 So, what we will do we will I will just start on assuming that you know basics about the normal random variable. What a normal random variable is what its applications are and I will start from something we call the standard normal random variable. Did he go through that? Standard normal? No, right? So, I think that is good then ok. So, you have normal random variable or the Gaussian which is the most common type of random variable has a huge amount of applicability in real life. You know normal random variable is very very good at modeling phenomena which occur in real life, various phenomena like natural phenomena, synthetic phenomena everything. Normal random variable is very good at that. The standard normal is a specific type of normal random variable as you can see from the definition is the normal random variable and instead of x now you are putting z with mean equal to 0 and standard deviation equal to 1 ok. So, normal random variables can have any mean and any standard deviation. Now, standard normal is the normal random variable with mean equal to 0 and sigma equal to that means standard deviation equal to 1. You will see that I have emphasized the because there is only one standard normal random variable ok. You cannot have two standard normal there is only a unique standard normal random variable. So, we continue with that ok. So, first you look at the functions that means the density function and the distribution function and you can of course derive them from the definitions of these same functions that we know about the normal random variable. So, for normal random variable we had this big equation and we continue with that we just put sigma equal to 1 over there mu equal to 0 over there that is the basic idea. So, now for pdf and here you will see we are using the sign the Greek letter phi ok the small phi over here this is typical for standard normal. So, whenever you see that phi you know that it is a standard normal random variable that we are talking about ok. So, phi evaluated at that z is equal to just like what you had for the normal random variable except that we are putting sigma equal to 0 and mu equal to 1 ok. Similarly, if we go ahead we can also find out the cdf the cumulative distribution function for the standard normal random variable and for that we use the capital letter capital phi which means the cdf. So, instead of f we are using the Greek letter phi here ok and it is a it is usual it is customary that instead of x we usually denote it with z ok. So, how do we get that just by integrating the pdf from minus infinity to z ok and this z over here is a dummy variable like we have used for the definition of cdf of normal random variable. So, overall the basic idea is that the standard normal random variable is a special case of normal random variable where the mean is 0 and the standard deviation is 1 of course, the variance is also equal to 1. Now, in general not only for the normal distribution, but for any continuous random variable we say that there is a standard form for that normal random variable. What is the standard form? For any given random variable x we have this standard form given by x minus mean of x divided by sigma that means standard deviation of x ok that is the standard form. So, even for normal distribution if we use the standard form we get the standard normal distribution. Similarly, if you take any other let us say uniform you can also define the standard form of a uniform random variable just by applying let us say if the random variable is u then u minus mu everything divided by sigma u that would be the standard form for a uniform normal uniform random variable ok. So, that is what we say that you can even define it for a non normal random variable not only for normal random variable, but standard normal is the most important one because we use it a lot use it to compute the values of pdf and cdf not only for the standard, but also for the regular normal random variable with mu not equal to 0 sigma not equal to 1. We will see how ok. Let us discuss this property of the standard normal random variable if we have this standard normal random variable z then you can write that cdf at minus z is equal to 1 minus cdf at z and it is very easy to see how it takes place I had the time. So, I have drawn it already. So, this is the pdf small phi let us write pdf over here of the standard normal random variable. So, it is mean is at 0 this is 0 ok and normal random variable is symmetric about the mean. So, we have a symmetry about the mean now what we have is cdf at minus z 1. So, here we go to z 1 this area beneath the pdf up to z 1 is the cdf at z 1 right. So, this is phi at minus z 1 and symmetrically opposite we have plus z 1 equidistant from the point 0. So, this area would be equal to 1 minus phi of z 1 is that ok everybody I think it is pretty easy to see how it is. So, this part the left of z 1 if you take the area of that that will be phi z 1. So, this is 1 minus phi z 1 and you can see it is symmetric. So, this area will be equal to this area that is why we can write that phi of minus z is equal to 1 minus phi of z very important property of the standard normal distribution and we use it a lot for finding out values of the cdf ok. The cdf is also treated as an operator like any mathematical operator you operate it on any value it gives you another value and you will see how we do that. So, we apply this phi on value 2.25 we get some value we apply this phi on minus 1.244 we get