 I will continue with the example, I was discussing at the end of lecture 17. So, this was the conditional pdf of x 1 given x 2 equal to x 2, and we were given the marginal of x 2 also, then we had to find constant C 1 and C 2. Of course, the criteria is that they should they are pdf. So, the integral in the specified region must be 1. So, we computed we made this computations in the last lecture. So, by saying that this integral 0 to x 2, because given an x 2 when you do not draw this your x 1 will vary from 0 to x 2, because this is the region of integration. So, 0 to x 2 integral this equal to 1 implies C 1 is 2. Similarly, by integrating this from 0 to 1 because this is the marginal of x 2, then this comes out to be this gives us C 2 equal to 5. So, the joint pdf of x 1 and x 2 will be the conditional of x 1 given x 2 into the marginal of x 2. So, the product of the 2 which we have already have now with us since we have computed C 1 and C 2. So, that will be the joint pdf of x 1 and x 2 and that we also computed as 10 x 1 x 2 square, where the range for x 1 is from 0 to x 2 and x 2 varies from 0 to 1. Now, is this a pdf surely, because it is a product of 2, this we have verified is a pdf, this we have verified is a pdf. So, the product must also be a pdf. So, there is no need to verify this again though if you want you can integrate for respect to x 1 from here to here and for x 2 from 0 to 1 and you can show that this integrates. So, double integral will come out to be equal to 1. So, now we have to find in the number 3 is compute the probability of x 1 between 0.25 and 0.5. So, to do this I need to compute the marginal of x 1 marginal pdf of x 1 and that will be taking the joint pdf and here you will be integrating with respect to x 2. So, given an x 1 this is the line and therefore, your x 2 will vary from x 1 to 1. So, this is the range for x 2. So, x 1 to 1 you integrate the joint pdf to obtain the marginal pdf of x 1 and so this comes out to be. So, you integrate in respect to x 2. So, x 2 cube by 3 10 x 1 and so this is the expression for the marginal. So, once you have the marginal for x 1 then you want to integrate again the marginal from 1 by 4 to 1 by 2 to obtain the probability and the number that I get is this. So, may be you need to simplify this further and get the right answer. Then fourth one required you to find the conditional probability of x 1 given x 2 is 5 by 8. So, this is from again 0.25 to 0.5. So, the conditional density function we already have which is C x 1 upon x 2 square. So, x 2 is given to be 5 by 8. So, we have to integrate this from 1 by 4 to half to x 1 upon 5 by 8 whole square d x 1. So, here this is a simple this thing x 1 square by 2 this is 1 by 4 to half and what you get is 12 by 25. But now, since we have also talked of conditional expectation I thought we will include that part also here. So, for example, you are asked to find the expected value of x 1 given x 2 is equal to x 2. And so you will find out the expectation here that means it will be the joint p d f it will be the conditional p d f of x 1 given x 2 equal to x 2 which is 2 times x 1 upon x 2 square and since you are finding the expectation here with respect to x 1. So, it will be x 1 into this. So, this is the integral and of course, x 1 varies from 0 to x 2 and you are integrating with respect to x 1. So, 2 by x 2 square comes outside and then this will be x 1 square. So, which is x 1 cube by 3 and 0 to x 2. So, you simplify and this is the expression 2 by 3 x 2 which is the function of x 2. So, which is to be expected because in this integral you are integrating with respect to x 1 and the limits are from 0 to x 2. So, this and since x 2 is given to be x 2. So, this will turn out to always be a function of x 2 when you are finding expectation of x 1 given x 2 equal to x 2 which I mentioned yesterday also when we defined and now this is a straight line passing through the origin. So, the expectation that comes out is a function of x 2 and this is a straight line passing through the origin. So, that means here the relationship is linear that means the expectation of x 1 given x 2 equal to x 2 is a linear function of x 2. Then therefore, follows that expected value of x 1 given x 2 capital x 2 will be 2 by 3 x 2 will be a random variable. See the whole idea here is and this was repeated earlier also what we are trying to say here is that for a particular value of x 2 this is what you get the expected value of x 1 slash conditional x 2 equal to small x 2. Now, as values of x 2 vary then this becomes the expected value of x 1 condition on the random variable x 2 and so the notation is that we call it 2 by 3 capital x 2. So, this is a random variable and so for different particular values of x 2 we will get the expected value from this formula this is the whole idea right. And therefore, since this is now a random variable we can again talk about the expected value of here and this I was I had done this even earlier in the last lecture also the same thing. So, therefore, expectation of expectation of x 1 conditioned on x 2 will be just 2 by 3 into expectation of x 2 which will by the formula would be you know integral 2 by 3 0 to 1 x 2 f x 2 x 2 d x and this will turn out to be the expected value of x 1. So, when you take first expectation of x 1 condition on x 2 and then again and that comes out to be a function of x 2 then you take expectation again and then you will get the expectation of x 1. So, this is idea been repeated in the last lecture and I have asked you to verify. So, that