For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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## Transcription

### Independent Random Variables

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Intuition
- Definition and Formulas
- Theorem
- For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 1
- For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 2
- Example I: Use the Definition to Determine if Y₁ and Y₂ are Independent
- Example II: Use the Definition to Determine if Y₁ and Y₂ are Independent
- Example III: Are Y₁ and Y₂ Independent?
- Example IV: Are Y₁ and Y₂ Independent?
- Example V: Are Y₁ and Y₂ Independent?

- Intro 0:00
- Intuition 0:55
- Experiment with Two Random Variables
- Intuition Formula
- Definition and Formulas 4:43
- Definition
- Short Version: Discrete
- Short Version: Continuous
- Theorem 9:33
- For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 1
- For Continuous Random Variables, Y₁ & Y₂ are Independent If & Only If: Condition 2
- Example I: Use the Definition to Determine if Y₁ and Y₂ are Independent 12:49
- Example II: Use the Definition to Determine if Y₁ and Y₂ are Independent 21:33
- Example III: Are Y₁ and Y₂ Independent? 27:01
- Example IV: Are Y₁ and Y₂ Independent? 34:51
- Example V: Are Y₁ and Y₂ Independent? 43:44

### Introduction to Probability Online Course

### Transcription: Independent Random Variables

*Hi there, these are the probability videos here on www.educator.com, my name is Will Murray.*0000

*We are working through a chapter on Bivariate densities and Bivariate distributions*0006

*which means we will have two variables, a Y1 and Y2.*0012

*We have just been looking at some videos on marginal probabilities and also, on conditional probability.*0016

*We are going to be using some of those ideas in this video.*0023

*Today's video is on independent random variables.*0027

*I will be using the notion of marginal density function.*0031

*If you do not remember anything about marginal probability or marginal density functions,*0036

*what you might want to do is just go back and just briefly review the idea of marginal density functions.*0041

*Because, we are going to use that, we will use the definition of those,*0047

*in this video today on independent random variables.*0050

*Having said that, let us jump in.*0055

*The intuition of independent random variables is sort of one thing.*0057

*And then, there is a definition and then there is a theorem about independent random variables.*0062

*There are three different ways to think about it.*0067

*There is intuition, there is a definition, and then there is a theorem which is also very useful.*0069

*I will spell out each one, I got a side on the intuition, and then the next side will be the formal definition,*0075

*and then the next side will be the theorem that you sometimes want to use.*0082

*The idea here is that, we have an experiment with two random variables, Y1 and Y2.*0086

*Intuitively, independence means that if I tell you the value of Y2, I tell you the value of one of the variables,*0092

*you really have no new information about the distribution of the other variable Y1.*0101

*Maybe, you can make a prediction about Y1 and then I would tell you the value of Y2,*0109

*and say do you want to change your prediction about Y1?*0115

*If they are independent then no, you would not change your prediction of Y1*0117

*because the new information about the value of Y2 does not tell you anything new about the value of Y1.*0122

*That is the intuitive idea of independence.*0130

*If we spell that out, in terms of equations, what we have here, F1 of Y1 is the marginal density function of Y1.*0134

*What we have on the right here, F of Y1 condition on Y2 is the conditional density function of Y1.*0158

*The idea is that on the left, this marginal density function of Y1, this is how you would describe Y1,*0176

*if you have no information at all about Y2.*0186

*If you knew nothing about Y2, this is how you think the density of Y1 behaves.*0190

*On the right, we have the conditional density function.*0199

*This is, if I told you a particular value of Y2, how would you describe the density of Y1 with that extra information.*0201

*On the right is, how you would make predictions with the extra information about Y2.*0212

*On the left is, how you would make predictions with no information at all about Y2.*0218

*The idea of independence is that, those should be the same.*0223

*The extra information about Y2, does not change what you know about Y1.*0226

*That should be kind of intuitively why this formula makes sense.*0233

*This is not actually the formula that we will use to check whether variables are independent.*0238

*I’m going to give you a different definition on the next slide but I think this is sort of the more intuitive formula.*0245

*After I gave you the new definition on the next slide, I will try to connect it back to this formula*0251

*so that you see how the two ideas are related.*0257

*I know that is a bit confusing to have two different ways of approaching something.*0261

*I'm going to try to persuade you that they do both make sense, and that,*0266

*you can get back and forth from one to the other.*0269

*In the next slide, we are going to look at the formal definition of independence,*0274

*which I think is a little less obvious but I will connect it back to this intuitive idea.*0279

*The formal definition of independence for random variables, in terms of probability,*0285

