 Assalamu alaikum, welcome to lecture number 25 of the course on statistics and probability. Students you will recall that in the last lecture I discussed with you in some detail the concept of continuous probability distributions and towards the end of the lecture we were discussing one particular example. Let us pick up the same example and understand how we will compute any conditional probability in the case of a continuous probability distribution. As you now see on the screen the example that we were considering was find the value of k so that the function f of x defined as follows may be a density function f of x is equal to k x in the range 0 to 2 and it is equal to 0 elsewhere. Compute the probability that x is equal to 1 as well as the probability that x is greater than 1 also compute the distribution function f of x. Students you will recall that all the questions we solved last time but there was one more question and as you see on the screen we wanted to compute the probability that x is less than half given the information that x does lie somewhere between 1 by 3 and 2 by 3. So now the question arises that how will we tackle this problem? We will apply the same definition that we did when we dealt with this concept in the very beginning. As you will recall the probability that B occurs given that A has already occurred is equal to the probability that both A and B occur divided by the probability that A occurs. So in this problem we will look at the situation in which both A and B occur. As you see on the screen this may be that is x less than or equal to half or A that is that x is lying between 1 by 3 and 2 by 3 students. If you consider the range x lying between 1 by 3 and half you will realize that both these requirements are being fulfilled. If x lies between 1 by 3 and half as you see in the numerator you will realize that x is less than half and x is somewhere between 1 by 3 and 2 by 3. If it is falling in the range 1 by 3 to half both of those events are occurring. Now the denominator has to be the probability of A and as I said a few minutes ago in this problem slash ke baad jo chi is hai that is A and that is that x lies between 1 by 3 and 2 by 3. So we divide probability that x lies between 1 by 3 and half by the probability that x lies between 1 by 3 and 2 by 3. In other words we divide the integral of x by 2 from 1 by 3 to half by the integral of x by 2 from 1 by 3 to 2 by 3. We had found that k was equal to half and hence our probability density function was f of x is equal to x by 2. Now computing the integrals we obtain x square over 4 within the limits 1 by 3 to half divided by x square by 4 within the limits 1 by 3 to 2 by 3 and computing this quantity our final result is 5 by 12. Students this is the way we can compute any conditional probability in the case of a continuous probability distribution. I mean this probability 5 by 12 which is case may I is occurring in the range 1 by 3 to 2 by 3. I A, abhum ek aur example karte hain which may appear to be a slightly more complicated situation. As you now see on the screen if we have the statement that a continuous random variable x has the distribution function capital F of x as follows. Capital F of x is equal to 0 for the range x is less than 0. Capital F of x is equal to 2 x square over 5 for the range 0 to 1 and for the range 1 to 2 it is a much more complicated expression minus 3 over 5 plus 2 over 5 multiplied by 3 x minus x square over 2. Also capital F of x is equal to 1 for the range x is greater than 2. Given this entire information we are required to find the probability density function of this continuous random variable and the probability that the absolute value of x is less than 1.5. Students abto aap kainge ke bas bahothi jada sakht problem ho gaiye because now we do not have available to us small f of x jis ka hum integral le le te rather we have capital F of x. How do we proceed to solve this question? Aapko yaadhe na many last time aapko ek bara important rule tia tha that the relationship between the small f of x and the capital F of x is that the derivative of capital F of x is equal to small f of x. Of course, this tally is with what you study in pure mathematics and now if we apply this in this problem we will be in a position to solve the question. As you now see on the screen small f of x will be the derivative of capital F of x and for the range x lying between 0 and 1 this derivative comes out to be 4 x over 5. For the range 1 to 2 this derivative comes out to be 2 over 5 multiplied by 3 minus x and for the range x less than 0 or x greater than 2 small f of x comes out to be 0 and students I would like to advise you to compute all these derivatives on your own. Jaisa ke aap sab jaanthi hain it is the general perception that it is much easier to find the derivatives as compared with the integrals and I am sure that you should be able to compute all these derivatives conveniently. Aacha is ke baad next hum kya chis compute kar na chata hain the probability that the absolute value of x is less than 1.5. Students iss silsle mein sabse ahem baad sabhajne ki ye hain ke if we are saying that the absolute value of x is less than 1.5 it means that x itself lies anywhere between minus 1.5 and 1.5 dekhe fars ki jay ke x is equal to minus 1.3 uska absolute value uske aap jab leenge what will you get 1.3 and that is less than 1.5. So, in order to compute this particular probability as you now see on the screen it is equal to the probability that x lies between minus 1.5 and 1.5 and this means that we have to find the integral of our probability