 So, I will start by talking about independence of random variables when they are jointly distributed. So, just again an extension of the same concept that we talked about for single random variable. Now, here one can write it in 2, 3 ways x belongs to A and y belongs to B. So, there are two subsets A B of the real line and then probability this should be equal to probability x belonging to A into probability y belonging to B and this should hold for all possible subsets A B of R. Now, in other words if the events. So, another way of saying is that this event and this event are independent for all A B subsets of R. Another way of saying it then an equivalent definition of independence of x y would be that you know now you take real numbers A B and then you are saying that x is less than or equal to A y is less than or equal to B then probability is equal to probability x less than or equal to A into this should hold for all A and B and this is if and only if. This or you then you know by using the three axioms you can show that this and this are the same definitions that you can then take two real numbers and you can describe your events in that way. So, then so this will immediately imply that your distribution function cumulative distribution function for the joint cumulative distribution function can be written as a product of the individual cumulative distribution functions. Now, if x and y are discrete then we say that we can you know equivalent way of writing it is or this is how we can if p x comma y is p x x into p y y that means individual probabilities for all x y and surely this definition and this definition are the same because this is the general one which covers both continuous and discrete. Now, you can if you just take A as singleton x and B as a singleton y then this follows from here right because x equal to A. So, A comma B that means x comma y if A is simply a singleton B is also singleton y then this becomes x comma y and this is probability x into probability y of course with respect to x and respect to y right. So, two I mean two follows from one when you choose the sets A B to be singletons right and the other way also it is valid because if star is valid for all x y then you can you can write this definition as the first part of star as summation p x y x belonging to A y belonging to P that is our definition right. But since this is x and y are said to be independent then this can be written as p x x into p y y where x summing over all x is in A and summing over all y in B then you can separate out the summations you can write out this way and therefore, this is probability x equal to I did not yeah. So, probability x belonging to A into probability y belonging to B which is the same as this. So, if this is valid two follows star follows and if star is valid then double star follows. So, this is the whole idea now and of course, an equivalent condition for this for the continuous case would be that the joint pdf can be written as the product of the individual pdf. So, now you have so many equivalent ways of expressing the same concept I will take this example from Sheldon Ross interesting to you know again show the computations and the how we make use of independence of random variables. So, here a man and a woman decide to meet at a certain location if each person independently arrives at a time which is uniformly distributed between 12 noon and 1 p m. So, the arrival time of both of them is a I mean for the man and the woman both is random variable are random variables and this is the time is between 12 noon and 1 p m. Now, you have to find the probability that the first to arrive has to wait longer than 10 minutes. So, therefore, that is why I converted the pdf as 1 by 60 for each x. So, here let me define the random variable x as time in minutes past 12 noon when the man arrives and similarly, time in minutes past 12 noon when the woman arrives at the fixed spot. So, both are random variables and they are uniformly distributed between 0 and 60. So, that means, the pdf will be 1 by 60 for each. So, when you take the joint pdf that will be 1 by 60 square and we will say that x and y vary between 0 and 60. So, the probability that you are asking for is x is the probability for the man to arrive. So, if the man arrives first then x plus 10 should be less than or equal to y that means this is the time when the woman arrives. So, the man has to wait because he is the first to arrive he has to wait for longer than 10 that is probability x plus 10 less than y, but since x and y are continuous random variables computation of the required probability is alright. And similarly, if the woman arrives first then y plus 10 should be less than or equal to x, but since x and y are identically distributed independence assumes. So, therefore, the two events are the same. So, that means the probability will be exactly the same from symmetry. So, twice probability x plus 10 less than or equal to y this is what we have to compute. And so, you write down the limits you see for x the limits are 0 to y minus 10 x is less than or equal to y minus 10 x varies from because the arrival the man can arrive at 12 noon itself. So, then x will be 0. So, x actually varies from 0 to 60. So, here it is 0 to y minus 10 and the man woman can arrive has to because the man has