means you will have to compute the marginal of x 1 and then which I think you have already here yes that is a conditional pdf. So, you will have to compute the marginal of x 1 and then find out the expected value of x 1 independently and verify that this it comes out to be the same what you will get from here. And will further I mean I will continue with this concept and take another example to make this thing things clear try to do it. So, this was an example I got from the net roll a die until we get a 6. So, this is the experiment you continue rolling a die until you get a 6 and let y be the random variable which is equal to the total number of rolls till a 6 comes up. So, you continue rolling the die till a 6 shows up and then you stop right. So, and x is the number of once we get in this process. So, while you are rolling the die you keep noting also the number of times 1 appears. So, and then of course, the experiment stops the moment a 6 comes up. So, x is the number of once you get in this process right. Now, you are asked to compute expectation of f x given y is equal to y. That means, the number of dies number of times you had to roll the die is small y. And then this process you want to compute the expectation of x given that you had to roll the die y number of times. So, y capital y equal to small y means that there were y minus 1 rolls of the die which were not a 6. All other numbers appeared, but the 6 did not appear. Because the experiment will stop the moment a 6 appears right. So, y minus 1 rolls did not show a 6. Now, therefore, you see an x is the number of once that come up before the experiment ends. So, that means, in the string that you get the y minus 1 rolls that you made of the die x is the number of once that show up. So, when we can treat x as a binomial random variable with the number of experiments of y minus 1 and the probability of occurrence of 1 is 1 by 5. Because remember up to y minus 1 rolls 6 is not appearing right. So, and the other 5 numbers are equally likely. So, the probability of a 1 showing up is 1 by 5 and the number of experiments that you make is y minus 1 right. And so, we will treat occurrence of 1 as a success and occurrence of 2, 3, 4, 5 as a failure. And therefore, x can be treated as a binomial random variable with n equal to y minus 1 and p equal to 1 by 5. So, you see the whole idea is that you can how you can compute certain quantity that you require by just realizing how the experiment has been conducted. So, therefore, immediately you can say expected value of x when y is given to be small y is similarly, because for a binomial random variable the expected value is n p. So, n here is y minus 1 and p is 1 by 5. So, without any hassle we get the expected value the conditional expected value of x given y is equal to y. Now, this is again a function of y that is as y takes values possible values 1, 2, 3 and so on. The points 1 by 5 y minus 1 will lie on a straight line expected value of x by y that means conditional expectation of x given y capital y is a random variable. Now, conduct the experiment and get an outcome omega that is omega is a string of 1, 2, 3, 4, 5 ending with a 6. So, like we conduct the experiment we keep rolling the die till we get a 6 and we record the outcomes at each roll of the die. So, it will be a string of these numbers ending with a 6. So, omega is that then you compute how many numbers are there in the string. So, that means how many times we had to roll the die. So, now you can say that y small y is the value of capital y at omega because omega represents a string and y counts the number of elements in the string. So, this is this is actually our relationship small y is capital y at omega. So, y is the number of rolls of the die to get a 6. So, y omega is a random variable surely because this can go on depending I mean when the 6 shows that is not a certain event. So, this is a chance element here and therefore, y omega is a random variable and then you compute the expectation of x given y equal to y which we did right now. And so that means you are relating omega with this because given an omega you computed the y and then you compute the expected value of x given y equal to y. So, that is expectation of x given y equal to y is a mapping that maps omega to 1 by 5 y minus 1 and remember I defined random variable also as a mapping because I said random variables associate with sample space real numbers. And so here also what you have happening is that this expected value of x given y equal to y is a mapping that maps omega to this where y is capital y omega. So, therefore, this is therefore, omega is map to 1 by 5 y omega minus 1 which is a random variable. So, now when you write capital y omega so this is I mean the idea was to explain to you again in a different way why we are saying that this is a random variable though it should be clear because as values of y change there is probability associated with what value y takes right. And therefore, this is again a random variable you can see you can go ahead and make some more computations for example, if you look at variance. So, again the conditional variance of x given y equal to small y I know because I have said that x is a binomial random variable. So, this number I know and for this number also I know. So, if you want to compute expectation of x square given y equal to y then this minus this is equal to the variance right. And so and therefore, I can say that yes and this is equal to n p q by the binomial formula. So, p q and n so 1 by 5 into 4 by 