 So, while talking of the Poisson process, I said we have to first talk about a counting process. And so, we said that there should be some norms for the, have to be followed for the counting process. And so, the first one is that it should have independent increments. I should underline increment, independent increments. So, that means that the number of events that occur in disjoint intervals are independent. So, that means say for example, if you are saying that up to 8 a.m. you are counting the time, then n 8 would be the number of events that have occurred up to time 8. And so, that n 8 will be independent of say for example, the number of events that have occurred in the, from between at 8 and 10 a.m. So, that means n 10 minus n 8. So, both these are random variables. And so, we will assume because the intervals are disjoint. This is up to 8, this is 0 to 8 and then this is from 8 to 10. So, the two intervals are disjoint and therefore, we expect, we want that these two, the number of events that means the corresponding random variables must be independent. Now, for the, for example, a that I see I gave you three examples of counting processes. First one was you know arrivals at a post office. And so, example a, this may be a reasonable assumption because we may assume that people as long as the, for the time that the post office is open, people will come in any time. And so, people coming say between say up to from, if it is open from 8 a.m. then 8 a.m. to 10 a.m. the number of arrivals and the random variable indicating the number of arrivals and the say between 10 and 12, the arrival, the random variables, the random variables giving the number of arrivals in these disjoint intervals are independent. Now, for b, which is the number of births in a particular town or this number of births between time t and t plus delta t, suppose we are taking, will be large. If n t is large, so that means that if you taking it over a long span, then at a particular time in the population is large, then the number of births will be large. So, here it will depend that means it does not seem reasonable that n t is independent of n t plus delta t minus n t for any t. So, it will depend, see this will depend on the number of births that occur if you are taking up to or in the sense that we are saying that the random variable n t and the random variable n t plus delta t minus n t may not be independent, if you are. Now, for example c, it will be a reasonable assumption, so this is example c referred to the number of goals that are hit by a hockey player. So, here again if you take the time span to be one season of hockey tournaments, so during one particular season we expect the hockey player to be either to have continue to have a good form or not have a good form. So, if he has a good form, then he will hit number of goals hit by him up to time t or from t to s t to s plus t should be the two random variables should be independent. So, the time span is important, if I take the time span to be two years, then certainly it will matter because one cannot maintain a form for let us say up to two years or five years, so the time span. So, if you restrict your time span, then c would be a reasonable assumption, for c the independent increment assumption would be reasonable. The other important assumption for counting process is stationary increments. Now, here what we are saying is that the number of increments that occur should depend only on the length of the interval. So, here for example, if you have s comma s plus t, then the length of the interval is t. So, it will not matter what value s takes, as long as the interval length has time t, then the number of increments that occur during this interval is just dependent on the length of the interval. Now, here we can again see whether the counting processes that we wrote down is that reasonable assumption for all those counting processes. For example, for a it is not a reasonable assumption, why because for a post office and even similarly you can consider a bank, there may be a rush hour. If there is a rush hour, then certainly you cannot say that this and this are independent, say if that this is the rush hour, then you know that there will be more arrivals and so the random variables n 8 and n 10 minus n 8 would not be independent. So, if you have the concept of rush hour, but if you sort of ignore the rush hour and then you look at the counting process for a post office, then this may be assumption of stationary increments may be a reasonable assumption. Now, again for b it may not be a reasonable assumption, in this case some sort of pattern has been observed in the number of births. See some of our time may be more babies may be born during winters and so again if some pattern has been observed in the town that you are considering, then again b may not be because it will not follow the assumption of stationary increments. So, it will matter if your time span is during summer, so then for the same period more babies may be born as opposed to the time span this is during winters. So, for