to wait 10 minutes or more then it should be 10 to 60. So, the arrival time of the woman would be from 10 to 60. So, this is what is important. So, once you fix the limits and you know the joint pdf then computing the probabilities is not a problem. So, one has to spend time and thinking as to how to so the event is described very clearly. So, here the limits for y for the woman. So, arrival time would have to be from 10 to 60 and this is from 0 to y minus 10. So, therefore, you integrate with respect to x and the so it is simply x. So, y minus 10 by this thing and then when you integrate with respect to y it will be y square by 2 minus 10 y from 10 to 60. So, this will be 2 by 60 square and this will be 1 by 2 60 square minus 10 square and then minus 10 into 60. So, will continue from here this was y square by 2 minus 10 y from 10 to 60. So, then this becomes 60 square minus 10 square this one by 2 and minus 10 into 60 plus 10 into here this is 10 into 10. So, when you compute this from 10 to 60. So, this is minus 10 into 60 plus 10 into 10 and so when you simplify the numbers this here this will become 25 100 upon 60 square and this is 25 by 36. So, desired probability is this much that means, whoever arrives first at the meeting place will have to wait for more than 10 minutes. The probability of that is 20 upon 36. So, let me continue with the examples of jointly distributed independent random wearables. Now, here is an interesting example it is called the Buffon's needle problem. Buffon was a French naturalist and so he formulated this problem. It says that table has parallel lines drawn on it and the distance between two consecutive parallel lines is d. Now, you drop a needle on the table and of course, one possibility is that the needle lies like this and so it does not intersect one of the lines, but suppose it does intersect one of the lines. So, that is what we have to find out. So, a needle of length capital L where L is less than or equal to d the distance between two consecutive parallel lines is randomly thrown on the table. What is the probability that the needle will intersect one of the lines? So, we have to compute this probability and now here I have drawn this diagram. So, this is the needle and b is the middle point of the needle and you drop a perpendicular from here to the nearest line. So, that is the x. So, that will be a random wearable because you do not know what the position. So, that means the two random wearables that describe the position of the needle is the distance of the center of the needle from the closest nearest line and the angle it makes with the nearest line. So, that is that angle I am calling as theta. So, this theta and this is the distance which is x. So, these are the two this thing and we are saying that you see this thing is if this is the center point then this is L by 2 the length of the needle is capital L. So, this length is L by 2 and if your O B is less than L by 2 then the needle will intersect the one of the lines. So, that way geometrically this is described. So, this is you know making use of geometry to also explain things in probability theory. So, here x is a random wearable equal to the distance of the center of the needle from one of the lines and so now you can see that x upon in the right angle triangle O A B x upon O B will be sin theta. So, sin theta in the right angle triangle O A B x upon O B sin theta or x is O B sin theta that is O B is x upon sin theta and this has to be less than L by 2. So, this is our condition for the needle to intersect one of the parallel lines drawn on the table. So, and x varies from 0 to d by 2 yes because it can come up to here because after that it is the same thing then if the position of the line will become like this. So, if this can up to d by 2 and theta can vary from 0 to that means either the needle almost lies on the parallel line or it makes you know it stands like this these are the two. So, theta varies from 0 to pi by 2. So, these are the ranges for the two and of course since any position is equally likely we will say that both the random wearables x and theta are uniformly distributed. So, x is uniformly distributed in the interval 0 to d by 2 and theta is uniformly distributed from 0 to pi by 2. So, you see then one of the density p d f for a uniform random wearable is the one upon length of the interval in which it is defined. So, here it is pi by 2. So, 2 by pi and here it is 2 by d. So, this is the joint p d f that is 4 upon pi d. So, now you want to integrate you want to find out the probability x is less than l by 2 sin theta. This is our condition. So, therefore, theta varies from 0 to pi by 2 and x will vary from 0 to l by 2 sin theta. Then this is to be 4 upon pi d d x d theta. So, this is the integral which will give you the required probability and so that is simple enough because with respect to x see this is independent of x and theta both. So, here you get integral x and so that will be l by 2 sin theta into 4 by pi d and then you integrate respect to d theta here. So, that will be minus cos theta 0 to pi by 2 which is equal to 1. So, the required probability is 2 l upon pi d. So, you know one can construct interesting examples. So, here we use the independence of the two random variables to compute their joint p d f and then which which