 अस्लामलेकुम, welcome to lecture number 26 of the course on statistics and probability. students अप याद होगा के पिछले लेक्चमे हमने univariate continuous probability distributions के बारी में काफी detailed discussion की ती अर उसके बात है याद याद बारी येट प्रवबेलिटी दिस्टॉष्याज़त् है You will recall that we started an example and i encouraged you to compute the various probabilities on your own. से निस बॉओन ते दिखुचन कीक्क्रशावे 숫िल, से कीस्सिया Indigenous दोस्मद किक्श्चशना किन्चा bare सा से निस साक्ँच्छा sound कर और सशा स्फ्थ free साकटी कना, सी निस मोहदा, बोल्स and y is the number of red balls selected then find number one the joint probability function f of x y number two the probability that f of x plus y is less than or equal to one number three the marginal probability distributions number four the conditional probability distribution f of x given one also we want to find the probability that x is equal to zero given that y is equal to one and last but not the least we are interested in determining whether or not the random variables x and y are independent students you will remember that I told you last time that in this particular problem we have nine possible situations zero zero zero one zero two one zero one one one two and two zero two one and two two I told you everything that the value that I said earlier that is the value for x and that is the value for y as you see on the screen for example the combination two zero means that x is equal to two the number of black balls that you draw is two and y the number of red balls is zero you will agree that it is not possible to have x equal to two and y equal to one because we are drawing only two balls so how is it possible that two balls are black or one is red since it is an impossible event you can see on the screen that the probability for the combination two one that is x is equal to two and y is equal to one the probability of this combination is zero as far as all the other probabilities are concerned I encouraged you to try to compute them on your own and in that there is a rule of combinations because as I told you earlier that when there is no importance of order we are drawing two balls in the same order the first and the second balls do not matter so whenever there is such a situation we apply the rule of combinations as you once again see on the screen the probability for the combination zero zero is given by three c zero into two c zero into three c two divided by eight c two or because there are three c zero ways of drawing zero black ball out of three black balls and two c zero ways of drawing zero red ball out of two red balls and at the same time three c two ways of drawing two green balls out of the three green ones that we have in our bag. ये जो कुच मेंने कहा this forms the numerator और इसको आपको याद है के multiply हमने इन सब को इसले किया के हम and की बाद कर रहें we are doing this and the next thing and the third thing. और denominator जो है that is eight c two the total number of ways in which we can draw two balls out of eight students. जा थक इन फरमूलाय की detail का ताल लुख है जाहर है के उनको बाद बार तो हम नहीं दो रहा सकेंगे I am sure that you must have revised the lectures that you have had earlier and you will recall the detailed अब नहीं ताल वो लाग है याद का थाप ओड़े रहा है नहीं ती कारूट नें लागा है थकुढ़त को रहाुर लेग लीग इस भी तुछ नहीं नहीं च़े करै रहा है लगु़ of ncr. ताल रहा आपको देंगे ड़ाई बाद का ताल बाद ऴाद लिए हूँ. अगर अप जीदो जीदो की है, मquisite of 0 and 0 is 0 which is less than 1 and the sum of 0 and 1 gives you 1 and the sum of 1 and 0 is also 1. इं तीन के अलावा अगर अगर अप ओभी कोईभी 바꿋क्टी देखेगे तोधाद, अप इं तीन के अईखाँ के अब आपवर सब पक hearing of the entire भीि ज़ादि, ज़ादि, उआ फ मार, ख़ूग़ूऐ वीि। घा रेंशि देखँएं तो आब देखऑेंगें से उजं ज़ादि, औह घ्ये न제를सON ग़्दि, इक आथ शनधो क canned दख़्दि, गया वी। भींій सन चिक जं शुरेंं भीा पतिका todavía ड़िए म अपland अज आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आद आ� करि और सोग मेंट देखात नहींके लेगा। तब गाई सोग मूझादित्न लेगा करनें जीवाए थ्तादाथ बचन्तगे हता सच्पनी लिए ती बचागा गो विlardan चहता तब प्रशाजी सेदाने की औत लिए तिटने करेचिना. बॉपोगतु म्वाियले मही रीझर नहीं परतािथ। हीि जेत से तो रए लकशों तुतसे लिमلام थारिएन की पॉवेलीह fading is not served in. मिक्सा क代消icit SEO सो इस लिरे तेई में में औग US आज of how see in the signs to garlicuchi. करी लिन में में उड़ा ड़ों। less than or equal to one, if we conduct this particular experiment. All right. What was the next question? Students, we were required to also find the marginal probabilities. और आप याद होगा के मैंने आप से कहाता, के वो जो बाई वेर्येट टेबल है, उसके मारजेंस में जो प्राभबिलिती कर रही है, they are called the marginal probabilities. As you now see on the screen, the marginal probability distribution of the random variable X is 10 by 28, 15 by 28 and 3 by 28 against the values 0, 1 and 2 and similarly we have the marginal distribution of Y. Next, we are interested in computing a conditional probability और शाएद अभी तक आपको मैंसुस होता है, के कनदिशनल प्राभबिलिती तो बहुती दिफिकल्त चीज है, जब के मैं बार-बार आप से ये बाद कहती हूँ, के it is actually quite simple. Let us now see the situation on the slide. The probability that X is