the same length of time the 2 may not be the number of events in the interval has the same distribution for all s, so it may be different distributions. Then again for c it will be reasonable here again if you are saying that if the person if the player if the hockey player is in form, then surely the number of goals that he hits will depend on the length of the time that he has played. And so it will have the same distribution and will depend on the distribution of the number of goals that he hits would be dependent on the length of the interval and not on the when he hits. If I am again restricting myself to let us say 1 season or may be 2 seasons if that is considered to be reasonable that a person will continue to be informed for a player will be continue to be informed for 1 season or 2 seasons may be sometimes it depends whatever the way to look at it in that case the stationarity increment assumption would be a reasonable one for c. And so this is what I am trying to say is that you have to before you start applying a modeling a situation for example, a particular counting process you have to see that certain basic assumptions are satisfied and in that case you can you know then we will see that based on these 2 assumptions we can now talk about the poison process. So these are the 2 basic assumptions under which we will now formulate our this probabilistic model for the for counting for the counting process and which is which we will define as the Poisson process. So after having defined a counting process now I make a definition of a Poisson process. So what we are saying is that the counting process n t t greater than or equal to 0 is said to be a Poisson process if it satisfies the following conditions. So as we said that we start the counting from time t equal to 0 so n 0 is 0 then and this has a independent has independent increments which we have already said that well we define the conditions for a counting process and then we said additional properties were has independent increments. Then 3 is that number of events that occur in any interval of length t has Poisson distribution see this is the thing. So therefore, we are saying that it will be a Poisson process the counting process will be a Poisson process if it the number of events that occur in any interval of length t has Poisson distribution with mean lambda t where lambda is some constant positive constant. So that means what we are saying is that because the interval of length is t time interval is the length t. So therefore, probability of n s plus t minus n s equal to n because what we are saying is the number of events that occur in this interval s plus t comma s you can say that the length of the interval is t equal to n will be e raise to minus lambda t lambda t raise to n upon n factorial and n varying from 0. So this is your probability for any value of n here integer value of n. So now since we are since this probability is only dependent on the length of the time interval that is t it is not dependent on when the counting process you started the beginning of the interval. So therefore, you can immediately conclude that the process that we have defined also has the stationary increments property which we said that it does not matter it is only the number of events that occur with depend on the length of the interval and not on when you started counting. So therefore, this would be a Poisson process that is what our definition is. So now, let us just look at this definition and see you can of course, condition 1 is fine that simply says that counting begins from time t equal to 0 which we have been saying repeatedly. Then 2 can be verified from the knowledge of the process that it has independent increments whether this is a valid assumption or not for the process that you are trying to model then you can tell from the knowledge of the process itself. Now 3 says that the number of events that occur in any interval of length t has Poisson distribution. So it is not clear how to verify 3 and this is the whole thing I mean if I am calling a process a Poisson process then I am just saying that the number of events that occur here follows a Poisson distribution. So it is not clear and therefore, this definition is not a very implementable decision or you know you cannot really verify this condition because right. So therefore, an alternate definition is needed to determine whether a counting process is a Poisson process or not we would be wanting to have another definition which hopefully would be more easily implementable or verifiable that a given counting process is Poisson or not. So let us see so here we will say that again the same mean lambda t if it satisfies the following conditions. So counting process is said to be Poisson with mean lambda t if it satisfies the following conditions. So this is same as for the first definition n 0 is 0. Now what we are saying here is that probability n h equal to 1 that means in the time interval h your probability that n h is equal to only one arrival takes place or one event occurs it should be in