I think I will probably be now. So, we have already shown this that the joint p d f will be the product of the two marginal p d f's if the random variables are independent. Now, another proposition and this sometimes makes life easy and you can immediately find out the independence of the random variables. So, this is if and it holds for discrete and continuous both. So, what we are saying is that random variable x and y are independent if and only if their joint density function f x of x comma y can be written as a product of two functions h x and g y. So, this is solely a function of x, this is function of y and you see the range the limits are also independent. So, x varies from minus infinity to infinity, y varies from minus infinity to infinity and since. So, what we are saying is that since the integral double integral here is equal to 1 because this is a p d f. So, this implies that when you substitute when you replace this by h x into g y the integral separates into two single integrals and. So, this is this will have to be something like c 1 and this is c 2. So, this product is 1. So, therefore, you see 1 upon c 1 h x will be p d f and 1 upon c 2 g y will also be a p d f because by definition this is c 1. So, 1 upon c 1 will be 1 and here 1 upon c 2 will be 1. So, therefore, this is the thing that means you will always able to convert these two since this is the p d f and if it is a product of two such functions single valued single variable functions then you will be able to convert both of them into p d f this is the idea. So, let us just look at a few examples if your joint density function is given as 12 e raise to minus 3 x into e raise to minus 4 x this is minus 4 x and x between 0 infinity y between 0 and infinity then you see I can multiply this by 3 and this by 4. So, this now is your exponential with parameter 3 you immediately recognize it this also is exponential with parameter 4 and. So, both are p d f's and. So, therefore, you conclude because it says if and only if see this is a part. So, we have told you already that if they are independent then the joint p d f will be a product of the individual p d f's and here I am saying that if f x y can be written like this then again x and y are independent. So, therefore, I can immediately conclude that x and y are independent because your f x y can be written as a product of two p d f's. Then if you look at this function here x e raise to minus x plus y now here again I can break it up into two functions x into e raise to minus x and this e raise to minus y. So, this I know immediately is exponential with parameter as 1 lambda equal to 1 this 1 and here I can quickly verify that this is also a p d f which means that integral of x e raise to minus x from 0 to infinity d x and if you now integrate this by parts then you get minus e raise to minus x x take this as the first function. So, 0 to infinity this gives you 0 then you have plus 0 to infinity e raise to minus x d x which integrates to 1. So, therefore, this is a p d f and this also is a p d f. So, the proposition again tells us that x and y have to be independent random variables. So, let me now show you take another example see f x y is 24 x y and you see here you can decompose it into a function of x and a function of y. So, let us now, but the thing is that the area of integration. So, this is x between 0 and 1 y between 0 and 1, but x plus y less than 1. So, you see here the connection is there and therefore, the suspicion is that the two random variables are not independent and we will see. So, I have shown here the area and over which the valid region. So, this is x plus y less than 1 this is the line. So, here in the square 1 1 you this is the area on which you have to concentrate. So, now let me first verify that this is a p d f. So, therefore, see your range for x will be from 0 to that means given a value of y. If I am integrating with respect to x then I fix a value of y and then I draw a line. So, therefore, I will be. So, my range for x is then from 0 to see when you fix 1 minus y. So, x varies from fixing a y then my range for x will be 0 to 1 minus y and that is what I have written here 0 to 1 minus y 24 d x d y and. So, even you integrate with respect to x this is x square by 2 0 to 1 minus y and so the integral is the value of the integral is 1 minus y whole square by 2 and here just open it up take y inside and then you integrate you this is what y square by 2 minus 2 y cube by 3 and y 4 by 4 from this is 0 to 1. So, and so this will be when you substitute the values at 1 you get this which is equal to 1. So, we have verified that the p d f this is a p d f and now you compute the marginals and we will show that the product of the marginals is not equal to the joint density function. So, because that condition was if and only if remember. So, here when you integrate with respect to y. So, again this will be now 0 to 1 minus x that means you have fixed x. So, once you fix an x then your y will vary like this. So, from 0 to you are fixing x. So, y varies from 1 minus x. So, this will be the length of the range for y. So, therefore, 0 to 1 minus x x y d y and again this will be x into 1 minus x whole square the same integral that we did and so this will be 12 into x 1 minus x whole square x varies now from 0 to 1. Similarly, the