*is the probability of both of them taking a particular value is the same as, if you evaluate them separately*0292

*and find the probability that each one of them takes that value separately and then multiply those probabilities.*0301

*That is actually the discreet version of independence.*0307

*The short version of that same formula is the probability of Y1, Y2 is the probability of Y1,*0310

*the marginal probability of Y1 × the marginal probability of Y2.*0320

*Let me stress here that this P1P2, those are marginal probabilities.*0326

*On the left, we have the joint probability function.*0340

*That was the discrete case and the continuous case is the analogue of that.*0346

*On the left, we have the joint probability function.*0353

*On the right, we have the two marginal probability functions, marginal density functions.*0365

*The idea of independence is that the joint probability density function,*0375

*maybe it would be better if I said joint density function instead of joint probability.*0386

*Let me write that down.*0391

*The idea of independence is that the joint density function is equal to the product of the two marginal density functions.*0393

*Let me write that a little more clearly, densities.*0404

*The joint density function factors into a product of the two marginal density functions.*0412

*They sort of split apart and they are independent there.*0418

*Let me try to connect this, this is the formal definition of independence.*0422

*This is the one that we are going to use for most of the problems.*0429

*Let me try to connect this up with the intuitive formula that I gave you back on the previous slide.*0432

*The way you can make those match is, you have to remember that F, the joint density function F of Y1 Y2,*0441

*one way to think about that is to sort of first evaluate the marginal density function of Y2.*0450

*And then, once you know what Y2 is, evaluate the conditional density function of Y1 condition on Y2.*0457

*This is that old conditional probability formula.*0466

*If you remember this and then, you kind of plug this into the formula for independence,*0472

*if you plug that in right there, F of Y1 Y2 is equal to F2 of Y2 × F of Y1 condition on Y2.*0480

*What you notice is that from both sides, you can cancel out an F2 of Y2.*0493

*We could cancel and if we canceled F2 of Y2 from both sides, we get on the left F of Y1 condition on Y2.*0503

*We cancel the F2 of Y2.*0519

*On the right, we would get just F1 of Y1.*0521

*That is exactly the intuitive formula that I showed you on the previous side.*0527

*That is how this formula, this definition connects up to the intuitive formula from the previous side.*0532

*This is intuition from the previous slide.*0539

*That is where you can derive the intuitive formula, if you like to have a formal justification*0550

*and how it connects up to the formal definition here.*0562

*There is one last way to think about independence and that comes from a theorem.*0567

*Let me go ahead and show that to you.*0572

*The theorem says that, for continuous random variables Y1 and Y2, they are independent if and only if,*0574

*the domain where the joint density function is defined a non 0, is a rectangle.*0582

*And, the joint density function can be factored into a product of a function of Y1 only and a function of Y2 only.*0592

*Let me expand that a little bit.*0601

*Condition one here means that, when you are graphing the domain, you would have some kind of square or rectangle.*0603

*It could be infinite, you can have something like this, you could have something that goes on forever.*0614

*A rectangle but it goes on forever, maybe something like this where it goes on forever,*0621

*in terms of one variable or it could also go on forever, in terms of the other variable.*0627

*These would all be considered rectangles, even though they extend infinitely far.*0638

*Or even, you can have something that goes on forever in both directions,*0643

*that is still considered to be a rectangle, for the purposes of this theorem.*0650

*What you could not have is some of these triangular domains that we have been looking at, in some of these examples.*0654

*I think we had one example where there was a triangular domain like that.*0661

*That was the triangular domain and that was automatically not independent.*0666

*All these others, at least as far as condition one is concern, would qualify as being independent.*0675

*The second condition means that, you can factor F of the joint density function F of Y1, Y2.*0683

*You can factor that into a function of Y1 × a function of Y2.*0691

*It is okay for either of these functions to be constants meaning, you do not have to see the variables in these functions.*0699

*Sometimes, you might just have a function of Y1 and you would say that other function is just 1,*0704

*and that is still okay to be constant.*0712

*It is okay, if either one of these functions are constants.*0722

*You would have to be able to factor it and separate it into a function of Y1 and a function of Y2 separately.*0725

*Inextricably, next in the density function that you cannot factor it then it is not independent.*0734

*We will work through the examples.*0741

*I’m going to try to solve most of the examples using the definition but then,*0742

*in a lot of them we will come back and applying this theorem.*0745

*We will see that, if we can use this theorem, we could have gotten the answer a lot more quickly,*0750

*just by kind of glancing at the region of definition, or just trying to factor the density function.*0755

*We will try to do the examples both ways, that you can get a feel for both of them.*0763