5 into y minus 1. So, 4 by 25 y minus 1 is the variance and expectation x given y equal to y we have already computed as 1 by 5 of y minus 1. So, therefore, expectation of x square given y equal to y is variance plus this which is 4 by 25 y minus 1 plus 1 by 5 y minus 1 whole square which is 1 by 25 into y minus 1 whole square. And so when you simplify this expression you get this y minus 1. So, this is a quadratic function of I should say here quadratic function of small y and so random variable represented by expectation x square given capital Y is a random variable which is this. So, this becomes a quadratic function of y. So, I hope this gives you a little insight into what we mean by and so and now another role that the conditional expectation plays is as a best approximation. Conditional expectation as a best approximation and I thought we should talk about this to give you some more feeling and this is value of a random variable x is observed. So, suppose a value of a random variable x is observed and based on this observed value an attempt is made to predict the value of a second random variable y. See sometimes it may be easier to observe value of a certain random variable and then if based on that observed value you can make some attempt to predict value of another random variable that helps you because then you do not have to you know conduct another experiment to obtain the value of y. But of course, the value that you predict from knowing the value of x may not be the exact one. So, let g x denote this predictor. So, suppose so it is some function of the random variable x. So, this denotes this predictor. So, that is x equal to x is observed and then g x. So, g of small x is our predictor for the value of y. So, for one value of y and then you observe another value of capital x then g of that value of x will give you another predictor for the value of y. So, this is the idea. Now, how to choose g because you have to have some concept as to what is the g that is acceptable to you. And so of course, the quality that g must possess is that it should be as close as possible to y. So, now the whole idea is what do you mean by closeness here and how can we define and of course, later on also when we talk of limits and so on and convergence and probability and law these things will become more clear. But right now let me say that our criterion is to minimize expectation of g x minus y whole square. So, if this is my criterion then I want to choose that g which minimizes this expression this expectation g x minus y whole square. So, expectation of this should be as small as possible. Now, we will show that g x equal to expectation y conditioned on x is the best choice. So, this is the whole idea and therefore, you see another role that the conditional expectation plays. And the proposition is so we want to show that expectation of y minus g x whole square is greater than or equal to expectation of y minus expectation y given x whole square. So, then this will establish that this is the smallest value of this and therefore, the best choice for g is expectation of y given conditional expectation of y given x. So, we will just prove this proposition for you. . So, before we prove this proposition I will like to state a theorem. In fact, the proof is also straight forward. So, the theorem says that if x and y are to jointly distributed random variables and h x is a function of x whose expectation exists. So, then if you have the conditional expectation of f x given y and then we take the expectation again this would come out to be expectation of h x. Now, this is on the same lines as if you say that expectation of x given y and then you take the expectation again. So, this is E x remember we have already shown this result. So, on the same lines we are trying to show that expected value of h x conditioned on y and when you take the expectation again it will come out to be the expectation of h x provided of course, the expectation of h x exists. And so I have just written down for the continuous case I have just written down the expression for expectation of the expectation of conditional of h x on y. Then it will be minus infinity to infinity minus infinity to infinity h x. The conditional p d f would be f x y x y divided by f y because this is conditioned on y and then when since you have this turn out to be a function of y. So, when I take the expectation again it will be f y y d y d x and this you can see that this will cancel out and it will get expectation of h x. Now, similarly the result can be stated that if you condition you take a function g of y and then condition on x then just reverse the roles of x and y and then you will get here expected value of g y. That means your expected g of y conditioned on x this will turn out to be expected value of g y. So, provided of course, expectation g y exists. So, this is the proof I mean in case x and y are discrete random variables you can just imitate the proof for the continuous case. So, once this theorem is there we will be using it to prove this proposition. And so let me now consider the proof of the proposition. So, this is I am considering y minus g x whole square conditioned on x and because in my proposition I am taking condition on x. So, now add and subtract expected value of y by x within the brackets. So, then this is expected value of y by x plus expected value of y by x minus g x whole square conditioned on x open up the brackets. So, then this will be y minus e y by x whole square conditioned on x the expectation of that