density function from minus 1.5 to 1.5, but since our probability density function has different forms in different regions of the x axis therefore we have to split the integral into different parts and then we are able to solve this problem. So, the probability that modulus of x is less than 1.5 is equal to the integral from minus 1.5 to 0 of 0 plus the integral from 0 to 1 of 4 x over 5 plus the integral from 1 to 1.5 of 2 times 3 minus x over 5. Solving this integral our answer is 0.75. In other words it is 75 percent probability that for this particular density function our random variable x lies somewhere between minus 1.5 and 1.5. Students aap ko yad hoga ke pishli martaba hamne detail me mathematical expectation ka concept diskas kia. Now of course the concept of mathematical expectation mean variance moments and moment ratios it holds equally for the continuous probability distributions. The only difference is that in the earlier situation we were finding various sums and in this situation the summation sign is replaced by the integration sign. Let me explain this to you with the help of the following example. Find the expected value of the random variable x having the pdf f of x is equal to 2 times 1 minus x in the range 0 to 1 and it is equal to 0 elsewhere. In order to solve this question we compute the expected value of x which is given by the integral from minus infinity to infinity of x multiplied by f of x. Students ye hamne expression jo istimal kia hai integral of x into f of x I hope you realize that this is nothing new actually it is very very similar to the formula that we had for expected value of x in the case of a discrete random variable. Yad hena in that situation we had e of x is equal to sigma x into f of x where f of x of course represented the probability. To jasa main e kuch der pehle kaha tha ab jab ke ham continuous random variable si rather than having sigma yani summation we will have integration. And so as you now see on the slide again expected value of x is the integral of x into f of x and in this problem it is equal to the integral from 0 to 1 of 2 x into 1 minus x and that is equal to 2 times x square over 2 minus x cubed over 3 and the limits are 0 to 1 and solving this expression the final result is 1 over 3. Now as you just noticed the answer the mean the expected value of x is equal to 0 to x 1 times x square over 3 minus x square over 3 minus x square over 3 plus 1 over 3 minus x square over 3 plus x square over 3 minus x square over 3 minus x square over 3. 2 in this particular problem has come out to be 1 over 3 which is equal to 0.3333. So, let us try to visualize this situation and if we draw the graph of this particular probability distribution, we have the picture that you now see on the screen. As you can see this one is again a straight line, but because of the negative sign that we had in the equation of f of x, this is a downward going line because the slope is negative. And the total area under the curve is equal to 1 because of the fact as stated before that for any probability distribution the total area under the curve is always 1. As far as the mean is concerned as I said it is 0.33 and so if we want to represent it graphically you can see that it occurs a little before the number 0.5 and as such it is not in the exact middle of the distribution, but it is obvious that you can get the mean in the exact center of the distribution only if it is an absolutely symmetric distribution. So, if we have a problem in which we get a triangular distribution which is absolutely symmetric then of course, our mean will be in the exact center of the distribution also. If we have the uniform distribution, the uniform distribution being the one for which we have a horizontal line and that is why we say uniform students that is also absolutely symmetric if you stand in the mirror between the left hand side will be the mirror image of the right hand side and so in this case too your mean will be in the exact center of the distribution. And then of course, we have the normal distribution, the beautiful bell shaped distribution that I have already talked about, but about which I will talk in much more detail in a coming lecture and in the case of the absolutely symmetric normal distribution again the expected value of x that is the mean will lie in the exact center of the distribution. You will remember that in the last lecture I discussed with you two properties of mathematical expectation, the expected value of a constant is equal to the constant and the other one was expected value of A x plus B is equal to A times the expected value of x plus B. I A is property we try to verify in this situation and you will realize that because we are dealing with a continuous situation this time all these expected values will have to be computed by the use of integration. So, as you now see on the screen suppose that A is equal to 3 and B is equal to 5 then we wish to verify that expected value of 3 x plus 5 is equal to 3 times expected value of x plus 5. Now the right hand side of the equation is very simple to find because we have already found that the expected value of x for this particular problem is 1 over 3 and therefore, 3 times E of x plus 5 comes out to be equal to 1 plus 5 and that is 6. In order to compute the left hand side of this equation we have to proceed as follows. Expected value of 3 x plus 5 is equal to the integral from 0 to 1 of 3 x plus 5 multiplied by 2 times 1 minus x and this is equal to 2 times integral from 0 to 1 of 5 minus 2 x minus 3 x square. Solving this integral students you find