 As Salaamu Alaykum, welcome to lecture number 23 of the course on statistics and probability. Students, you will recall that in the last lecture, I discussed with you the Bayes theorem, quite an interesting theorem in which we deal with conditional probabilities. After that, I began the discussion of discrete probability distributions. A very important topic and in this regard, I discussed with you the graph of the discrete probability distribution, the mean, the standard deviation and the coefficient of variation for a discrete probability distribution. After that, we talked about the concept of the distribution function. Today, I will elaborate upon the concept of the distribution function and I will explain to you the graph of the distribution function of a discrete random variable. Let us do this with the help of the example that you now see on the screen. Find the probability distribution and the distribution function for the number of heads, when three balanced coins are tossed, depict both the probability distribution and the distribution function graphically. In order to solve this problem, the first thing to note is that because the coins are balanced, therefore, the equi-probable sample space for this experiment is head head head, head head tail, head tail head, tail head head, head tail tail, tail head tail, tail tail head and tail tail tail. Students, if we let x denote the number of heads that we obtain in this experiment, then obviously the values of x are 0, 1, 2 and 3 and the probabilities are 1 by 8, 3 by 8, 3 by 8 and 1 by 8. For the same reasons that we have discussed in earlier lectures, the classical definition of probability m over n gives us all these probabilities. And of course, the sum of these four probabilities is 1, so that we have a proper discrete probability distribution as you now see on the screen. If we are interested in drawing the line chart of this distribution, we get a graph of the form that you now see and it is very clear that this particular line chart is absolutely symmetric. Therefore, if we are interested in finding the mean of this distribution, students, I hope you realize that you do not even have to go for the formula. So, in this example, the x values are 0, 1, 2, 3 or the exact center that is 1.2, 0.5. If you use the formula, which we have discussed last time, that mu is equal to sigma x into p of x, then you will get exactly the same result. If you remember, I told you last time that instead of p of x, probabilities can also be denoted by small f of x. And the formula for mu can be written as sigma x into f of x. Today, we want to talk about distribution function. So, as you now see on the screen, if we cumulate the column of probabilities, we get first value 1 over 8 as it is, 1 over 8 plus 3 by 8 gives us 4 by 8, 4 by 8 plus 3 by 8 gives us 7 by 8 and 7 by 8 plus 1 by 8 is 1. Hence, the distribution function of this particular probability distribution is capital F of x is equal to 0 for all those x values, which are less than 0. It is equal to 1 over 8 for all those x values, which lie between 0 and 1 including 0, but excluding 1. It is 4 by 8 for x lying between 1 and 2, including 1, but excluding 2. It is 7 by 8 for x lying between 2 and 3, 2 included, but 3 excluded and it is equal to 1 for all values of x greater than or equal to 3. The question arises that, the simple table, which was cumulated simply the first value as it is and then added and added further, why is it that I have expressed that simple thing in such a complicated form students. I will try to explain it to you in a very, very simple manner and I hope that you will concentrate. I have said that capital F of x is equal to 0 for all values of x, which are less than 0. From the basic definition that I discussed with you last time, what was it? That it is the probability that my random variable x takes a value less than or equal to some specified value x, small x or capital X. Now, when I am saying that capital F of x is 0 for all those x values, which are less than 0, it means that probability is 0 for x to be less than 0. Any individual probability is 0 and if we accumulate it is 0. Well, that is obvious. When we are tossing 3 coins, the minimum number of heads that can come is 0. We cannot have minus 1 heads, minus 2 heads, minus 4 heads and even if we write these values in our column which we had written as 0, 1, 2, 3, we would have written minus 1, minus 2, minus 3 and so on. If we want to write probabilities against them, then what was it? 0, 0, 0, because minus 4 heads cannot be written. It is an impossible event and the probability of an impossible event has to be 0. If we make that 0's column for probabilities, then against the capital F of x column, which we had to accumulate, then students, if we write 0, then 0 plus 0 would have given us 0 and again 0 plus 0 would have given us 0 for all those x values which are in the negative part of the real line. So, this is the reason why we express capital F of x in this manner. Now, as you have noticed, I would like you to look at the screen once again and try to concentrate