this form that means lambda h plus order h. Now this of course means that your function here is of values of a higher order than h and you say this is higher order of h. So this is linear that means the terms here would be square cube powers of h. In other words you can say that what this means is that limit o h upon h as h goes to 0 is 0. So therefore, you know this is of higher order than h square cube this for example, if you take o h to be h square then h square upon h limit of this as h goes to 0 is 0. So this is the idea that means higher order terms here. So this and so what we are trying to say is that you can always discretize the occurrence of the events that means if you take if you make the interval h small enough then you know there will be the probability of an occurrence of an event is positive in a small interval. So therefore, you can discretize and the second one and so of course, this is the third condition the second of course, is the same as that one that process has stationary and independent increments. So these two properties we require for the counting process to be able to to be able to to be able to say that it is Poisson process. So this is has to be satisfied. Now this one tells you that the probability of the occurrence of event in a small interval is dependent on lambda and that you can separate out the. So in other words the you know bunching of occurrences of events is not permitted here that probability n h greater than or equal to t is of order h. So that means when h is small this probability is really is very small of the two or more events occurring in a very small interval of time h. So that is of order h and since as I have told you that this means that the higher powers of h then linear. So therefore, this would be you know when h is very small this also be very small. So given lambda positive the same lambda we are saying here when I say this that means it is understood that lambda is greater than 0 here. So this is the alternate definition and let us now see that obviously when we are saying that this also describes a Poisson process that also describes a Poisson process that should be we should be able to show that the two definitions are equivalent. So I will do it one both ways I will first show that definition 1 implies definition 2 and then show you the definition 2 implies definition 1 and this is very interesting and nice simple. So here see we start with this definition 1 says that probability n of t plus s minus n s equal to n is e raise to minus lambda t and lambda t raise to n upon n factorial and n values like this. Now put s equal to 0 t equal to h because this is for all s t and n equal to 1 in this equation then you obtain that probability n h equal to 1 because this is 0 t is h s is 0. So n h equal to 1 is given by e raise to minus lambda h then lambda h raise to n is 1 and this is it. Now just expand e raise to minus lambda h that only 1 minus lambda h plus lambda h whole square by factorial 2 and so on multiplied by lambda h. So when you bring lambda h inside this will be lambda h plus all terms will be of higher order because this will be square h square this will be h cube and so on. So I can write it this way and which satisfies. So that means your condition 3 implies condition 3 here for the definition second definition and similarly it is also implied from here because when you want to compute this probability n h greater than or equal to 2 this will be 1 minus probability n h equal to 0 minus probability n h equal to 1. So n h equal to 0 again since your definition is saying that when n is 0 the probability is e raise to minus lambda h that is all because your n is 0. So that is e raise to minus lambda h and this is probability n h equal to 1 we have just computed is this thing. So minus lambda h in order minus of plus does not matter whatever and now you see when you expand this because this will be 1 minus of 1 minus lambda h plus lambda h whole square by 2 factorial minus lambda h plus o h. So then 1 so you have a minus sign here. So this becomes plus because minus and then 1 minus lambda h. So therefore this is 1 1 cancels out lambda h minus lambda h cancels out and you are left with something because this is order h. So order higher than h and this is also o h. So therefore the whole thing is o h. So you can see that definition 1 implies definition 2 and now we will show you that definition 2 implies definition 1 and you will see that this then we would most of the time be working with this definition of the Poisson process. So let us see we will try to now show that definition 2 implies definition 1 and of course when then we will talk about inter arrival times later but you see here probability n t plus h. So this I am saying is the probability of n t plus h is equal to n this is what we are saying what we mean by this. So this is a new notation I have started. So p n t plus h that means number of arrivals are n in up to time t plus h. Now this can be thought of as n minus 1 arrivals up to time t and then 1 arrival at within the interval h. That means