 Before jumping into the result of finding the optimal mechanism for any number of agents, we will first focus on the optimal mechanism designed for a single agent. So, let us assume that there is only one agent. How can we design the auction such that we get the maximum revenue out of it? So, the motivation is very simple. We first want to find a solution of a simpler problem and then we will generalize. So, the setup here is that there is only one agent. So, therefore, type set is just given by an interval 0 to beta and the mechanism f comma p is just mapping that 0 to beta to 0 to this interval 0 comma 1, which is the probability of that agent getting allocated. So, there is still a chance that the object will not be sold at all. So, and the payment is again taking this values between 0 and beta, which is the type and mapping it to the set of real numbers. Now, in this setup, because there is exactly one agent, so there is no belief system for the other agents, the BIC and DSIC definitions become the same and therefore, we can write the DSIC condition as we know it. So, player is getting better payoff, at least, weekly better payoff when it is reporting it truthfully versus when it is reporting, misreporting to S, that is it. Similarly, individual rationality is just the expected utility of this agent, which is non-negative. Now, the expected revenue earned by a mechanism is given by the payment that it makes and the probability of, so the prior probability of this specific type of that individual. So, because the expected revenue is calculated from the point of view of the mechanism designer, even though the type is deterministically known by the player, it is not known to the mechanism designer and that is the reason it is taking the expectation. So, the problem of optimal mechanism is to find a mechanism M star in this class of IC and IR mechanism such that the revenue earned, so this is the expected revenue, that revenue earned by that mechanism M star is going to outperform all the other mechanisms and we will call this M star to be the optimal mechanism. Now, what is the structure of the optimal mechanism? That is the question that we are interested in. So, if we are considering only IC and IR mechanisms, which are the kind of the basic requirements of a mechanism, we use the characterization theorems and lemmas as we have seen before. So, we know that this payment formula has to follow this integral form for incentive compatibility and for individual rationality, this constant of this payment should be non-positive and because we are trying to maximize the revenue, so there is no reason why we should choose anything smaller than 0, we should be exactly matching it to 0. So, P 0 will be exactly equal to 0, so this is the condition. So, it makes our expression of finding the optimal mechanism a little easier because now we have the payment which reduces to this form, only the remaining part of the payment formula. So, therefore, the expected revenue is given by this term which we have already defined and notice that this is only given by the allocation rule. The moment you fix the allocation rule in this optimal mechanism, you exactly know what is the payment formula and therefore, this is only a function of this F which is the allocation rule and we can expand this out by replacing this PT appropriately to find out what is the expected revenue. This is what we have to maximize with respect to F with the condition that F is non-decreasing because that is the requirement for incentive compatibility on this allocation rule and as long as that is satisfied, we have a DSIC or IC in this case because they are equivalent, we have this IC mechanism and we are trying to maximize the revenue. So, let us look at how we can do that. So, the first thing we are going to do is we are going to kind of get a little simpler expression, we will do this integral in steps so that we can reduce this expression to a smaller form and this lemma is telling you how you can reduce it. So, the expected revenue for any implementable allocation rule F is given by this formula where you have T minus this strange looking expression which is a function of that prior distribution, the distribution here and the density here and then you are taking the expectation with respect to GT. So, how can we find this? We can just start from the very the first principle. So, we had this expected revenue of this player, expected revenue from this mechanism for the auctioneer in this original form that is what we have started with. Now, if we just expand this out it is doing this. So, this term is getting multiplied here, we are just using the distributive form and the second part has this component. Now, we are going to use a kind of a transformation of this limits. So, we are doing the change of limits and here is how we are doing it. So, what we have here is we are first integrating with respect to X and then integrating with respect to T and the X is running from 0 to T and T is running from 0 to beta. So, what we are going to do in next is actually changing the order of integration and that will help us in understanding this integral better. So, in this so how do we do this change of order in the integration? So, this is standard limit switching. So, we have X on this X axis and T on the other axis in the original form X was running from 0 to T and T was running from 0 to beta. So, we will have to look at which space does this double integral satisfy. So, the first part is when X is moving from 0 to T. So, if you look at this part here this is essentially where T the X is starting from 0 and going to the value of T. So, this is the first part 0 to T and then the second component that is T is getting a swept over from 0 to beta. So, therefore, this strip this yellow strip is getting up and down the spanning the whole space here. So, that is the space on which this integration is happening. Now, we can switch these sides. So, instead of going from 0 to T first and then sweeping it over between 0 to beta we are first looking at T which is going from X. So, this is X to beta and that is the order in which T is going and then sweeping