this 0.10675 what is this this is again this is nothing but the cdf of the standard normal random variable, but for every value of the standard normal random variable z you have a value of the cdf right and that is what we treat as an operator. So, we call it the standard normal operator and you have standard normal tables you open up any book you open up your text book by Ross or open up any book on probability and statistics which discuss which discusses about this distribution there you will see a table of standard normal distribution. Where does that table give you? That table will give you values of the cdf for standard normal variable z for different values of z you will get the value of the cdf and also software packages will have built in functions using which you can determine what is the value of the standard normal random variable at a given value of the standard normal random variable ok. You can easily obtain the cdf software packages by mean we will see some examples using Sylab. If you are familiar with spreadsheet operations like numeric spreadsheet or open office calc those things will also have some built in functions ok and they are very useful for finding out the cdf of standard normal random variable. Now, so far I have been talking why should we find out the cdf of standard normal random variable because it is useful for finding out the cdf of a normal random variable. Do you see how if we know the cdf of a standard normal random variable can you find out the cdf of a normal random variable? I leave that to you ok think about it ok. This is what basically it is cdf of any random any normal random variable x with mean equal to mu x and standard deviation is sigma x for that the cdf which we denote by f of x at x should be equal to this is the cdf of the standard normal random variable at x minus mu x by sigma x. What is this? This is the standard form of the normal random variable we already showed that z equal to x minus mu x divided by sigma x. So, f x at x will be equal to f z at z or phi at z capital phi at z ok. So, you use that property to find out the value of any normal random variable with a non-zero mean and non-one standard deviation ok. So, if we have the standard normal tables in your book and you have to find the cdf of a normal random variable with mean equal to 3 and standard deviation equal to 0.2. What we do? First we have to find out let us say we have to find out it at a value of x equal to 5. So, 5 minus 3 divided by 0.2 we find out the value of the standard form and then we go to the table find out the value of phi at that standard form very simple task. We also define the inverse normal operator or the inverse standard normal operator which is the inverse of this capital phi function. So, it is the inverse of the cdf of the standard normal operator and if you remember the previous examples these examples are exactly opposite of what we discussed then we said earlier that phi at 2.25 is equal to 0.98778 or so. Now, you turn it around you say that phi inverse at 0.98778 is equal to 2.25 and similarly the other one ok. So, what is this? This is the cdf of the standard normal z at 2.25 ok. So, we also have the inverse normal operator it has its applications I will see when you go through some examples we will see when you use the inverse of the standard normal operator and of course, one can also define the pdf's and treat them also as operators. So, this is the pdf of the standard normal this is the inverse pdf of the standard normal, but these are not used as commonly as operator as the other two the cdf ones are very very important. Now, we come to Sylab and the function that we have in Sylab to find out the cdf of the normal random variable it is very easy to remember it is cdf not finds the cdf of any normal random variable having any mu and any standard deviation ok. You have to know the mean and standard deviation of course, unless you have the mean and standard deviation we cannot define the normal random variable. So, definitely you need to know those two parameters those two moments of the distribution. Once you know this then this is the format cdf not then within quotes you write pq in capital then the value at which you are finding out the cdf of the normal random variable then the mean and then the standard deviation ok enter this function you get the value of the cdf at value x for any normal random variable we go to examples ok. So, these what we type in the Sylab prompt cdf nor within quotes pq we are trying to find out the value of the cdf at 2.25 for a normal random variable which has mean equal to 0 and sigma equal to 1 that means this is the standard normal random variable. So, using the same function cdf we can also find the cdf for a standard normal random variable. So, for that again is the previous example we can see that let me go back we found the cdf of the standard normal at 2.25. This what you are doing cdf of the standard normal at 2.25 and you get the same value this pq over here tells you was the value that we are trying to find out which is the unknown over here ok. The cdf nor function has multiple applications it can be applied to find out the cdf it can also be applied to find out the inverse of the cdf ok. The inverse normal we will see how this first argument tells you what you are trying to find out using the cdf nor command. So, that is what I said you can also use it for nonstandard normal variables putting any mu any sigma instead of 0 and 1 you can put 3 and