see from t plus up to t plus h you want n arrivals or n occurrences. So up to t if there are n minus 1 occurrences then in the interval time length of h you want 1 arrival and according to our definition 2 this is the probability of 1 arrival in the time interval h length of interval is h and since we are talking of independent increments. So this will be product that means I can say this plus there is no that means there n arrivals up to time t and there is no arrival in time length h this is what and this will be probability n h equal to 0 which follows from 3 and 4. See 3 said that probability of 1 arrival is lambda h plus order h and probability n h greater than or equal to 2 was of order h. Now when you say that there is no arrival then that means you want 1 minus probability n h because you want this to be. So if you take the complement so then n h greater than or equal to 1 would that be okay. See if you want n h equal to 0 in time so that means arrivals are 1 2 and so on in the interval h. So 1 minus of that will be right. So therefore yes so this would be then 1 minus lambda h minus because this is o h and this is lambda h plus o h. So therefore this is what you have. So therefore probability of n h equal to 0 is 1 minus lambda h minus order h. So again independence of events independent increments so this will be p n t into this right. So this is how you can describe n events in time up to t plus h by breaking up the event into number of arrivals up to time t and then n minus 1 1 arrival in time h or n arrivals up to time t and no arrival in time n h. So this is how we will write it down right. And therefore this will be if you simplify this expression p n t is coming from here plus lambda h p n minus 1 t minus p n t right. So this is it plus our terms of higher order of h right. Now just rewrite this this is p n of t plus h minus p n t divided by h. So that will become I have divided by h here. So lambda times p n minus 1 t minus p n t and this is order h upon h. Now when you take the limit you can immediately see that when you take the limit as h goes to 0 this will be limit as h goes to 0 of this and here this will be lambda this is independent of h and this we have seen will go to 0 right order higher order of h means that this limit is going to 0 right. So this is what you have and then this is nothing but the derivative of p n t. So this will be the derivative and this is equal to lambda p n minus 1 t minus p n t. Now so this is valid for n varying from 1 because you have this n minus 1 and d p 0 t upon d t is minus lambda p 0 t because when you are looking at n equal to 0 see from here then you do not have this you simply have this right p n t because I should have n equal to 0 and right. So it will be if you want me to write it down separately see here becomes crowded. So this will be p 0 t plus h is p 0 t. So no arrival up to time t and no arrival in the time interval. So will be 1 minus lambda h remember I can ignore that term because that will anyway go to 0. So then this will be p 0 t plus h minus p 0 t divided by h. So limit of this as h goes to 0 it is not legible but you can I am talking loudly. So you can hear. So this is equal to p 0 t minus into lambda. So this is therefore this is derivative d p 0 t upon d t is equal to minus lambda p 0 t and so you this much knowledge you have about the differential equations. So here from here it follows that p 0 2 is e raise to minus lambda t and so you have these two sets of differential equations. So here the solution is immediate and then this is n from 1 to t. So if you put like for example n equal to 1 it will be d p 1 t upon d t which will be from here minus lambda p 1 t plus lambda e raise to minus lambda t because I am putting substituting for p 0 t. See when n is 1 this is p 0 t which is e raise to minus lambda t. So therefore this is it. Now of course there are methods to solve these differential equations are difficult but what we are saying is that you if you just try p 1 t equal to lambda t e raise to minus lambda t it will satisfy this equation. Just differentiate and substitute here the two sides will be equal. So p 1 t is a solution here and then in general the solution would be p n t you can very easily verify that you know for all values of n this is the solution to your general differential equation that you obtained here. And this is nothing but as I said that my definition is that p n t is nothing but probability of n t equal to n. So n arrivals up to time t. So therefore you see all other conditions remain the same in the definition 2 and 1. This was the only thing because we were not sure how we would go about verifying in definition 1 that the number of arrivals in time interval of length t will follow Poisson distribution. So now you know using these properties you know a probability of definition of number of occurrences in the time interval h given those the third and fourth condition of definition 2 help us to show that the number of probability of number of arrivals up to time t would be follow a Poisson distribution. So a nice