marginal of y same integral because the symmetric the function is symmetric with respect to x and y both and the limits are also. So, therefore, this will be this and y between 0 and 1. So, you see that f x y is not equal to f x into f the marginals. So, therefore, the two random variables are not independent. So, when you say that you can break up the joint p d f into individual you know functions of the single variables then make sure that the ranges are also independent. Otherwise, the random variables will not be independent. So, we will continue and then maybe we will keep coming back to. So, but independence is a very important concept it does simplify lot of things in probability theory. Now, let me start talking about sums of random variables I have already told you earlier that how we can you know get the distribution for sums of independent random variables. So, here again the catch word is independent we need that. So, now, for example, if x and y are two random variables which are independent and if f x comma y is the joint p d f then if you want to write the distribution function for the random variable x plus y this will be given by this which will be. So, now, here again the same thing that we used in this case. So, since x plus y less than or equal to t. So, if you are integrating with respect to x then y is your y gets fixed. So, this will be minus infinity to t minus y to range for x and then y the range for y will be minus infinity to infinity. But since this is x and y are independent. So, this can be written as a product of the individual p d f's the marginal p d f's. So, therefore, I can separate out my integrals minus infinity to infinity f y y d y and minus infinity to t minus y f x x d x and. So, this you know it gives you what this is the probability of capital x less than or equal to t minus y. So, therefore, this is your cumulative distribution function for x at t minus y this integral and this is the integral minus infinity to infinity f y y d y. So, this is called convolution of because here this is the distribution function for x at t minus y minus infinity to y. Well not exactly you are well I am writing small f y d y. So, anyway this will be called as a convolution because this is at y and this is at t minus y. But I will make things more clear here see now if you want to if I differentiate this with respect to t then I get the p d f for x plus y. And you see that here this is this and remember we have done it already we can exchange the integral and the derivative sign provided this thing here is a differentiable. And so I take the derivative inside this gives me this and this now f x t minus y d d t will give me the p d f of x at t minus y and this is f y y d y minus. So, now this is also this you can see better as a convolution of the two p d fs f x and f y. So, f y y then f x is t minus y where we are looking at the event this less than or equal to t. So, this is called the convolution and sometimes it comes in handy, but I will try to show you again that means my philosophy is that you should try to use geometry and direct methods as much as possible to get answers. The formulae are important and sometimes they are very helpful and I have shown you that at times it has helped us to use the derived formulae to get the result, but sometimes it also helps to do things directly. So, let us take this example sum of two uniformly distributed independent random variables x and y on the interval 0 1. So, now I want to find out the p d f of the random variable x plus y where both x and y are uniformly distributed on the interval 0 1 and they are independent. So, then the variable z which represents the sum will now vary from 0 to 2 because this varies from 0 to 1 this varies from 0 to 1. So, the sum can vary from 0 to 2. So, as I am saying we will do it ab initio without using a formula directly I will try to compute the cumulative distribution function and see. So, therefore, the diagram is here 0 to 1 0 to 1 x axis y axis. Now, you see what happens is when you if your t is less than or equal to 1 see t equal to 1 gives you this line x plus y equal to 1 and this portion is covered by t greater than 1. So, as long as t is less than 1 you see this is uniform. So, that means, uniform mass over this region. So, therefore, to get this probability I just need to compute the area of this triangle. So, to get probability x plus y less than or equal to t it is simply the area of this this because the joint density function is what your joint density function is 1 1 into 1 which is 1 because both are uniform on the interval 0 1. So, the mass spread over is you know same uniform unit and so, the area of the valid region would be your probability. So, here as long as t is less than or equal to 1 the area. So, therefore, if you take this or you take this then it will always be the area of this triangle. So, therefore, this is half t square because this side is t and this side t. So, base and height are both t. So, half t square for 0 less than t less than 1. Now, when t becomes more than 1 then you see it is no longer because it is this region which is a triangle, but you need this area. So, therefore, what I will do is from the total area 1 I will subtract the area of this triangle and that will give me the required probability. So, now, so this