 So, now, we are going to talk about jointly distributed random variables. See, there may be events which need to be described by more than one random variable. So, there may be you know the phenomena, the random phenomena concerned with the event are more than one dimension. And therefore, we need to be able to of course, then we from two dimensions we go to multi dimensions also, but right now to keep it simple, we will first talk about two dimensional random variables and then may be extend the notion to more than two. So, if x and y are two random variables, we define their joint cumulative distribution function as follows. So, probability f of x less than or equal to a comma y less than or equal to b is actually the probability that x is less than or equal to a and y less than or equal to b a and b two real numbers between minus infinity and infinity. So, essentially in this diagram, if this is b then you are asking for y less than or equal to b and then x less than or. So, it will be this whole region, if you can see this line and this line. So, whole of r 2 extending from here to this end, that will be of. So, you are talking of the probability over this region. Then the moment you define the joint CDF, the cumulative distribution function then you can talk of the marginal CDF. So, for example, of marginal CDF of x can be obtained from f a b as follows. Sorry, this should be a b. The probability part comes here x less than or equal to a y less than or equal to b. So, therefore, from f a b we can obtain the marginal CDF of x as follows. So, this is f x a which is probability x less than or equal to a. So, this will be actually therefore, that means this means that y is allowed to take any possible value. So, therefore, y probability x less than or equal to a and y less than infinity. So, now, let me just define it in a proper way in the sense that this is also the same as b going to infinity. So, limit that means if I take the event x less than or equal to a y less than or equal to b and then I take the limit. So, you see what will happen as b goes to infinity. So, x less than or equal to a y less than or equal to b you are talking of this then when say say b 1 then you talk of this y less than or equal to b 2 where b 2 is greater than b 1 then this is a bigger event right this event is contained. So, therefore, now you have a sequence of these events which are increasing. So, increasing sequence of events as b goes to infinity and if you remember my definition of probability as a continuous function, continuous set function. So, then in that case I told you that we can in that case extend I mean when you have a increasing sequence of sets then and you are computing the probability you want to take the limit then the probability of the limit is the limit of the probability. So, therefore, I can take limit outside and this will be limit as b goes to infinity of the this event x less than or equal to a y less than or equal to b. So, you take the probability here and therefore, this will be limit and this is now your f a b by definition. So, this limit b going to infinity of f a comma b and this will be then f a comma infinity. So, f a and f a comma infinity are the same. Similarly, if you want to compute the marginal of y then we will take the limit f a comma b a go to infinity which will be f infinity comma b. So, exactly in the same way we will argue out that the b remains the same and then a keeps increasing. So, again you have increasing sequence of sets and so when you take the probability I can you know I can exchange the limit and the probability and get the answer. .. So, now similarly you want to compute and then of course, you see the properties that we had defined for that cumulative distribution function must have for a single variable. So, the same will apply and you can apply them to f x and f y. So, through those you can get the property for the joint. Because, the property that x f x and f y possess combine them and you can you know then put together the properties that your joint cumulative distribution function must have. Now, suppose you want to compute the probability of x greater than a and y greater than b. So, this we can write as 1 minus probability of the complement of this event which is x greater than a y greater than b complement. And then since x and y are independent in the sense that you know I can write this as. Now, this can be written as x less than or equal to a union y less than or equal to b right from if you recall your De Marben's law and so on. So, then this will be 1 minus probability x less than or equal to a minus probability y less than or equal to b. And then you add probability x less than or equal to a comma y less than or equal to b and this diagrammatically also you can immediately see that this is how from here we have to get here. Because, you are computing the probability x less than or equal to a and y less than or equal to b. So, this will be you see x less than or equal to a would give you this region right and y less than or equal to b will give you this region right. So, you see the region extending from here x less than a y less than b is this region which you have subtracted twice. Because, once when you do it for x less than a probability x less than or equal to a. So, it is this. So, then this region is coming into it then when you subtract probability y less than or equal to b then again this whole region is coming right. And therefore, you add this again. So, in words of your I should have written this that means your. So, therefore, I will simply write