way of showing that and because therefore definition 2 is easier to verify we feel because that you know to digitize and so on you can sort of approximate the condition 3 actually that is the important one in definition 2 and then that is probability of one occurrence in time interval length h is of this order that you can supposedly easily verify there are methods to do it. So therefore most of the time we would be but now once we have established the equivalence of definition 1 and 2 it does not matter whichever you feel when needed you can use it in your you know when you are trying to analyze certain results and or obtain certain probabilities. Now as I had written there inter arrival times we would now like to look at the probability of that means the time. So here time between 2 occurrences because again the occurrences are all chance events are unpredictable. So we want to now look at the if it is possible to determine the distribution of these inter arrival times. We have seen that probability n t equal to 0 is e raise to minus lambda t. So no arrival from 0 to time t this is your. So that means up to and if you define x 1 as a time of the first arrival if the time of the first arrival and then if I want to compute the probability that x 1 is greater than t that means that there has been no arrival in the interval 0 to t. So no arrival from 0 to t is the event that n t is 0. So the 2 events are the same that means if the first arrival has not occurred the time for the first arrival see x 1 is the time of the first arrival. So if the first arrival time is greater than t that means in the interval 0 to t no arrival has taken place which implies that n t is 0. So the 2 events are the same it is just that here we have defined the random variable n t and here is the random variable x 1. So therefore and this probability is e raise to minus lambda t. So if somewhere I have written that this event can be taken as x 1 greater than or equal to t then this is not correct because you are saying that the time of the first arrival is greater than t. So it cannot be or if you are saying that n t is 0 then this is equivalent to the event that x 1 is greater than t it cannot be greater than or equal to t because in the time 0 t no arrival has taken place. So any arrival that takes place will be after t. So this is the important thing and therefore 1 minus f of x 1 t. So this is 1 minus of f x 1 t if f is the cumulative distribution function for x 1. So then this is equal to e raise to minus lambda t and so f x 1 t is 1 minus e raise to minus lambda t. So this gives you the and so that shows you that therefore your f x 1 will be lambda e raise to minus lambda t. So if you differentiate this equation on both sides you will get lambda e raise to minus lambda t. Now this is exponential lambda. So the distribution of x 1 is exponential lambda and therefore the expectation is expected value of x 1 is 1 upon lambda. So this tells you what that 1 upon lambda is the mean time of first arrival. But actually now if you look at x 2, x 2 is the elapsed time between the first and the second arrival that means here the first arrival occurred here and the second arrival occurred here. So I am calling this as x 2. So that means if this is x 1 so the time of the first arrival is x 1 then time of the second arrival will be x 1 plus x 2 because this is the inter arrival time. So x 2 is the time between the first and the second arrival. So for example, so now if you look at that is why I have defined here this is the elapsed time between the first and the second arrival or this is the occurrence I keep calling arrival but which also means the occurrence. Now if you want to look at this probably t 2 greater than t condition that t 1 is s. So the first arrival took place we are writing t 2 t 1 but here I have been writing x 1 x 2. So it does not matter you can make it x 1 and we can make this as x 2 because this is x 2. So x 2 so therefore this will be probability n s plus t minus n s is 0 because the second arrival has not taken place up to I mean this length is not is bigger than t. So therefore s plus t will be this so n s plus t minus n s is 0 given t 1 is s. So therefore probability n s plus t minus n s 0 is 0 to minus lambda t. So this is again so you see this will also be 1 minus f x 2 t and therefore you can again see that it will be exponential. So we will continue with the you know so that means the inter arrival times are all exponential and that you can relate it with the memory less property of the exponential distribution which comes to your independent increments. So the two things are related. In this exercise 8 I will be going over some problems related with the limit theorem that we had done and also the Poisson process that we had talked about after that. So x 1 and x 2 are question 1 x 1 and x 2 are independent random variables x 1 is binomial and i comma p i 1 to 2. So x 1 that means is binomial n 1 comma p and x 2 is binomial n 2 comma p. So use m g f to find the distribution of x 1 minus x 2 plus n 2. So here you see all that you