line is x plus y equal to t and therefore, it intersects x equal to 1 at t minus. So, therefore, the y point is t minus 1 because t is greater than 1. Similarly, this is t minus 1 1. So, then this length both these lengths this point is 1 1. So, when you do it 1 minus t of minus 1 plus 1. So, 2 minus t. So, this length is 2 minus t this is length is also 2 minus t. So, the area of that triangle is half 2 minus t whole square and therefore, the required probability is 1 minus 1 by 2 into 2 minus t whole square when t lies between 1 and 2. So, see looking at the diagram things become really simple and therefore, when you differentiate this with respect to t in this case it becomes t 0 less than t less than 1 and for this because there is a minus sign a minus sign. So, 2 2 cancels 2 minus t as t varies between 1 and 2. So, now, if you draw the picture graph of f 2 t between 0 and 1 it is given by this line and between 1 and 2 it is given by this line. So, it is a triangle. So, therefore, this is also known as triangular distribution. Now, I leave it to you to try the try out the convolution formula and then see that you should get the same answer. So, you can sit down and verify for yourself the formula is clear here for different values of t you will have to. So, here of course, your this thing will be from 0 to 2 because your t can vary from, but looking at the diagram things really become simple and you can. So, wherever possible use geometry or direct methods. Now, trying to look at this example where you know you have x is uniform 0 1 and y is exponential with parameter 1 and x and y are independent. So, now, we want to look at the. So, the joint density function of x and y will be just the product of the individual p d f's. So, this is for the uniform it is 1 and for the exponential it is e raise to minus should have written e minus y. So, this is here I just wrote here e minus y and so your y varies from 0 to infinity x is between 0 and 1. So, now, you want to find out the probability that x plus y is less than or equal to t. That means, this is your f of x plus y t is what you want to find out and so you see now here what I have right now written is when 0 is less than t less than or equal to 1. So, you see the region of integration as I had told you is x between 0 and 1 and y non negative. So, going to infinity. So, this is the region for integration and I was trying to tell you that we have to separate out the integration into two parts because you see for x plus y less than or equal to t as long as t is between 0 and 1 then this is this area which will has to be covered. So, therefore, for example, if you take the line x plus y equal to t if this is the line then the limits for x are from 0 to t and for y will be from 0 to t minus x. So, therefore, for t between 0 and 1 these are the limits and so you can therefore, integrate with respect to y first. So, that will be 1 minus e raise to minus t minus x because integral of this will be minus e raise to minus y and so from 0 to 1 minus e of minus of t minus x d x and then you integrate with respect to x the limits are 0 to t and so this will be x and because this is plus x. So, this will remain minus e raise to minus t of minus x from 0 to t and therefore, this is your cumulative density function for t between 0 and 1. Now, for t greater than 1 the lines would be like this x plus y equal to t which is greater than 1. So, in that case you are for a given x your y will vary like this from 0 to t minus x and x will vary from 0 to 1 because when you are talking of t greater than 1 this is this line then x will vary from 0 to 1 and t will vary from 0 to t minus x and therefore, for t between 1 and infinity it will be 0 to 1 0 to t minus x e raise to minus y d y d x then again the same integration this 1 is exactly the same and then from 0 to 1 with respect to x. So, this will be minus e raise to minus t minus 1 plus e raise to minus t plus 1. So, please just verify that these are valid cumulative density functions of course, here as t goes to 0 this goes to 0 right t goes to 0 this is 0 and this is 1. So, 1 minus 1 is 0 and as t and from here as t goes to infinity you can see that this will go to 1 because this will go to 0 this will go to 0 and you will be left with 1. Have I written it correctly here let us just make sure that as t goes to infinity then this goes to infinity also and so e raise to minus of t minus 1 also goes to 0. So, therefore, the limiting value is 1. So, this is the valid cumulative density function and therefore, now we want to compute the p d f then just differentiate. So, this will be 1 minus e raise to minus t as t is between 0 and 1 and this one here would be minus e raise to minus t and then this will be plus e raise to minus of t minus 1. So, this is your p d f for. So, therefore, you know all these things as you work out and get experience you can get a feeling how to go about and the diagram in course, you can do it in 2 or 3 dimensions. So, therefore, diagram helps to tell you that you have to break up the integration into 2 parts. So, for between 0 and 1 for t between 0 and 1 the limits for x are different and when t is greater than 1 the limits become different. So, therefore, this can only you can get the feeling by looking at the figure and then you know deciding accordingly. So, we