that expected value of 3 x plus 5 is equal to 6 exactly the same result that you had for the right hand side. I would like to draw your attention now to a special case of this particular property. As you now see on the screen if we put B equal to 0 then our equation E of A x plus B is equal to A times E of x plus B takes the very simple form that expected value of A x is equal to A times expected value of x. This means that whenever you have to get 2 x or 4 x or 2 or 4 that will come out and it will simply become 2 times E of x. So, this is a very simple rule and a very convenient formula to remember. Expected value of basic concept or formula in the case of continuous distributions, we are interested in computing the moments and the moment ratios of any continuous probability distribution. So, let us compute the moments and the moment ratios for the example that you now see on the screen. A continuous random variable x has the pdf f of x is equal to 3 over 4 times x into 2 minus x in the range 0 to 2 and it is equal to 0 otherwise. Find the first four moments about the mean and the moment ratios. In order to solve this question, students, we have two options. One is that we directly apply the moment about the mean will be given by expected value of x minus mu whole raise to r. But you will realize that in this situation we have to compute integrals and it is possible that if we use this formula our integral becomes a very complicated expression. So, we have the other alternative that I will initially find the first four moments about 0 that is expected value of x minus 0 raise to r. In other words expected value of x raise to r that is not perhaps as difficult as the one I mentioned earlier. Or, when we take these moments, then the same relationships that you have discussed, if we apply them we will find the moments about the mean. So, as you now see on the screen, the first moment about 0 is denoted by mu 1 dash and it is expected value of x minus 0 whole raise to 1 which is exactly the same thing as expected value of x and applying the same formula as before our answer comes out to be 1. The second moment about 0 is expected value of x square and this is equal to the integral from 0 to 2 of x square multiplied by f of x and solving this expression the answer is 6 over 5. Then the third moment about 0 is mu 3 dash equal to expected value of x cubed and that comes out to be 8 over 5. In a similar way the fourth moment is found to be 16 divided by 7. Students, now we will apply those relationships that we discussed earlier when we were dealing with frequency distributions. You remember that there were moments being denoted by m 1, m 2, m 1 dash, etc. But here of course, we are using mu 1, mu 2, etc. and hence as you now see on the screen, the relationships connecting the moments about the mean with the moments about 0 are mu 1 is equal to 0 which is always true regardless of whatever distribution we are dealing with. Mu 2 is equal to mu 2 dash minus mu 1 dash whole square and applying this formula in this particular problem mu 2 comes out to be 1 over 5. Mu 3 is equal to mu 3 dash minus 3 times mu 1 dash into mu 2 dash plus 2 times mu 1 dash cubed and substituting all the values of mu 1 dash, mu 2 dash and mu 3 dash. The third moment about the mean mu 3 comes out to be 0. Finally, mu 4 is given by a lengthier expression similar to the one that we did earlier and applying this formula mu 4, the fourth moment about the mean comes out to be 3 divided by 35. Students, ye chur results abhi hamne hasil kye hain, inki interpretation pe jara gaur ki jaye. Hamne dekhaha kye mu 2 is equal to 1 over 5 aur aapko yaad hain a, ke second moment about the mean is exactly the same thing as the variance. If you remember that, then of course you can find the standard deviation of this particular probability distribution by taking the square root of 1 over 5 and that is very convenient. And as soon as you found the standard deviation, of course you are also in a position to compute the coefficient of variation which as you remember is given by the formula sigma over mu into 100 and which gives us the dispersion of our distribution relative to the mean of the distribution. The answer is in percentage form, so that we can compare the variability of this particular distribution with the variability of any other similar distribution. Iske baad zara mu 3 pe gaur ki jaye mu 3 ki ka jo answer aaya hai that is 0 and do you not remember that we have discussed in a lot of detail that for an absolutely symmetric distribution, the third moment about the mean is 0. Iske matlab yeh hua ke yeh jo distribution hai, it is absolutely symmetric. To aap ke zehen me to phir aap shahit kuch confusion payda hoge hoge hi, ke hamne iske graph to draw bhi nahi kia and just because of this third moment we are making such a big statement that this distribution is absolutely symmetric. Students why do not you draw the graph of this particular distribution? It is extremely simple aap ka jo aap ki jo equation hai f of x is equal to something aap uske andar x ki wo jo muhtalif possible values hai un me se kuch values uske andar rakhiye and compute f of x and then draw a simple graph of f of x against x and find out for yourself is the distribution actually symmetric or not. Now the next thing that I would like to discuss is the two moment ratios that also we have discussed when we were talking about frequency distributions. As you will recall in that situation we used to say that b 1 and b 2 are the two moment ratios that give us indication regarding the skewness and the kurtosis of the