have to show is so basically these two problems are on the you know m g f of the joint density functions of random variables and then I had also introduced the concept of m g f for more than one variable. So now here you see you can rewrite this x 1 minus x 2 plus n 2 as x 1 plus n 2 minus x 2. So since x 2 is binomial n 2 p your n 2 minus x 2 will become binomial n 2 1 minus p. Is that clear? Because you see when you consider n 2 minus x 2. So if x 2 has r successes then n 2 minus x 2 will have n 2 minus r successes. So therefore for n 2 minus x 2 the successes of x 2 would be failures here and therefore your you see the the probability for a failure is 1 minus p. So that is all. So once you recognize that you can write this as x 1 plus n 2 minus x 2 then n 2 minus x 2 is binomial n 2 and 1 minus p. So once you know this then you can immediately and then they will be independent. So therefore because x 1 and x 2 are independent. So x 1 and n 2 minus x 2 are independent and therefore you can write down the joint m g f. So you should be able to do it right. Now question 2 x 1 and x 2 form a bivariate normal random variable I mean they are bivariate random normal variables with parameters mu 1 mu 2 sigma 1 square sigma 2 square and rho. So rho is your correlation coefficient show using m g f that y and z given by. So y is defined as x 1 minus mu 1 and z is defined as x minus mu 2 minus rho into sigma 2 upon sigma 1 into x 1 minus mu 1. So you have to show that these two random variables are independent and that y and z are each normal random variables. Here the idea is to use m g f's but actually there is a easier way and surely to be able to do this problem using the m g f's I leave it as a challenge and maybe when I am discussing you know set of miscellaneous examples we will revisit this. But right now the way to do it is see because y is x 1 minus mu 1. So this will continue to be normal remember because x 1 minus mu 1 the mean of y will be 0 variance will be the same because by shifting the mean of a random variable the variance does not change. So therefore y is again normal with mean 0 and variance sigma 1 and similarly z is also this should be x 2 here that is missing. So x 2 minus mu 2. So here again you have shifted the mean of x 2 by this quantity mu 2 plus rho into sigma 2 and of course this is the whole idea when you write z like this x 1 minus mu 1 upon sigma 1 that is that because you are computing the. So this is the whole idea. So here again the mean has been shifted and therefore z is also normal with the same variance sigma 2 and the mean would be. So by definition of z we see that z is the random variable which is given by a conditioning x 2 on x 1 and we had while computing the conditional p d f's we have seen that this will also turn out to be normally distributed random variable and you can immediately see that the mean of z that is the expected value of z is 0 because expected value of x 2 is mu 2 and expected value of x 1 is mu 1. So when you take the expected value of z it will turn out to be 0. So therefore we need to compute the variance and the variance will simply be expectation of z square. So which I have done here by writing out the expectation of the square term and then taking expectation inside as we can do it and therefore it will turn out to be 1 minus rho square upon sigma 2 square. So this is the variance. Now when you want to define the covariance then it will be covariance of y comma z. So that will be expectation of x 1 minus mu 1 into x 2 minus mu 2 minus rho sigma 2 upon sigma 1 into x 1 minus mu 1. And again here we can take expectation inside. So this will be expectation of x 1 minus mu 1 into x 2 minus mu 2 and then minus rho sigma 2 upon sigma 1 expectation of x 1 minus mu 1 whole square. And expectation x 1 minus mu 1 into x 2 minus mu 2 is the correlation coefficient rho into sigma 2 sigma 1 because expectation of x 1 minus mu 1 into x 2 minus mu 2 is the covariance and so that can be written as rho into sigma 2 sigma 1. Then minus rho sigma 2 upon sigma 1 into expectation of x 1 minus mu 1 whole square is sigma 1 square. So therefore when you substitute that value you turn out turns out to be 0. So y and z are uncorrelated and now we use the result that for if y and z are normally distributed then being uncorrelated is equivalent to their being independent. So therefore y and z are independent. Then 3 is from an earned containing 10 identical balls numbered 0 1 to 9 and n balls are drawn with replacement. So the draw a ball and then put it back we just notice the number. In the following let occurrence of 0 on a draw mean that the draw yields a ball with number 0. So a draw whenever the occurrence of 0 means that you draw a ball you notice the number or you wrote it down somewhere. So if it is a 0 then it is a draw means that the draw yields a ball with number 0 and then we put the ball back. Now what does the weak law of large numbers assert about the occurrence of 0s in n drawings. So the mean is 1 by 10 because they are identical balls. So probability of drawing any ball