 So, in the previous lecture we discussed the symmetric or unbiased random walk extensively and derived it is a symptotic form and then we also arrived at an expression in terms of real space and time variables by introducing the concept of diffusion coefficient. Subsequently, we gave a very brief introduction to the problem of what is called as random walk with the bias or it is also called asymmetric random walk which is basically a natural extension to the symmetric random walk that we discussed. There are many areas where asymmetric random walk is of considerable importance. For example, areas of importance could be or areas of relevance, transport of atoms or particles under external fields a simple example would be for example, the falling of a tiny dust particles. So, this is a problem of asymmetric if assume that there is a collision and random walk then the external force would introduce a certain asymmetry at each step. Some sense of direction there are other equally important cases such as for example, trends in the environmental variability of environmental parameters which is a matter of hot interest due to climate forcing for example, one expects temperature trends and all. And this is always superimposed with the other variabilities and randomness. So, in some ways this can also be although it will be too simplistic, but in a idealized way it can be called as a random walk with bias. There are other situations price of commodities and in physical sciences for example, growth and evaporation kinetics. There is always a forcing because of external conditions and that could again leads it to a problem of biased random walk. Then of course, in what is called as the gambler's ruin problems here a lucky gambler or unlucky gambler could have a certain trend say lucky or unlucky gambler. If it is totally neutral then it is equivalent to a symmetric random walk, but one has a streak of luck or misfortune then it will be a trend. Before any application one must first understand and formulate it in a way which will yield some results that can be discussed or held as a benchmark. The real problems could be much more complex. So, we take up a simple idealized problem and that is what I have shown here. You can see that there is a real line in which states are represented by sites indicated by numbers. So, there is a walker as earlier he starts from the origin and then executes random walk, but when he executes the random walk he he or she the random walker let us say he has landed at the m side. Now, we call it a symmetric if the transition probability from m to m plus 1 is not the same as the transition probability from m to m minus 1. So, if the former is p the latter is q all that we know is that p plus q equal to 1 and relatively p minus q need not be 0, but it could have some what is called as a bias factor. So, we now take up this problem in detail. So, let us therefore, postulate that the transition probability from state m to state let us say m plus 1 is not equal to same transition probability from m to m minus 1. So, let us denote this is origin here this is a state m minus 1 this is state m and this is state m plus 1 and this is p and this is q. So, p is the upper transition probability and q is the downwards transition probability and p plus q equal to 1 and we can say that p minus q is not equal to 0. We can call a factor p minus q equal to gamma in denote the gamma as a bias factor gamma equal to 0 it is a symmetric random work corresponds to symmetric random work. We can express also p and q in terms of now gamma because if you add the 2 by add we get also implies we can say p 2 p equal to 1 plus gamma or p will be 1 plus gamma by 2 and q will be similarly if you subtract 2 q will be 1 minus gamma so, q will be 1 minus gamma by 2. So, once we have bias factor we can express both the upward and the downward transition probabilities in terms of gamma. Now, it stands to reason by the same arguments that we have put forward in the case of symmetric random work that the occupancy probability w at a site m for an n plus 1th step should have come from transitions from m minus 1 with the factor p and transition from m plus 1 with the probability q and of course, from nowhere else because we continue to use the nearest neighbor transitions only. So, nearest neighbor model nearest neighbor transition probabilities transitions. So, we continue to use that then we can write them as earlier the equation for the occupancy probability. So, let us denote w n m as earlier, but if you want we can keep a superscript to be b for bias. So, right now I will write it as bias for the bias case as a occupancy probability at site m superscript bias just denotes for distinguishing from the previous case. Then as from following the argument that I mentioned the occupancy probability at the same site m at the n plus 1th step would come due to transition to the right from the at the n from the occupancy probability at the nth step at site m minus 1 and also would come by a transition from the right corresponding to the occupancy probability at the nth step at from the site and m plus 1. So, this is quite obvious now. So, all that we have to do is to solve it and we need an initial condition and as earlier we say that w to begin with thus walker was at the origin where delta now represents the chronicle delta. We can as earlier use the concept of generating functions define generating function g f or g m z will be since all states are accessible in principle. So, the sum will be over all states and it will be summed over the state or the site index and not the step index. So, this is site sum generating function where z now is a kind of a conjugate parameter that replaces the site, but keeps the memory of the site there and z as we mentioned could be any real number and the function g n z should exist in a small strips for some values continuous values of z. So, this is the definition. So, with this definition we can now work with the equation this equation here where the transition probabilities are given as a recurrence relation. So, we can multiply both sites by z to the power m and sum over multiply by z to the power m and sum over m let us call this as equation 1 and this as equation 2. So, then we we can easily see that the left hand side will become g of n plus 1 since we have multiplied by z to the power m it is only the index that changes g n plus 1 and the right hand side we will have now p here and we can easily see by following the same strategy that we adopted earlier that it can be written as m prime equal to minus infinity to infinity w n m z to the power m prime plus 1 plus q m prime equal to again minus infinity to infinity w n again here of course we will call m prime m prime, but z to the power m prime minus 1. So, this m prime was just new dummy index replacing m in the first equation on the right hand side the m prime was actually m minus 1 on the second equation the m prime was defined as m plus 1. So, that in both the cases we