continue with some more of these examples. . So, expectation of x plus y is equal to expectation x plus expectation y. Now, let us just try to first show this under independence of x and y. So, then I can write the joint density function as the product of individual density functions and so, e of x plus y this should be x plus y also. So, you are integrating x plus y integral minus infinity to infinity minus infinity to infinity x plus y into f x x into f y y d x d y. So, I have written the joint density function as the product of individual density functions and so, therefore, one can then separate out the integral. So, this will be x f x x f y y d x d y and here it will be y. Now, then since if you integrate respect to y then integral f y of y d y will be 1 because this is a PDF of y and so, you will be left with x f x x d x minus infinity to infinity and similarly, here integration respect to x will result in 1 and so, you will get integral minus infinity to infinity y f y y d y and this is e x plus e y. But, I want to point out that it is not necessary to show this result you do not need independence of x and y actually it is always true and that you can immediately verify because if x y is the joint density PDF of x y joint PDF of x and y then e of x plus y we write in this way integral double integral minus infinity to infinity x plus y f x y yeah this is this of x comma y then again I write this as sum of two integrals which is this and this, but you know making the same argument that minus infinity to infinity of f x y x y d y will give you f x the marginal density of x and then so, it will come out to be integral x into marginal of x density function of x plus similarly, y you first integrate with respect to x here to get the marginal of y and then you get this right and therefore, this is e x plus e y. So, for showing that e of x plus y is e x plus e y you do not need independence of x and y, but under independence we can show another result which is that expectation of x y product of x and y is actually product of the expectations. So, this is e x into e y and this for this I will need independence because now I will write e x y as this double integration x y and the joint can be written as the product of the marginals. So, f x x into f y y d x d y and now here again I can separate out the integrals. So, this is integral x f x d x minus infinity to infinity into minus infinity to infinity y f y d y and now each of this this is e x and this is e y and so, I get this as the product. And this result now I will prove in while showing that the under independence variance of x plus y is equal to variance of x plus variance of y which holds only when x and y are independent. So, therefore, once we have shown this result we can now derive that result to compute the variance of x plus y we will require the result that I am giving right now we will use independence of the variables x and y. So, let me just write down variance x plus y is expected value of x plus y minus e x minus e y whole square. So, I open up the square term then this will be x minus e x whole square plus sorry this should not be there plus y minus e y whole square plus twice x minus e x into y minus e y. Now, from linearity of the expectation function we have seen it earlier that the expected function is a linear function. So, it follows that the expected value of the whole expression I can take expectation inside and therefore, this will be expectation of x minus e x whole square plus expectation of y minus e y whole square plus twice expectation of x minus e x into y minus e y. Now, this is the place where I will use the linearity of x and y because if x and y are independent sorry the independence of x and y if x and y are independent then x minus e x into y minus e y are also independent this being a constant. So, it can immediately be concluded that if x and y are independent then x minus e x and y minus e y are also independent. Therefore, the expectation of the product can be the is equal to the product of the expectations and therefore, this product expectation of this product is equal to expectation of x minus e x into expectation of y minus e y. So, now here you see that this will be e x minus e x which is 0 into e y minus e y which is also 0. So, the whole thing is 0 and so this expression reduces to simply expectation of x minus e x whole square plus expectation of y minus e y whole square which is equal to variance x plus variance y. So, when trying to compute the variance of sum of two variables two random variables then I need to be able to write it like this I need x and y to be independent. Now, let me show you what independence means to us and of course, interesting relationships between various p d f's that we have discussed or special kinds of random variables. Now, if x is gamma s lambda that means parameters s and lambda and y is gamma with parameters t and lambda. So, lambda is same, but these two numbers differ then x and y and x and y are independent then x plus y is gamma s plus t comma lambda that means the first parameters is the second parameter is the same then the first one get added up when you add up the two random variables. And this I will show you using the m g f because I missed out on the m g f. So, m g f is also straight forward because this is expectation