this probability in terms of 1 minus f x a minus f y b. So, this will be equal to 1 minus f x a minus f y b plus f plus f a b. So, therefore, to make the equation correct I add this once and then because I need to subtract only see I need this region. So, the valid region that I require is this. And therefore, from 1 I subtract this whole and so this got subtracted twice for I added once to make it proper. So, this will be your that means now one can compute whatever probabilities you want related to these two random variables it can be done once we have made this definition. Now, in general if x is lying in the interval a 1 comma a 2 and y in the interval b 1 comma b 2 then this can be written as f a 2 comma b 2 plus f a 1 comma b 1 minus f a 1 comma b 2 minus f a 2 comma b 1. So, here again we will just simply diagrammatically look at the equality that we have here. So, x is varying between a 1 and a 2 and y is varying between b 1 and b 2 then f a 2 b 2 is this whole area. In fact, if it is not if it is only non negative variable then it is this region. So, just assume for argument sake that this is both the variables are non negative, but otherwise it will extend to infinity and this will extend to infinity. So, anyway because this is f a 2 b 2 is probability x less than a 2 and y less than b 2. So, in the case of non negative variables this is the total region then f a 1 comma b 1 is this region let me this here then f a 1 comma b 2 a 1 comma b 2 is this whole thing and f a 2 comma b 1 is this region here. So, you see that here you are subtracting this region twice and in this a 2 plus b a 2 comma b 2 you are adding it and then you are adding this. So, that gets cancelled out and this region also gets cancelled out. So, from the whole of f a 2 comma b 2 you are finally, left with this particular region this is the idea I am equating the probability with the area in a sense and that is how I am trying to explain. So, if you keep this picture in mind and you can always make the right computation now in case x and y are discrete random variables the joint mass function can be written easily because here you are simply wanting to compute these probabilities probability x equal to a y equal to b is b a comma b this is how we will define. So, if we can compute this probability then that is the probability a comma b and this for all possible values the pairs a comma b taken by x comma y and then we can also compute the marginal probability mass function of x which will be simply probability x equal to a and this will be your summing these probabilities over. So, fix the x at a and then you are summing it over all possible values of y. So, you are summing up p a comma y for all possible values of y that will give you the probability that x attains the value a because here the value of y is not is a material. So, therefore, you sum it up over all possible values that y will take similarly p y b will be summing up these probabilities x comma b when p x b is positive. So, in this case you want to sum up over all possible values of x to get the probability that y will take the value b and we will go through an example here. So, let us consider this three balls are randomly selected from an urn containing two red three white and four blue balls. So, the total number of balls is 9 and now you want to find the joint distribution function of x and y when x and y where three balls are chosen from the urn and x represents number of red balls and y number of. So, I pick up three balls from the urn then I note the number of red balls present in those three balls and the number of white balls. So, now we want to write down the joint mass function or distribution function for x and y. So, let us now continue with the example with which there the urn contains two red balls three white and four blue total number of balls is 9. So, you want to compute the cumulative mass function for the variables x comma y where x is the number of red balls and y is the number of white balls when you pick up three balls from the urn. So, let us I will just compute a few and then you can try to complete the table I think may be I have completed it already. So, when you want to compute the probability of 0 0 that means x is equal to 0 and y is 0. So, which is no red ball and no white ball. So, that means all the three balls that you pick up from the urn must be blue. So, the probability of all three balls being blue is 4 c 3 and total number of ways in which you can pick up three balls from 9 balls is 9 choose 3 and. So, this number comes out to be 1 upon 21 which is here. Then I should have computed p 0 1 and I think I have written it here as. So, it is this number is actually. So, it is got mixed up this is 1 by 7. So, that is probability 1 0 is 1 by 7. So, if you want to compute have I done it somewhere here no. So, therefore, let us just quickly compute this number p 0 1. So, 0 1 is probability no red ball and 1 white ball. So, this will be no red ball that means 1 white ball and 2 blue balls. So, blue is 4 c 2 and 1 white out of 3. So, this is 3 choose 1 divided by 9 choose 3. So, let us quickly compute this. This should be 6 3 and then from here you will get a 6 and this will be 9 into 8 into 7. So, 6 3's are 18 and this will be. So, twice twice 4 then 2 and 3. So, this is 3 by 14. So, 3 by 14 is this. So, 3 by 14 is this number and this way I have made some computations. So, 0 2 you have to compute and so on. So, I have shown you some computations here p 1 2 then p 1 0 p 1 1 we have computed and so you go on. So, now the whole idea is that once you