 We will go through exercises on discrete random variables and I will try to give you small hints so that you can work them out yourself. Now the question one says that each voter is for proposition that patrol price should be slashed by rupees 5. So this is the answer people are supposed to vote for this with probability 0.9. So probability that a person will vote for this that the prices should be slashed by rupees 5 is 0.9. What is the probability that exactly 7 of 10 voters are for this proposition? So you should be able to guess what distribution you have to use here. At least one half of an airplane's engines are required to function in order airplanes there is an apostrophe there. So at least one half of an airplane's engines are required to function in order for it to operate. If there are obviously we are assuming that even number of engines are there and so half of them have to function in order for the plane to be able to fly. So if each engine independently functions with probability p for what values of p is a four engine plane more likely to operate than a two engine plane. So please first try down the probability of two or more engines working for a four engine plane and for a two engine plane it will be one or two engines working for a two engine plane in terms of p and then write down the inequality that you want the probability for the four engine plane to be higher and so what would be the values of p for which this inequality would be satisfied. In 3 a news boy purchases papers at rupees 2.50 and sells them at rupees 3 however he is not allowed to return unsold papers. If his daily demand is a binomial random variable with n equal to 25 and p equal to 1 by 3 approximately how many papers should he purchase so as to maximize his expected profit. So now see you start with you do not know exactly how many newspapers he buys so let us say the number is r that means r success is binomial random variable. So you know the probability what is the probability when he has when he this is daily demand is a binomial random variable with this right how many papers should he purchase right. So essentially what you have to show here is that the expected value would be a function of r the number of newspapers he buys right and that you have to then maximize with respect to r. So that will tell you what the optimum value of r will be because here see when you take the expected value depending on the demand it will tell you that see when he is able to sell a newspaper he earns 50 paisa if he is not able to sell a paper then he loses 3 rupees. So accordingly you have to write down the expected profit for this news boy and then maximize it and find out the optimum value of the number of papers he must order. If x has a distribution function f what is the distribution function of e raise to x so you have to find that out again apply the definition what is the distribution function of alpha x plus beta where alpha and beta are constants alpha not equal to beta. Then question 5 let n be a positive integer valued random variable show that expected value of n so n takes positive integer values so then expected n is sigma i varying from 1 to infinity probability n greater than or equal to i. So it is a matter of you know writing out the expression for e n and then rearranging the terms so that you can get at this answer. Six problem if x is a Poisson random variable with parameter lambda show that e raise to x n is lambda e raise to x plus 1 raise to n plus 1 now use this result to compute e expected value of x raise to 4. This is straight forward from a set of n randomly chosen people this problem I have already discussed with you in one of the lectures I explained the notation e i j that person i and j have the same birthday assume that each person is equally likely to have any of 365 days of the year as his or her birthday. So then you have to find these conditional probabilities that I have written out here and then I am asking the question are e 3 4 and e 1 2 independent then what can you say about the independence of the events e 1 3 and e 1 2 now you can almost guess what the answers would be but anyway you have to work it work out and show that they are that means you have to show that the probability of e 3 4 intersection e 1 2 that means all these 4 people having their same birthday will be product of the probabilities of e 3 4 into e 1 2 and so on. Question 8 prove the recursion formula for x Poisson which is probability x equal to i plus 1 is lambda upon i plus 1 into probability x equal to i with lambda as its parameters. So Poisson random variable with lambda as the parameter then you have to show this and if you start with probability x equal to 0 equal to e raise to minus lambda then you can compute probability x equal to 1 so on from this recursion formula then you also I want you to compute probability x less than or equal to 100 when the mean of the Poisson distribution is 100 and you can compare your results because Poisson tables are not easily available so in the website you can go and look at the tables. For a hyper geometric random variable determine probability x equal to k plus 1 upon probability x equal to k so here also