of e raise to t x plus y. So, then this because I will write it as e into t x and into e into t y this function I can write as this then again I just told you that yeah. So, you might say that y are these independent that can also be shown that if two functions are independent then their functions are also independent of course, under certain condition, but e raise to t x and e raise to t y are also independent random variables. So, when I write expectation of the product that will be the product of the expectations. See there is nothing much to you can actually use the again write the joint this thing because the joint density function will again with the product of individual p d f. So, even when you write this function here as an integral you separate out the integrals and again just as we prove this result exactly you can show that this expectation is equal to the product of the expectations. So, then that means if x and y are independent then their m g f of the sum is the product of the individual moment generating functions. So, this is the important result and that is what I am going to use here. So, I am showing you that we know that when x is gamma s lambda its moment generating function is given by lambda upon lambda minus t raise to s and the moment generating function of g t comma lambda is lambda upon lambda minus t raise to t. And so, the sum would be the product of the m g f. So, the m g f of the sum is the product of the m g f and so, this becomes lambda upon lambda minus t raise to s plus t. So, now I am concluding immediately that this that means the p d f of x plus y is gamma with parameters s plus t comma lambda. While defining m g f for you and looking at its properties I had mentioned that m g f uniquely characterizes the p d f of the random variable. If you know the m g f of a random variable then you can tell from what is the p d f of that random variable. This can also be proved, but I will simply use the result here and so therefore, once I know that the m g f of x plus y is of this form then I will conclude that this that means the p d f of x plus y must be s plus t comma lambda. And now let me this I had already mentioned this relationship between the exponential and gamma. I had said that exponential lambda is gamma 1 comma lambda that means the first parameter is 1 and this is the common thing. So, when you have a exponential random variable it is also gamma 1 comma lambda. Now, if x and y are independent exponential lambda random variables both having the same parameter then by this result x plus y will be gamma 2 comma lambda. So, I am using this result here that the first parameters get added up if the second parameter is the same since the exponential x and y both are exponential with. So, they are both gamma 1 comma lambda. So, their sum will be gamma 2 comma lambda and so on. So, and now what I am trying to say here is that this concept of independence of two variables can be extended to many more. In the discrete case I had already shown you that characterizing the independence of more than two variables becomes tedious, but we can anyway use the end result. So, here see the thing is that you can sequentially use this result you know about m g f's about variance and expectation. So, what we are saying is that if x 2 first of all you have two variables. So, x 2 and x 1 are independent. Then once you have this then you can apply that x 3 is independent of x 1 comma x 2. So, then you can apply the result that means you will go on adding. So, what I am trying to say here is that if you have two exponential random variables both with the same parameter I add them. So, I get a gamma 2 comma lambda then I add x 3 to it. So, that means I will be talking of x 1 plus x 2 plus x 3. So, this will become 3 comma lambda. So, the concept can be recursively used. So, here then you will say that x 4 is independent of these three and so on. So, finally, x 1 is independent of x 1 x 2 up to x n minus 1. So, this is the whole idea. So, therefore, now I can say that when you have gamma s comma lambda where s is an integer. So, if you look at this result then s is an integer. So, gamma s comma lambda is the sum of s independent exponential lambda random variables. So, when s is an integer gamma distribution has been built up by adding exponential distributions s of them. And if you remember when I was talking of the gamma distribution and then we were looking at the. So, we had said that suppose there is a service counter and there are people ahead of you who are n minus 1 you are the n th person then gamma n lambda you can. So, that means the time that you have to spend and we had said that if these time taken to service each customer is exponential with parameter lambda then the total time till the n th person gets serviced. That means this is this includes the servicing of n minus 1 people ahead of him and then the n th person is this person. So, when then total n people get serviced that will become gamma n lambda. So, this was the connection right and so we I had used this and I had told you that we will be able to show this prove this result also that the n exponential distributions with the same parameter lambda will add up to gamma distribution with parameters n comma lambda. So, I will continue this exploration mode in the coming lectures.