have completed this then as I was saying that if you want the marginal distribution for x then. So, for example, if you are looking for the probability, let me just take it from here. So, if you add up the numbers here this is what probability x equal to 0 then you are giving values to 0 1 2 and 3. In other words you are adding up probability. Let me write it as 0 0 plus probability 0 1 plus probability 0 2 plus probability 0 3. So, there can be 3 white balls because total number of white balls is 3. So, all this will give you the probability that x is equal to 0 because when you pick up the 3 balls if it does not contain any red balls then those 3 balls will either contain no white ball 1 white ball 2 white balls or 3 white balls. So, this all these probabilities add up to the probability x equal to 0 and so this will be right. Similarly, when you add up the second row that will give you probability x equal to 1 second third row will give you probability x equal to 2 and this will be probability x equal to 3, but since this is all 0 because there are no the number of red balls is only 2. So, you cannot have number of red balls in the sample that you pick up as equal to 3. So, therefore, this is all 0s similarly here you cannot have the combination 3 comma 3 because you are picking up only 3 balls and similarly 3 comma 2 3 comma 1 is also not possible from here 2 comma 2 and 2 comma 3 is also not possible. So, these are the only numbers and you see this now finally, because this is probability x equal to 0 probability x equal to 1 probability x equal to 2 which are all the possible values that x can take in your sample of 3 balls because there are only at most 2 white balls that can appear in a sample. So, this must add up to 1 and similarly when you add up see when I add up these probabilities this will give me the probability of the y is equal to 0. That means there is no white ball in the sample when you add up these numbers this will give you the probability that y is equal to 1 and similarly this. So, all these 4 when you add up that should also add up to 1 and that will give you a good check. So, again that means when we are defining the marginal pdf's and so on then we must continue to check the validity and so make sure that your calculations are because otherwise you will know that you have made a mistake somewhere. So, you can go back and check. So, all these numbers so the numbers here will give you the marginal distribution of y here the numbers will be marginal distribution of x and so on. So, this is how you write down the joint probability distribution of 2 random variables. Now, take another example so that I want to make sure that you understand how the calculations are being done. Now, this is an example in which there is a community in which 15 percent of the families have no children 20 percent of the families have one child then 35 percent of the families have 2 children and 30 percent have 3 children. Now, the probability of the children being a boy or a girl is same that means it is half half. So, once you pick up a family at random. So, the experiment that we are doing here is that you pick up a family from the at random that means any family is equally likely then you want to find out the number of boys and girls in that family. So, both of them are random variables that means number of boys in the family and the number of girls in the family the family that you have picked up at random. So, here again let us see I have made a few calculations. So, if you want to compute P 0 0 then that means no children. So, of course, this is straight forward you know that 15 percent of the families have no children. So, this is simply 0.15. So, which we write here then when you want to compute P 0 1 that means no boy just one girl. So, this is the event probability of one girl child now I will write this as a conditional probability saying that probability one child into probability one girl given that there is one child in the family. So, that will come out to be 0.20 into half because there is only one child in the family P 0 1. So, that means only one girl and no boy. So, this the total number of children in the family is 1. So, that probability is 0.20 then it being a girl or a boy the probability is the same therefore, it will be into half. So, that is 0.10 that is the number you enter here. Then similarly, if you want to compute P 0 2 then it will be two children and so I am writing the conditional straight away. So, two children probability of that and then two girls given there are two children. So, this is 0.35 from here families having two children that is 0.35 and then two girls. So, 1 by 2 into 1 by 2. So, 1 divided by 4 and so that is 0.0875. So, 0.0875 then when you want to compute I have done it here it is not very this thing, but anyway it is coming up may be I can just show the calculation in a better way. So, when you want to have 0 3 that means all the children are girls. So, this again will be probability 3 children which is 0.30 then into 1 by 8 because all the three are girls. So, 1 by 2 raise to 3 and therefore, you divide by this by 8. So, this is 0.0 then 3 will be 24 then 6 then 8 7's are 56 and then 40 and 8 5's are 40. So, 0.3 0.0 3 7 5 0.3 7 5. So, when you add up this numbers this will here will be the probability that x is 0. So, that means no probability when you pick up a family at random from the community it there are no boys in that family. So, that number will come out to be this you can add it up. Similarly, this will give you the probability that x is equal to 1 that means 