is equally likely with any of the numbers 10 numbers is equally likely and therefore the weak law of large numbers will say that if s n is the number see we are denoting by s n the number that number of balls that showed up with number 0. So then s n by n is the relative frequency. So out of the n draws you have s n balls have shown up with number 0. So s n by n as n goes as becomes large will converge to 0 1 by 10 the mean is 1 by 10. So will converge to 1 by 10 in probability that is your weak law of large numbers. So s n by n will converge to 1 by 10 in probability. Now b part is how many drawings must be made in order that with probability at least 0.95 the relative frequency of occurrence of 0s will be between 0.09 and 0.11. So that means you are wanting to know. So s n by n is the relative frequency you want to compute this probability that the relative frequency is between 0.09 and 0.11 and this probability should be at least 0.95. So therefore now again I standardize the whole thing. So this is s n by n minus 0.1. 0.1 is the expected value here and so subtract 0.1 from either side and so this finally, because the difference is 0.01. So this is what you have. Now by Shebyshev's inequality this number this probability is greater than or equal to 1 minus variance of this s n by n which is because this is now binomial. So 1 by 10 into 9 by 10 so 0 and non-zero that is how I am treating this is. So p q p q into 1 by n because s n by n the variance. So variance of s n is n p q and that divided by n square. So this is 1 by n and so this into divided by. So Shebyshev's inequality 0.01. So therefore we will compute the value of n when this is equal to 0.95 and for higher values of n it will be higher. So therefore now you have the equality you can now get a equation for n and you will get the value of n for which this probability would be between 0.019 and 0.10. So you can do the rest of the problem. Then part c is use a central limit theorem to find the probability that among n numbers thus chosen 0 will appear between n minus 3 root n by 10 and n plus 3 root n by 10. So here again you want this probability using central limit theorem n minus 3 root n by 10 less than s n less than n plus 3 root n by 10. So then I see n by 10 is the mean of s n. So I write it here this and then you see the variance variance of s n will be n into 1 by 10 into 9 by 10 because s n is now binomial with mean 1 by 10 and sorry mean n by 10 and variance n p q. So this is this and so under root of this will be 3 root n by 10. So 3 root n by 10 so when you divide so now this becomes a standard normal variant and so the central limit theorem says that probability this less than or equal to I mean for n large enough less than or equal to 1 you want to compute. So this is approximate probability which is 2 phi 1 minus 1. So phi 1 the value is given to you at the end of the problem and so you can compute this. So this was a good use of central limit theorem. So now let us go to problem 4 an employee in a call center works from 8 a m until 5 p m with breaks between 10 30 to 10 45 then from 12 30 to 11 30 and 40 14.45 to 15 hours. Assume that calls come in according to a Poisson process with expected number of calls per hour equal to 6. So you will have the probability that there are at most 10 calls during the breaks. So here the whole idea is because you see it does not matter for the when the arrivals when the phone calls are coming by as a Poisson process then the inter arrival times between the phone calls is exponential which is memory less. So that is what you are using here. So then what we can do is we can just add up the total break up the break time which is 15 1 hour and 15. So 1 hour 30 minutes. So therefore, 3 by 2 hours and you want 10 you want at most 10 calls during this 3 by 2 hours. So therefore, you will have to write the probability. So you sum up that means the calls can be 0 1 2 up to 10. So your lambda is 6 into 3 by 2 6 by 6 into 3 by 2 because 3 by 2 hours. So lambda t lambda t becomes your parameter now and within this time within the time 3 by 2 hours you want at most 10 calls. So you can write down the Poisson probability. That is the probability that first call of the day is after 8 10 a.m. So that means for 10 minutes you do not want any call. So 0 call and therefore, again this will be. So your time would be you have to see since your arrival rate is given per hour. So you have to convert the 10 minutes to the fraction of an hour which is 1 by 6. So therefore, it will be e raise to minus lambda t probability that no call comes during the time 8 to 10 a.m. So that means 1 by 10 1 by 6 into 6 which becomes e raise to minus 1 will be the probability. My idea is not to really give you all the answers but just to give you hints. What is the probability that the employee can do something else for 45 minutes without being disturbed by a call. So here again we are repeatedly using the memory less property. So 45 minutes can be anywhere and therefore you want the break time. That means now your time is 3 by 4 hour. So you do not