I am asking you to find out the recursion formula. Question 10 the number of eggs laid on a tree leaf by an insect of a certain type is a Poisson random variable with parameter lambda. However such a random variable can only be observed if it is positive since if it is 0 then we cannot know that such an insect was on the leaf because there will be no eggs present on that leaf if we let y denote the observed number of eggs then probability y equal to i is equal to probability x equal to i given that x is greater than 0 where x is Poisson with parameter lambda so find E y so I am asking you to find the conditional expectation of x. Question 11 each game you play is a win with probability p you plan to play 5 games but if you win the fifth game then you will keep on playing until you lose. So here negative binomial would be used find the expected number of games that you play find the expected number of games that you lose. So I hope you enjoy doing this exercise. I will continue with the you know examples and some more results about the cumulative distribution function. It is a very important concept and also as we go along with special distributions and so on we will be able to get more familiar with the whole idea. So let us consider the function g x which is defined here for x less than 0 by this and for x non negative by this. Then we want to show that limit of g x as x goes to minus infinity must tend to 0 which it does because e raise to x goes to 0 as x goes to minus infinity. Similarly limit of g x as x goes to plus infinity should go to 1 and that also you can see because this is e raise to minus x. So as x goes to infinity this portion goes to 0 you are left with 1 so that is fine. Then you want to show that g x is monotone and g x is monotonically increasing. So we take the derivative here because g is a continuous function differentiable also. So I can take the derivative and if the first derivative is non negative which it is right e raise to x half this is non negative for all x therefore g x is monotonically increasing. So g prime x will be half e raise to x for x less than 0 and for and it will be see there is a minus sign and so there is a minus 1 will come from here. So it will become plus so that will be half e raise to minus x for x greater than or equal to 0. So we see that g prime x is non negative for all x this is a non negative function this is a non negative function. So therefore g prime x is non negative for all x which implies that g x is monotonically increasing which is again a property of g that for a cumulative distribution function. So we just verifying that this qualifies to be a cumulative distribution function and then other properties we will check. G satisfies all the conditions for being a CDF. Now we also have this result that if f is continuous in an interval then f the integral of f from minus infinity to x this is differentiable on that interval. So wherever in whichever region f is continuous capital f the cumulative density function would be differentiable and so then therefore that means if you are given the CDF and then it is differentiable then you can differentiate it and say that it will be equal to the PDF wherever the function is differentiable. And also we have seen it already that since this is equal to probability x less than or equal to x then and this is integral minus infinity to x of f y d y and this defines again the area under the curve from minus infinity to x. So wherever the mass is and so now given this since we have verified that this is a valid CDF that is now find out the probability density function and which we said that we get by differentiating the CDF. So here you see g prime x is half e raise to x if x is less than 0 and when you differentiate this part the minus sign minus sign becomes plus. So this is half e raise to minus x which is for x non negative. So actually this is very important that when you define the PDF you have to specify where it is defined because that means where it is non zero and where it is zero because you have to say where the mass is of the random variable. So it is very important sometimes people just forget this and you write this only which is not correct because you must specify the region in which it is defined that means where the mass exists. Now so therefore this and this is a non negative function you see that and now you can verify that this is a valid PDF by integrating it from minus infinity to infinity. So that will be minus infinity to 0 e raise to x and then plus half 0 to infinity e raise to minus x dx and you can see that the integral would be half e raise to x minus infinity to 0 that will give you. So at 0 it will be 1 so that is half and minus half so when you integrate this it will minus sign half e raise to minus x 0 to infinity this again gives you half. So half plus half is 1. So therefore we have verified that your small fx is a PDF and also that this is a CDF. Let us take another example of this CDF now here. An electronic component functioning time and electronic components functioning time before it fails can be considered a continuous random variable x and suppose it is PDF is given by this. So therefore you want the