equal to small x, given that Y is equal to 1, is equal to the joint probability of capital X being equal to small x and capital Y being equal to 1, divided by the marginal probability that Y is equal to 1. Students, you remember that I told you that capital X and small x have a difference. Capital X represents the random variable that we are talking about and small x represents any particular value of this variable. So when we say that probability that capital X is equal to small x, given that Y is equal to 1, then there is no confusion in this. You know that capital X has three possible values. It can be 0, it can be 1 or it can be 2. So now we should say that we are saying this again and again. Why don't we say it in a summarized form that we are interested in the probability that capital X is equal to small x, given that Y is equal to 1, where small x can be 0, 1 or 2. So as you now see on the screen, f of 0 given 1 is equal to f 0, 1 over h of 1 and that is equal to 6 over 28 divided by 12 by 28 and that is equal to half. In a similar way we find f of 1 given 1, which also comes out to be equal to half and f of 2 given 1 comes out to be 0. Students, that has come out to be 0. See what is possible that X is equal to 2 given that Y is equal to 1. Try to interpret this. Y is equal to 1 means that the number of red balls that we have is 1 and this information we have achieved that we have one red ball and now we are trying to take out this conditional probability that we have two black balls and that we have one red ball. So this cannot happen. Because you have to take out these two balls so how can two black balls come from one red ball? In this way this is an impossible event and hence appropriately the conditional probability f of 2 given 1 is equal to 0. And after this our next question was what is the probability that X is equal to 0 given that Y is equal to 1? तो आप तो किसी किसम का कोई मस्लाही नहीं है ये तीन हमने जो निकाली है इनही में से एक हमारा आनसर है as you now see on the screen the probability that X is equal to 0 given that Y is equal to 1 is equal to 1 over 2 that is 50%. The last part of the question was are X and Y statistically independent students according to what we did few lectures ago two events A and B are said to be independent if the probability of A intersection B is equal to the probability of A into the probability of B and when we apply this concept in case of a bivariate probability district distribution as I said last time this formula becomes F of X Y is equal to G of X into H of Y so now in this example all we have to do is to determine whether this equation holds or does not hold if all these 9 cells which we have if they hold for all then we can say that X and Y are independent but if any of these for one cell this equation is violated then of course we say that X and Y are not statistically independent so in this example let us take one particular cell the one that you now see on the screen for the cell 01 meaning that X is equal to 0 and Y is equal to 1 that is the number of black balls to be 0 and the number of red balls to be 1 in the two that I have drawn students we have already found that F of 01 is equal to 6 by 28 but the marginal probability of X corresponding to this particular cell in other words G of X for X equal to 0 which can also be called G of 0 is equal to 10 over 28 because when you add the three probabilities in that particular row you do obtain 10 by 28 similarly the marginal probability of Y equal to 1 in other words H of 1 comes out to be 12 by 28 now when I multiply H of 1 students I obtain a value which is not equal to 6 by 28 the joint probability of X equal to 0 and Y equal to 1 since the equation of independence is not holding in this situation therefore we conclude that the random variables X and Y in this particular example are not statistically independent alright this was the discussion of the bivariate distribution in the case of discrete random variables we now proceed to bivariate situations in the case of continuous random variables as we now see on the screen the bivariate probability density function of continuous random variables X and Y is an integrable function F of XY satisfying the following properties number one F of XY is greater than or equal to 0 number two the double integral of XY with respect to X and Y must equal to 1 and also the probability that X lies between A and B and Y lies between C and D is obtained from the formula F of XY and the limits have to be according to the limits of X and Y so if you concentrate on the formula that you see students if you first integrate according to Y then your limits for that particular integral will be C or D will come under that expression and because you will integrate according to X therefore the limits of X A to B will come out from this point I have just shared this is a vitally important point and of course pure mathematics and here this is very important because the double integral is the formula through which we can compute any by-variate continuous probability density function in the situation required probability so let us visualize what you now see on the screen we can conceive the X Y