on all those parts that we have for capital F of x. As you can see, for the range x lying between 0 and 1, capital F of x is equal to 1 by 8. We are saying that 1 by 8 is the value of capital F of x, not just for one isolated x value, but for all those x values which are lying between 0 and 0.99999. Come, let us try to understand it. I have written a few values between the values 0 and 1. I have written 0.2, 0.4, 0.6 and 0.8. Obviously, I could have written any value between 0 and 1, but the point to note is that against all these values, the probabilities are 0. Now, if you concentrate on the third column, students, that is the column of cumulative probabilities. The first value 1 by 8 as it is 1 by 8 plus 0 gives you 1 by 8, 1 by 8 plus 0 again gives you 1 by 8, 1 by 8 plus 0 yet again gives you 1 by 8. You can see that between 0 and 0.99999, you can write any value against capital F of x will come out to be 1 over 8, but the moment you arrive at the value x equal to 1, your cumulative probability becomes 1 by 8 plus 3 by 8 and that is 4 by 8. This is exactly the reason why the graph of the distribution function of a discrete random variable is of a very interesting shape. As you now see on the screen, this particular graph is like a staircase and because of this fact, it is also called a jump function or a step function. The graph of the distribution function starts from minus infinity and goes up to plus infinity and as you can see on the slide, the level of the graph is level 0 for all values of x less than 0. As soon as we arrive at the point x equal to 0, the graph takes a jump and it reaches level 1 by 8 and it maintains this level for all values of x from 0 to 0.99999 and so on. But the moment we arrive at the point x equal to 1, the graph takes a jump again and it reaches the level 4 by 8. It maintains this level until the value 1.99999, but then as soon as we arrive at the point x equal to 2, our graph jumps again and reaches the level 7 by 8 and lastly, it takes the jump when we arrive at x equal to 3 and now it reaches the level 8 by 8 and that is 1. Yani, saari baat ka lubbe lubab yeh ke minus infinity se shuru hua or plus infinity tak jaega aur darmean mein steps ki tara yeh chalega aur jumps lagaega at all those x values where we actually have non-zero probabilities. Students, iske baraks, the graph of the distribution function of a continuous random variable is a smooth curve which rises from level 0 to the level 1. Iske esme bhi yeh hamesha yad rakhe ke jo lowest level hoga that will be 0 or jo highest level hamara distribution function ka graph attain kar sakta hai that is the value that is the level 1. Once yopar yeh jaan nahi sakta. After all, have we not discussed so many times that the sum of the probabilities is 1 and it can never be more than 1? Students yeh jo graph mein abhi aapko dikhaya jaisa ke aapne dekhah ke wo jumps to hai unke darmean koi hamne line vagara draw nahi ki thi. Lekin, in certain books you will find a dotted line in the vertical direction as you now see on the screen. These vertical dotted line segments do not have any significance from the mathematical point of view. In fact, from my point of view it is better not to draw them. Aayi aap yeh dekhte hain ke ham iss distribution function ko real life perspective ke hi saap se kista se interpret karenge. Zahirei ke iss ka yehi muffoom bane ga, ke if we toss three balanced coins the probability of obtaining at the most one head is 4 by 8, the probability of obtaining at the most two heads is 7 by 8. All right, let us now consider another interesting example as you now see on the screen. A large store places its last 15 clock radios in a clearance sale. Unknown to anyone five of the radios are defective. If a customer tests three different clock radios selected at random, what is the probability distribution of X where X represents the number of defective radios in the sample? In order to solve this question, the first thing is to realize that out of a total of 15 clock radios, 10 are all right and 5 are defective. The total number of ways of selecting three radios out of 15 is 15 C 3 according to the discussion that we have already had in an earlier lecture. Also the total number of ways of selecting three good radios out of 10 is 10 C 3. And total number of ways of selecting no defective radio out of five defective radios is 5 C 0. And hence the total number of ways in which the event which is favorable to what I want is 10 C 3 into 5 C 0. Hence the probability of obtaining no defective radio is 10 C 3 into 5 C 0 divided by 15 C 3 and this is equal to 0.26. So, the first is the rule of combinations. And we are only interested in the combination of the radios that we are that we are drawing. Then of course, we will apply the rule of combinations. And the second rule which is applied in this of course, that is the rule of the classical definition of probability M over N favorable over the total. So, the statement of the question is that the radios we are selecting that selection is not deliberate it is at random. Now, what I have just computed is the