first question is for what value of lambda is fx a PDF and obviously we apply the condition because it is a non-negative function. So the second condition is that from 0 to infinity since here again you see the mass is for x non-negative for x less than 0 it is 0. So I will integrate the function from 0 to infinity and e raise to minus x by 50 so this becomes minus 1 by 50 comes to the top to the numerator minus 50 lambda e raise to minus x by 50 from 0 to infinity. So at infinity this is 0 and at 0 it will be 1 so 50 lambda is equal to 1. This is the condition we need so that this is a valid PDF and therefore we get that lambda is 1 by 50. So the first question has been answered. Now I want you to compute probability x less than or equal to 150 so you compute 0 to 150 1 upon 50 e raise to minus x upon 50 dx which comes out to be this and then at 150 so at 0 it is 1 and then this is minus e raise to minus 150 by 50 minus 3. So look up the value for this from your calculator and you get the answer. Now as I said that there are discrete random variables, continuous random variables and mixed kind of random variables. So I will take up this example for a mixed random variable. Suppose you are given this function capital F x so I plot it for x less than 0 it is 0 so this is the portion then you see for and see note that here this is x less than 0 for x equal to 0 and greater than 0 then this part operates and therefore the value jumps to 1 by 3. So and then it continues to be 1 by 3 till x reaches 1 but it is not equal to 1. So therefore at this point it is 1 by 3 and then it is like this and here at 1 here so you see the right continuity part is satisfied there is a jump here again at x equal to 1 because at x less than or equal to 1 less than 1 it is 1 by 3 the moment you attain the value x equal to 1 it jumps to 2 by 3. So here again the jump is of 1 by 3 and then after that see what happens this is x upon 3. So as x varies from 2 to 3 at 2 see this is 2 by 3 and this is also 2 by 3 x equal to 2. So therefore and then it continues like this the slope is of this line is 1 by 3 and then of course at 3 as long as this is the line and then at 3 it becomes 1. So you see here you have both kinds that the function take has jumps so it is a step function for some values and then after that it is a continuous function. So therefore that is what I have said that here and we have already seen that you know how we compute this difference but I am saying that probability x equal to 0 this will be f 0 minus f 0 minus. So f 0 minus is 0 f 0 is 1 by 3 and therefore the value is 1 by 3 probability x equal to 0 and probability x equal to 1 will be f 1 minus f 1 minus and so that will be f 1 f 1 will be 2 by 3 and f 1 minus is 1 by 3. So again the difference is 1 by 3 so the two jumps are of 1 by 3. So this is what will happen and here again I want to point out see the result that we have shown that probability x less than or equal to x or of course even for an interval same thing will hold I had shown you last time that the area under that strip would be the probability when x lies in an interval. So here this thing is valid only when f is you know f is continuous and then so this is a x is a continuous random variable. Now so therefore for discrete and mixed kind we will not apply this result. So only when the random variable is completely continuous you can apply that result. So here this is this result so one has to be careful and keep this in mind that you cannot compute this by area under the curve because the p d f would not be this is your cumulative density function. So p d f would be for these two values will be a bar chart and then it will be this thing. So therefore the concept of area under the curve does not apply for mixed random variable or for discrete random variable it only applies to continuous random variable. You know computation of c d f is very important you should get the idea very clear in your mind how you can go about doing it and the validity you must make sure that when you consider a function to be a c d f it must satisfy the it must have all the properties that we have you know said that the c d f must have and so on. So I mean you know even this is not enough you should work out many more problems to get familiar with this concept. We will talk about uniform random variable now one of the simplest and very widely used continuous random variable and try to get the get a feeling about this particular distribution. So now see suppose we pick a number at random between 0 and 1. So this is important we pick a number at random which means that any number is equally likely and using this term very often describing events and so on. So we pick a number at random between 0 and 1 the probability that it lies in the interval 0 x should be proportionate the proportion that the length of the interval 0 x is of the length of the whole interval 0 1 this is what we mean by picking a number at random. I will repeat that if we are saying that I pick a number here and I say that it is lying in this interval 0 comma x x is of course a number which is less than 1 then the probability that the number lies in this interval should be the proportion that