use become linear functions of linear functions of psi. So, once all the once the once all players other than player i play linear strategies the the strategies of the other players become linear functions of psi and the optimal action for player i then is to pick is is to do some sort of a mean square estimation based on the cost that is involved and that then also becomes a linear function of its own information. So, the which means that if the others if if the others play a linear strategy it becomes optimal for player i to also play a linear strategy and then this is this from this if one can argue that the that a particular type of tuple of linear strategies is actually team is actually person by person optimal. And once it is person by person optimal we know from the previous result that it is also it is also team optimal. So, one basically argues that if u j star is equal equal to lambda j times times y j for all j not equal to i then u i star then the optimal action for player i is to also play a linear strategy i. So, if then u i star equal to this is is optimal. So, a linear strategy is is optimal in response to lambda j for j not equal to i. So, in a so therefore it becomes optimal for all the other players for for the other players to play linear strategies. So, this ensures therefore that that person by person optimal linear strategies are person by person optimal and therefore linear strategies are also team optimal. Now this this actually is one of the once one set of results which are which are rather where it is once one problem class in which the static team problem has actually been solved very cleanly not too many other results like this are known the LQG static team is therefore also one of the most widely applied structures in in team decision theory. So, with this we have conclude with this we have concluded our study of static team problems and come to the conclusion of with with a study of linear quadratic linear quadratic Gaussian teams. So, from here now we will go into problems that are of a dynamic nature and we will study those we will study those problems in in a little bit more depth and we will see how a version of the Wittson-hausen problem actually corresponds to corresponds to what looks like a communication problem. Let us now cast our mind back to the Wittson-hausen problem. So, the Wittson-hausen problem remember the what what do we know about it we know that it is a linear quadratic Gaussian problem we know that it has non classical information pattern we know that in that the optimal controller is for that problem are not linear. So, some kind of non-linear controllers are optimal we also know that we have that the that there is a dual effect because of the kind of information structure that we have there is a dual effect in this problem and finally what we do not know about the problem is that we do not know the form of the optimal controller we just know it is something non-linear but we do not know what it is. Now, let us let us cast think back about this and let us recall again the role that the first controller plays in the in the Wittson-hausen problem which is what we have understood by the term dual effect. So, in the in the Wittson-hausen problem you have so this is Wittson-hausen. So, this is the Wittson-hausen problem. So, you have a a source or we have an the initial state as x 0 this is being seen by the first controller who takes a who applies a control gamma 1 and produces an action right that is his action. Now, this action we denoted by u 1 now this action had two different effects this action was present in the cost function of the controller. So, it is present here in the cost and therefore it has this is what we consider as the direct effect. So, this is the direct effect of this of this action and because that is what directly affects the affect the cost function. Now, the other indirect effect let us say the indirect effect is that gamma gamma 1 came is that gamma gamma 1 was also present was also impacting the information of the second controller right. So, because through through the information of the second controller. So, so this was the indirect effect. So, when you came to the second controller the the information of of gamma 2 was affected was dependent on gamma 1 right and then when and then gamma 2 chose its action as a function of this function gamma 1 right and this is what we called the dual effect this was a manifestation of the dual effect. Now, the we have seen before that therefore as a result of this dual effect there is a dilemma for the choice as far as the choice of gamma 1 is concerned. Gamma 1 can say well I want to minimize the direct cost by choosing the right kind of action u 1 that affects the cost directly or I can say well I want to also focus I want to minimize the indirect cost which is that affects the information of gamma 2 which then will affect the action of gamma 2. So, so the dilemma for gamma 1 is to choose the right balance between the direct cost and the indirect cost right. So, the dilemma for gamma 1 is striking in striking the right balance between the direct cost and indirect cost. Now, in addition to what I just mentioned that we know about the Wittson-Auser problem we also know a few other things. We have also seen variants of this problem that look that we were able to solve right. So, we have seen variants which we have been able to solve rather easily. So, what was one of those variants one of those variants was wherein was one where the second controller did not have a non classical this is one of the variants was where the problem did not have non classical information where the we somehow gave access of x 0 to the second controller. So, this was one of the variants we considered one of our one variant was one variant was gamma 2 as access has access to x 0 as access to the initial state. In other words it has access to the information that the first controller has when it takes its action. So, as a result this becomes the which makes it a classical information structure. Now, because this is