*Let us jump into those.*0767

*Example 1, this is an example we have seen before in some of the previous videos.*0770

*But, we have not looked at it in quite this light before, in terms of independence.*0776

*We have F of Y1 Y2 is 6 × 1- Y2.*0784

*The region there is a Y1 there, Y2 there, and our region is from 0 to 1 on both variables.*0790

*But, we are only looking at the region where Y2 is bigger than Y1.*0799

*That is this triangular region and we are looking at that color blue region.*0803

*The question is, whether Y1 and Y2 are independent.*0811

*If you are on top of your game right now, you already know the answer because there is a shortcut to the answer.*0815

*I’m going to go ahead and use the definition because that is what the example asks me to do.*0821

*I will use the definition, I will be very honest, I will work it out.*0826

*But kind of secretly that people who already know have already glance of that region and*0829

*there is a shortcut to the answer, that hopefully you already figured it out.*0834

*Let us go ahead and work it out.*0838

*What we are trying to figure out to test our definition of independence is whether F of Y1 Y2,*0841

*the joint density function separates into the two marginal density functions F1 of Y1, F2 of Y2.*0847

*It is a question there, whether those two were equal.*0858

*We will work it out, we will see if they are equal, and then we use that to determine if they are independent.*0862

*That is the definition of independence.*0866

*You got to remember the marginal density function F1 of Y1.*0869

*We did calculate this, this is one of the examples in the previous lecture.*0872

*I will go ahead and calculate it again.*0878

*We always have this variable switch, you always integrate over the other variables.*0879

*This is Y2, and Y2 in this case, goes from the line Y1 = Y2 or Y2 = Y1 to Y2 = 1.*0884

*In this case, we are integrating from Y2 = Y1 to Y2 = 1.*0899

*My joint density function is given in the problem, 6 × 1- Y2.*0905

*We are integrating with respect to Y2.*0914

*I forgot my D in there, that is very important, DY2.*0919

*I will just go ahead and integrate that.*0924

*The integral of 6 is 6Y2 -, the integral of 6Y2 is 3Y2²,*0927

*Y2² integrate that from Y2 = Y1 to Y2 = 1, which is 6 -3, -6Y1.*0936

*That is a Y2, it looks like it did not show up there.*0950

*I forgot my 3Y2², that is very important there.*0954

*-6Y1, - -, + 3Y1² and that simplifies a bit to 3Y1² - 6Y1 + 3.*0959

*Fair enough, that is my marginal density function for Y1.*0974

*F2 of Y2 is, we will switch the roles of the variables there.*0978

*We are integrating over Y1, Y1 goes from 0 up to Y1 = Y2.*0985

*Y1 = 0 to Y1 = Y2.*0994

*I’m doing this a little bit faster than I did in the previous videos.*0998

*We did figure out both of these marginal density functions, as examples in the previous video.*1002

*You can go back and check them out, if you want to see this work out a little more slowly.*1007

*6 × 1- Y2 DY1, I’m integrating with respect to Y1.*1011

*Be very careful here, not with respect to Y2.*1020

*6 × 1- Y2 × Y1, that is because Y2 is just a constant.*1023

*When we integrate with respect to Y1, evaluate that from Y1 = 0 to Y1 = Y2.*1030

*I get 6 × 1- Y2 × Y2.*1038

*Let me look at my condition that I'm trying to check here.*1044

*That is whether the joint density function splits apart into the two marginal density functions.*1048

*That is 6 × 1- Y2 is that equal to F1 of Y1 was this, 3Y1² - 6Y1.*1055

*It looks like I forgot a Y1 up there, + 3.*1068

*That is multiplied by 6 × 1- Y2 × Y2 F2 of Y2.*1073

*Clearly, if we expand out all that mess on the right, we are not going to get the equivalent expression on the left.*1080

*This does not work.*1087

*We can say that the Y1 and Y2 here, these variables Y1 and Y2 are not independent.*1091

*That is a formal check of how to determine whether or not these variables are independent.*1108

*Let me show you the secret shortcut that hopefully you had in mind, even before we started.*1116

*Without doing any calculus at all, I knew that as soon as I graphed this region that this was not independent.*1123

*That is by the theorem, Y1 and Y2 are not independent.*1134

*That is because the theorem said that the variables are independent if and only if, the region is a rectangle.*1152

*Another condition which I do not even have to check because I already know the region is not a rectangle.*1163

*Because the region is not a rectangle, it is a triangle.*1168

*That is another and much quicker way of solving this problem, is to invoke that theorem there.*1181