want any call to come in between for 3 by 4 hours and your lambda is 6. So we can do this now. Now consider a Poisson process with parameter lambda. What is the conditional probability that n 1 is n given that n 3 is n. So here again this is a good question in the sense that it will help you to understand the Poisson processes. Do you understand why this probability does not depend on lambda. So now here you see what is it that you have to find. You have to find probability n 1 is n given that n 3 is n. So that means a probability n 1 is n and see actually you can interpret this as that n 1 is n and then n 2 is 0 because n 1 is n and n 3 is also n. So that means for 2 prime periods 3 minus 1 there has been n. So therefore this is this and then divided by a probability n 3 equal to n. This is what you have. So your arrival rate is lambda. So therefore you have to write this. So n 1 is the number of arrivals which is equal to n n time 1 period and in time 3 period also you have given that n arrivals are there. So that means all the n arrivals have taken place between 0 and 1 and so there are no arrivals in the time 2 units of time 0. And now if you write out this probability you see that the answer will be independent of lambda and again I want you to work out the details. Then question 5 is over. Question 6 jobs to be performed on a particular machine arrive according to a Poisson input process with a mean rate of 2 per hour. Suppose that the machine breaks down and will require 1 hour to be repaired. What is the probability that the number of new jobs that will arrive during this time is 0 2. So that means essentially you are asking for 1 hour gap that means the machine is takes 1 hour to be repaired. So then in this 1 hour you want to know the probability of 0 arrival, 2 arrival or 5 arrivals. And this is the Poisson process and the mean rate is lambda which is equal to 2 per hour. So this you can say is a simple problem. But again I just want you to familiarize yourself with all these concepts and therefore I have done it. Now question 7 suppose you arrive at a single teller bank to find 4 other customers in the bank. One being served and the other 4 waiting in line. The statement is a little this thing because here it should have said 5 people. But anyway 4 refers to the 4 people waiting in the queue one is being served and you join the end of the line. So you are the 6th person. So if the service times are all exponential with rate mu what is the expected time you will spend in the bank. So therefore you see this is 6 service times because 5 are already there and because of the memory as property the person who is being served again the probability of it is completing. So whatever service time is over is immaterial. So therefore 6 services have to be completed and therefore the completion time for the 6th person is a gamma distribution with mean 6 mu or to be 6 comma mu. It will be gamma 6 comma mu. So it will be 6 comma mu and therefore you know what is the mean. The mean would be 6 by mu. I think the mean should be n by mu if it is gamma n comma mu. Anyway just verify that so you will find out the expected waiting time of the 6 person when he joins the queue in the bank. So cars pass a certain street location according to a Poisson process with rate lambda. So certain location in the street the cars are passing at the rate of lambda and a woman who wants to cross the street at that location waits until she can see that no cars will come by in the next t time units. She has some idea that she is standing at this particular place and then she looks at the side and sees that now for quite a distance she cannot see any car coming. Then she will feel free to cross the street and that time according to her t time units something like that. So find the probability that her waiting time is 0. So waiting time is 0 means that means she comes to the location and then the probability that there is no car coming for the next t units. So here again because your arrival rate is Poisson with rate lambda. So your arrival times are also exponential with parameter lambda and therefore you can find out the probability that her waiting time is 0. That means no arrival in the time 0 to t. So find out the Poisson probability that which is e raise to minus lambda capital T. No arrivals and find her expected waiting time. So now you have this distribution e raise to minus lambda t as the probability that her waiting time is 0. So you find out her expected waiting time. So this you can now do by yourself. To answer the second part that is what is the expected waiting time of the woman who waits at the crossing for the cars to come. So you will have to compute the expected value of the waiting time by using conditional probabilities. So I will just write down the solution for you on the board. So I hope with all these hints and almost some of the problems have solved almost completely. So anyway I hope you will enjoy doing it and I definitely will try to come up with a large list of interesting and challenging problems at the end of the course.