plus the square of this expected value of y by x minus g x whole square and conditioned on x expectation plus twice the product of these two terms. And so we would say the conditioned on x separately and then expectation that. So, twice e of that. So, let me call this equation as the star denoted by star. Now, see for a given value of x equal to x this will be a function of x remember we have shown you that the expectation of y conditioned on x. So, for every fixed value of x it will be a function of x and so for the purpose of you know taking this expectation because that this expectation will be respect to this will be. So, therefore, I can treat this as constant and so this will come out of the expected sign. So, this is a function of x and hence can be treated as a constant. So, because I will be computing it for different values of x this thing and so for every fixed value of x this will be a constant. So, I can take it out of my expectation and so therefore, y minus e of y by x conditioned on x into expectation of this I can take this outside and then this will be expectation of y minus expectation y by x conditioned on x. Now, here again you see when you bring x inside it will be y by x y conditioned on x and this does not change because this is already conditioned on x. So, this will become expectation of y by x I mean this of course, is outside. So, this portion will be expectation of y by x minus expectation of y by x which is 0 and so therefore, the in the star expression this portion has no contribution this is equal to 0 and so the right hand side this reduces to simply the expectation of y minus expectation y by x whole square conditioned on x plus expectation of y by x minus g x whole square conditioned on x. So, this is what you have and since for a given value of x y minus g x is a function of y therefore, by the theorem above see this will be a function of y only because this will be a constant for every fixed value of x and so by our theorem expected value of expected value of y minus g x whole square conditioned on x. So, for different values of x this will be a function of y only. So, for each fixed value of x it will turn out to be expected value of y minus g x whole square and then so for different values of x the whole thing will become y minus g x expectation of y minus g x whole square. This is the same concept is getting over and similarly, expectation of y minus expectation y by x whole square conditioned on x will turn out to be expectation of y minus expectation of y by x whole square. I mean this whole square and then the expectation is what we are doing and so you get that when you know you have this because this is equal to 0. So, I got yeah so where is the greater right. So, the right hand side when I take the expectation I get see we started out from here we started out from here and then since this got 0 then I took expectation of both the sides. So, this became expectation of y minus g x whole square and this became expectation of y minus expectation of y by x whole square plus expectation of y by x minus g x whole square. So, the conditioned thing disappeared and since these expectations of squares are non-negative. So, therefore, I can this because this portion the expected value of y by x minus g x whole square this will be non-negative. So, therefore, my equality will convert into an inequality and is what you have. So, this is what we are trying to prove right that you know for any function of g x for any function g of x if you take this expectation y minus g x whole square here again if you want you can put this then this will always be greater than or equal to expectation of y minus expectation y by x whole square expectation y by and this whole square expectation of that. I should first say y minus expectation y by x whole square and expectation of that. So, this is what we wanted to prove. So, in other words you see the process that you are observing value of x and then computing the expectation y given x and in the earlier lecture when I had considered defined this conditioned expectation and conditional expectation and taken this discrete case. So, I showed you how you change keep changing values of x and then you will compute the expectation of y given x right. So, this in a way we are treating as a good approximation for the value of y and when we are when our criteria criterion is in terms of minimizing this expression then obviously no other function of g x will qualify to be the best predictor except for the expectation y given x. So, in this way we are treating or showing that this can also be looked upon as a very good approximation for the value of x. So, for different values of x we will compute this I mean we will observe possible values of x all possible values compute this and this will give us the different values for of y. So, that means you can compute this expectation for well yes there is still a few questions which are not answered, but and hopefully I will continue looking at this again. Now, take this example you can show that for any real number a in fact what we have shown here can be also you know this you can show very easily that expectation of x minus e x whole square is always less than or equal to expectation of x minus a whole square for any real number a. And you will do the same thing here you will write x minus e x plus e x minus a open of the square and so on and then again one part when you look at this thing e x minus a. So, this is the constant and therefore, when you are taking the expectation of the