*That is the two different ways you could solve this problem.*1189

*The problem did ask you to use the definitions, that is why I worked it out from scratch.*1193

*I started with the joint density function and I want to see if it could be,*1198

*if it was really the product of the two marginal density functions.*1202

*I calculated the marginal density function F1 of Y1 and F2 of Y2.*1206

*Each one, you have to switch the variable that you are integrating with respect to.*1211

*F1 of Y1, we integrate with respect to Y2.*1215

*F2 of Y2, we integrate with respect to Y1.*1218

*And then, I describe this region separately, in terms of Y2 or in terms of Y1.*1222

*I ran it through these integral, did a little multivariable calculus.*1233

*And then, I multiply those two marginal density functions together to see whether*1237

*I would get back the joint density function that I started with.*1242

*Actually, I did not even bother to work out the multiplication because I could see that,*1245

*there is no way this is going to come out to be 6 × 1- Y2.*1249

*It is definitely, if you multiply out all this mess on the right, it is not going to work.*1254

*Therefore, by the definition of independence, Y1 and Y2 are not independent.*1258

*A quicker way that we could figure that out is, to use the theorem that I gave you on the third slide of this lecture.*1266

*It just says, first of all, look at the region and see if you got a rectangle.*1272

*If you have not got a rectangle then immediately, you know they are not independent.*1276

*If you have got a rectangle, there is another condition you need to check.*1281

*But, we could have stop as soon as we saw that region was a triangle, we know that they are not independent.*1284

*Let us move on and we are going to look at example 2 now.*1292

*F of Y1 Y2 is defined to be Y1 + Y2 and our region is a square.*1297

*Let me go ahead and graph that out.*1305

*We do not have the easy shortcut that we had on the previous example,*1307

*where we knew that they were not independent because the region was not a rectangle.*1313

*Here, a square counts as being a rectangle.*1318

*Here is Y2, here is Y1, there is 0 for both of them.*1321

*There is 1, there is 1, and so our region is just this very nice square*1325

*which means maybe they are independent because, at least the region is a square.*1330

*But, we are going to use the definition to calculate it out.*1335

*That means, we are going to need to find the two marginal density functions.*1338

*F1 of Y1 means you integrate over Y2.*1342

*It looks like I just integrate for Y2 = 0 to 1, Y2 = 1 of Y1 + Y2 DY2.*1348

*Be careful when you integrate, because you have to integrate keeping in mind that the variable is Y2.*1359

*Y1 is a constant, very common mistake that my students make when they are doing their probability homework,*1366

*is they cannot keep the variable straight, which one you are integrating.*1373

*We integrate Y1, that is a constant, the integral is just Y1 Y2.*1377

*Y2 is the variable, the integral is Y2²/2.*1384

*We want to evaluate all that from Y2 = 0 to Y2 = 1.*1389

*Let us see, when Y2 is 1, we will get Y1 + ½.*1397

*When Y2 is 0, it looks like both the terms dropout.*1407

*I found the marginal density function F1 of Y1.*1410

*F2 of Y2, if you look at the function that we started with, Y1 + Y2 is totally symmetric between Y1 and Y2.*1414

*The region is symmetric too, it is going to be the exact same calculation.*1423

*Just switch the roles of Y1 and Y2, you will end up with Y2 + ½ that is because everything is symmetric in this problem.*1428

*Let me make that a little more clear here, clear that I'm skipping a few steps because*1439

*I can tell that they are going to be the same, as the previous one.*1442

*You are just switching the roles of Y1 and Y2.*1446

*I want to check if Y1 and Y2 are independent.*1450

*I want to check if F of Y1 Y2, the joint density function, is equal to the product*1453

*of the two marginal density functions F1 of Y1 × F2 of Y2.*1462

*In this case, Y1 + Y2 is that equal to Y1 + ½ × Y2 + ½.*1470

*Now, if you multiply those out, no way that is going to be equal.*1480

*It is definitely not going to be equal.*1484

*Y1 and Y2 are not independent, that is the conclusion we have to draw from this.*1491

*Y1 and Y2 are not independent, by the original definition of independence.*1500

*The way I calculated that was, I really wanted to check, here is the definition of independence right here.*1523

*It says that the joint density function is equal to the product of the marginal density functions.*1528

*But I worked out the marginal density functions, F1 of Y1 means you integrate with respect to Y2.*1533

*I integrated the joint density function, I had to be careful there that Y2 was the variable and Y1 was just a constant.*1540