plane and on the X axis we can determine two points X1 and X2 and on the Y axis two points Y1 and Y2 such that we wish to compute the probability that X lies between X1 and X2 and at the same time Y lies between as you see on the figure that you have in front of you these four values they give us four points on the X Y plane the points are X1 Y1 X1 Y2 X2 Y2 and X2 Y1 and these four points are the rectangle now students these are the rectangle this is not the probability that we are trying to compute this is a very important thing that I am sharing with you when you were doing a univariate situation you had an X axis there was a curve and under it you were trying to find a certain area here as you can see this is not the area that will give you the probability that you want but here now we have two dimensional situations three dimensional situations have come now we have two variables X and Y the X axis is for the Y variable but students the X axis the Z axis is the axis for F of X Y and in this situation this X Y plane you understand as if that is the floor the X Y floor and against this vertically Z axis F of X Y the various values when we plot so instead of getting a curve which you have in the case of a two dimensional situation F of X against X now you will have a surface and you want to compute not the area but the volume of the region on the floor that you just saw on the slide if you look at the slide once again you now I hope will realize that you are looking at the floor or on this the volume which is three dimensional volume you want to compute and double integral provides you that particular volume hence in this particular situation students it is the volume under the surface which gives you any required probability or if you compute the common volume then it has to be equal to one hence if volume is representing probability then the total probability is always equal to one we have said this so many times so students you have seen mathematics world of its own or if you study in this or are not trying to understand it then you will feel that it is very interesting statistics becomes even more interesting because we are able to apply these mathematical intricacies world situations and in our everyday life we deal with these phenomena we can interpret them through these mathematical models so you have seen that summation or integration is not just a formula but they have a lot of significance alright let us now apply all these concepts to an example given the following joint probability density function f of x y is equal to 1 over 8 multiplied by 6 minus x minus y for the region x lying between 0 and 2 and y lying between 2 and 4 f of x y is equal to 0 elsewhere number 1 we would like to verify that f of x y is a joint density function number 2 we would like to calculate the probability that x is less than 3 by 2 and y is less than 5 by 2 we are also interested in finding the marginal pdf g of x and h of y also we would like to compute the conditional probability density function f of x given y as you have seen in this also we are interested in what we are doing in the discrete situation we were solving so let us proceed step by step the first thing is that we would like to verify that this is indeed a proper probability density function and how do we do that we have to verify that the two basic probability density function f of x is greater than or equal to 0 for the entire region and it cannot be negative anywhere and number 2 the total volume under that surface which is generated by this particular equation that total volume has to be 1 so 1 over 8 multiplied by 6 minus x minus y now you will say that it is also possible that x is coming with minus and y is also minus sign so our f of x y this may come out to be negative but students you also see that x is going from 0 to 2 that is going from 2 to 4 if you substitute their maximum values in this expression then what do you get 1 over 8 multiplied by 6 minus 2 minus 4 in other words f of x y comes out to be 0 in this particular case if you keep any value of x which is less than 2 or any value of y which is less than 4 or bigger than 2 or equal to 2 then you will see that f of x y comes out to be greater than 0 now as far as the other condition is concerned we will have to compute the double integral of f of x y and as you see on the screen the double integral of 1 over 8 multiplied by 6 minus x minus y within the intervals 2 to 4 as far as y is concerned and 0 to 2 as far as x is concerned comes out to be equal to 1 I would like you to once again note since we are writing y inside therefore the first integration will occur with respect to y and x will act as a constant after that when we have applied the limits of y we obtain an expression that contains only x which in this case over 8 multiplied by 6 minus 2x and when we integrate this particular expression with respect to x and apply the limits 0 to 2 the answer is 1 exactly as what we required students the next part of the question was find the probability that x is less than 3 by 2 and y is less than 5 by 2 and we proceed in a very similar fashion and as you now see on the screen the probability that x is less