1 boy in the family probability x equal to 2 and this will be probability x equal to 3. And here also you can see that this these combinations and of course, because the boys and girls are equally likely. So, the tables so once you have computed this part then this is symmetric because whether having a 1 boy and 2 girls or 1 1 when you have 2 children. So, it is whether girl or boy is same therefore, this will be the same then 1 2 see this one we are having yeah this is 2 0 I am sorry 0 1 and 1 0. So, that was the number then this number for example, 2 1 and 1 2 1 2. So, these two numbers so this is symmetric because boys and girls are equally likely similarly here whether all the 3 children in the family are boys all the 3 children are girls is the probability must be the. So, I am sorry this should be 0.375 so make that correction. So, this will be 0.0 375 and then again as we said when you add up all these probabilities they should be add they should add up to 1 and here also you will get the marginals. So, this will be probability y equal to 0 probability y equal to 1 probability y equal to 2 and probability y equal to 3. So, I would like you to complete the tables I have made some computations for you this. So, this is the same now you have to make just these two computations and then you can complete the table. So, this is this should give you an idea as to how to go about computing joint when the variables are discrete how to compute the joint distribution function for discrete random variables. So, let us continue with the jointly distributed random variables suppose x and y are continuous. So, I will now define another way in other way in the sense that now I will define this through your joint pdf and so we are saying that if there exists a function f x comma y real valued function which is defined for real x and y having the property that for every set c a subset of r 2 that means all pairs of values the pairs of values which are there in c then x y belonging to c that means you see the convention is that the first these are pair of coordinates then the first value is for x and the second value is for y. So, then probability x y belongs to c is this. So, now this is different from the way we define the cumulative distribution function right in the beginning. So, of course, the two will merge to the same thing. So, you have the probability should be f x y d x d y for a continuously distributed joint you know jointly distributed continuous random variable. So, f x y is called the joint pdf of x comma y and if a and b are two subsets in r square then probability that x belongs to a and y belongs to b will be integration over a with respect to d x and actually see what you should make a convention that this must refer to the later one here. So, I should have said this is d y and d x because just a convention. So, therefore, when you are writing the limits you know the ranges then it is nice to remember that this one refers to the second integral and this is the first. So, the order has to be maintained. So, therefore, this is a b. So, I should have written d y d x of f x comma y. So, the same thing is being sort of repeated and then for if your f a b as we defined earlier is now you know x lying between minus infinity and a y lying between minus infinity and b then the partial derivatives here delta square delta a delta b f a comma b will be f x y a comma b. So, this is the relationship between the cumulative joint cumulative distribution function and joint pdf and then of course, the marginal distributions here. So, this will be x belonging to a and y is from minus infinity to infinity. So, in that case you are integrating with respect to y from minus infinity to infinity and this integral when you are integrating with respect to y will be the marginal. Just as I showed you in the discrete case that you add up for all the possible values of y to get the marginal and the probability of you know x for a certain value. So, same here the marginal with respect to x will be you integrate the joint pdf from minus infinity to infinity and for the marginal of y pdf of marginal y this would be integrating with respect to x from minus infinity to infinity and of course, depending on whatever the region of actual definition is fine. . Let us take up this example. So, if you have this function defined here which is defined for all values of x is from 0 to infinity and for all values of y from 0 to infinity and it is 0 otherwise. So, it is in the first quadrant that the function is defined then verify that this is joint pdf. So, therefore, the integral total integral from 0 to infinity and this should be 1. So, suppose I just integrate with respect to y first. So, the derivative here the integral is minus e raise to minus y from 0 to infinity that gives you 1 because at infinity this is 0 at 0 this is 1 plus 1. So, then this is this now you differentiate integrate with respect to x. So, this is minus 3 upon 3 e raise to minus 3 x going from 0 to infinity which is equal to 1. Now, if you want to compute this particular probability then your limits will be. So, this time I have taken care of the order. So, this is 2 to infinity for x and for y it is from 0 to 1 and. So, again the same integral you do it with respect to y first take the limits and then this is 1 minus e raise to minus 1 then you integrate with respect to x and this will be. So, the final answer will be this and so you can go on. So, this is nothing new except that your dimension has increased and the same concepts are there the same axioms will be followed. So, we