*That is why I got Y1 × Y2 here and Y2².*1548

*Worked out to Y1 + ½, it is a function of Y1.*1552

*Y2 works the exact same way, it gives you Y2 + ½.*1557

*When I plug those in, Y1 + ½ × Y2 + ½, if you multiply those together,*1562

*will you get the original joint density function that we start out with, no you do not get that.*1567

*They are not independent.*1573

*By the way, it is less obvious than it was in example 1.*1575

*In example 1, we had our region that was a triangle.*1579

*Immediately, the theorem told you that it was not independent.*1582

*In this case, our region was a square and that condition did not make it obvious anymore.*1587

*Maybe, you could have looked at Y1 + Y2 and said, can I factor that into a function of Y1 × a function of Y2.*1592

*And said, you cannot factor that, then you would have known that they are not independent.*1600

*The safest way is actually to check this definition and to calculate the marginal density functions,*1606

*and see if they multiply to the joint density function.*1612

*They do not, in this case, the variables are not independent.*1616

*In example 3, we have a discreet situation.*1623

*We are going to roll two dice, a red dice and a blue dice.*1627

*We are going to define the variables Y1 is what shows on the red dice and Y2 is the total.*1632

*You might think, Y1 be what shows on the red dice and Y2 is the blue dice.*1639

*We mixed them up a little bit to make it a little more interesting.*1643

*The question is, whether Y1 and Y2 are independent.*1648

*Let me just mention that there is an intuitive answer to this, which should make sense to you.*1652

*Intuition here is that, remember the intuition of independence is that if I tell you the value of one of the variables,*1659

*you will have some new information about the other.*1667

*In particular, this Y1, we know it is going to be somewhere between 1 and 6.*1672

*Y2 is going to be somewhere between 2 and 12 because it is the total showing on both dice.*1677

*The intuition is, if I tell you what is showing on one of the dice, or if I tell you what one of the variables is,*1684

*does it change what you might expect about the others, about the other one?*1697

*In this case, yes, it does.*1701

*Intuition is, it does change your prediction, that means these variables are dependent on each other,*1703

*they are not independent.*1713

*Let me write that down to make it clear.*1716

*No, they are not independent because, let me just give an example value here.*1718

*Let me say, suppose you roll these two dice and you are wondering what kind of roll you are going to get.*1729

*Suppose, you peeked at the red dice, if you get a 6, if Y1 = 6 that means you peek at the red dice and*1735

*say what dice came out to be a 6?*1746

*My prediction is, I’m more likely to have a high total.*1751

*That is going to change what I expect about Y2.*1755

*Then, Y2 is more likely to be large.*1759

*If I know that one die rolled very high, then, it is more likely that I got a large total.*1775

*In particular, if I just say I'm rolling two dice, I could get anything from 2 to 12.*1780

*But if you tell me that Y1 is 6, I know I'm not going to get 2 as a total, not anymore.*1786

*I know that I’m going to get at least 6 as the total.*1792

*That is a very strong intuitive hint that these variables do depend on each other, they are not independent.*1794

*Let me check it using the formulas as well.*1805

*I'm going to check P of Y1 Y2, this is the definition of independent.*1808

*It should be equal to P1 of Y1 × P2 of Y2.*1814

*I do not know whether that is true, if they are independent then they should be true.*1826

*I'm going to take some values of Y1 and Y2, I will go ahead and take those values that I mentioned.*1831

*Y1 is equal to 6, I’m going to pick those.*1837

*This formula should be true for all values.*1841

*If it is not going to be true, I can pick whatever values I want to illustrate that it is not true.*1844

*Y2, I will pick 12 just because I think that, if I know the red dice is 6,*1850

*I think that is going to change my probability of getting a 12.*1857

*Just think about whether the probability of 6/12 is equal to P1 of 6 × P2 of 12.*1860

*The probability of 6/12 means that I got a 6 on the red dice and a 12 total.*1873

*In order to get that, I have to get a 6 on the red and a 6 on the blue.*1880

*This is really the probability of 6-6.*1885

*The probability of rolling double 6 is 1/36, 1/6 × 1/6.*1890

*P1 of 6, what is my probability that the red dice is equal to 6, that is 1/6.*1898

*P2 of 12, what is my probability that my total is 12?*1907

*Again, to get 12, I have to get 6 on both dice, that is 1/36.*1912

*Now, is 1/36 equal to 1/6 × 1/36, sure is not.*1920

*That does not work out.*1928

*Since, we found some values for which that equality did not hold, we can say for sure,*1931

*that Y1 and Y2 are not independent.*1938

*That agrees with the intuitive answer that we already gave.*1944

*That does agree with the intuition, that is quite reassuring that our intuition is not completely off base*1951