probability of x equal to 0, where x represents the number of defective radios in my sample of size 3. We will compute the probabilities of x equal to 1, x equal to 2 and x equal to 3. And as you now see on the screen these probabilities come out to be 0.49, 0.22 and 0.02. The sum of these probabilities comes out to be 0.99 or this is only because of rounding error. The line chart of this distribution is as you now see on the screen and students I hope that by looking at this line chart it is clear to you that it is not at all necessary for a discrete probability distribution to be absolutely symmetric. As you can see on the slide the shape of this particular probability distribution is slightly positively skewed. Now, let us talk about the distribution function. Just as I explained in the previous example, we can construct a column of cumulative probabilities. And as you see on the slide we have capital f of x is equal to 0.26 against the value x equal to 0 or us ke baad 0.26 plus 0.49 is 0.75, 0.75 plus 0.22 is 0.97 and so on. If we wish to interpret this third column that we have just constructed the interpretation will be something like this. The probability that I will have at the most one defective radio in my sample is 0.75 and the probability that I will have at the most two defective radios in my sample is 0.97. Alright now that we have discussed in detail the concept of the distribution function students I now begin the discussion of the concept of mathematical expectation. As you now see on the screen, let a discrete random variable x have possible values x 1, x 2, so on up to x n with corresponding probabilities f of x 1, f of x 2, so on up to f of x n such that sigma f x is equal to 1. Then the mathematical expectation or simply the expectation or the expected value of x denoted by E of x is defined as sigma x into f of x. E of x is also called the mean of x and it is usually denoted by the letter mu. You have seen that this is not a new concept. This is exactly what we discussed in the last lecture when we tried to find the mean of our probability distribution. And today also we have discussed the mean of the first example. Yes certainly expected value mean and the question is why it is called as expected value. This is why students that this is that value that we can expect to have if we repeat our random experiment again and again and again and again a very very large number of times. What I am trying to say is take that same example of tossing the three coins. You toss it again, you might get two heads and then toss it again and you might get one head. If you do this experiment a very very very large number of times then on the average in the long run you expect to have per toss 1.5 heads. Now you will say that what is 1.5 heads? But students I have already explained that on the average dare head per toss of three coins come a fume ye hai, ke agar aap in three coins ko ek tafa na toss karein, balki das tafa toss karein to aap expect kar sakte hain ke dare nahi, pandra head aapko milenge in all. Gohi baat ke aap us decimal ko shift karne ek step aage. So, you get 15 heads in 10 tosses or 150 heads in 100 tosses. So, this is the concept of mathematical expectation. The value that we expect to have on the average if we repeat the experiment a very very large number of times. Let me explain this to you with the help of a very interesting example. As you now see on the screen suppose we have a situation in which we can say that if it rains an umbrella salesman can earn thirty dollars per day. If it is fair he can lose six dollars per day. What is his expectation? If the probability of rain is 0.3, students aap uske andar doteen points dekhne baale hain. Pehli baat to ye ke shayad aap haraan ho rahe ho ke how can he lose six dollars? Well it is obvious agar wo investment kar chuka hain and he has already spent a certain amount of money on buying those umbrellas that he is wanting to sell. To agar wo suraj nikla wa hain aur bilkul baar chhe nahi ho rahe hain to zair hai ke wo chuke wo spend kar chuka hain. Usse saab se we can say that he will lose a certain amount of money. Dhusri baat hi hai ke humne isme abhi padha ke hum kair hain ke what is his expectation if the probability of rain is 0.3. Zair hai ke ish shakhs ke point of view se to expectation joh hai wo paeson ke hi ke hawale se hain hain. Ke how much can he expect to earn from his sale selling his umbrellas and the information that we have is that the probability that it will rain on any day is 0.3 and the probability that it will not rain is obviously 0.7 by the rule of complementation 1 minus 0.3 will give us 0.7. To hai ye dekthe hain aap screen par ke ish ke mathematical picture ke abanti hain. As you can see the two mutually exclusive events are rain and no rain and the random variable that we are interested in the amount earned has only two possible values. If it rains x is equal to 30 and if it does not rain x is equal to minus 6. Ye minus 6 maine isli likha ke losing six dollars is equivalent to earning minus six dollars. Now as I have just said the probability of rain or