 So, we showed that the survival probability of the random walker in the presence of an absorber is so, 1 D absorber unbiased continued. So, we showed that the survival probability S T in the presence of an absorber is equal to basically an error function of x naught by 4 D T. So, this as a T tends to infinity we saw that the because the error function of an argument 0 is a function itself it tends to 0. So, S T tends to 0. We can now evaluate some other quantities of interest for example, the loss rate loss rate of the particle or of the walker random walker. More precisely it can be defined as the probability current of adsorption. So, we call it as J equal to minus D S by D T is the probability current or absorption current upon differentiating the error function. We can show that this becomes x naught by 2 root pi root pi D 1 by T to the power 3 by 2 e to the power minus x naught square by 4 D T that is because the derivative of the error function is the Gaussian itself from its definition and then you have to differentiate 1 by root T and you get this. Interestingly this function now clearly shows that the current is 0 at T equal to 0 J equal to 0 when T equal to 0. How? Because when T equal to 0 e to the power minus x naught square by 4 D T becomes e to the power minus infinity which tends to 0 much faster than T to the power 3 by 2. Although it is a 0 by 0 form, but this 0 goes much faster. So, eventually J becomes 0 and J tends to T to the power minus 3 by 2 as T tends to infinity. So, the current slowly it tends to 0 as a power law, but as T to the power minus 3 by 2 law. One can plot the current distribution therefore, as it is 0 at T equal to 0 the absorption current to be 0 high and then decays like a power law. So, approximately the reason why it is 0 is after all the walker is at a distance x naught and he takes some time to reach the absorber and hence initial times is quite far away from the absorber and therefore, current is 0. It goes to 0 sufficiently far away in time because eventually one day absorber gives no chance of escape for a random walker. No matter how far you have moved eventually you might have to come back to the origin visit at least once and one visit at the absorber is sufficient to kill the absorber from the from the side space. So, this perception this way of looking that an absorber therefore, is basically a counter and absorbing barrier is basically a keeps a count of the very first visit that the random walker has made. Since there is no return quite often this concept is now adopted in defining what is called as first passage time distribution that is the distribution the probability distribution that the random walker has crossed the point to 0 for the first time. So, that is equivalent to the quantity minus d s by d t or alternatively the j t that we have given. So, the if you define that probability distribution as f t then f t is j t and it is given by the expression that we wrote it is x naught by square root of 4 pi d t to the power 3 by 2 e to the power minus x naught square by 4 d t. So, although mathematically it is equivalent to the current seeing it as a survival time distribution or first passage or first contact time distribution gives a an additional ah interpretation a additional strength in handling particle transport problems when problem is of course, much more complex than this it may not be a free space it may be a space in which particle is moving under some specific forces then you need not solve the entire problem it is sufficient to obtain a obtain a formula for this ah first passage time distribution which is equivalent to calculating the currents. Interestingly this distribution while it is normalized because we know that the integral f t d t it will be normalized to unity because it will be it is a derivative of d s by d t and hence it is as infinity minus as 0 which is 1. So, as a result the function exists, but however other moments for example, if you want to know the mean time of contacting the absorber of a random walker if you ask the question ok random walker is eventually probability is going to go to 0, but what is the mean time that quantity very easily you can show that that will be infinity. That means, this is a distribution function which is normalizable, but which does not have a first moment. In fact, it will not have any other moment now t square also will go to 0 ah quote infinity and why does it go to infinity it is easy to see because this integral 0 to infinity if you multiply by t it is going to be essentially t to the power minus half e to the power minus x naught square by 4 d t d t and this integral goes as t to the power minus half as t tends to infinity and integral of t to the power minus half is divergent. So, this will tend to infinity in fact, this integral is infinity. So, we we started some time about central limit theorem the kind of restrictions we put on the mean and the ah standard deviation existence of these two moments. That time we saw that there is distribution called a Cauchy distribution Cauchy distribution which does not have higher moments and we have got one more example of such distributions arising naturally in the context of random walk. Now that we have handled the problem of 1 d walker with undergoing a symmetric random walk and we found that eventually the random walker is going to be absorbed on the absorber. The question arises will there be a possibility of random walker surviving eventually if you have a biased random walk this question is of great practical interest. Therefore, we explore the problem of biased random walk in the presence of an absorber. So, 1 d absorber unbiased random walk it is not unbiased random random walk with bias random walk with bias. So, we have the same situation there is an absorber at site 0 and the random walker starts from lattice points discrete lattice points like this and let us say from a point M naught and it is of interest to find his survival probability at some lattice point M or his occupancy probability. So, the same old question. So, to find the occupancy probability W n let us say well we can call it as biased. So, we can call it as bias at a point M starting from some point M naught. So, this is the M. So, hence we postulate that since it is a problem with bias the bias probabilities are P to the right and Q to the left P naught equal to Q yeah, but P plus Q equal to 1 P transition probability from some site M to M plus 1 Q transition probability from M to M minus 1 only nearest neighbors transitions P plus Q equal to 1. With this we can set up the random walk equation like we had done earlier that is the occupancy probability at the n plus 1 step of being found at M with the absorb with the bias. So, it is an absorber plus bias problem we will we will put this this is a superscript later at the end when we summarize a formula, but right now in order to keep up the speed of writing we will not introduce it here, but always keep it in our mind. So, this probability comes from the probability that he was at M plus 1 side. So, he was at let us say at the previous step at the nth step he was at M minus 1 side and he jumped to the right plus at the nth step he was at M plus 1 side and jumped to the left