will now you know continue developing this theory and then talk of independent random variables jointly distributed independent random variables then we will talk of some of random variables. So, of course, the thing is that this concept can be extended to more than 2 and except that the writing part is a little tedious, but the same thing will follow. So, I will just revisit this I had shown you this probability try to well computing this probability I had drawn the diagram. So, let me just make it more clear because I had a feeling that I did not do a very good job last time. So, let us see probability x greater than a and y greater than b then if this is my origin this is x equal to a this is y equal to b then this is the area that you want to compute the probability on your region x greater than a y greater than b is this event represented by this area. So, now we said we will write it as x less than or equal to a. So, x less than or equal to a is this whole area extending on this side whole area and then y less than or equal to b will be this area. So, therefore, all this and all this gets covered and you see that this area the area that is x less than a y less than b this is. So, this portion because when you are covering x less than a then it is all of this y less than b then it is all this. So, therefore, x less than a y less than b this particular area gets covered twice. So, you add here and therefore, you get this. So, therefore, 1 minus of all this that means if I write probability x less than or equal to a plus probability y less than or equal to b minus this then I will get the region corresponding to this and therefore, so 1 minus of that and that will be the probability for this. So, just wanted to revisit this thing and here also in the other example that we had taken when we were talking of a community in which there are families and there were probabilities associated with the families having no children, one child two children and three children. So, I thought that though I had asked you to compute I had computed some probabilities for you and I asked you to compute the remaining one, but I realize that probably you need a little more working out and so I thought I will give you the hints here. For example, when you are computing 1 comma 1 then it is the probability two children into the conditional probability that it is either a boy girl or girl boy. You see when you are saying 1 1, so then it can be you know the first child is boy and the second is girl or the first child is girl and the second child is a boy. So, therefore, this will be 0.35 into because the probability of two children families having two children is 0.35 then into actually boy girl will be 1 by 4 because each of them are equally likely, but since there are two possible cases. So, 2 into 1 by 4 therefore, this is half and so this number comes out to be 1 1 this is 0.175 now for 1 2 when you want one boy and two girls then the that means three children. So, the probability of having three children is 0.3 30 percent of the families have three children. So, this is 0.3 then the conditional probability of having a boy and two girls and then with three children having three children and then one boy and two girls. So, now here again you need to say that see it could be the first child who is the boy other two are girls then it is first girl then boy then girl and girl girl and boy. So, these three possibilities are also there therefore, it will become 3 by 8 because each of them is half. So, then when you have these three possible cases you know favorable for your event then this is 3 by 8. So, 0.3 into 3 by 8 that comes out to be 0.1 1 2 5 and then the rest we had computed and then these are the marginal. So, that means this I told you is probability x equal to or b equal to 0 and probability b equal to 1 probability b equal to 2 and probability b equal to 3 and similarly here this is the marginal for probability the girl the probability girl is 0 no girl in the family of course. So, probability no there is one girl in the family two girls and three girls and then you see that this adds up to 1 because they are marginal pdf's or probability mass functions and so on. So, once you find out the joint then you find out the marginals and then you can do all other computations that are required the expectation variance and everything you can compute as we have done it for you like even for the joint also you can do it expectation x y you will do it and so you will multiply I think I have given you the expression p x y this we worked it out this into your the values that x and y take some y is positive and so on. Now, I will continue with some more examples of the continuous random variable. So, here let us see you take a circle of radius r and let us say it is centered at the origin. So, the equation of the circle is x square plus y square equal to r square. So, when you are taking the inside of the circle then it is less than or equal to r square. So, this is the circle given to you then let us and we just pick a point on inside the circle then that is a random because and we are saying that all points are equally likely to be picked up in the inside the circle then x let x represent the x coordinate of the point that you have picked up y is the capital y represents the y coordinate. So, now both x and y are random variables and since any point in the circle is equally likely therefore it will be a constant the p d f of the x comma y that joint p d f of x comma y will be a constant inside the circle and 0 outside and this is an example