point 2 and evaluate it at 5 and also you can use cdf nor for finding out the inverse ok. And here I am not writing capital phi, but I am writing capital F because this is not the standard normal inverse right. This is inverse for any normal it can have any mu any sigma and you can still take the inverse. Now, the format changes a little bit from the previous one the first argument is x, x is the value at which you know the cdf right. What you get? What you get if you take the inverse of the cdf you get the value at which the cdf is taken that means x. So, that is what you are trying to find out we are trying to find out x we supply the value of mean the value of standard deviation the cdf p is the cdf or probability q is also the probability this is 1 minus p for some reason sila needs both p and 1 minus p ok. On the probability axioms that we know that q is always 1 minus p, but it needs both. So, we will supply both that is ok. So, here is a program here is a command to find out the inverse of the cdf of a normal random variable. So, we are trying to find out the inverse 0 mean standard deviation equal to 1 and I took the value remember it was 0.98778 or something I just rounded it off and for q you can write 1 minus this or you can find out the value of q and write it over here either way both works you can write it recursively that is the good part of it. And this is the answer sila gives you ok you can check this what we had p inverse of 0.98778 equal to 2.25 same thing from the sila command cdf now. The important part is over here the first one x signifies that we are finding the value x for which we know the cdf. Instead of 0 and 1 we could have used any other value of mean any other value of sigma. We go to an example this what we have daily rainfall in months and months in Mumbai can be described with this normal random variable x which has a mean of 55 and a standard deviation of 15.5. These are in millimeters, but if you remember I said that for random variables we do not need to put the units, but as engineers we need to keep a track of what units are there. What is the probability that it rains more than 70 millimeter ok. So, you want to find out the value 1 minus the cdf at 70 is that ok. What is the rain is less than 70 millimeter less than equal to 70 millimeter that is the cdf at 70. So, the probability that it is more than 70 millimeter is 1 minus cdf at 70. So, you find 1 minus cdf nor we are trying to find out the cdf. So, you put pq over here at 70 for a mean 55 for a standard deviation 15.5. This is the answer we get which means 16.7 percent ok. Very simple just go to Sylab apply a function like that and you get the answer. What is the probability of a moderate rainfall and by moderate let us say we define moderate by 30 to 70 millimeter rainfall. So, that is the cdf at 70 minus cdf at 30 for the same thing and this is the answer you get 78 percent. And you can see 55 is the mean 30 to 70 is around the mean. So, there is some concentration. So, most of the data 78 percent of the data is in the range around the mean. Mean is a measure of what is a measure of central tendency of a distribution. So, this one tells you that you have more data around the mean. Now, you look at the other way what is the 84.6 percentile rainfall. For design purpose many a times it is defined that you have to design for a 90 percentile rainfall. That means it will be exceeded only 10 percent of times. Usually design guidelines tell us that way. So, let us say the design guidelines ask us that you have to design for a 84.6 percentile rainfall. So, you find the inverse now. So, again you use the cdf NOR command and you are finding out the inverse for a mean 55 sigma equal to 15.5 cdf is 8.846 that is 84.6 percentile and this is q this is p this is q equal to 1 minus p you put that you get a value 70.8 millimeters. So, now you can design with this intensity of rainfall. We go to a related random variable log normal which is kind of close to the normal and it also has a lot of application and the primary use of log normal is probably that it is very good when you are dealing with multiplication of various random variables. Then log normal becomes very useful. See the overall goal is whenever you are dealing with real life data you want to fit some known distribution to that data. Why? Because once we have that description in terms of a known distribution then it is very easy to incorporate the uncertainty the variability in a given random variable in your calculation. Because for a normal random variable we know the cdf we know the pdf we know the mean sigma everything. So, that is what we always try to do for a given data set we try to fit a known random variable like normal random variable like log normal like uniform and so on. So, log normal is another option most of the times we try that. How do you define it? X is a log normal random variable if its logarithm is normally distributed. So, y defined as ln x is normally distributed then x is a log normal random variable. It is valid in the range of 0 to infinity. Now if you go to the theoretical things the base of the logarithm it is not important. So, instead of the natural logarithm you can take it to base 10, base 2 whatever you want. Log normal random variable sometimes you will find they are very close to normal distribution. So, we will see some pictures. This is the definition of the pdf. Does it look similar to the pdf of a normal random variable? Does it look similar to you? It has