product term this will come out and see the second factor will be expectation of x minus e x which will be 0 and suppose the same reasoning you can show that this is always less than or equal to this. So, you can do this for any real number a. Now, look at the example of a bivariate normal distribution of x and y and we have already looked at this p d f which is you know in terms of sigma x sigma y and rho. So, these are correlated because I mean I am just taking the general case x and y are not necessarily independent. So, this expression in an exercise 10 in exercise 5 of and question 10 I ask you to show that the conditional distribution of x given y equal to y is the is again a normal p d f with mean this and variance equal to rho square you know why am I writing rho square they should be sigma square sorry. So, this will be sigma x square into 1 minus rho square. So, this is your mean and that is your variance. Now, therefore, expectation of x given y equal to y will be this mean because this is for capital Y equal to small y. So, this is mu x plus rho into sigma x upon sigma y y minus mu y which is again a linear function of y. So, what we have found is that when x and y have a bivariate normal distribution the best overall predictor because according to our proposition of x with respect to y turns out to be mu x expected value of x condition on y is mu x plus rho sigma x upon sigma y into capital Y minus mu y and this turns out to be linear in y. So, that is all we can say that the best overall predictor in case x and y are bivariate normally distributed then the best overall predictor of x with respect to y is this expression which is the conditional expectation of x given y it turns out to be this and this is linear in y. So, this is all what we can say to talk of best linear product linear predictor and so on that is another thing that we have to go about it in a different way and I have to first define what we mean by a linear predictor and so on. So, right now all we are saying is that the best overall predictor in case x and y are both normally distributed that means they have a bivariate normal distribution then the best overall predictor would be linear in y. And similarly, if you were talking of the best overall predictor of y with respect to x then that will be linear in x. Now, let us just further continue talking about conditional expectation. So, now we will show that expectation x y is expectation x into expectation of x expectation of y given x. So, this is what a simple calculation we will do now here of course, I have not written out all the steps. So, for example, when you want to write expectation x y. So, actually the starting expression will be x y into the joint d x d y. So, this is the thing, but as we have seen that the joint pdf of x and y can be written as product of marginal of x into the conditional of y given x. So, therefore, this is what I am writing here. So, expectation x y I am writing as x y f x x into f of y given x d x d y. Now, the thing is that so y integral y into f y by x conditional pdf of y given x. So, this I can separate out and I can write this as y because this is not of when I integrate this is given value of x is fixed here f y given x. So, x capital x equal to small x. So, I integrate this with respect to y and then so I can just separate out this double integral into minus infinity to infinity x f x x and then this will come out to be a function of x. So, then the whole thing I will integrate as a function of d x. So, you can immediately see that this is expectation of y given x and then x times f x this gives you expectation of x into expectation of y by x. So, simple calculation, but so again just emphasizing the fact that this is a random variable and therefore, becomes a function of x and so you compute this expectation again and you get this. So, you know you are using conditional expectation to compute expectation. Now, let us go through this exercise of computing row and of course, for different situations you can use different techniques to handle it. So, now here if you are given that x and y have a bivariate normal distribution. So, this is mu x mu y sigma x square sigma y square and row. So, this is your bivariate normal distribution and you are given that row is greater than 0. So, you have to find the conditional expectation of y between 4 and 16 given x is equal to 5. So, I mean you are given that this probability is equal to 0.954 you have to determine row. So, the row is the unknown here and therefore, you want to determine that and now from our this result that x given y is this. So, what will be this thing? So, from here only or we can write here that expectation of y given x this will be mu y. So, just replace x by y mu y plus row sigma y by sigma x and then this will be x minus mu x. So, this is the formula which I have written it here already. So, this is mu y plus row times sigma y upon sigma x into x minus mu x that will be this and then since x is given to be 5 and mu x is also 5. So, therefore, this part is 0 and so expectation y when x is equal to 5 is 10. So, the conditional expectation of y or the mean is 10. So, y by x is normally distributed with mean 10 and the formula for the variance would be yeah. So, variance formula for the variance y given x equal to 5 is row y square into 1 minus row square which is 25 times 1 minus row square. So, this is the variance therefore, this is what I have written here. So, this is normally distributed with mean 10 and variance 25 into 1 minus row square. So, therefore, I