in other words the probability of x being equal to 30 is 0.3 whereas the probability of x equal to minus 6 is 0.7. In order to compute the expected value of x all we have to do is to multiply the column of x with the column of probabilities and doing so we obtain 13 to 0.3 equal to 9.0 and minus 6 into 0.7 equal to minus 4.2 adding the two numbers we obtain e of x equal to 4.8. In other words students hain yeh kare hain ke yeh jo shaks hain jo umbrella salesman hain he can expect on the average to earn 4.8 dollars per day. Lekin yad hain ke yeh concept bohot zyada repetition. So iska yeh matlab hain ke agar wo selling ka yeh process bohot din tak jari rakhe aise bohot se dino tak gin me probability of rain 0.3 he maintain ho to phir at the end of that period he will have earned that much amount which is equivalent to 4.8 dollars per day. So iska matlab yeh hua ke jab kabhi baarish hoti hai he is earning 30 dollars jab baarish nahi hoti he is losing 6 dollars. Lekin at the end of it on the average he is still earning and he is earning 4.8 dollars per day. Eis saari discussion ke andar ek aur point bhi baar important hain. Me aapse yeh pushna chaati hu ki yeh jo earning ki baat hum kare hain x is the number of dollars he earns and we are saying that either he earns 30 dollars or he earns minus 6. Yeh jo variable hain students is it a discrete variable or is it a continuous variable? Eiska jawab yeh hain ke agar hum question ko iss tara se post karein jis tara ke hum ne iss example mein kiya hain then it is a discrete variable. Eis liye ke iss tara ke scenario mein either it is 30 or it is minus 6. Yeh nahi baarish ho rahi hai to 30 hai baarish nahi ho rahi to minus 6. So, we cannot have minus 5 minus 4.5 plus 1.7 or iss haale se the discontinuity is there and therefore, it is a discrete variable. Of course, you can argue ke in a real life scenario aise toh nahi ho satta maybe other values are possible and so on. But of course, the point is that here we only wanted to explain the concept of mathematical expectation in a very simple manner. All right. Now that we have discussed the mathematics, we will now move on to the next slide. So, we will now move on to the next slide. So, we will now move on to the next slide. All right. Now that we have discussed the mathematical expectation of x, I would like to draw your attention to the fact that for many purposes, we are also interested in computing the mathematical expectation of some function of x. And as you now see on the screen, if capital H of x is a function of the random variable x, then capital H of x is also a random variable and it also has an expected value, because any function of a random variable is also a random variable. Students ab iss me hoi itne confuse ho nahi baat nahi. Dekhye H of x to ek symbol hain a, the function can be any function. It may be x square, x plus 1, 2 minus x or any such function. To saab zahir hai, ke agar x veri karega, to x square bhi to veri karega na. In the example of the three coins that we have been discussing, x ki values thi 0, 1, 2, 3. Aur agar mai x square ki baat karun, to saab zahir hai, ke x square ki values hain 0, 1, 4, 9. Ab x khud jo hain that is a random variable, because it is associated with the random experiment ke jo hain toss kar rahe hain usme kitte head hasil ho rahe. To usse tara x square jo hain 0, 1, 4, 9 that is also a random variable, because that is also associated with the same random experiment. Agar hain ek head mil rahe hain to hamara variable x square jo hain that takes the value 1. Aur agar hain 2 heads mil rahe hain to hamara jo variable x square hain uski value 4 hain. Is saab se I hope you understand that the probability of x equal to 1 is the same as the probability of x square equal to 1. Kyuke probability to students x ya x square pe depend nahi karthi. Probability to us event pe depend karthi hain jis ke tahit x ki value 1 ho rahe hain aur x square ki bhi is situation me 1 ho rahe hain. Wo event kya hain ke jab hamne toss kia we obtained only 1 head and 2 tails. Isi tara the probability that x is equal to 2 is the same as the probability that x square is equal to 4. Kyuke isli a ke ye dono jo values hain pehli x ki aur dosri x square ki, they depend on that event whose probability is 3 over 8, the event that I will get 2 heads and 1 tail. Is havale se we are able to understand how to compute the expected value of h of x as you see on the screen e of h of x is equal to sigma h of x into f of x. In particular if h of x is equal to x square then e of x square is equal to sigma x square into f of x. It is important to note that e of x square is not the same as e of x whole square. E of x is sigma x into f of x. Uska jo square keringe to wo us sum ka square ho rahe jo us column ka sum hai. Lekin e of x square is sigma x square into f of x. Yani ap ek column konstruct keringe of x square into f of x aur uska sum keringe, which is not the same thing as we just discussed. Isli ye jo differentiation hai that is very important. Also if h of x is equal to x minus mu whole square then e of x minus mu whole square is equal to sigma