some similarity. At least in this part you can see this half is there denominator of sigma y square and so on. y over here is the corresponding normal random variable y is log of x. We said x is a log normal random variable is if ln x that is y is a normal random variable. So, y is a normal random variable with mean equal to mu y and standard deviation equal to sigma y. Then you can write that the pdf of the log normal random variable x is this. We have not put the mean and standard deviation of the log normal random variable x in this equation. Do you see that? We do not have mu x or sigma x in that equation or in that expression for pdf. So, definitely we need to know the mean and standard deviation of the corresponding normal random variable when you are trying to find out the pdf of a standard normal random variable. I am sorry when you are trying to find the pdf of a log normal random variable. This is how it looks like. Log normal random variable pdf for different values of sigma. Sigma here is the sigma of the corresponding normal ln x or y is the normal random variable. Remember that? So, for different values of sigma and constant value of mu you can see how the log normal random variables pdf change. You can see if you go by this one, the red one which has a very low standard deviation 1 over 4 kind of looks like the normal random variable. If you go even further down sigma equal to 1 over 8, it is almost the pink one, almost the normal random distributions pdf. So, sometimes log normal is very close to normal, but not always. For example, the blue one or the green one. Now, go to the cdf. So, you want to find out the cdf of the log normal random variable x. That means capital X less than equal to the value small x. You can write that in a form that the probability of capital X being less than equal to small x is nothing but the probability that ln of x should be less than equal to ln of small x. Do you agree to this? Is ln x a non-decreasing function with x? Yes? If it is so, then you can write this. That means ln x is nothing but y. So, this is the definition of cdf of y. So, f y at y and y is a normal random distribution. If it is a normal random distribution, then its cdf can be obtained from the standard normal operator. This is the cdf of the standard normal random variable and how do you obtain using the standard form? So, y minus min of y, everything divided by standard deviation of y. u y and sigma y are not the mean as standard deviation of the log normal random variable, but they are the mean as standard deviation of the corresponding normal random variable for a log normal random variable x. This is important. So, basically what you see that we need to find out those two and here how the cdf looks like. So, again if you go to very low value of sigma like 1 over 4 and 1 over 8, the pink and the red curves, the s curves like you have seen for normal distribution and if sigma is very high like 10, the black one or the blue or the green, it is very different from the s curve, but even this black one at one point of time has to be asymptotic to cdf equal to 1 has to be. Finally, it has to reach cdf equal to 1. Maybe it takes equal to infinity, but it has to reach there. So, all the expressions of pdf and cdf is based on the mean and sigma of the normal random variable. So, we need to find out the mean and standard deviation of the corresponding normal random variable given the mean and sigma for the log normal random variable and these are the equations. You can note them down. So, if we know random variable x which is log normal, if we know the mean and variance, we can find out the mean and variance of the corresponding normal random variable which is ln x. And once we find these two, we can find out the cdf and pdf of the log normal random variable x. And again we take an example, we will continue this similar example. Again the same monsoon rainfall, instead of considering it to be a normal random variable, we consider it to be a log normal random variable with the same mean and standard deviation. And the question you ask is what is the 84.6 percentile rainfall. So, this is what we are trying to find out. We know that the cdf of the log normal random variable is 0.846. We want to find out the value at which it obtains that cdf. And we know this expression f x of x is equal to phi of ln x minus mu ln x divided by sigma ln x. I just wrote that a few slides back. This is what we wrote f x of x is equal to this y is nothing but ln x. So, we know this we are going to use this. So, we find the mu of ln x, we have the expression, we know mu of x and variance of x. So, you get the value and the variance of the corresponding normal. Again we know the mean this is 55 standard deviation is 15.5. We put those, we get the variance. If you know the variance, we can also obtain the square root which is the standard deviation of the corresponding normal random variable. Once you have these two values, then you can plug it in into the basic expression. And this is the answer that you are trying to find out at which value it obtains the cdf of 0.846. Now, does this expression look ok to you? Let me get back. So, x will be phi inverse of this thing from that we get this. Is that ok? Well, you can do it for yourself and see what answer you get. You can use standard tables to find out phi inverse. How do you use standard tables to find out phi inverse? Tables have values for given x, what is the cdf of the standard normal? You calculate