of a two dimensional two variable uniform distribution. So, this represents a uniform distribution because any point in the region in the inside and on the circle is equally likely right and then now you want to find out the value of c. So, we will have to say that this must integrate to 1 the whole thing and since this is simply minus infinity to infinity and minus infinity to infinity d x d y this is an element of area. So, when you integrate over the whole of the circle you will get the area of the circle by everybody. So, this is you have already done it in your class 12 and so on. So, then c into pi r square is equal to 1 therefore your constant is 1 upon pi r square which is expected right for in one dimension also I told you that when you had a uniform random variable then this was the length of the. So, if this was point a this was b then the probability density function for any for this uniform random variable was 1 upon b minus a which is 1 upon length of the interval in which the random variable is defined. Here the random where the pair x comma y is defined inside the circle when it is uniformly distributed this pair and that means that the p d f must be 1 by pi r square the area of the region. In this case it is the area of the circle now you want to compute the marginal of x and marginal p d f of x and marginal p d f of y. So, for x you will integrate from minus infinity to infinity f x y d y and this is what I want to point out that. So, you have fixed value of x if you have fixed a value of x then how does your y vary y varies along this chord right and the length of the chord is because this length is x this is r the radius of the circle. So, this length is under root r square minus x square. So, therefore, this point has the coordinates x comma under root r square minus x square and this point has coordinates x comma minus under root r square minus x square. So, your y varies from minus under root r square minus x square to under root r square minus x square. So, this is it this reduce because there is no other mass outside this. So, this is c into d y and the c is 1 by pi r square. So, this becomes twice under root r square minus x square and this is defined for all x between minus r and r right because your x varies from this point to this point along this and as your x varies along this your y varies along this. So, the whole circle gets covered right similarly because there is symmetry. So, your marginal of y will be twice under root r square minus y square upon pi r square where again y varies sorry this is mixed up this is minus r less than or equal to y less than or equal to r 0 otherwise. Now, suppose you want to find out the distribution of random variable d which is the distance of the point from the origin. So, this is under root x square plus y square. So, you take any point here and then this is the distance the length would be under root of x square plus y square. So, we find out the distribution function of d which will be f d r. So, that means probability d less than or equal to r. Now, since everything is non-negative distance is a non-negative number r is non-negative. So, this event is the same as squaring up both the sides that means x square plus y square less than or equal to r square. The two events are the same and again by the same argument that for x square plus y square less than or equal to r square. So, all points inside the circle having radius r all these points will satisfy this or define this event. Every point inside the circle of radius small r and therefore, the area here is pi r square small r square and divided by the p d f which is pi r square. So, you can you know want you can do it from by integrating and everything, but this is a straight forward because everything is uniformly distributed. So, therefore, the probability of a point lying inside the circle small r is pi r square just as we said that the you know this is the that means this is the area which is favorable to our event and then the density of the function is of the two pair is 1 upon pi square r square. So, therefore, the probability the distribution function of d is r square upon capital r square. So, now, if you want to find out the p d f of d p d f of d would be differentiating this with respect to small r. So, 2 r upon r square and small r varying from 0 to capital r because now d is a non negative random variable and the value of d will vary from 0 to r because the point is here then this way and this way. So, it can go up to capital r and similarly you can find out the expectation of d which will be 2 by 3 r. So, you know just by you know geometry helps you draw pictures then you can you know get a good idea as to how to go about solving a problem. So, in the moment you have more than one you know if you consider a pair of random variables and you can see that the complexity will increase the moment you consider higher dimension you know that means 3 random variables together joint density function of 3. But as I go along I will try to solve some more problems relating to 2 random variables joint distributions of 2 random variables. So, another example on joint distribution of random variables suppose f x y is a function given like this 6 by 7 x square plus x y by 2 x is between 0 and 1 and y is between 0 and 2. So, then you want to verify that this is this represents a pdf of the variables x y. So, I will just integrate from 0 to 1 and 0 to 2 and the working out shows that this is x square y plus x y square by 2 if you are differentiating with respect to y because I have written the limits