this green part from left to right which is from 0 to this maximum value which is also beta. So, that sweeping changes this order of integration the integrand that is gt times fx remains the same. So, what advantage does it give us? So, we can write this explicitly in this form so this is giving us the integral with respect to this density and we know that this density is nothing but g of T and if you look at that so it will be at beta it is going to be 1. So, 1 minus gt is going to be the inner integral and the other part we have ft. Now, this ft integral over 0 to beta is the same integral which is also happening here. So, therefore, we can actually take this ft and dt outside and we can look at the rest of the integral together. So, this is going to be so you just do some amount of free arrangement to find out that this is going to be T minus 1 minus gt divided by gt multiplied by this gt and ft and that is exactly what we wanted to satisfy. So, this integral formula will have this expression here and that is the expected revenue. So, essentially here this ft was missing so that that should be here. So, the expected revenue is given by this rather simpler formula and now the mechanism the optimal mechanism that we are trying to find solves this problem of maximizing this quantity subject to the condition that f is non-decreasing. Now, in order to do that we are going to assume some regularity conditions on this distribution g and that regularity condition is known as the monotone hazard rate condition. So, what does it say? It is saying that this ratio of gx by 1 minus gx that is just the inverse of this expression here is monotone non-decreasing in x. So, that is the monotone hazard rate condition and this is not an arbitrary condition that we are imposing various standard distributions like uniform distribution and exponential distribution. They satisfy this monotone hazard rate condition and for those kind of conditions those kind of distributions or priors we can use this fact that this quantity is non-decreasing. So, how can we make use of this fact? So, what advantage does it give? So, if we look at this expression that x is equal to 1 minus gx capital gx by lowercase gx. If g satisfies monotone hazard rate condition then we know that gx by 1 minus gx is non-decreasing in x. So, the inverse of that is non-increasing in x. So, you can sort of get a pictorial idea. So, this is 1 by gx by gx that is monotone non-increasing and x is exactly a monotone increasing function. So, this function will have a unique solution and that is the fact that we are going to state here. So, there exists a unique solution to this expression. We will make use of this. So, if we will now define a term called wx and this wx is nothing but this x minus 1 minus gx by gx. So, essentially in this term we are just replacing this part with w of t and looking at under monotone hazard rate condition what happens to this w of t. So, we observe that this expression. So, let us assume that this x star is the unique value at which this wx becomes equal to 0. So, we already know that this is going to be 0, the uniquely 0 and we know that wx is going to be positive because this term together. So, we know that this term is monotone non-increasing. So, minus of that is monotone non-decreasing and x is increasing. So, therefore, wx is something a function which is in increasing function. So, this is w of x with respect to x and we know that it hits this 0 point. So, assuming that this is the the y axis is 0 here, then it hits this at the point x star and anything beyond that any x which is above x star this is going to be positive, wx is positive and anything below that it is going to be negative and then that gives us a very clear idea of how optimization 1. So, because our objective is to maximize this quantity wt gtft its integral between 0 to beta, we should not be picking when we should not be allocating the object when this wt is negative because that is going to give us something negative integral. So, for a while if we just keep this condition of f being non-decreasing aside, we can see that whenever t is less than x star there is no reason to allocate that object. So, fd should be equal to 0. If it is more than x star then we know this is positive we should completely allocate this object and at the point where it is exactly equal to x star we can allocate it in whatever proportion it does not really matter. But what we observe interestingly is that this f is non-decreasing. So, we have a threshold point this is very similar to that second price option that we have discussed earlier. So, if we plot f of x in this case then we know that up to x star this is exactly equal to 0 and then from x star onwards it is getting fully allocated and at the x star point it can have any value in between. But this is certainly a non-decreasing f and that satisfies the condition for the optimal one. So, even though we did not care about the non-decreasingness the result that we got is actually non-decreasing. So, therefore it is an optimal solution to opt one. So, we can state all these sub-results in the form of a theorem a mechanism under the monotone hazard rate condition is optimally if and only if f is given by this equation 1 the equation that we have just mentioned where x star is the unique solution of this expression x equal to 1 minus gx by lowercase gx and then we can do calculate the payment formula from the standard integral formula and that will you will find that this is going to be x star if t is actually it should be t is strictly greater than x star for all the other cases that is t is less than or equal to x star this is going to be exactly equal to 0. So, that is very nice. So, whenever we find some t which is strictly greater than x star we know how much payment to charge and how to allocate. So, we will have to allocate it fully and anything below that we are not going to charge anything and we are not selling that object. So, this is this is a very interesting result. This is the optimal mechanism which maximizes the revenue for the seller for one player. Now, we are going to generalize it for multiple players.