than 1.5 and y is less than 2.5 is the double integral of our pdf 1 over 8 into 6 minus x minus y and the limits for y are 1.5 but the limits of x are 0 to 1.5 integrating in the same manner as before first integrating the expression with respect to y and applying the limits of y and later integrating the expression containing only the x variable with respect to x and applying the limits of x the final answer is 9 over 32 in other words there are 9 chances in 32 that if some particular bivariate phenomenon follows this particular probability distribution then x will assume a value not greater than 1.5 and y will assume a value at the same time not greater than 2.5 alright now next question is that we want to compute marginal distributions students actually this is easier because now you don't have to do double integration but you will simply have one integration you remember that in a discrete situation when we were trying to remove the marginal distribution of x then we took sum that was over the y values similarly when we were computing the marginal distribution of y then we summed over the x values now exactly the same logic applies here if we want g of x which is the marginal distribution of x then we will not find the sum but the integral of our pdf with respect to y and if we want to compute the marginal distribution of y then we will take the integral with respect to x as you now see on the slide g of x is equal to the integral of 1 over 8 into 6 minus x minus y with respect to y and the limits are 2 to 4 and solving this integral the answer is equal to 1 over 4 multiplied by 3 minus x students it is very important to note that after we have obtained this integral we have written that this equation is valid for the range x lying between 0 and 2 and we have added the statement that g of x is equal to 0 for x less than 0 or x greater than 2 what I have told you in the end what does this mean students actually it is very very important that we write that g of x is equal to so and so where x lies between certain limits and these are the same limits which were our initial limits for x in this problem the first thing I have told you region is determined according to x and according to y then our marginal distribution that will also be according to that region for g of x x will have the same limits 0 to 2 and similarly when we compute h of y then we will have the limits of y as you now see on the screen the integral of 1 over 8 multiplied by 6 minus x minus y the integral of this expression with respect to x within the limits 0 to 2 and when we solve this expression the answer is 1 over 4 multiplied by 5 minus y and you note that now we are writing along with this equation that this equation is valid for the case y lying between 2 to 4 and h of y is equal to 0 if y is less than 2 or y is greater than 4 we can also say that h of y is equal to 0 elsewhere the next concept is of the conditional probability density function is of bivariate continuous situation or then the same thing that there is no new thing everything is the same as before we have to divide the joint probability function by the marginal one in order to obtain the conditional probability density function so as you now see on the slide the conditional pdf of x given y equal to small y is equal to f of x y divided by h of y and substituting the expressions for f of x y as well as h of y that we have just obtained the final answer is that f of x given y is equal to 6 minus x minus y divided by 2 times 5 minus y or you note students that with this expression I have not written that y lies between 2 and 4 rather I have written that x lies between 0 and 2 so I have done this on purpose I want to convey that right now we are computing f of x given y so some value of y is fixed available we have got this information that y is equal to a certain value and while it is there the conditional distribution of x that will have to have the range x lying between 0 and 2 i.e. x's any value between 0 and 2 we take and we want to get a conditional probability for any fixed y so we will compute like this another thing that even y of course you can give different values so that after slash the y that is written its different values are possible now in a very similar fashion as you now see on the screen the conditional distribution of given x is equal to f of x y divided by g of x which is the marginal distribution of x and substituting the expressions the final answer is f of y given x is equal to 6 minus x minus y divided by and in this y can have any value from 2 to 4 here another point is also very important see we have computed the marginal distributions or any distribution if it is a proper probability density function then it will have to satisfy the basic properties of a proper PDF so it is very interesting to note that for the marginal as well as the conditional distributions that you have just obtained if you apply that same basic principle that we have to check whether it is positive or zero at the most in the entire range and it can never be negative and the other condition that for these univariate distributions that you have just found the total area under the curve