with respect to y first 0 to 2. So, let us integrate with respect to y first and therefore, this is what you get and then you this you look at this then integration with respect to x 0 to 1. So, that will give you this function and finally, you get the integral to be equal to 1. So, therefore, the function does represent because it is a non-negative function since x and y are both non-negative. So, this is a non-negative function the integral in the defined in the specified area integrates to 1. So, it represents a pdf. Now, to compute the marginal or the pdf of x or the marginal of x, you see I have already done it here because you have to integrate this integral from 0 to 2 dy. So, this integral gives me the marginal of x therefore, this is 2 x square plus x 0 less than or equal to 1. And you see that if I now integrate this from 0 to 1, it will give me the answer 1 and hence this is also a pdf. It is non-negative in the specified region and of course, I should also say 0 otherwise. See it is very important that your definition of the pdf must be complete in the sense that you must specify the region on which it is defined. So, here it is in between x 0 and 1, this is the pdf is defined by this function and it is 0 otherwise. Then you are asked to compute x greater than y. So, if you have to compute x greater than y, then see I have written down this is the line which represents x equal to y. So, this is the region over which you want to compute this probability because you have to specify the event. So, then it will be see here if you are integrating with respect to x, then you fix your y that means when I am integrating with respect to x, my x is varying. So, y is fixed. So, when you fix a y in this region, how will your x vary from y to 1 because this represents 1 here is this line. So, that means you are integrating from here to here for a fixed y and as y changes from 0 to 1, you will cover the whole area like this. Therefore, range for the variable x is from y to 1 and y varies from 0 to 1. So, this is very important and again I will keep repeating this that draw the diagram we can two dimensions you can always do it and then you get a feeling for how the what are the ranges of integration and so on. So, integration with respect to x gives me x square x cube by 3 plus x square y by 4 from y to 1 and that is important because I must integrate with respect to x first since the limits for x are in terms of y. So, then I will first integrate with respect to x and then get the function as a function of y and then I integrate with respect to y. So, this has to be the order and you must keep this in mind and so finally, this comes out to be 15 upon 56. So, the arithmetic should be correct. Now, here what I was saying is that suppose you had to compute the event probability y greater than x. So, in that case you see what will happen is it is this region over which you have to do it. So, then you see you will have to break it up into or what you can do is maybe no it is not necessary because then if y is greater than x you will want to write the limits for that means for a given x how will your y vary that is no problem. The y will vary from here to here that means it will vary from x to 2. So, I will integrate with respect to that means if y is varying from x to 2. So, we will have to integrate with respect to y here. So, the limits would be x to 2 and then x varies from 0 to 1. So, this will be the this will give you this probability. If you integrate first with respect to y x to 2 the same function and then you integrate with respect to x because for a given x. So, this is how when you have the diagram you can immediately see that given a value of x the corresponding value of y can range from x to 2 and that is it. You can compute this way right. Now, just look at another example and maybe we do not need to compute it fully, but I will give you an idea here. This is f x y is defined as c into y square minus x square into e raise to minus y. Now, the limits for x are from minus y to y and y varies from 0 to infinity. So, therefore, to draw the diagram see what I said is that when y is fixed then my x is varying from minus y to y. So, it has to be between the lines this is x equal to y and this is x equal to minus y right or minus x equal to y whatever whichever way. So, this is this and this is this this line. So, your x is in between these two lines and your y varies from 0 to infinity. So, it is this region extending to infinity right this is the region. So, once you know this then there is no problem because anyway you was first want to find out the value of c. So, find value of c such that this defines a pdf. So, then you can integrate and you see the what will happen is that you come up to here. Now, this is y cube e raise to minus y. So, you will have to you know do integration by parts and it will have to be done three times because you know you this will be your first function this is second. So, you will have to in the first iteration you will get y square then you will have to do it again to get y and then get rid of the y. So, therefore, it will be three times you will have to repeatedly apply integration by parts to get the value of c. So, that this integral finally, has to be equal to 1 and this yeah. So, once you have the diagram in front of you you cannot go wrong you can always find out the correct limits and then decide how to you know like what I was doing is I was dividing the region in a certain way to do the integration which you have already done in your while you were doing your calculus course.