For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

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### Marginal Probability

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 0:00
- Discrete Case 0:48
- Marginal Probability Functions
- Continuous Case 3:07
- Marginal Density Functions
- Example I: Compute the Marginal Probability Function 5:58
- Example II: Compute the Marginal Probability Function 14:07
- Example III: Marginal Density Function 24:01
- Example IV: Marginal Density Function 30:47
- Example V: Marginal Density Function 36:05

### Introduction to Probability Online Course

### Transcription: Marginal Probability

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

*We are working through a series of videos on experiments involving two variables.*0005

*Right now, we will always have a Y1 and a Y2 in all of our experiments.*0011

*We are talking about joint density functions and things like that.*0016

*Today, we will talk about marginal probability.*0020

*Marginal probability is kind of a tool that you use in the service*0024

*of calculating conditional probability and conditional expectation.*0028

*We do have another video that comes after this one where you will see this being used*0033

*in the service of conditional probability and conditional expectation.*0038

*In this video, we are just going to learn what marginal probability is.*0042

*We will learn how to calculate it, we will practice it with some examples.*0046

*Marginal probability is something you can talk about with discrete probability or with continuous probability.*0052

*As I said, we have an experiment with two random variables Y1 and Y2.*0058

*We are going to talk about the marginal probability function.*0064

*There is a marginal probability function for Y1 and then there is a separate marginal probability function for Y2.*0068

*What they mean is, the marginal probability function of Y1, we call it P1 of Y sub 1.*0075

*By definition, the := that means it is defined to be.*0082

*It is defined to be the probability that Y1 will be a particular value of y1.*0087

*The way you would find that is, you would look at all the possible values of Y2.*0094

*Add up all the possible values of Y2 and then find the probability of each combination*0100

*of that particular fixed value of Y1 with all the different possible values of Y2.*0106

*Let me emphasize here that, that Y1 there is fixed and we are adding up over all the possible values of Y2.*0113

*And then, you can also talk about the marginal probability function of Y2 which means you have a fixed value of Y2.*0123

*You are trying to find the probability of getting that particular value of Y2.*0130

*The way you find that, is you add up all the probabilities of combinations with all the different possible y1s.*0135

*This is really quite confusing for students because there is a sort of a subscript change here.*0143

*We are finding the marginal probability function for Y1, notice that there are 1’s in the subscripts there.*0148

*What we will do is we add up over all the possible values of Y2, and then, vice versa.*0155

*When we are finding the marginal probability function for Y2, we add up over all the possible values of Y1.*0161

*That gets a little confusing, we will do some practice with this.*0168

*You will see in the examples how it works out.*0170

*That is just the discrete case, the continuous case is very much analogous.*0173

*We are going to take a look at that with the continuous case but you will see it is kind of the same thing.*0179

*Except that, we are going to change the summation signs to integral signs.*0183

*The continuous case, we will talk about the marginal density function of Y1 and Y2.*0189

*For Y1, we talk about F sub 1 of Y1.*0195

*Again, we add up, instead of P of Y1 Y2, it is F of Y1 Y2, the joint density function.*0200

*We take the integral over all possible values of Y2.*0207

*To make it most general here, I have written from Y2 goes from -infinity to infinity.*0213

*For the marginal density function of Y2, we add up or we take the integral over all possible values of Y1.*0221

*I wrote Y1 goes from -infinity to infinity.*0231

*It is true in many of our experiments and many of our problems,*0237

*but we do not actually have distributions covering an infinite range.*0242

*You would not actually have to integrate from -infinity to infinity.*0246

*The point here is, you just integrate over whatever the range is for that particular variable.*0250

*Here, you just integrate over the full range for Y1.*0257

*Here, you just integrate over the full range for Y2.*0263

*Whatever those full ranges are, for whatever that variable is, that is what you integrate over.*0272

*Again, we had that same subscript change that I mentioned with the discrete case,*0277