*and the formulas do back it up.*1957

*Let me recap that.*1960

*We are rolling two dice, we have, what shows on the red dice and the total.*1962

*The question is, whether those are independent.*1967

*I do not think they are going to be independent because I think, if you to tell me what is going to come up on the red dice,*1969

*then I can probably say a little more about what the total is likely to be.*1976

*I will not be able to say exactly, but if you tell me that I get a 6 on the red dice,*1980

*then I know the total is somewhere between 7 and 12.*1984

*If you tell me that I get a 1 on the red dice, then I know the total is somewhere between 2 and 7.*1988

*It is really going to change, what I expect the total to be.*1993

*Similarly, if you tell me what the total is, maybe, I know a little more about what the red dice might be showing.*1997

*Like, if you tell me that the total is 12, I know the red dice is a 6 now.*2003

*That is the intuition there, which is that knowing one variable will influence what you predict for the other variable.*2007

*That means they are dependent on each other, which means they are not independent.*2016

*That is why I made that intuitive prediction.*2022

*In order to back it up, I checked it with the formula.*2025

*I just grabbed two values of Y1 and Y2, if they are independent then this formula should hold for all values of Y1 and Y2.*2027

*That is my definition of independence.*2038

*I'm going to check it out just with these two values, the Y1 and Y2, 6 and 12.*2042

*On the left, I’m finding the probability that the red dice is 6 and the total is 12 which means,*2048

*we must have rolled double 6.*2055

*The chance of getting a double 6 is 1/36.*2058

*On the right, P1 of 6 means what is the probability that the red dice is a 6, it is 1/6.*2061

*What is the probability that the total is 12?*2068

*Again, you would take double 6, that is 1/36.*2070

*I just check if the arithmetic works out and it does not.*2075

*1/36 is not equal to 1/6 × 1/36.*2078

*Y1 and Y2 are not independent and that confirms the intuitively that I made, at the beginning of the example.*2082

*In example 4, we have the joint density function F of Y1 Y2 is E ⁻Y1 + Y2.*2093

*My region here is the region on Y1 and Y2 both going from 0 to infinity.*2103

*There is Y1 and there is Y2, both regions go from 0 to infinity.*2116

*The question is, are Y1 and Y2 independent?*2123

*Again, there is sort of two ways I can check this.*2127

*One, is by using the original definition of independence.*2130

*And one, is by using the theorem that we got.*2136

*The faster way will actually be the theorem.*2139

*I'm going to check it from the definition first, just so that you understand that method.*2142

*And then, we will see how the theorem would actually be much faster.*2147

*We will check it out using the definition.*2151

*Remember, the definition of independence was that F of Y1 Y2 should be equal to,*2153

*should factor into the marginal density functions F1 of Y1 and F2 of Y2.*2161

*Let me work out what those are, those marginal density functions.*2170

*F1 of Y1, by definition, that means you switch the variable.*2174

*F1 of Y1, I was already thinking ahead to the integral that I’m about to solve.*2179

*I have to integrate over Y2, Y2= 0 to infinity, Y2 goes to infinity.*2184

*I will take a limit for that, of the joint density function E ^ (-Y1 + Y2) DY2.*2192

*When I solve that out, I’m integrating with respect to Y2.*2204

*If I think about that, I can factor out an E ^- Y2 and an E ^- Y1.*2210

*E ⁻Y1 will just be a constant, factor that right on out.*2219

*That is E ⁻Y1, I got the integral of E ⁻Y2 DY2.*2225

*And that is not such a bad integral, it is E ⁻Y1 × –E ⁻Y2.*2233

*I want to evaluate that from Y to = 0, and then take the limit as Y2 goes to infinity, that is E ⁻Y1.*2241

*Y2 going to infinity means, we are talking about E ⁻infinity, that is 1/E ⁺infinity or 1/infinity which is just 0.*2252

*-E⁰, -E⁰ which is 1, those two negatives cancel and I just get a +1.*2265

*I get E ⁻Y1, notice that I get a function of Y1 which is what I'm supposed to get,*2277

*when I take the marginal density function.*2283

*This is actually symmetric, F2 of Y2 was going to behave the exact same way.*2289

*I’m not going to belabor the details there, it is going to be E ⁻Y2.*2294

*I will check out my definition of independence, F of Y1 Y2.*2300

*Y2 is equal to, or possibly equal to F1 of Y1 × F2 of Y2.*2306

*E ⁻Y1 + Y2, that is my joint density function that I have been given.*2316

*F1 of Y1, I worked out was E ⁻Y1.*2323

*F2 of Y2, I worked out was E ⁻Y2.*2327

*Those could combine, we get E ^- Y1 + Y2.*2334

*In fact, it does work, that equality really holds true.*2338

*By the definition, yes, they are independent.*2346

*Y1 and Y2 are independent, checking from the definitions.*2360

*That is really very reassuring there.*2364

*Let me show you another way you could have done this problem, which is to use the theorem, two parts of the theorem.*2368