the length of this interval is of the length of the interval 0 comma 1. And if you can recall that we in the discrete case I had told you number of favorable cases divided by the total number of cases is sort of an extension of the same concept that if I am saying that the probability that the number I pick up lies in this then length of this interval divided by the length of the total interval should be the probability that the number that I have picked lies in this interval and this in a sense captures the concept of a uniform random variable or what we keep saying a number is equally likely in this interval and so on. So this will be when we are saying that probability that the number lies here that means this actually describes that probability because we are saying that number that I pick between 0 and 1 lies in the interval 0 x that means the number is smaller than or equal to and so it is between 0 and small x. So therefore this is fx and what I am saying is that the cumulative density function is x upon 1 which is equal to x if x is between small x is between 0 and 1. So this I immediately get the cumulative density function of this random variable and if you draw the picture you see it is like this because as x goes from 0 to 1 and then it stays at 1. So you see this is there is no discontinuity here, no jumps here. Therefore this represents a continuous this is the graph of a cumulative distribution function for a continuous random variable. The pdf of course as we have now can use this fact that this is differentiable from 0 to 1 and so the derivative of this function will give you the pdf fx equal to 1 in the interval 0 1 and 0 otherwise. So simple right. So this is one special case when I said that the random variable is defined on the interval 0 1 and so it is uniformly distributed in that interval. In general we say that x is uniform random variable on the interval alpha comma beta if its probability density function is given by 1 upon beta minus alpha x lying between alpha and beta and 0 otherwise. So you see I will again repeat that whenever I am defining a pdf or a cumulative density function I have to define the region on which it is specified. So of course for the cdf it goes up to infinity because after whatever the values are over then it stays at 1. So for and therefore see here again this is proportionate to the length of this. So the length of the interval is beta minus alpha. So 1 upon beta so any so here for example if you draw the graph of this pdf of a random uniform random variable this will be alpha and let us say this is beta then this is the length. So this height is 1 upon beta minus alpha which is greater than 0 and you see if you look at the area under this curve this is a rectangle height is so the area is 1 upon beta minus alpha into the length length is also beta minus alpha so this is equal to 1. So this is a valid pdf and the area under the graph you know that concept. So now if you are for example if you if you are wanting the probability that x is less than or equal to some gamma where gamma is some number here then it will be this area under the curve. So this is how you can very simple to picturize not go wrong with it but just make sure that you write the probability correctly and always validate always make sure that you have the right numbers. Then you want to compute the expected value of general uniform random variable this will be 1 upon beta minus alpha integral alpha to beta x dx which comes out to be x square by 2 alpha to beta. So 1 upon beta minus alpha into beta square minus alpha square by 2 that leaves alpha plus beta by 2. So very simple way to remember whatever the interval for the uniform you just add the two end points divide by 2 that gives you the expectation. So it is a middle point the expectation yeah so what will it be yes just see here this is alpha and so alpha plus beta minus alpha by 2 right because the length of the interval the midpoint of this interval is beta minus alpha by 2 which you add to alpha to get to this point right. The length here is beta minus alpha so half of the length this length is beta minus alpha by 2 added to alpha that gives you alpha plus beta by 2 right. So that is the midpoint the expectation and expectation x square will simply this make it beta cube so x cube by 3 and from alpha to beta this will be beta cube minus alpha cube by 3. And you know expand this multiple divide by beta minus alpha you get this and then for the variance it will be expectation x x square minus expectation x whole square which when you simplify comes out to be 1 by 12 beta minus alpha whole square. So again here it is easy to remember the formula for the variance length of the interval square divided by 12. So the length of the interval is beta minus alpha square it up divide by 12 that gives you the variance of the uniform random variable. Now let us look at this example buses arrive at a specified stop at 15 minute interval starting at 7 am so that is the first bus arrives at 7 then the next bus will arrive at 7.15 then 7.30 7.45 and so on. So at interval of 15 minutes the buses keep arriving. If a passenger arrives at this stop at a time that is uniformly distributed