 to complete all aspects of the random work with bias. We can also derive the asymptotic form of the occupancy probability as n and m tends to infinity. For that we invoke the asymptotic form we obtained earlier for the symmetric case that is W n 0 m as n comma m tends to infinity such that m square by n fixed. So, under this assumption the asymptotic form was square root of 2 by by n e to the power minus n square by 2 n refer to the symmetric case or unbiased random work. So, therefore, the present expression W n biased m equal to W n unbiased m 1 plus gamma to the power n plus m by 2 1 minus gamma to the power n minus m by 2 takes the following form if you note that any function for example, any function say a to the power x is equal to e to the power x log a we can always write like this. So, using this property we can write W n bias equal to the asymptotic form. So, we straight away go to the asymptotic form let us say n tends to infinity and also m tends to infinity at the under the condition. So, this will be square root of 2 by by n e to the power minus m square by 2 n and regarding 1 plus gamma functions we use this exponentiating logic that is we write it as e to the power n plus m by 2 ln of 1 plus gamma plus n minus m by 2 ln of 1 minus gamma. We further assume that gamma is small then ln of 1 plus gamma to the first order we keep it as gamma only and to the order gamma square which we are going to neglect. So, then working out the detail we are going to have W n with bias m is going to go over to square root of 2 by pi n e to the power minus n square by 2 n that term will be there and it will be n plus n plus m by 2 ln of 1 plus gamma is just gamma similarly the next term n minus n by 2 it will be minus gamma. So, I change the sign here and write gamma then of course, of the order of gamma square and we decide to neglect those terms on simplifying we see that this will be root of 2 by pi n e to the power minus m square by 2 n and now here n by 2 minus n by 2 gamma which cancels. So, m by 2 gamma plus m by 2 gamma. So, that leads to simply m gamma m by 2 plus m by 2. So, plus m gamma I make it a complete square now e to the power minus 1 by 2 n I take out then it is going to be m square here and minus 2 m n gamma. We can add for tentatively we can call it as we can add a order of gamma square term or subtract because we are going to neglect it. So, you can make it a perfect square now e to the power minus m minus n gamma whole square and and here of course, we will have an order of gamma square term which we have decided to neglect. So, it will be just this upon neglecting terms of the order of gamma square and higher. So, this is a symptotic form for the occupancy probability in the limit of large n and m. You can note that here as compared to the symmetric random walk this term m square has been replaced with m minus n gamma whole square that is consistent with our derivation that the peak m bar is n gamma and of course, variance sigma square will be n. It is basically now a Gaussian solution Gaussian approximation for problem corresponding to random walk with bias. We will encounter this bias problem as we move forward especially when we discuss gambler's ruin problems and revisit some of these results, but with a very new approach and a new perspective. In the category of random walk on lattices next important problem that needs some attention is random walk with the pause. So, let us say we have a very general problem with pause or pausing or waiting can use the underlying concept is just this. So, as earlier there is a random walker who starts from 0 and let us say it means after several steps he is at site m that he has we work with the nearest neighbor transition probabilities only. However, unlike the cases we discussed thus far when you toss a coin or when you decide to take a next step it is not necessary that he transits either to the left or to the right, but he could just stay where he is with some probability. So, we have now 3 possible transitions possibly q to the left and let us say a probability delta to himself and p to the right at each step. So, a step is decided by a coin tossing let us say, but it is not necessary that he takes either left or right. So, of course, coin tossing is dichotomous. So, one must have 3 probabilities the probability of to the right to the left and to himself or to the site state itself. So, this you know earlier we were representing it via for example, transition probabilities ending on themselves. So, transition to this transition from let us say we have 3 states a b c then certain probability that we assign to transition to the state itself is what we call as delta. Random walk with the pausing is of a practical relevance because many a time if you are talking of atoms which execute diffusion on a lattice they many a time need not get released after landing on a site immediately. There could be some waiting time before it jumps either to the left or to the right. Let us see how to formulate this problem within the framework of what we discussed thus far. So, let us take a most general case because we have so far discussed a symmetric random walk asymmetric random walk, but without pauses. Now, let us generalize it to the case of pausing and bias together. So, that all the cases can be later seen as special cases of one of them. So, general case most general situation random walk with bias plus pause. So, to get back to our problem R w bias plus pause it is modeled as I mentioned probabilities m to m minus 1 to m plus 1 with the 3 probabilities p q and probability delta such that p plus q plus delta equal to 1. That basically implies that p and q are not themselves conserved p plus q will be 1 minus delta where delta is let us define p transition probability to the right q transition probability to the left and delta transition probability in fact it is a probability of pausing probability of pausing or no transition to neighboring sites. So, that is the most important part to remember. We can redefine the left and the right transition probabilities by normalizing with respect to 1 minus delta like for example, the here for example, if you call this as equation 1 and this is equation 2 then 2 may be rewritten as as p by 1 minus delta plus q by 1 minus delta equal to q or we may denote normalized left and right transition probabilities p prime plus q prime equal to 1 where p prime now is the original transition