will standardize usual thing that we do. So, now computing this probability I will standardize the variate here which means I will subtract 10 and divide by the oh yeah. So, divide by the standard deviation. So, standard deviation is 5 under root 1 minus row square is what you have and so this becomes 6 by 5 under root 1 minus row square. So, this is the CDF for this probability less than or equal to this. So, therefore, I mean the cumulative density function for the standard normal. So, this minus this is minus 6 this and again from symmetry of the standard normal distribution around the origin we can write this as twice 5 of 6 upon 5 under root 1 minus row square minus 1 because this can be written as 1 minus of 5 of 6 upon 5 under root 1 minus row square minus sign outside. So, therefore, is what you get and you have given this is equal to 0.954 or 6 upon this is equal to 1.954 by 2 which is 0.977. So, if you look up these standard normal tables the value of z which corresponds to 0.977 is 2 that means 5 of 2 is 0.977. So, this you can obtain from the tables for the standard normal and so this means that this number 6 by 5 under root 1 minus row square should be equal to 2 and so from here you do a simpler arithmetic square everything. So, you get 100 by 36 or 1 minus row square is 36 upon 100. So, row square is this so absolute row is 0.8, but we are given row to be positive and therefore, we will take the positive value from here which gives you row is 0.8. So, that means x and y are positively correlated if row was minus 9 row was minus 0.8 then we would say that x and y are negatively. So, the relationship between this of course, also throughout this we have also been able to establish that your correlation coefficient of the covariance can measure effectively linear relationship between the variables, but it fails to show you quadratic relationships and so on. So, we saw that right. So, this comes to sort of our treatment of conditional expectation and how this can be used for computing various things. In fact, for computing your actual this thing because we have shown that this result also expectation is expectation x and so on. So, and then here this result we have obtained for you and then I have also shown you the role of conditional expectation as a best predictor. So, this is it and let me now discuss exercise 6 with you which is related to what all we have discussed in the last 2 lectures last 3 lectures. So, a fair die is rolled 3 times fine it should be a full stop instead of comma. Let the random variable x be let the random variable x i be equal to the number that appears on the ith trial for i varying from 1, 2 to 3 and then we define y as the max of x 1, x 2, x 3. So, the largest number so up when the die has been rolled 3 times then you notice you record the numbers that show up and then you take the maximum one. So, y is the value of the maximum of x 1, x 2 and x 3 find distribution function and probability density function of y. So, distribution function means cumulative distribution function and probability density function of y. So, this question 1, question 7 consider the joint probability density function of x comma y given by f x y of x comma y. So, here it should be x and y as the suffixes are the capital 1 and small x y is equal to 2 minus x minus y x between 0 and 1 and y between 0 and 1 0 otherwise. Now, here part 3 I want you to find. So, part 1 says find the conditional probability density function of y given capital x equal to small x and then in part 2 I want you to find the expected value of y given capital x equal to x and e y. Then 3 I want you to find out e y through I want you to show that e y is actually equal to e expectation of, conditional expectation of y given x. So, what we have been doing in a lecture also we have been verifying you know we have been computing e y independently by first computing the marginal of y and then it is expectation and secondly by breaking it up into first the conditional probability a conditional expectation of y given x and then we take the expectation again. So, you have to say that the two processes lead you to the same answer. Now, exercise question 2 I already discussed with you in the lecture. So, you can have an alternate expression for the correlation coefficient when x and y are given to be two random variables. Question 3 is that x and y have the joint probability density function f x y equal to 1. So, y lying between minus x and x and x between 0 and 1. So, please be careful when you draw the boundaries for y because you see y 1 boundaries y equal to minus x and the other is y equal to x. So, that means your y varies from I hope you can just make sure that. So, you can see along the x axis you will have one line y equal to x and the other will be in the fourth quadrant y equal to minus x and therefore, your y will vary from minus x to x. So, this is what you have to be careful and then you have to draw the graph of. So, given the joint density function you will compute the conditional because you have to draw the graph of expectation of y given x equal to x as a function of x. So, you know how to do it right you have to compute the conditional problem pdf and then compute the expectation of y given x equal to x and also you have to draw the graph of. So, find out both the conditional pdf y given x and conditional pdf x given y as a function of y. So, you get some feeling about the question 4. So, suppose the conditional probability density function of x 2 given x 1 equal to x 1 is given by this function