between 7 and 7.30 am. So this is again an example because you cannot the time of the arrival of a passenger can be sort of treated as a continuous random variable you might say that you know your clock gives you discrete time but essentially the concept is that you will treat this as a continuous random variable. So here the distribution of his arrival time is a uniform random variable between 7 and 7.30 am which equally likely what time he arrives from 7 to 7.30 am. So then you have to find the probability that he waits less than 5 minutes for a bus. So he waits for less than 5 minutes for a bus. See since the person should have to wait less than 5 minutes therefore the event that we have to the probability of the event that we are asking for is 10 less than x and less than or equal to 15 it should be not less than or equal to because we want the time waiting time to be less than 5 minutes. So equality would have been valid if we had said that the time can be waiting time is 5 or less but here we are saying the time is waiting time is less than 5 minutes. So therefore it should be 10 less than x. So make the correction because while computing the probability I think I have said 10 less than or equal to x. Similarly here it will be 25 less than x. So see here he would if he has to wait less than 5 minutes then it should he should be arriving. So his arrival time should be greater than 25 and less than or equal to 30. So please make that correction in the computation but as I said the probability part it does not make a difference but here it will the event has to be described correctly. Then you see he should arrive between 10 and 15. See his time is between 7 and 7.30. So the first bus arrived at 7. The next is going to arrive at 7.15. So if he has to wait for less than 5 minutes then he should arrive after 10. And then it should be less than 15 because he has to get the bus. So this 500 interval if he arrives in this interval then he will have to wait or less than 5 minutes. Similarly if he arrives in this time interval 25 and 30 so this is in minutes then again he will have to wait for at most 5 minutes. So the event that he has to wait less than 5 minutes for a bus. So these two events capture this event and since they are disjoint I can add up the probabilities. This and these are disjoint because a person cannot arrive at both the times. Both time intervals either he arrives here or here. So therefore these are the two events and so the probability is add up. And here the probability as I told you this is simply when you divide I was drawing the figure for you somewhere here. So it is the length of the interval divided by the interval in which your wearable lies. Yes what example we looked at and so this length interval is 5 and 1 by 30 because his distribution is uniform distribution and the length of the interval is 30. So 1 by 30 is the probability of being I mean the pdf is 1 by 30. So therefore this is 5 by 30 plus 5 by 30 which is 1 by 3. Now the second part is he has to wait more than 10 minutes. So if he has to wait for more than 10 minutes then he should either arrive in this interval because if he arrives any time between 0 and 5 minutes see x. So that is what important I should have specified. X is the minutes past 7 am when the passenger arrives at the stop. So x is the random wearable because his arrival itself is random wearable. So therefore random phenomena. So x the number of minutes which are past 7 am when the passenger arrives at the stop. So therefore if you want to describe this event that he is he will have to wait for the bus for more than 10 minutes then he should either arrive that means his x should lie between 0 and 5 or between 15 and 20. Because if he arrives at 15 minutes the bus has just left and so he will have to wait for the next bus which will come at this thing 7. If he comes at 7.15 then the other one will come 30. So he will have to wait for more than 10 minutes and up to 20 because if like he arrives at 7.20 then he will have to wait for 10 minutes because the next arrival would be at 7.30. I hope that this event is described by these two again these two are disjoint. So therefore they add up to we can add up the probabilities and this will be 5 by 30 plus 5 by 30 which is 1 by 3 again. Now an important usefulness of random wearables of course there will be many occasions when we will see how uniform distribution is used. For example in simulation you need to generate random numbers from a particular pdf and simulation is the order of the day. Sometimes you cannot get physical data you try to generate it. So here what we will do is you want to generate random numbers and we can use this concept of cdf very nicely here. So take x to be a random wearable and capital F is the corresponding cumulative distribution function and now define the random wearable y which is f x because now whatever the function f I just place capital x there. So this becomes a random wearable again and this is uniform 0 1. So this is the wearable we will use the property of the cumulative distribution function. Now we can show this immediately this property that because if you consider the probability of y less than or