it back and most of the times we do linear interpolation between two cdf values or two x values. Same thing can also be done using psi lab. So, we write the same expression, exponential of this. We are trying to find out the inverse. This is the inverse over here. So, that is what we have. This is what you get. We go to the last one to discuss today. This is gamma distribution. There is a pdf like this. It is defined only for positive x values. P and k are parameters to describe the pdf of gamma distribution. And this gamma k is called the gamma function. Are you familiar with the gamma function? So, you know what the gamma function is. I would not go into the details of it. This is the pdf. There is a specific reason why I am discussing gamma. When we go through some examples, we will see. So, the gamma function is kind of like the factorial. Even you can calculate it for non integers. Factorial is only for non integers. But, gamma function can be calculated for non integers as well. Are you familiar with this? The incomplete gamma function? Well, if not, then this is the similar expression. You have the gamma function in the denominator and the numerator is a little changed one. You can find out the differences. It is defined for parameters u and k. k is the same k as the parameter in the gamma distribution also. u is where you are evaluating it. That is why you have integration from 0 to u. The incomplete gamma function is what we use to find out the cdf of the gamma function. How do you find the cdf of the gamma function? On the basic definition, we can use the pdf. So, probability that x is between a to b, that means cdf at b minus cdf at a. So, that probability you integrate the pdf from a to b. Is that ok? I will go back to the pdf expression. This is the expression for the pdf. It is valid only from 0 upwards. This is the expression for the pdf. You can note that v and k are constants. So, of course, gamma k is also a constant. So, you can take out that v raised to k and gamma k outside the integral. We do an integration of the other part of the pdf from a to b. You can divide it up if you expand it. We take the vk inside the integral. Then, the limits change to 0 to vb and 0 to va instead of 0 to b and 0 to a. Since you have taken the v raised to k inside, we change the integration limits. What are these? If you look at closely, these are nothing but the incomplete gamma functions. These are the incomplete gamma functions evaluated at vb. Earlier, we had u over here. These are the incomplete gamma functions evaluated at va. So, you can write the probability that x is between a to b is the incomplete gamma function at b minus incomplete gamma function at a. So, the incomplete gamma function is kind of like the cdf of the gamma distribution. See, probability that x is between a to b is the cdf at b minus cdf at a, which is same as i at vb minus i at va. So, not exactly cdf, but very similar to cdf. You can easily find out tables for this i so that you can evaluate the cdf of the gamma distribution. These are the two moments. Mean of gamma is given by k over b and the variance of gamma is given by k over v square. k is called the shape parameter and v is called the scale parameter for the gamma distribution. So, what do they do? They change how the cdf looks like. They change how the cdf looks like. So, these are again descriptors. Instead of mu and sigma, you can probably use k and v. You remember similar things that we use for Poisson distribution. What is the parameter that you use for Poisson distribution? Do you remember what is the mean of a Poisson distribution? Lambda, standard deviation also lambda. There was only one parameter to describe the pdf and cdf of the Poisson distribution. That was lambda. For the gamma distribution, we have k and v, which is the shape and scale parameters. Using those, you can find out the mean and standard deviation. Now, this is why I have discussed the gamma distribution. It has a specific relation with the Poisson process. If you remember what the Poisson process is, you remember we discussed that when we discussed the discrete random variable called the Poisson distribution. What is the Poisson distribution? When do we apply Poisson distribution? We apply it to describe the occurrence of an event over time or over space. For example, for this table, if a crack is going to appear over the length, where is the crack is going to appear? That will have a Poisson distribution. Again, I gave examples like number of earthquakes of such and such magnitude over next 10 years. How many? That usually follows a Poisson distribution. We call that a Poisson process having a specific mean occurrence time, which we denoted by nu and then we had t and we had a relation with lambda, nu and t and so on. If you look back, you will find those things out. Now, we are looking at the relation of the gamma distribution, which is a continuous distribution with the Poisson process, which deals with a discrete random variable. I hope you remember those things. So, what we say that if the occurrences of an event constitute a Poisson process in time, then the time till the kth occurrence of that event follows a gamma distribution. At the first hearing, it may sound a little strange, how are we relating these two, but later on you will see it is not very complicated and you can even prove that. So, if the occurrences of an event follow a Poisson process, that means how many earthquakes over the next 10 years, then the time