 So, let us do an example of a Bayesian Bayesian game and let us see how the how a Bayesian Nash equilibrium we will compute a Bayesian Nash equilibrium of this game in mixed strategies ok. So, the game here has two players let us denote these players by 1 and 2 ok. Now player 1 has player 1 here as has two types let us denote these types by I 1 and I 2 and player player 2 has just one type and that is just denoted by is just this ok all right. Now, these types get realized with probability half. So, probability that of this is equal to probability of this is half ok all right. Now, there are two possible actions for the players ok so, let us player 2 player 1 there are two possible actions for the players. So, player 2 has actions L and R and these actions are the same regardless of are the same in the I mean since player 2 has only one type ok he has just this one set of actions which is L and R, but for player 1 because he has two possible types the action set also depends on his type ok. So, here he has actions so, when his type is when the type profile is 11 comma 2 when that is the type profile he has two actions let us call them up 1 and down 1 and the payoffs are 1 0 0 2 0 3 and 1 0 and when the when the type profile is 1 2 comma 2 you have the actions up to down to and what are the actions for player 2 it is still L and R right because he has he is actually has only one type ok and the payoffs then are 0 2 1 1 1 0 and 0 2 and both players are maximizing ok both maximizing players all right. So, let us write out this game in extensive form so, you have nature which is playing with so, nature chooses a type 1 1 for player 1 that is one other type 1 2 for player 1 ok. Now then player 1 has two choices which is up 1 down 1 up 2 down 2 player 3 player then comes this is player 1 then player 2 will play player 2 has two actions left and right now what is the information set of player 2 here what does player 2 know so, it is a simultaneous move game after the types have been realized so, what does the player know player knows only his own type right and plus it is a simultaneous move game so, what is his information set everything in one this so, he has this information set right so, you so, the information set of player 2 is this now let us write go back here and we need to write out mixed strategies for the player ok. Now player 2 has only one information set so, he can randomize so, you think of it as not mixed strategy behavioral strategy so, for this information set he is going to choose he is going to randomize between L and R ok suppose he chooses L with probability Q and R with probability 1 minus Q so, it is going to so, remember this so, you have L with write this in a different color so, L is chosen with probability Q R is chosen with probability 1 minus Q what about L and R here it would have to be the same because there is only one information set so, this Q and 1 minus Q ok now, what about player 1 player 1 can be allowed to randomize differently in each matrix right so, here he can he let us say chooses X and 1 minus X and Y and 1 minus Y ok so, if you remember what we what we discussed about these for these games what happens effectively is that this becomes a game in a Bayesian Nash equilibrium becomes equivalent to a Nash equilibrium in which every type of every player is a distinct player so, because now there are there are a total of 3 types one for player 2 and 2 for player 1 is this effectively now looking for a Nash equilibrium as a for a 3 player game this is clear that is what we are effectively looking for ok so, you can imagine this is going to be quite complicated because it is anyway 2 player game is itself complicated 3 player game means that we have to now solve for these 3 variables X Y and Q simultaneously ok and so, this is just a way to just to give you a flavor of how you can go about doing some calculations like this so, I will just go through these arguments one by one so, first let us argue to find this now so, first let us argue that Q has to be Q cannot be either 0 or 1 in an equilibrium so, first we show that Q cannot be either 0 or 1 so, what happens if let us say what happens if Q is 1 so, let us come back here ok if Q is 1 then basically player 2 is choosing L he is choosing L here and choosing L here also ok now if he is choosing L here what is the best response of player 1 in this type he would choose so, players are maximizing so, he would choose X equal to 1 right he would get 1 then ok now, but if he is choosing X equal and then his and what about the best response in this one he would choose Y equal to 0 right he would choose D 2 is this clear ok in that case if player 1 is choosing U 1 here and D 2 here ok now, what is the payoff that player 2 can get from this so, this type gets realized with probability half this type he gets realized with probability half right so, he is getting half times 0 plus half times 0, but if player 1 is going to be choosing U 1 here and D 2 here what should player 2 actually instead do player 2 can in fact play R instead of playing L right because then he will get 2 here so, in other words L is not if L is chosen by player 2 then player 1 would choose U 1 here and D 2 here ok in which case player 2 would get half would get 0, but then player 2 can shift to R and get something more than 0 for sure ok so, in other words playing Q equal to 1 is not a you cannot be an equilibrium alright likewise you can argue that Q equal to 0 also cannot be an equilibrium ok for the same you can come up with the same sort of argument so, if Q is equal to 0 then basically player 2 is playing player 2 is playing this R alright if player 2 is playing R then the best response for player 1 in this type is to play D 1 and in this type it is to play U 2 alright, but the best response to that is not R anymore ok so, you can check that ok. So, what this means is that so, if Q is equal to 0 must play U 1 comma D 2 which gives player 2 half of half of what half of 0 plus half of 0 which is 0 so, if Q equal to 1 this is with Q equal to 1 right similarly, if Q equal 0 player 1 must play must play what D 1 comma U 2 it gives what does it give player 2 half into 3 plus half into 1 which is 2 ok, but if player 1 is playing D 1 here and U 2 here so, if player 1 is playing D 1 so, instead in Q equal to 1 would have given him Q equal to 0 sorry I have been looking at the wrong block so, if Q is equal to 0 player 1 must play D 2 so, that gives him which gives player 2 what does it give him half into 0 plus half into 1 right yeah which is half instead playing Q equal to 1 would give him would give player 2 half into now half into 3 plus half into 2 right is that right half into into 2 which would be 5 by 2 ok. So, this shows that Q which means that Q is Q equal to Q is between 0 at a Bayesian Nash equilibrium alright