equal to small y then this is the probability of f x less than or equal to y which will be 0 if y is less than 0 and will be f x inverse y if y is between 0 and 1 and this will be 1 when y is greater than or equal to 1. Now I have used the concept of f inverse here which I did not mention earlier. Now the whole idea is that here let us take x to be continuous random wearable in that case your f x is monotonically increasing which we have seen. So through so many examples also that f x and we proved it also as one of the properties that f x is an increasing function and therefore the f x inverse will exist. So there is no problem in case x so I am now doing the giving you this property of generating of random numbers for a continuous random wearable. But there are certainly even when x is a discrete random wearable this may not be unique but you can very easily you know the ways of determining a unique value for the f x inverse which is possible. So we see whenever x is not a continuous random wearable and is discrete then for certain interval as we saw the value of the function f x will remain constant. So we can decide that the inverse when we take the inverse will take the smallest value of y so that is possible. So we can define we can determine the inverse in a unique way whether the function random wearable x is discrete or continuous and so this will be valid for both of them. But here I am just now talking about x being continuous random wearable. So therefore this is this and since now again this thing can be written as f x inverse y and therefore from the definition of f x and f x inverse this it comes out to be y. So which is that means your capital Y has a uniform distribution 0 1. So the idea is that you generate random numbers u 1, u 2, u n random numbers from the uniform. So of course you might say how do you do that and there are methods there are computer methods for generating random numbers but which are which are actually pseudo random numbers. So there are whole lot of techniques and lot of software available for generating these random numbers from which are actually pseudo random numbers from the uniform 0 1. Once you do that then you will say that the x i's given by f inverse of u i are random numbers from the distribution of the original random wearable that you started. So the process is that you generate random numbers from the uniform 0 1 and then take the x i's which are given by f inverse x u i and these will be the random numbers from the distribution of x. So now you see you can immediate application of your uniform distribution. So you can generate any number of values from a given from a specified pdf to you know do all kinds of analysis that you want to do about that data. So I would like to revisit the concept of expectation of a random wearable now and for the discrete case we saw that it is defined as summation x i p x i over all x i for which the probability is positive. And if these values x i's that the random wearable x takes are finite in number then this sum will is a finite number because you are adding up p x i's are all between 0 and 1 these are finite in number then this will add up to a finite number. So in that case whenever the random wearable takes a finite number of values the expectation always exists. But we also saw that in the case of Poisson a random wearable where the values taken by the wearable are countably infinite. In that case the expectation which is the sum of the series you could add up and show that it is actually equal to the parameter of the Poisson. And in fact it is right so this is also called the mean of the parameter Poisson distribution. So for this case where the values taken by the random wearable are infinite countably infinite the expectation exists. So therefore there always has to be when we define the function e x we will say that this is the expectation provided it exists. So now look at another random wearable which takes a countably infinite numbers for example take the probability of x equal to n as c of 1 n square. So x takes the values 1 to infinity. Now since you want this to be a valid p of f so the summation n to 1 to infinity of probability x equal to this should be capital X capital X equal to n is this summation. And therefore this is equal to and we know this series is a convergent series and it is known that the sum of the series 1 to infinity 1 upon n square is actually pi square by 6. And so your c must be 6 by pi square. So once I define my c to be 6 by pi square this is a valid p m f. But when you want to compute the expectation the expectation would be c sigma n into 1 by n square n varying from 1 to infinity. And then this some series 1 by n summation 1 to infinity we all know is a divergent series and therefore expectation does not exist. So therefore one has to be cautious and careful and make sure that your I mean e x is defined only if it exists. Similarly in the case of continuous random variables this integral may not always exist even if your f is a valid p d f the probability density function. And I will give you an example here this is known as the p d f where f x is equal to 1 upon pi into 1 upon x square x varying from minus infinity