now a classical information structure now we can say that we were actually able to solve the problem rather easily. We found there that we found what the optimal controllers there were and in fact found the optimal cost as well under those controller. So, now what do we learn from this particular this particular example what this we learn is that if this information here of x 0 is supplied to gamma 2 then in that case the optimal choice the finding the optimal controllers is rather easy to do. In other words this dilemma is this dilemma between striking the right balance between direct and indirect cost is not that hard dilemma. So, it is rather it is quite easy to know what the optimal action then should be. So, what we find is therefore that if the information of the initial state there actually is no dilemma between direct and indirect cost. So, if the information of x 0 is present with gamma 2 then there is no dilemma between direct and indirect cost. Now, what this means is that the root of this dilemma is the lack of knowledge of x 0. So, if this x 0 could somehow be communicated to gamma 2 then this dilemma goes away and therefore in that case the problem becomes rather easy to solve. So, in other words the core of the dilemma between for controller 1 is really the dilemma between taking the optimal control action and all and between doing a some sort of a communication which is communicating enough about x 0. So, that gamma 2 can then take its action rather it can take its next action and together these two actions turn out to be optimal. So, the crux of the dilemma in the Wittson-hausen problem is the dilemma between communication between control and communication. This is what we find. So, thanks to this we can now ask ourselves we can ask an interesting question. So, if this is really the dilemma between control and communication then there ought to be extreme cases of this particular problem in which either one of the effects is present or the other effect is present. So, what are the extreme cases then of the Wittson-hausen problem? So, there are one can talk of two extreme cases. The first is the an extreme first extreme case is when you have classical information structure. First extreme case is what is called the classical information structure. Now, why is this an extreme case? This is an extreme case because in this case the information of the initial state is already available with the second controller. So, therefore there is no need for communication. So, in this when one has classical information structure the information of the initial state is already available with the second controller and therefore there is no need for communication. The other extreme case is one where you have pure communication. See the other extreme case is one when you have pure communication. Now, why is this the other extreme case? See notice that the dilemma in the Wittson-hausen problem is the dilemma between control and communication. When we take the classical information pattern we have taken communication out of the picture and the first controller can focus only on control that means choosing the right control action without bothering about how it is affecting the information of the other agent. Because the information of the other agent is independent of its choice of policy. So, the classical in this classical information pattern the here in this case the the here the communication aspect is irrelevant communication is irrelevant and only control matters. So, the controller can focus on only control. In the other case the other extreme case which is the case of a pure communication problem in that case we what will happen this control is control has no importance has no relevance and only communication matters. So, thanks to this what we see is that actually Wittson-hausen problem lies somewhere in between these two the these two extremes the one extreme in which communication is irrelevant and only control matters which is the classical information pattern which is where MDP is reside OOMDP is reside and so on. So, this is let us say P also the domain of POMDP is MDP is and so on. The other extreme is a pure communication problem where control has no relevance and only in the issue the issue at hand is only communication. Now, I have not yet defined for you what a communication problem really is but I will I will come to that in the following lectures. But nonetheless remember that in spirit at least this is what is this is what this is where the Wittson-hausen problem finds itself that it finds itself somewhere tied somewhere between these two extreme cases and what is fascinating is not only is it some is the Wittson-hausen somewhere between these two extreme cases what is also fascinating is that these two extreme cases are actually well studied problems. The classical information structure is really what we have studied already as Markov decision processes are partially observed Markov decision processes. The other extreme which is a which is the communication problem that is an that is also well studied and that is an instance of that then is a that is a that is and that is a subject of communication theory. So, what Wittson-hausen problem finds itself in between is is sort of in a regime where there is no really no good theory it finds itself in a regime where which is somewhere in between a communication between communication theory which is developed solely for the purpose of communication and and Markov decision theory which is which is developed assuming a perfect information structure we are assuming a classical information structure. So, this is where the Wittson-hausen problem finds itself. Now for the remaining part of the course what we will see is we will look at since we have already looked at the classical information pattern in depth for the remaining part of the course we will focus on this this aspect which is the communication aspect and we will see what what how exactly is a communication problem can how can that be thought of as an extreme case of the Wittson-hausen problem.