*By the way, I gave you this theorem in the third slide of these lectures.*2379

*Just check back and you will see the third slide.*2384

*The domain is a rectangle, it is an infinite rectangle.*2386

*But, that does check out here.*2396

*Remember, it is okay for it to be infinite, according to the theorem, it is okay if the rectangle is infinite.*2399

*I'm looking at this domain here, it is an infinite rectangle.*2406

*That is condition one of the theorem, that is satisfied.*2410

*Condition two of the theorem was that, the joint density function F of Y1 Y2*2415

*had to factor into a function of Y1 only × the function of Y2 only.*2423

*Let us check that out here.*2428

*E ⁻Y1 + Y2, yes, I can factor that into E ⁻Y1 × E ⁻Y2 which is a function of, let me just write that as G of Y1.*2431

*Y1 × H of Y2 because I do have Y1 only in the first function and Y2 only in the second function.*2453

*That second edition of the theorem is satisfied.*2464

*Once, I have checked both of those conditions, I can go to that same conclusion and say yes, they are independent.*2468

*I could have saved myself doing a lot of integration there, if I had used the theorem.*2475

*I want to make sure that you are comfortable using the definition.*2481

*But also, using the theorem which can save you lots of time, if you know how to use it.*2484

*Let me recap the steps here.*2489

*The first way that I want to check this problem was to look at the definition.*2491

*Does the joint density function factor into the two marginal density functions?*2495

*I had to calculate the two marginal density functions F1 of Y1, you switch the variables and*2501

*you integrate over Y2 with respect to Y2.*2508

*I looked at my range on Y2, that goes from 0 to infinity.*2512

*That is where I got these limits right here.*2517

*I integrated the joint density function that I was given, E ⁻Y1 + Y2.*2520

*That is where that came from.*2526

*That factors, it is really nice that if factors because we are integrating with respect to Y2, that means the term with Y1,*2529

*pulls right out of the integral and then I'm just doing an integral on Y2, pretty easy one.*2538

*Plug in my limits, infinity and 0, and I just simplify down to E ⁻Y1.*2544

*That was the marginal density function on Y1.*2553

*The exact same arithmetic occurs with Y2, except you are just switching the two variables.*2558

*I'm going to check that definition, does the joint density function separate into the two marginal density functions.*2566

*When I plugged everything in there, it look like I had a true equation.*2578

*Just by the definition, I get that the two variables are independent.*2582

*A quicker way to do that would have been, both the theorem from the third slide of this lecture.*2587

*But we look at the domain, that is looking at this domain right here.*2593

*It is an infinite rectangle, that does satisfy it is not a triangular region or anything like that.*2597

*If you look at the joint density function, we can factor it into a function of Y1 × a function of Y2.*2605

*They separate the variables there.*2612

*And that, right out there, would have been enough to confirm to me that these variables really are independent.*2615

*In our last example here on independent random variables,*2625

*we are given a joint density function of 4Y1 Y2 and our region, I will go ahead and graph it.*2629

*Y1 and Y2 both going from 0 to 1, here is Y2 and here is Y1.*2637

*We want to figure out, I seem to switch my variables for some reason, I certainly would not want do that.*2644

*There is Y1 and there is Y2, here is 0 and 1 on both axis.*2652

*There is my region, a very nice square.*2660

*By the way, if you are really on your game right now,*2663

*if you have been paying close attention to everything in this lecture, you already know the answer to this problem.*2666

*If you really know what is going on.*2672

*I'm going to take it a little slowly and we will work it through.*2674

*We will find an answer and then I will kind of comeback at the end, and show you how you could have done it very quickly.*2680

*If you knew quickly, what you are doing.*2685

*We are going to check the definition of independence which it asks us*2690

*whether the joint density function factors into the product of the marginal densities.*2695

*Let me find the marginal densities, F1 of Y1 is equal to, we integrate on Y2 here.*2700

*Y2 goes from 0 to 1, my joint density function is 4Y1 Y2, and we are integrating DY2.*2708