x minus mu whole square into f of x. This particular expected value is called the variance and it is denoted by sigma square as indicated in the last lecture. The shortcut formula for the variance as indicated before is e of x square minus e of x whole square. And if we are interested in computing the standard deviation, we will find the positive square root of the variance. Students, aap recall keringe ke pichle lecture me ham ne mean aur variance ke ilawa coefficient of variation bhi compute kia tha. And that is very, very easy as soon as you have found the mean and the standard deviation. You can divide the standard deviation by the mean and multiplied by 100 in order to get the coefficient of variation. Aur iska fayda kya hai jaisa ke pehle pataya gaya aap compare kr sakte hai variability of one particular probability distribution with the variability of another probability distribution. Mean or variance dono expected value ke under ham compute kr sakte hai. Toh students jo higher moments hai, the third moment, the fourth moment and so on. Of course, they can also be computed in the same manner. As you now see on the screen, if h of x is equal to x raise to k, where k is 1, 2, 3 and so on, then expected value of x raise to k is equal to sigma x raise to k into f of x. And this is called the keth moment about the origin. Ab isski andar aap note karein ke jab bahut pehle hamne sample data ke liye moments discuss kie the tab bhi to hamne kuch issi tara se define kia tha na? Ke, for example, the first moment about 0 was sigma x minus 0 whole raise to 1 over n aur yaha peh, we can say expected value of x minus 0 whole raise to 1 issliye ke waha peh jo sigma over n, yani sum over n hum karte the, woh bhi to mean hi tha na? Kisi bhi quantity ka sum karein aur usko number of terms se divide kardein, to woh mean hi hota hain na? Aur yaha peh expected value ka matlab hai ke mean? So, you must try to understand that they are very, very similar. Jab bhi aap sample data se yeh cheez hai compute karein ke kuch iss karein ke aap sum over n karein ke aur jab bhi aap probability distribution ke liye yeh cheez hai nikalne ke kuch iss karein ke aap expected value compute karein ke. They are basically one and the same thing. To, yaha jasme kahi thi ke first moment about 0 yaha first moment about the origin ka matlab yaha peh kya hoga? Expected value of x minus 0 raise to 1 ab minus 0 likne ki to koi zaroora thi nahi hai. So, we can write expected value of x raise to 1 ya simply expected value of x. So, this is the first moment about 0 ya first moment about the origin. Likne agar second moment mein interested hai toh kya mo jahega? Expected value of x raise to 2, third moment about 0, expected value of x raise to 3 and so generally kth moment about 0 is expected value of x raise to k. Aur jo formula hai jyaisa ke kahi defa kaha e of anything is sigma of that thing into f of x. Bahotsi baate jabhi tak ki aur bahotsi slides jo apne dekhi unka jo lubbe lubab hai nahi yehi hai. E of anything is sigma of that thing into f of x and of course, yeh jo formula mein aapko deh rahin hoon, yeh jo tip mein aapko deh rahin hoon. This is in the case of a discrete probability distribution. Agar ham continuous distribution ki baat karein, then the summation sign will be replaced by integration aur pure maths mein integrals toh apne padhe hi hai. Likne aap varee na karein, abhi usme kuch deer hai. Fil hal, we are concentrating on the discrete situation. Yeh toh the moments about 0, but then of course, we can also compute the moments about the mean as you now see on the screen. If h of x is equal to x minus mu whole raise to k, then we get an expected value which is called the kth moment about the mean and it is denoted by mu k. That is mu k is equal to sigma x minus mu whole raise to k into f of x. Ab iss ke andar bhi, of course, hum kth values change kar sakte hain. We can put k equal to 1, 2, 3, 4 and we will get mu 1, mu 2, mu 3 and mu 4. The first 4 moments about the mean. Please yeh note ke je agar mu ke saath subscript laga hai, then we are talking about moments aur agar mu ke saath koi subscript nahi hai, we are talking about the mean aur yeh bhi aapko yaad hoga ke first moment about the mean, yani mu 1 that is always 0. Isliye jo 3 moments hume compute karne honge, they are mu 2, mu 3 and mu 4. Either we do them by this formula that you just saw or we first find the moments about 0 and then use those relationships that I discussed with you in an earlier lecture that connect the moments about the mean with the moments about 0 or about any arbitrary origin a. Yeh moments nikalne ke baad hume kya nikalna hota hai? As you will recall, we would like to compute the two moment ratios that will enable us to talk about the skewness and the ketosis of our probability distribution. As you now see on the screen, the first moment ratio is beta 1 is equal to mu 3 square over mu 2 cubed and the second one is beta 2 equal to mu 4 over mu 2 square. Aap note kari once again that the formulas are exactly similar to the formulas that you had before. Pehle hum b 1 or b 2 krehte or m 3 or m 4 