get w n m prime only. So, this now becomes a easy to move forward. So, we see that we get an equation g n plus 1 z equal to it will just come to p z plus q by z multiplied by g n z. We can easily see this equation because now the right hand side if you see you can take out the z out. So, the rest is simply g n of z similarly here also 1 by z if you take out the sum becomes just again g n z. So, it becomes p z plus q by z that is how we get this equation you can call it 3. Now, if we iterate this equation the following exactly the same steps using the initial condition iterate with n equal to 0 1 etcetera and use the initial condition which amounts to which implies g 0 z equal to 1. So, we get g n we get g n of z simply alpha z plus beta by z whole to the power n. So, since the g not equal to 1 is quite obvious because it was present only at the origin. So, all other z the powers they it does not exist all other coefficients do not exist that is why it becomes 1. So, you have now the general solution you can call it 4. Now, this is the generating function which is quite useful. Many results can be gotten as we have been seeing in many cases by just using the generating function alone. For example, we can easily note that at z equal to 1 g n 1 should be 1 because it is sum of all the probabilities which is true here. So, sorry it should be p and q not alpha beta p z plus q by z and when z equal to 1 p plus q is 1. So, g n 1 is 1. So, that sort of confirms that we are on the right track. So, we can now explicitly obtain the occupancy probabilities by expanding the generating function in powers of z using the binomial theorem as we did earlier. So, we will not repeat that derivation you can refer back to the method adopted in the case of unbiased random walk where we used a plus b to the power n equal to n c r a to the power r b to the power n minus r formula. So, similarly when we do this becomes sum say m equal to minus n to n n c n plus m by 2 p to the power n plus m by 2 then q to the power n minus m by 2 then z to the power m. So, this is the general expression for g and z expanded in terms of powers of z. So, as earlier we identify the coefficients of z to the power m identify the coefficients of z to the power m with w the occupancy probabilities since the subscript is n this will be w n and since it is z to the power m it will be at the site m as per the definition this as per the definition of g n. Then we get the formula w n with bias m is n c n plus m by 2 p to the power n plus m by 2 q to the power n minus m by 2 this is a actually formally the problem is solved. We can give it a slightly more transparent form in terms of gamma replace p and q with gamma using the relation p equal to 1 plus gamma by 2 and q was 1 minus gamma by 2 then we get w n with bias will be an expression which will have the form w n let us call it 0 without bias m and 1 plus gamma to the power n plus m by 2 into 1 minus gamma to the power n minus m by 2 where w n 0 this is occupancy probability for unbiased case and this we have derived an expression as 1 by 2 to the power n n factorial divided by n plus m by 2 factorial into n minus m by 2 factorial where these factorials have to be for integers which means if n is even m also should be even I mean if yes when n is even m will be even and when n is odd m will be odd that will be maintained this expression was our starting expression for symmetric random walk. So, the problem is fully and clearly obtained which makes it very clear that if gamma is 0 if there is no bias factor then the occupancy probability is same as for the unbiased case. So, that recoveries immediate in this. So, the next natural thing to do is to see the mean and the variance for this case for. So, mean and variance follows from the generating function as we saw or generating function g n z was p z plus q by z whole to the power n and we have seen it now several times that the mean what we call as m bar at nth step the mean at the nth step is given by g n prime to the evaluated at z equal to 1 and if you differentiate for example, here what we will get what is g n prime g n prime z is going to be n of p z plus q by z to the power n minus 1 and if you differentiate again this is going to be p and this will be minus q by z square. So, now if we set z equal to 1. So, g m prime 1 turns out to be if you put z equal to 1 and note that p plus q equal to 1 then the first term the n remains the first term becomes 1 and the second term becomes p minus q and in terms of the bias factor this is n gamma. So, we have the expression m bar mean equal to m bar is n gamma is an interesting result, but quite obvious now that is how the bias factor operates. In fact, another definition of the bias factor could be a measure of the mean displacement after several steps two things to be noted here the mean displacement is proportional to the number of steps gamma is a system parameter. So, it does not depend on n. So, it is proportional to the number of steps unlike the standard deviation which was going as square root of n and it is of course proportional to the bias factor. So, what it means is over superimposed by the random walk there is however a net trend in the direction of gamma positive. So, if gamma is positive here it is p minus q. So, p is greater than q that is if the forward jump is little more probable than the backward jump then on the whole the random walker tends to move in the forward direction towards the forward direction and gamma could be negative if q is more than p and then he would be moving in the opposite direction. So, that expected trend is reproduced by the mathematical formalism that we followed. Similarly, we can determine the variance which we simply mean short notation variance var which is defined as second derivative plus g prime 1 minus g prime 1 square. Here now one has to differentiate g n prime z once again it is a bit tedious, but very simple procedure to do. We will only give the results here it just turns out that this variance will become n into 1 minus gamma square or the standard deviation sigma equal to square root of 1 minus gamma square into n to the power half. If gamma is much less than 1 this will tend to n to the power half if gamma is far smaller than 1. So, if the bias factor is very weak then the standard deviation is almost undisturbed it remains the same at least to the order of gamma square same as that for the symmetric random walk. So, we can visualize this whole process as shifting of. So, the originally our random walker was here and slowly at some and after some steps the random walker will sort of would have moved with the mean which is now shifted and his standard deviation measure of the standard deviation would have been still this is after. So, I will join it for guiding the i after n steps this is m bar this could be sigma. So, there is a peak movement is seen unlike the symmetric case where only broadening takes place, but more or less the peak would remain at the origin itself. Thank you.