till the kth occurrence, let us say the second earthquake will follow a gamma distribution. Now, this time till the kth occurrence, what kind of variable is it, discrete or a continuous one? Continuous, right? So, no problem in that. It can have a gamma distribution and it will have a gamma distribution, which we can prove. Before we go into an example, again we will look at psi lab, what are the functions that we have? Again, just like normal random variable, we had CDF NOR for normal random variable. For gamma, we have CDF GAM and the expression is very similar. Again, within quotes, you put PQ, then you put the value at which you are evaluating the CDF, then you put the shape parameter and then you put the scale parameter. These are non mean and standard deviation. These are the two parameters K and V. This will give you the CDF and you can also, of course, obtain the inverse of the CDF. For that, instead of PQ, we write x over here and then the shape parameter, scale parameter, CDF and 1 minus CDF. Using that, we can find it out. Example, we do the same thing as we did for the normal random. So, CDF gamma, that means we are trying to find out the CDF using this PQ argument at 2.25 for K equal to 2 and V equal to 0.5 is this. So, instead of mean and standard deviation, we use K and V, which are equivalent. If we know K and V, we also fix mean and standard deviation. If we know K and V, we fix the PDF and we also fix the CDF. If the CDF is fixed and the PDF is fixed, you can obtain mean and standard deviation from those. So, those are also fixed. So, you put mu or mu and sigma or K or V is the same thing, but silo requires you to put K and V over here and the inverse. I think you are doing the inverse of the same thing. We obtain the CDF at 2.25, we got a value of 0.310 and we turn it around. We find the inverse for K equal to 2, V equal to 0.5 of CDF 0.310 and this is Q. Q is always 1 minus P and you get back at where you obtain this CDF of 2. So, you get the value of 2.25. Salib also has this function gamma K, using which you can obtain the gamma function for any value K, which can be integer or non-integer. Now finally, the example and with this we will stop. So, this is known an earthquake larger than M5. M is magnitude moment magnitude in Richter scale. So, larger than M5 hits Mumbai once every six years. That is known to us and you also know that it follows a positive process over time. And you are trying to find out what is the PDF of the time of occurrence of the third earthquake. And by earthquake we mean earthquake having a magnitude larger than M5. So, you can see it is following a positive process and we have a occurrence rate once every six years. So, use that information and this the PDF for the time of occurrence of the third earthquake. We said that if this follows a positive process, then the time to the third earthquake will follow a gamma distribution. So, use the gamma distribution for this. So, based on the information we know the annual occurrence rate nu is 1 over 6 once in every six years. And the occurrence time would follow a gamma distribution with K equal to 3 that is the third earthquake and V the shape parameter equal to the occurrence rate equal to 1 over 6. K is the kth event and V is the occurrence rate nu. So, the expression for the PDF. So, T 3 we put. So, time to third earthquake that is the random variable. This is V and this is V x raised to the k minus 1 this factorial or gamma function of k minus k and this is e raised to minus V x. So, that is the PDF that we wanted to obtain. What is the PDF for? This is the PDF for the time of occurrence of the third earthquake. And what is the mean for that? E of the time of occurrence of the third earthquake from now is 18 years. Where is the 18 coming from? This is K equal to 3 over V. I will get back mean or expected value of x is K over V. So, K over V, K is 3, V is 1 over 6. So, you have mean equal to 18 years. Make sense right? The occurrence of earthquake has a rate once in every six years. So, when do you expect the third one to happen on an average at 18 years from now on? That is what you have. Also, you can find out the probability that the third earthquake will occur between the 16th and the 18 years from now. So, we find out the CDF at 18 years minus CDF at 16 years. This is for T 3. We are trying to find out that T 3 occurs between 18 and 16. So, CDF of T 3 at 18 minus CDF of T 3 at 16. With K equal to 3 and V equal to 1 over 6, you get this 0.0786 as well. So, that means 7.86 percent. So, we relate the Poisson process which basically describes the number of occurrences over time or space of a discrete random variable like number of earthquakes, number of floods, number of defective pieces in a given production and so on. We are trying to find out when is it going to occur, the k-th event that will follow gamma distribution and you can find out the probability of that occurrence for any given instant. That is what we do over here. So, this is what we covered. Standard normal, standard normal operator and the inverse, the silo function for it. Example using a normal random variable. Then, we discussed quickly the log normal distribution. It is CDF and PDF and how it is related to the normal random variable. LNX is the normal random variable. How to obtain the mean and sigma for that? Then, one example with log normal random variable. Then, the gamma distribution properties relation to Poisson, the silo function and example. That is it. Thank you.