*If I integrate that, the Y1 is a constant and the integral of 4Y2 is 2Y2².*2718

*2Y2², we still have the constant Y1.*2727

*We integrate that or we evaluate that from Y2 = 0 to Y2 = 1.*2732

*If I plug in Y2 = 1, I just get 2Y1.*2739

*Y2 = 0 does mean nothing.*2745

*I have figured out my marginal density function F1 of Y1.*2747

*F2 of Y2, if I do this exact same arithmetic, everything is symmetric here.*2750

*We will just swap the value of the variables.*2759

*It is going to work out to be 2Y2, that would be the marginal density function for Y2.*2761

*I want to check using the definition of independence, is F of Y1 Y2 is it equal to F1 of Y1,*2768

*the thing that I just calculated × F2 of Y2.*2778

*Is that going to work out?*2783

*I will plug in everything and I will see if it works out.*2784

*F4 × Y1 Y2 is that equal to 2Y1, that is what I just calculated, × 2Y2.*2787

*That was still a question, but when I look at them, it is really is, 4Y1 Y2 is 2 × Y1 × 2 × Y2.*2798

*Low and behold, it does check.*2808

*I really confirmed by the definition that, yes, Y1 and Y2 are independent.*2811

*That is very reassuring, that they do come out to be independent.*2826

*However, I hope that some of you watching the video were kind of chuckling to yourselves all along,*2830

*because you knew this answer in advance.*2836

*Here is how you knew the answer in advance.*2839

*You remembered that theorem that I gave you on the third slide of the lecture.*2841

*You can go back and check that out.*2848

*There are two conditions you had to check.*2849

*The domain is a rectangle and a square, in this case, does qualify as a rectangle.*2852

*You checked right away that the domain is a rectangle, that is confirmed.*2861

*The second condition that you have to check is that the joint density function,*2867

*F of Y1 Y2 factors into a function of Y1 × Y2.*2871

*Let me go ahead and say it is 4Y1 Y2.*2883

*If you wanted, you could write that as 4Y1 × Y2.*2886

*Or you can put a 2 on each part, it really does not matter.*2893

*The important thing here is that it is a function of Y1 only, × a function of Y2 only.*2895

*It does indeed factor as it is supposed to, in order to be independent.*2903

*Both of those are probably things you could have checked in your head, if you really knew what was going on.*2911

*In that case, you knew right away from the beginning of the problem, by the theorem Y1 and Y2 are independent.*2918

*You could save yourself this integration and the tedious checking, and jump to the answer right away.*2925

*Let me recap that problem.*2933

*I wanted to check this definition of independence,*2935

*that the joint density function does occur as the product of the two marginal density functions.*2939

*That meant, I had to calculate the two marginal density functions.*2945

*F1 of Y1, to calculate the marginal density function, you integrate over the other variable.*2949

*I put my Y2 there and my region is Y2 goes from 0 to 1.*2956

*That is how I got those limits right there, Y2 goes from 0 to 1.*2963

*My joint density function is 4Y1 Y2, I got that from the stem of the problem there.*2967

*Integrate that with respect to Y2.*2976

*The Y1 just comes along as a constant, the integral of 4Y2 is 2Y2².*2979

*Plug in the values for Y2, I get 2Y1.*2986

*The F2 of Y2 is completely symmetric.*2990

*The function and the region are both symmetric.*2994

*It is going to work out to be 2Y2.*2997

*And then, if I plug those both in, the 2Y1 and the 2Y2, I multiply them together and look,*3001

*I really do get the original joint density function that we started with.*3006

*It really did work out, in this example.*3011

*This is kind of a special example, if you check back some of the previous examples, example 1 and 2,*3013

*it did not work out when we multiply those.*3018

*In this case, it did work out and we can say that the variables are independent.*3020

*The quicker way to do that is to use the theorem that we had in the third slide of this lecture,*3026

*is to just look at this domain right here, and say that is a square.*3032

*A square counts as being a rectangle.*3035

*And then, you look quickly at the joint density function and say can I factor it somehow,*3038

*with all the Y1 on one side and all the Y2 in the other factor.*3044

*Yes I can, I can factor it just like this, all the Y1 in one part and all the Y2 on the other part.*3049

*I have successfully separated it into a function of Y1 × the function of Y2.*3055

*And that was the second condition that you had to check with the theorem.*3062

*The theorem says that both of those conditions are met, then Y1 and Y2 really are independent.*3065

*That is by far, a faster way to know quickly whether your variables are independent.*3071

*That wraps up this lecture on independent random variables.*3078

*This is part of the larger chapter on Bivariate distributions and Bivariate density functions.*3083

*In turn, that is part of the probability videos here on www.educator.com.*3090

*My name is Will Murray and I thank you very much for joining me today, bye now.*3096

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