is 1 if you apply these conditions then you will see that they are satisfied note that I have just told you that the marginal distributions they are univariate continuous distributions or these conditional distributions they are also univariate distributions now why did I say this you should pay attention to this alright now that we have completed the discussion of some of the very fundamental concepts of bivariate distributions both discrete and continuous students the next point that I would like to convey to you is that there are two important properties of mathematical expectation that are valid in the case of bivariate distributions as you now see on the screen the first property is that the expected value of y is equal to the expected value of x plus the expected value of y and the second property is that if x and y are independent then expected value of x into y is equal to expected value of x into expected value of y regarding the sum of the expected values also holds for the difference of expected values and the equation becomes expected value of x minus y is equal to expected value of x minus expected value of y students here is an important point when I talked about the second property I said that if x and y are independent random variables then expected value of x, y is equal to expected value of x into expected value of y but when I gave the first property that expected value of x plus y is equal to expected value of x plus the expected value of y then I did not impose any condition that x and y must be independent in the second situation I imposed this condition and if you pay attention then the concept that we discussed before you can relate it to this you remember that we said that if a and b are two independent events then the probability of a intersection b is equal to the probability of a into the probability of b then if x or y are independent random variables then the expected value of x into y is equal to expected value of x into expected value of y yani a similarity you can see although I am not saying that the two things are identical let us now apply this concept to an example suppose that x and y are two discrete random variables with the following probability distribution the x variable takes the values 2 and 4 and the y variable takes the values 1, 3 and 5 and the joint probabilities are 0.10 0.20 and so on we would like to find the expected value of x the expected value of y and the expected value of the product x y in order to solve this question first of all students we will need to compute the marginal probability distributions g of x and h of y adding over the rows of this bivariate table h of y comes out to be 0.25 0.50 and 0.25 against the x values 1, 3 and 5 also adding over the columns of this bivariate table we obtain g of x is equal to 0.40 and 0.60 now you add the value of h of y or add the value of g of x you will see that the sum of the probabilities is equal to 1 do you remember that I told you that your own wish is that you want to keep the value in the first column that doesn't matter but the important thing is that you remember that if you have taken y in the column then h of y will also come in the column and similarly if you have placed in the top row then in the bottom row you will get now what is the formula you remember that in a univariate situation e of x is equal to sigma x into f of x or instead of f of x we have g of x and so as you now see on the screen e of x is equal to sigma x into g of x and in this problem it is equal to 2 times 0.40 plus 4 times 0.60 and that is equal to 3.2 similarly e of y is equal to sigma y into h of y and substituting the values of y and the corresponding values of h of y e of y comes out to be equal to 3.0 hence e of x plus e of y is equal to 6.2 now the real thing in this situation that is how do we compute e of x plus y that is your equation which you want to verify according to the property that I shared with you that is e of x plus e of y so this equation right hand side we have solved and we have just found that 3.2 plus 3.0 is equal to 6.2 the question is left hand side e of x plus y how do we compute this students since this is a situation therefore e of x plus y is given by sigma sigma x i plus y j multiplied by f of x i y j and the first summation is over i whereas the second summation is over j now there is no need to be confused regarding this formula it should be taken simply as an extension of the formula that is valid in the univariate situation students when we wish to compute the expected value of x into y our formula becomes double summation x i y j into f of x i y j we will discuss with you the details of this formula in the next lecture with reference to this very example and we will find that e of x plus y will come out to be exactly the same as e of x plus e of y also interestingly in this problem we will find that e of x y comes out to be e of x into e of y so that we can say that x and y are statistically independent in the meantime I would like to encourage you to practice with the various concepts that I have dealt with today and in particular with the concept of double integration best of luck and Allah Hafiz