 Let us illustrate the application of the convolution theorem by taking up a hypothetical example. This is not a realistic example, but just introduced merely to illustrate how to obtain the impulse function from information obtained on the output function. So, let us say that in some experiment one has obtained an output function phi t as t square by 2 e to the power minus t. This functional form has been decided by fitting the data or some other theoretical reasons for all t greater than 0. Also this output has been because of two things. One a certain form of the input function f t which is to be known, but what is known already is the response function r t. How the system behaves that transfer function or response function is known and let us say it is known to be e to the power minus t for t greater than 0. What is required is what was the signal function f t which gave the output phi t. So, we use the convolution theorem which says that phi tilde s should be equal to f tilde s and r tilde s and r tilde s is very simple. It is the Laplace transform of e to the power minus t that is it is basically 0 to infinity e to the power minus s t minus t d t which is going to be 1 by 1 plus s. Similarly, the output function will have its Laplace transform as 0 to infinity t square by 2 e to the power minus t d t e to the power minus s t and that will turn out to be 1 by 1 plus s cube. So, that easily you can see that it is a gamma function and when we factorize out s plus 1 will come out and t square d t e to the power minus s plus 1 t will be integrated out. So, from the knowledge of these two functions phi tilde and applying it in equation 1, we can see that the Laplace transform of the quantity of interest is nothing, but the ratio of the Laplace transform of the output function and the response function which is going to be 1 by 1 plus s whole square. If we go back to some of our previous presentations, you could see that this is nothing, but the Laplace transform of a function f t equal to t e to the power minus t because if you Laplace transform e t e to the power minus t one is going to get 1 by 1 plus s square and that inversion could be obtained by looking up at the table of Laplace transforms. So, this is a very idealized or a purely theoretical example of reconstruction of the signal by using convolution theorem. There will be several occasions where Laplace transforms will be required in our study and we will be discussing it as the case arises. We move over to another family of generating functions called generating functions for discrete variables. As we discussed on several occasions, when my sample space is when my sample space consists basically of discrete points, we assign probability mass functions on those points denoted by P n where n could be points on the real line. They can have both positive and negative values. So, this family of P n can also be handled via generating functions, but instead of the integral transforms it is more advantageous to construct sums. So, we have a family of generating functions constructed as sums of a function z to the power n P n summed over all n and this generating function is now a function of the variable z. So, z is a conjugate variable, it is conjugate to n the discrete point n and z of course, is a continuous function and it can be complex. It is a domain depends on the summability of this function n equal to n in the entire domain of n. So, it is quite useful in dealing with discrete variables and we can give some simple examples. For example, if we have a function f P n equal to q to the power n into p let us say we have some function and we can construct its generating function g with respect to variable z by following the definition and let us say this function P n is defined for 0 1 2 etcetera then it will be z to the power n q to the power n p n equal to 0 to infinity which is going to be p into n equal to 0 to infinity z q to the power n. Assuming that this quantity z q is less than 1 its sum is going to be p divided by 1 minus z q where we use the property that sigma x to the power n n equal to 0 to infinity will be 1 by 1 minus x for x less than 1. Similarly, this is valid for z q less than 1. So, we have transformed the p n function to a function of z which contains all the information that the p n carry like in the case of both characteristic function as well as the generating function in the integral transform method. The various moments of the p n distribution can be easily generated via derivatives of the generating function with respect to z. We will skip the details here. We can easily see that if we define the moments say m k kth moment by definition is going to be n is basically n to the power k p n n equal to let us say the domain is 0 to infinity its over the entire space of n then as earlier we see that m 1 equal to n p n which is n bar assuming that sigma p n is normalized. Similarly, we can define the second moment n square p n which is n square bar. So, all these moments can be also related to the derivatives of the generating function. So, we can see that for example, g prime evaluated at z equal to 1 is going to be m 1 and we can easily see that the g double prime 1 it will of course, be involving n square average minus n average that is m 2 minus m 1 it is not directly m 2. The reason is when it is differentiate second times you get n into n minus 1 and all that. In any case we can evaluate the variance sigma square as by suitably adding and subtracting m 1 and m 2 by definition this is nothing, but m 2 minus m 1 square and it is going to be g double prime 1 plus m 1 minus m 1 square where m 1 is g prime 1. So, we can obtain the mean as well as the variance of the distribution by taking the derivatives of the generating function. So, you do not need not sum the p n function all the time with all its moments. The in many random walk and other problems the sum based generating functions are very useful considerable time of this course we will be dealing with discrete discrete stochastic phenomena and generating functions based on the sum procedure using the z variable come very handy in obtaining the various properties of this probability densities associated with discrete processes. Another aspect of considerable importance in stochastic theory is the concept of probabilities and probability distributions. Probability distributions are central to description of stochastic phenomena because we are basically dealing with random variables. One way we can look at the entire methodology of stochastic study as finding the evolution of the probability distributions or finding what phenomena cause or lead to signatures of various probability distributions or what are the underlying processes which lead to a given distribution. So, in order to arrive at a better understanding it is important that we understand standard distributions which are known already. Although many of these distributions emerge with the new meanings when they come up in the theory of stochastic processes. So, we have a quick look over some of the very well known probability distributions which also occur centrally in the various processes that we are going to discuss. Coming to probability the first thing one has to define the problem define the variable appropriately and this becomes important. So, that one has a unique definition of probability associated with that random variable. There are examples that the problem may not be well formulated in which case one would end up with multiple solutions or multiple answers of probability thereby implying that some parameter or underline parameter has been missed in formulating the