distribution. Lekin jesa ke last lecture me bhi aap ne note kia hoge, in the case of a probability distribution these moment ratios are denoted by beta 1 and beta 2. And as you now see on the screen for this particular problem beta 1 which is mu 3 square over mu 2 cubed is equal to 0 square over 1 over 5 cubed and obviously that is equal to 0. As far as beta 2 is concerned it is given by mu 4 over mu 2 square and when we substitute the values our beta 2 comes out to be 2.14. Aap ke aad hoga ke humne kaha tha ke for a normal distribution the value of the second moment ratio is equal to 3 and for a leptocratic distribution it is greater than 3 and for a platikertic distribution it is less than 3. So the answer that we have just obtained students I leave it to you to interpret the shape of this particular probability distribution with regard to the peatness of the distribution. Alright now that we have discussed in quite some detail the concept of the univariate discrete probability distribution and the univariate continuous probability distribution students it is time for me to begin with you the discussion of bivariate probability distributions. Yaani aap hum us situation kisa deal karne wale hain jis me hum sif ek variable x me interested nahi hain balki we are interested in two variables simultaneously x and y. Aap kahange ke ek variable ki discussion hi kya kum thi ke aap hum 2 variables ki baat kare. Ispe mujhe ek hubsurat shere yaad aagyaa shahir ne kaha hain iptadai iskh hain rota hain kya aage aage dekhye hota hain kya. Students ilm to ek laa mutanahi samandar hain aur ne aur aap saari umar bhi lager hain to us samandar ka ek katra hi hasil kar sakte hain. But it's worth it. So, for the discussion of the bivariate probability distribution, let us begin with the concept of joint distributions. As you now see on the screen, the distribution of two or more random variables which are observed simultaneously when an experiment is performed is called their joint distribution. As indicated earlier, it is customary to call the distribution of a single random variable as univariate. Likewise, a distribution involving two, three or many random variables simultaneously is referred to as bivariate, trivariate or multivariate distribution. So, let us begin the discussion of the bivariate probability function in the discrete situation. As you now see on the screen, if we let x and y be two discrete random variables defined on the same sample space as x taking the values x 1, x 2, so on up to x m and y taking the values y 1, y 2, so on up to y n, then the probability that x takes on the value x i and at the same time y takes on the value y j is denoted by f of x i comma y j and this is defined to be the joint probability function or simply the joint distribution of x and y. Thus, the joint probability function also called the bivariate probability function f of x y is a function whose value at the point x i y j is given by f of x i y j is the probability that the random variable x takes the value x i and the random variable y takes the value y j and this equation is valid for all the i values from 1 to m and all the j values from 1 to n. This joint probability distribution is very conveniently represented in the form of the table that you now see on the screen. This time, we have a bivariate situation and we can write the x values in the first column as you can see and we can write the y values on the top so that inside the body of the table, we have all the probabilities of the form f of x i comma y j. The point to be understood is that the sum of all these joint probabilities is equal to 1 and this is exactly how we would want it to be because we have been studying for quite some time now that for any random experiment, the probability of the sure event is 1. When we do our experiment, it is clear that the possible ordered pairs x 1, y 1, x 1, y 2, x 3, y 4, etc. One of them has to occur. Therefore, the sample space which consists of all those outcomes that is a sure event has the probability 1. In other words, the sum of all those probabilities has to be 1. Because of the discussion that I have just had with you, we can say that a joint probability distribution has the following important properties. Number 1, f of x i comma y j is greater than or equal to 0 for all possible ordered pairs x i, y j. In other words, what we are saying is that none of these probabilities can be negative. The other property is that the sum of all such probabilities is 1. Students, you have noted that the expression you have seen on the slide, we have written that sigma sigma f of x i, y j is equal to 1. Pehle sigma ke saath subscript i attached hai aur ducksre summation sign ke saath subscript j attached hai. Aur ye double summation iss liye sонuri hai ke hamari jo table hai. That is a bi-variate table. Jab ham add karenge to ham, first we can add over the columns and then, we can add over the rows and that will give us the value 1. So, jeb kabhi bhi aap bivariate situation ke saath deal karenge, you will be involving double summation or agar continuous situation hai, then the two summation signs will be replaced by double integration. Je to table aap ne bhi deki, this brings us to the concept of marginal probability distributions. The probabilities that occur in the margins of that table are called marginal probabilities and as you now see on the screen, the formal definition of marginal probabilities is the probability distribution of the random variable x denoted by g of x is equal to sigma f of x i y j and the summation sign has j equal to 1 to n. In other words, g