to infinity this is known as the Cauchy's distribution p d f. And here again this is a valid p d f because integral minus infinity to infinity 1 upon pi 1 plus x square d x. So integral of 1 upon 1 plus x square is tan inverse x minus infinity to infinity 1 upon pi then tan inverse of infinity is pi by 2 tan inverse of minus infinity is minus pi by 2. So therefore it becomes plus and this is equal to 1. So this is again a valid probability density function but when you want to compute the expectation of this random variable you have to integrate this particular integral. And you can show that here you see if I make the substitution that x square is equal to so actually this is an improper integral and I will just consider the integral from 1 to infinity let us say. And if this is not existing then obviously 0 to infinity will also not exist and therefore this whole thing will also not exist. So let us put x square equal to t then your d x x d x is d t and so x d x you replace by there will be a 2 somewhere should have taken because this is this then your 2 x 2 x d x is equal to d t. So this will be 1 by 2 here x d x is 1 by 2 d t. So this and the integral of this is l n of 1 plus t 1 to infinity and you know that l n of infinity is infinity. So therefore this integral does not have a finite value so your expectation does not exist and hence your variance will also not exist. Now since variance of x would be e x square minus e x whole square and we have just seen that e x does not exist for this particular random variable. So therefore variance also will not exist because it has to be e x square minus this. So if this does not exist that means this is not finite then obviously variance will also not be finite and so we will say that it does not exist. So I just thought that I will put in this note here before we proceed with the other theory of probability theory. So that you can find you know you cannot always be sure that the expectation of a random variable will exist. Now another very important or widely used concept in probability theory is that of normal random variable and its probability density function. So the function is defined by 1 upon root 2 pi sigma e raise to minus 1 by 2 sigma square x minus mu square where mu and sigma are the parameters of the normal pdf and x varies from minus infinity to infinity. Now you can look at this thing here e raise to minus 1 by 2 sigma square x minus mu whole square. So it can be shown where you know people have already drawn the graph for different values of x and mu and sigma. So this is a bell shaped curve and it is symmetric about mu that you can see from here because x minus mu whole square. So it is symmetric that means on either side of mu you take the value this sign will not matter and therefore this is symmetric about the value x equal to mu and this was this distribution was discovered by was defined by let us say French mathematician Abraham de Maure in 1733. And can you believe that he was not a very he to make a living he used to spend time in a you know at that time in a dingy gambling house he would be sitting in the evenings spend whole evening there and trying to help people because he used this concept of the normal distribution to approximate binomial distribution and binomial distribution he was using to help people because it was a gambling house people would come to bet money and of course would want to win their bets and so he would give them the probability of you know winning which bet and so on. And so he actually use this concept for approximating binomial distribution and we will be discussing those approximations little later on. So this is how but the concept he introduced is a very very important and very widely used one and I think by the end of the course you will also see how important this concept is to the probability theory and you know for various estimations that we want to make about different events and their probabilities. So this and now let us see whether this is actually a valid PDF. So we want to integrate this function from minus infinity to infinity and show that the integral is equal to 1. So here what I do is I make the transformation yeah I have made the transformation y is equal to x minus mu by sigma. So then dy is dx upon sigma so dx upon sigma appears here which gets replaced by dy and 1 upon under root 2 pi is here then this is e raise to minus y square by 2 this remains from minus infinity to infinity because mu and sigma are finite numbers. So then we have to now integrate this and let me call this integral as i. So if I multiply this by another integral this is the notation. So i square now will become a double integral 1 upon 2 pi minus infinity to infinity minus infinity to infinity e raise to minus y square plus t square by 2 dy dt. See t and y are dummy variables so it does not matter and therefore I can say that it is this is also i minus infinity to infinity 1 upon 2 pi root 2 that becomes 2 pi because you are multiplying them. So e raise to minus t square by 2 dt. So this is what you have and now we want to be able to compute this integral and there this is where we will use polar coordinates. So let me show you the computations.