ok. So, which means that player 2 is playing a completely mixed strategy with something that is between 0 and 1 which means then that we as far as if you look at which means that the payoff that player 2 can get from each of his pure strategies must be equal right because he is mixing between between them is mixing between them. So, it cannot be that any one of them has higher payoff that is what we have basically shown here right. So, which what this means is that so thus payoff of player 2 from each pure strategy L or R must be must be the same ok. So, what is the payoff that he gets from when he plays when he plays L. So, firstly the the first box is first type is going to be realized with probability half times now the he gets a payoff of 0. So, here this is chosen with probability X the box itself is chosen with probability half then this action is chosen with probability X right by player 1 that is X here. So, you are going to get half into 0 into X plus half into 3 into 1 minus X ok. So, let us just write this 0 into X plus half into 3 into 1 minus X plus plus half. So, this is what he gets from this from this box and this and plus half of this what he would get from this box by by playing L ok. So, half into 2 Y plus half into 0 into 1 minus Y. So, this here is the from L remember what I am writing here is you of that the what the player would get when he knows his own type right. So, what you have to do is you have to sum over the types of all the other players times the probability that those types will occur given your own type times the payoff that will come and then there is also a mixing mixing mix strategy that is happening which is giving rise to these X's and Y's ok. So, you have also then the strategy the strategies that are chosen. So, that is what is written here ok. So, this is the payoff from L what is the payoff from R. So, this has to be equal to the payoff from R. So, this is half into half into 2 X plus half into 0 into 1 minus X plus now half into yeah half into 1 into Y plus half into 2 into 1 minus Y right. So, these 2 have to be equal now what this gives us is that X should be equal to 1 plus 3 Y by 5 ok. Now, this is a characteristic feature of these things that you get this kind of simultaneous equations across the types of players of the same player ok. So, which means that you must have that Q has to be between 0 and 1 and X has to be X is has to be equal to 1 plus 1 plus 3 Y by 5. Now, question is does every such is every such point actually an equilibrium is this enough. So, this is just we just know that this must be satisfied, but we do not know if this is you know this is if all such points are equilibrium ok. So, now, let us go through this let us let us drill down further into this. This is if we have narrowed it down to this now ok. Now, suppose so now let us take these cases. So, suppose Q is less than half if Q is less than half. So, we just know that Q has to be between 0 and 1 and and these which means that X and Y has to satisfy this. Now, whatever Q you choose X and Y should be a best response. Now, suppose Q is less than half ok what is player 1's best response. So, just go back to this if Q is less than half ok which means that Q is giving more weight to this ok. So, what is the what is player 1's best response? Yeah. So, is player 1's best response actually it turns out that X equal to if Q is less so you can write this out you can write out the exact thing ok. So, what is player 1 getting here? Here is getting Q plus 1 minus Q and here is getting sorry Q into 1 plus 0 into 1 minus Q here he is getting 0 into Q plus 1 into 1 minus Q right. So, if Q is less than half then player 1 would prefer this. So, which means X equal to 0 is his best response right and likewise what let us see here. So, if here he is get what is player 1 getting here he is getting 0 into 0 into Q plus 1 into 1 minus Q and here he is getting Q plus 0 into 1 minus Q. So, if Q is less than half here then what would player 1 prefer U 2 right. So, in other words if Q is less than half then in this type player 1 would prefer D 1 which is X equal to 0 and in this type he would prefer U 2 which is Y equal to 1 ok. So, if Q is less than half which means that player 1's best response is X equal to 0 and Y equal to 1. See I unlike for player 2 where I had to average over the two matrices player 1 actually knows which matrix he is in. So, there is nothing for him to average over because player 2 has only one type ok. So, that is become very simple here. So, player 1's best response now is just X equal to 0 and Y equal to 1. Now X equal to 0 and Y equal to 1 does not satisfy this equation here does not satisfy this equation ok. So, which means that Q less than half is not possible. We can similarly argue that Q. So, if Q is greater than half then what would happen the exact opposite would happen right because it is 0 into Q 1 plus 1 into 1 minus Q that type of thing. So, player 1's so then player 1's best response is X equal to 1 and Y equal to 0 which is again the same problem. So, X equal to 1 and Y equal to 0 does not satisfy this right this in this equation. So, neither of these satisfy X is equal to 1 plus 3 Y by 5 ok. Now what does this mean then? It has to be Q equal to half because we know that there is at least one equilibrium. So, which means that Q has to be equal to half ok. So, now if Q is equal to half ok then then if Q is equal to half then what you can see what happens to the payoff of the player of player 1. If Q is equal to half then both U 1 and D 1 will give him the same payoff here and U 2 and D 2 will give him the same payoff. In other words player 1 can play any X and any Y here, but then remember we were also constrained to satisfy this ok. So, in other words then what this means is Q is Q equal Q has to be half and X comma Y are such that X is equal to 1 plus 3 Y by 5 and X is in 0 1 and Y is in 0. Now you put this together actually this gives you this gives you some kind of a region you can say you can write that out. You can see actually that this it cannot X cannot because of this because X has to be between 0 and 1 and so on and Y has to be between 0 and 1. So, you can actually take Y to be anything between 0 and 1 the X will still remain in 0 and 1, but X cannot be too large because Y also has to be between 0 and 1. So, you can what you get is that Y is between 0 and 1 and X is between one-fifth and four-fifth. So, this is there is nothing more that needs to be said these all of these points are actually equilibrium. So, X, Y, Q, X, Y, Q such that this holds and Q equals half a region actually. So, in a three it is effectively you can think of this as a like a three player thing essentially two players collectively make the other guy indifferent between his pure strategies you know that is what is happening here ok.