probability divided by 1 minus delta. Similarly, q prime is the original left toward a transition probability divided by 1 minus delta and this is now normalized this of course, implies that p prime plus q prime equal to 1. Now, how do we address this problem of course, we can write this as a jump equation also. There would be then 3 terms corresponding to the left transition right transition and pausing probability because in the in the next step there is a certain probability delta that the random walker will land on the same site. But more elegant and a clever way is to make use of the existing results arrive at as an expression for the occupancy probability at the mth site. So, we may note that see of the n steps let us say that S is he has actually either jumped left or right in S steps the walker jumps or transits we will use jump as a simpler expression jumps either left or right in let us say S steps. So, in the remaining n minus S and does not and stays where he was or pauses in n minus S steps. So, we call those S steps as a successful steps steps where he walked moved n minus 1 where where he was stay put. So, he did not move. So, if we consider S as a successful events and n minus S the unsuccessful events he is net probability of being at any site M would be dictated by these S transitions only these are very important statement to digest. So, so, the so, his probability his occupancy probability say probability of being at M if S will be dictated by the S steps will be governed by the S steps. We will soon see what to do with S. Now, it is a new variable we have introduced let us say that S steps he has succeeded in undergoing transition at n minus S he has not succeeded. So, his being at site M will then be given by the biased random walk problem corresponding to the transition probabilities of p prime and q prime in the S steps. So, we can then say that if in S steps W SM with bias as we mentioned is the transition probability let us denote occupancy probability is the occupancy probability probability for site M in S steps. It will be including the bias case because our forward and backward transitions are not equal. If we denote this and if N S is the probability of executing S successful steps let us say S successful step by successful we mean the one with the transitions out of n steps out of a total of n steps which means N minus S were unsuccessful where with the pause married in jump. Then if these two probabilities are defined W SM is the occupancy probability the probability that in S steps he has occupied a site M and p is the probability that he has take he has executed S successful steps out of the total n steps then the required expression for the occupancy probability W in n steps with both pause and bias probability of being at M is nothing, but superposition of every possible value of the success number in each trial out of the n. So, the probability that he has taken S successful steps and the probability that he had he had he is present at site M with the of course bias. This is the full and the correct complete expression for the occupancy probability with the pause and bias. So, we have to now develop an expression for the p. So, what is PSN that is the probability that p probability that he has a executed S jumps in out of n total steps. So, S steps out of the total n were successful and n minus when were not successful. So, it becomes a binomial probability it is a binomial given that delta is the probability that is not successful given that 1 minus delta is the probability of success delta is the probability of non success or what you may call as in our present problem it is pausing. So, we can easily write the binomial expression for the probability that there have been S successes out of the total n as n C S 1 minus delta to the power S and delta to the power n minus S binomial distribution for S successes n steps in our case and for WS we have already derived an expression. So, we can the expression for WSM is going to be we also know that the probability of occupancy at site m with the bias will be given by S C S plus m by 2 p prime to the power S plus m by 2 because now it is p prime and q prime these are they add to unity. So, p prime to the power S plus m by 2 and q prime to the power S minus m by 2 this is a very important with the bias. So, upon combining the expression for the binomial expression and the occupancy probability expression with the bias case or partial occupancy probability expression we can explicitly write the problem of pause and bias as sigma n C S 1 minus delta to the power S and delta to the power n minus S then S C S plus m by 2 p prime to the power S plus m by 2 and q prime to the power S minus m by 2 where S varies from 0 to n all possibilities where we remind ourselves that p prime is related to the true probability as p divided by 1 minus delta and in terms of gamma it is much simpler it is 1 plus gamma by 2 the true bias factor and similarly q prime will be q by 1 minus delta which can be written as 1 minus gamma by 2 where gamma is the bias factor as earlier. The left to right jump bias factor does not depend on the weighting. So, we can replace p prime and q prime in terms of gammas and the expression then just we can repeat and write n pause plus bias m is going to be say it looks more complicated then either the bias case or the symmetric case obviously because we have included pausing probabilities. So, it will be n C S and one more S C S plus m by 2 1 minus delta to the power S delta to the power n minus S and 1 plus gamma to the power S plus m by 2 1 minus gamma to the power S minus m by 2. As easily you can note that if gamma equal to 0 special case unbiased, but random walk with pause. So, that is specially specifically when gamma equal to 0 that is random walk with pause, but no bias pause and no bias or symmetric random walk it will be a symmetric random walk for that we will have f n W n with the pause, but no bias will be just this sum S equal to 0 to n n C S S C S plus m by 2 1 minus delta to the power S delta to the power n minus S. We can rename it as symmetric random walk with the pause. In the next lecture we will explore some interesting aspects of how it transforms the normal random walk, how waiting introduces a new element in the problem of random walk and also we will develop an expression for its generating function understand the other parameters of the problem. Thank you.