of x i is equal to f of x i y 1 plus f of x i y 2 plus so on up to f of x i y n. Aap ye jo subscript i hai, this is itself a variable. So, if I put i equal to 1, I obtain g of x 1 is equal to f of x 1 y 1 plus f of x 1 y 2 plus so on up to f of x 1 y n. Iska mafoom kya hai, students, humari jo bivariate table hai aap ko yad hoga ki usme sabse pehli value x ki jo hai that is x 1 aur us ro mein uske against jo joint probabilities hai agar aap unko add karein to aap ko aakhir mein jo total value milti hai that is g of x 1. So, I would like to encourage you to study this table and the formula that I have just presented and you will realize that it is actually very simple. agar aap ne g of x i compute karna hai, yani x ki muhthulif values ke liye corresponding probabilities, to aap over the rows add karte jaaye and you will obtain the required probabilities. isi tara, if you want to find the probability distribution of the random variable y, you will do exactly the same procedure, but in the other direction aap ko yad hai ke us table mein y values are written on the top y 1, y 2, y 3 and so on. And if I want to find h of y 1 that is the probability that my random variable y takes the value y 1, what should I do aap uske against jo values column wise hai unko add karein je and what you get as the total that is what you want. And as you now see on the slide this can be written as h of y j is equal to sigma f of x i y j and the sigma sign now has with it i going from 1 to m. We will be applying all these concepts to an example in a short while, but before we do that students why do we not also define conditional probability in this kind of a situation? As you now see on the screen the conditional probability function for y given that x is equal to x i is f of y j given x i is the probability that y is equal to y j and x is equal to x i divided by the probability that x is equal to x i and this is equal to f of x i y j divided by g of x i. In a similar way we can define the conditional probability of x given y equal to y j. The next concept is that of independence and as before we can say that x and y are statistically independent if f of x y is equal to g of x into h of y. Now, the last expression I have shared with you, in that I said that f of x y should be equal to g of x into h of y. Of course, if we elaborate it and say that f of x i y j should be equal to g of x i into h of y j for all values of i and all values of j, but short k k we simply say that f of x y should be equal to g of x into h of y. Let us now apply all these points to an example. An urn contains three black, two red and three green balls and two balls are selected at random from this urn. If x represents the number of black balls and y represents the number of red balls out of the two that we select, then find number 1, the joint probability function f of x y. Number 2, the probability that x plus y is less than or equal to 1. Number 3, the marginal probability distribution g of x and h of y. Number 4, the conditional probability distribution f of x given 1. Number 5, the probability that x is equal to 0 given that y is equal to 1. And finally, we are interested in determining whether or not x and y are statistically independent. You have seen that there are many questions in this problem that we want to address. But first of all, let us see how we will construct this bivariate table. We will have to do it one by one. And first of all, let us consider the case when x is equal to 0 and y is also equal to 0. Now, in order to construct this table, let us consider the probability that x is equal to 0. Now, let us consider the probability that x is equal to 0. Now, let us consider the probability that x is equal to 0. Now, in order to have this particular scenario, we realize once again that we have to apply the rule of combinations and also we apply the classical definition of probability. So, to have the denominator of the classical definition, the total number of ways in which I can draw two balls out of 8, what will I compute? Obviously, 8 C 2. And to have the numerator, the number of ways in which I can draw two green balls out of this bag and no red ball and no black ball, so that x is 0 and y is 0. Students, how many ways of doing that? As you now see on the slide, this is equal to 3 C 0 into 2 C 0 into 3 C 2. You will remember that along with the rule of combinations, we also apply the rule of multiplication in this type of situation. Computing these values, our answer is 3. As far as the numerator is concerned and as far as the denominator is concerned, 8 C 2 is equal to 28. Therefore, the probability that x is equal to 0 and y is equal to 0 is equal to 3 by 28. Students, is it as I say, we will compute all the other probabilities and how many probabilities do we have to compute in all? You realize that if we draw two balls, then x variable can go from 0 to 2. We can have no black ball or one black ball or two black balls or y variable that can also go from 0 to 2. We may have no red ball, one red ball or both red balls. So, as you now see on the screen, we have a bivariate table in which there are nine cells and it is interesting to note that three of the probabilities are 0. Students, I will discuss with you this problem in further detail in the next lecture, but in the meantime, I would like to advise you to compute all the probabilities of this table doing your own calculations and in particular, think about the three probabilities that are equal to 0. The probability of an impossible event is 0. So, just think about it. Best of luck and until next time, Allah Hafiz.