this is the conditional pdf. So, x 2 positive and it is 0 otherwise. So, the region in which it is defined. So, x 1 equal to x and that f x 1 marginal of x 1 is also given by this function where x 1 is positive. So, both the variables are supposed to be positive take positive values and. So, again I want you to find out expectation of x 1 given x 2 equal to x 2 and then also find expectation x 1 and correlation x 1 comma x 2. So, should be able to do it because you have all the tools and this thing. Question 5 x 1 comma x 2 be a two dimensional discrete random variable or two discrete random variables with joint probability function. So, now you have to compute correlation x 1 x 2 and r x 1 and x 2 independent random variables. So, remember even if your correlation coefficient is 0 it will not necessarily imply that x 1 and x 2 are independent. So, to verify you will have to compute the you know show that for or find at least one pair of values of x 1 and x 2 for which the probability x 1 comma x 2 probability of x 1 equal to that number and x 2 equal to a particular number is not equal to the product of individual probabilities. If you can show that then you can conclude that x 1 and question 4 not independent, but otherwise you will have to go on verifying for all possible pairs which means you have 8 pairs. So, for 8 pairs if you can show that the product of the probability of the product is equal to individual product of the individual probabilities then you can conclude that they are independent for the for the discrete case this is the only way you can do it. Question 6 using the result that we just obtained this result for I just obtained it for you that expectation and sorry here they should be no comma. So, expectation of x y. So, please remove the comma expectation x y is equal to expectation x into expectation of y given x. So, I just prove this result for you now using this result show that covariance again here this is x comma expectation y given x is covariance x comma y. So, the comma in the last the two terms is, but here when you saying the result is expectation x y. So, comma is to be removed and then you can therefore, you can use this result show this result using what we have proved just now. Question 7 consider the joint probability density function of x comma y given by f x y of x comma y. So, here it should be x and y as the suffixes are the capital 1's and small x y is equal to 2 minus x minus y x between 0 and 1 and y between 0 and 1 0 otherwise. So, part 1 says find the conditional probability density function of y given capital x equal to small x and then in part 2 I want you to find the expected value of y given capital x equal to x and e y. Then 3 I want you to find out e y through I want you to show that e y is actually equal to e expectation of conditional expectation of y given x. So, what we have been doing in a lecture also we have been verifying you know we have been computing e y independently by first computing the marginal of y and then its expectation. And secondly by breaking it up into first the conditional probability a conditional expectation of y given x and then we take the expectation again. So, you have to say that the two processes lead you to the same answer. Question 8 is x y z are three random variables and n a and b are two constants prove that covariance of x comma a y plus b b is this. So, I had done it for you when the constant a was with x now you please do this should not be because remember in fact you can immediately do it because covariance x comma a y plus b is covariance a y plus b comma x. And therefore, from that result but then I have added a plus b here. So, please work it out and show that this result is true then the correlation coefficient of x comma a y plus b there will be no a because you see numerator there will be a from here. And then in the denominator also when you take the variance of a y plus b a will come out and so the a a will cancel. And of course, a has to be positive here because in the variance you will take out a only if a is positive otherwise you have to take out absolute of a. So, let x 1 x 2 x 3 be three independent random variables each with variance sigma square. So, they are three independent random variables with the same variance if we define new random variables. So, here w 1 is x 1 w 2 is root 3 minus 1 upon 2 of x 1 plus 3 minus root 3 upon 2 of x 2 and w 3 is a linear combination of x 2 and x 3 show that correlation coefficient of between w 1 and w 2 is equal to the correlation coefficient between w 2 and w 3 which is equal to half while w 1 and w 3 are uncorrelated. So, I just tried to I mean the purpose of giving this exercise was that you see that taking this linear combination some turn out to be correlated and some pair terms out to be uncorrelated. So, this is what you have to show. Now, tenth question is a fair die is successively rolled and let x and y denote respectively the numbers of rolls necessary to obtain a 6 and a 5. So, in fact the 6 part we discussed to at length and now so you are asking for. So, x is the number of rolls required when you till a 6 shows up and y is the number of rolls required till a 5 shows up. So, now find expectation x that means you will write down the probability mass functions for x and y and then compute e x compute conditional expectation of x given y is equal to 1 and conditional expectation of x given y is equal to 5. So, it should be easy computations once we have already handled the case when the you had to roll the die till a 6 showed up.