supposing this is this random walk equation given the fact that we are talking of we exclude of course, the absorber site in question and the absorber being defined as earlier at the absorber that is W n at the rather the transition probability the transition probability of the absorber property of the absorber is that the transition probability at M equal to 0 is 0 that is P equal to 0 at M equal to 0. So, particle which has landed on the site 0 does not come back to the walking space. So, that is the main criterion of absorption. So, we cannot straight away construct a virtual walker for this problem because there is no symmetry the left and the right jumps are not the same there is a for example, let us say that P is greater than q there is a slight bias towards movement away from the absorber. So, this means the virtual walker will not exactly be symmetrical with respect to the real walker starting from any point M naught. So, it is necessary that we symmetrize this problem first it is possible to symmetrize reduce it to a form which is then symmetric and apply the same virtual walker principle for that symmetrized equation and obtain a solution. So, in view of a symmetry a symmetric nature of the random walk we cannot set up a virtual walker on the left side of the absorber this is quite stands to reason. However, it is possible to symmetrize this equation equation. So, we see how to do that. So, for that it is actually it can be done in few steps, but we will skip the step, but confirm that what we are saying is true. So, supposing we define a new probability instead of W that is we transform W N M as P by q to the power M by 2 2 to the power N P q to the power N by 2 phi N M where phi N M is now the quantity of interest we have basically defined a new function phi N M such that the original occupancy probability factorizes out certain terms which involve P and q. So, when we substitute we can see that the you know the asymmetry gets removed. Now, the function phi N M is such that the initial condition phi 0 M this is W 0 M because if you put N equal to 0 it will be W 0 M P by q to the power minus M by 2 you know I am just putting N equal to 0 M putting. So, this terms go and it will be this and since W 0 M is the chronicle delta we want the walker to start from plus M naught. So, it will be delta M M naught P by q to the power minus now it is whether it is M by 2 or M naught by 2 does not matter because it exists only for M equal to M naught. So, I can as well make it a M naught. So, with this initial condition we can set up an equation for phi M itself and it actually comes to a completely symmetric equation of the form we will skip those steps it becomes phi N plus 1 M equal to half phi N M minus 1 plus half phi N M plus 1. Actually this almost follows straight forward when we substitute wherever W of N plus 1 comes it will be P q to the power N plus 1 by 2 and where M minus 1 will be M minus 1 by 2. Actually there will be lot of cancellation of terms and the choice has been so made that the P and q have been properly cancelled in each of the terms on the right hand side and leaving only half. We could have proceeded in 2 steps you could have arbitrarily taken some say lambda to the power N and alpha to the power M then substituted them then evaluated alpha and lambda 2 variables to be in such a way that everything else cancels except leaving half. So, once again you would have arrived at the same expression and in fact, that is how it is done and that is how these pre factors are gotten. So, this is symmetric random walk or symmetrized random walk with an initial condition that phi 0 which I have already evaluated M equal to delta M M naught P by q to the power minus M naught by 2. Now that this is a symmetrized random walk only difference being that instead of the original probability is just a Kronecker delta the intensity of the random walker or the strength of the random walker or the number of random walkers we can say has been changed from 1 to P by q to the power minus M by 2 some other number. So, my virtual walker also now will be at minus M naught, but with equal intensity P by q to the power minus M by 2. So, you have now the problem of a virtual walker and a real walker undergoing a symmetric random walk in the presence of a barrier and whatever the paths that are crossed by the true walker are virtually supplemented by the virtual walker same logic we use and we know that that leads to a solution which is the difference in the occupancy probabilities of a free random walker problem. So, with all that we can easily show that the probability of occupancy of the new variable in expressed in terms of the phi variable will involve it this factor which is actually the strength factor that we introduced and it will be W N 0 M minus M naught minus W N 0 N plus M naught. So, for all practical purposes the phi function has a same property as the W with the absorption absorber function, but for this P factor. So, with this now we can go back and obtain the full expression we can go back to this equation where we defined phi once phi is known we can obtain the W and we see that the solution comes out to be W N at M side. Now we can specify it with it is a bias plus absorber equal to P by Q to the power M minus M naught by 2 2 to the power N P Q to the power N by 2 and the free random walk probabilities. We can rewrite it again in terms of define the symmetry transition probability factor that is gamma such that P equal to half into 1 plus gamma Q equal to half into 1 minus gamma which implies that gamma equal to P minus Q it is basically asymmetric factor we have done it earlier the same gamma we are using then this expression turns out to be W M with bias plus absorber M equal to 1 plus gamma by 1 minus gamma to the power M minus M naught by 2 1 minus gamma square to the power N by 2 and the again free random walk probabilities. We can carry out the asymptotic forms now in order to explore the features of this distribution we have seen that the as N goes to infinity the occupancy probabilities have a Gaussian form, but we have now an additional factor in terms of gamma. So, we must explore how that varies as N tends to infinity. So, we can write for example 1 plus gamma by 1 minus gamma to the power M minus M naught as equivalent to exactly equivalent to M R M naught by 2. So, this will be M minus M naught by 2 it will be ln of 1 plus gamma minus ln of 1 minus gamma can always write like this. Now, it will go over to in small gamma approximation this will be gamma. So, it will be 2 gamma and gamma gamma cancels. So, we are going to be left with e to the power plus gamma into M minus M naught where we have assumed that gamma is now much less than 1 that is we have assumed log of 1 plus minus gamma equal to plus minus gamma we have made this use of this assumption. So, this part is fine now what about the term 1 minus gamma square to the power N by 2 and we can do a similar asymptotics this is actually e to the power N by 2 ln of 1 minus gamma square and again doing the Taylor expansion it is going to be e to the power minus of gamma square in by 2 into N. We continue with this and see the consequences of asymmetry on the survival probability. Thank you.