problem. So, the sum and substance of what I am saying is that we must first identify the right variable of the problem. So, in this connection some of the well identified statistical distribution become guiding factors. One such probability distribution often studied is the famous binomial distribution. This distribution refers to a sample space generated by throwing coins or tossing coins. So, a coin has a head and a tail and when you throw one coin when you throw coin at once it is assumed that there is a probability P that head turns up and there is a probability Q which will be 1 minus P that tail turns up. For a fair coin P equal to Q equal to half, but one can have coins with slight biases or if you do not want coin and still want to understand why P and Q are different one can imagine throwing dice and here I reduce the dice problem to a dichotomous problem by saying that dice has 6 sides any of the sides can come up, but however I state my problem like this what is the probability that the number 1 comes up that is of course is 1 by 6, but I say that I am interested in knowing it has a success if 1 comes up and failure if all other faces turn up. So, when I formulate it as success and failure problems my probability of success P is going to be only 1 by 6 and probability of failure all other faces are failures. So, it is going to be 5 by 6. So, it is now reduced to a binomial problem of P and Q type and very clearly with a physical basis for P and Q being different. So, let us revert back to the coin problem. Now, one has tossed this coin several times say 100 times and a natural question that arises is to find out what is the probability that R out of n tosses have turned up with head up and the remaining n minus R tosses have turned up with tail up. So, if you call head as a success and tail as failure the same question can also be asked that what is the probability of R successes and n minus R failures in a random trial involving n tosses of a fair coin. So, this probability is called as P of R probability of R successes and we can superscript it with the n to indicate that it is out of n trials. And this comes as one of the terms of binomial expansion this is n Cr the combinatorial term. Now, the probability that head turns up is P the probability that 2 heads turn up will be P square. So, the probability that R heads turn up is P to the power R and the remaining n minus R the tails turn up will be the AND statement. So, multiplied by Q to the power n minus R. And for us the order of appearance does not matter. So, number of distinct possibilities then turns out to be n Cr where n Cr is a well known term combinatorial term n factorial divided by R factorial into n minus R factorial. So, this is the probability of obtaining R heads and n minus R tails in a random trial involving n tosses of a fair coin. We can note this as the Rth term in the binomial expansion of P plus Q to the power n which is of course, 1 because P plus Q is 1 Q is 1 minus P. So, it and binomial expansion if you apply P plus Q to the power n is basically n Cr P to the power R Q to the power R R equal to 0 to n. So, if you sum all these probabilities it should come to 1 and which is what the probability is about that totally one of the sequences has to turn up. So, that is what it means. So, so the choice of our probability distribution function is consistent with its normalizability. The function Pr has many interesting properties it is a its mean mu which is obtained the mean success will be R equal to 0 to n R into Pr that is the definition of mean which is also we can call it as R expectation or R average and actually we can sum over this whole thing and it will turn out to be n into P. So, intuitively we can see that the probability of success for each coin is P on an average. So, on an average the probability of success of n coins it should be n into P because it is either the first one or the second one or any of the total n throes. So, we have the mean is n into P. Similarly, we can show that the variance around this mean that is n P which is obtained by definition definition of variance is m 2 minus m 1 square. So, take that second moment and then take the this is also m 1 that is the variance. So, binomial distribution is a two parameter distribution it has a mean and a variance and they are dependent on two variables n and P they are independent. We can also develop a generating function for this for example, g of since it is a discrete distribution our generating function d g z is going to be z to the power r Pr n R equal to 0 to n and it is going to be 0 to n z to the power r P to the power r q to the power n minus r n Cr which is going to be z P to the power r q to the power n minus r n Cr which is equal to q plus P z to the power n. Earlier it was P plus q to the power n which led to this series, but now P is replaced with z P and that is why we have this we have this. So, the generating function is easy to construct from this generating function and its derivatives at z equal to 1 you could have obtained expressions for the mean as well as the variance. Binomial distribution goes beyond coin tossing and there are many interesting examples where binomial distribution appears. For example, if we consider the case of infection transfer between a an infected individuals let us say an individual A and another healthy individual let us say an individual B. Supposing P is the probability of probability of infection transfer per encounter of a healthy individual with the infected individual and each time an encounter occurs let P be the probability that the infection is transferred from the infected person to the healthy person that is from A to B. So, A is infected so, I call it infected and this is healthy. So, we can now ask a question what is the probability. So, we can ask a question what is the probability that our infection transfer events will take place in n encounters. This problem is now identical to the binomial problem of the coin tossing and the probability that r infection transfer events have taken place out of n encounters. Given the probability of success of infection transfer is P and failure is Q equal to 1 minus P is given by P r n n C r P to the power r Q to the power n minus r. This distribution also efforts us to test whether a particular coin is fair or not. If it is a fair coin one expects that after sufficient number of thrown on an average 50 percent heads should have come and 50 percent tails would have come. But however, the question of what constitutes a sufficient number of throws remains open. So, if one decides for example, to test a coin with just 100 throws one would perform the 100 throws count the number of heads and come the number of tails and they will not be 50 50 there will be some asymmetry because of the fluctuations at what level the person can accept that it is a fair coin is ultimately left to a judgment. But to help that decision making it is the P n r gives us a probability of occurrence of spread around the mean value of 50 by let us say 1 sigma. So, for example, if there are 100 coin tosses for a fair coin the probability is I mean 45 and 55 for heads up and tails up lying between 1 sigma of the mean value. So, if this is constitutes about 73 percent of the area. So, if one is satisfied that the number lies between 45 to 755 one can sort of satisfy oneself that perhaps if you had done an experiment with the larger number of trials it would have come closer to 50 50 and the coin would have been fair coin. There are many other applications as we go along many other interpretations of these simple distributions which we will be using. Thank you.