krehte or aap b ki jaga the notation is beta or m ki jaga it is mu with a subscript. Beta 1 will enable us to determine the skewness distribution and beta 2 tells us about the ketosis. Aap ko yad hai ke abhi kuch ter pehle radios ke example me the line chart of our probability distribution was slightly positively skewed. Iska matlab yeh hua ke agar hum beta 1 nikalne for that example it should it will not come out to be 0 because as you know if the distribution is absolutely symmetric then beta 1 will be exactly equal to 0. So, I would like to encourage you students to solve that question on your own. Find a mu 2 mu 3 and also mu 4 find beta 1 and beta 2 and see what you get. Also you can do that for the first example the one of the tossing of the 3 coins in which case you remember we obtained an absolutely symmetric distribution. The next concept that I am going to discuss with you now is properties of mathematical expectation. There are certain properties which assist us in computing the various expected values in a very convenient manner. As you now see on the screen the first property is that if c is a constant then expected value of c is equal to c. In other words the expected value of a constant is the constant itself. I would like to explain this particular point to you not by deriving it in a mathematical way, but by giving you a very simple example. Everybody got 2 marks out of 20. Everybody got 2 marks out of 20. What is the mean of 2 marks out of 20? What is the mean mark? Obviously 2. And this variable in this case was not a variable. So, it is a constant and the mean value of this constant is equal to the constant it is equal to 2. The next property as you now see on the screen expected value of ax plus b is equal to a times expected value of x plus b where a and b are any 2 constants. As you see on the screen for the tossing of the 3 balanced coins x is equal to 0 1 2 3 and the probabilities are 1 by 8, 3 by 8, 3 by 8 and 1 by 8. Now when we multiply the x column with the column of probabilities the sum of that column is 1.5 as we have stated in the beginning of this lecture. The mean number of heads in a very large number of tosses of the 3 coins is 1.5 heads per toss. Now suppose that we are interested in finding the expected value of the random variable 2 x plus 3. Then we carry out the following computations. Subse pehle hum 2 x plus 3 column banayenge because that is the new variable that we are interested in. So, when we substitute x equal to 0 in the expression 2 x plus 3 our value comes out to be 3, the next one comes out to be 5, then we have 7 and 9. As I indicated earlier the probabilities remain the same and multiplying the column of 2 x plus 3 with the column of probabilities we obtain 3 by 8, 15 by 8, 21 by 8 and 9 by 8. Adding these we get 48 divided by 8 which is equal to 6. In other words the expected value of the random variable 2 x plus 3 is equal to 6. Ab aap note kanei ki ye ek neya random variable hai jo ke pehle random variable ki algebraic manipulation ke ze rye aaya hai aur is ney variable ki jo expected value hai that is 6. Jo property hum yaha verify karna cha rahe bo kya hai? Expected value of ax plus b is equal to a times expected value of x plus b. This example may obviously a is equal to 2 and b is equal to 3. And as you now see on the screen expected value of 2 x plus 3 which is equal to 6 can be written as 2 times 1.5 plus 3 and that is equal to 2 times e of x plus 3 and hence we have verified that expected value of ax plus b is indeed equal to a times expected value of x plus b. Students, the two properties that I have discussed with you are valid in the case of univariate situation. Yani ek hi random variable hai x aur uski manipulation aap kar rahe hain ya usse related bati aap kar rahe hain. But then we also have two very important properties which are expected value of x plus y is equal to expected value of x plus expected value of y and the other one that if x and y are independent then x expected value of x into y is equal to expected value of x into expected value of y. The point to understand is that we will be discussing those properties in detail but we can do that only in that situation when we deal with not a univariate situation but the bivariate situation. That will come later. In the next lecture the one immediately after today's I will first discuss with you the Shebyshev's inequality in case of a probability distribution and after that we will start the discussion of the univariate situation for a continuous random variable. Having discussed the univariate situation we will go to the bivariate situation and in that lecture we will be discussing those other two properties. To aap ne dekhah students ke aapki jo journey hai in the area of probability distributions actually abhi uska aapne sirf aagaz hi kia hai. Many exciting things are yet to come and in the meantime I would like to encourage you to practice all the concepts that you have done until now so that you feel prepared to address the concepts that will come after this. My very best wishes to you and until next time Allah Hafiz.