*which is that, when you are finding the marginal density function for Y1,*0281

*what you end up doing is you integrate over Y2.*0286

*Be very careful about that, when we are finding marginal density function for Y1, your variable of integration is Y2.*0290

*And of course, vice versa , when you are finding the marginal density function for Y2, your variable of integration is Y1.*0298

*That does get to be a little confusing, we will try to keep it straight.*0307

*As I said, the marginal density functions are, there are something that are used in computing conditional probability.*0311

*We will come back in the next video, the next day lecture, and we will see how these are used.*0322

*The purpose of this video is just to practice calculating the marginal probability functions.*0329

*It may not be that enlightening, why we are doing this at this point.*0333

*We are going to work through some examples, calculate the marginal probability functions, and just see what we get.*0337

*In the next video, we will come back and we will calculate conditional probability,*0345

*conditional expectation, using these marginal probability functions.*0349

*And you will see how we can actually apply these to real settings.*0353

*In example 1, we are going to roll to dice, a red dice and a blue dice.*0359

*The variables are not going to be the traditional ones you might think of.*0363

*Often, you would say Y1 is going to be the value of the red dice, Y2 is the value of the blue dice.*0366

*I changed that around a little bit to make it a little more interesting.*0371

*Y1 is going to be the value showing on the red dice.*0374

*Y1 could be anywhere from, when you roll a dice, you can get anywhere from 1 up to 6.*0380

*Y2 is the total showing on both dice, that can be anywhere from 2 up to 12.*0388

*They are not exactly symmetric, these two variables here.*0397

*We are going to calculate the marginal probability function of Y1.*0399

*By the way, we are going to do the same example again for example 2,*0404

*except that we will calculate the marginal probability function of Y2.*0407

*We will get to see both of them, in some kind of different behavior there.*0411

*Let us figure out the marginal probability function of Y1.*0416

*We showed two ways to calculate this.*0421

*Let me remind you the definition of the marginal probability function.*0423

*P1 of Y1, by definition, is the probability that Y1 is going to be equal to that particular value of Y1.*0427

*Then, the other way to think about it is, you add up over all the possible values of Y2,*0437

*of the probability of each combination Y1 and Y2.*0443

*I showed two ways to think about that.*0449

*I will calculate at least a couple of these using both ways and I will try to figure out which one is easier.*0453

*I’m going to look at all the possible values for Y1, and that is all the values from 1 to 6.*0459

*When Y1 is 1, P1 of 1, one way to think about it is, what is the probability that we are going to get that particular value of 1.*0466

*That is just the probability that the red dice comes up to be 1.*0477

*That is definitely P of Y1 = 1, that is definitely 1/6.*0482

*Another way to think about that is to say, it is the sum from Y2, all the possible values of Y2 which is 2 up to 12 of 1, Y2.*0489

*Then, we would add up P of 1, 2 + P of 1, 3, all way up to the probability of 1, 12.*0504

*That should not be 1/12, that should be 1, 12.*0519

*We asked ourselves, what the probability of each one of those combinations is?*0523

*Let us think about what the probability of 1, 2 is.*0528

*That means you got a 1 on the red dice and a total of 2, which the blue dice would have to be the 1 as well.*0531

*The probability of getting a 1,1 in the two dice is 1/36.*0539

*The probability of getting a 1 in the red dice and 3 total, means you have to get 1 red and 2 on the blue dice, that is also 1/36.*0546

*All the way up to here to the probability of getting a 1 on the red dice and a 7 total, which is still 1/36.*0558

*The probability of 1 on the red dice and getting 8 total.*0569

*What is the probability of getting a 1 on the red dice and 8 on the total?*0575

*In order to get that, you have to get a 7 on the blue dice.*0579

*You cannot get a 7 on a single roll of a dice, that probability is 0.*0582

*In turn, the probability of getting a 1 on the red dice and a 9 total is still 0.*0590

*All the way up to getting 1 on the red dice and 12 total, that probability is 0.*0597

*What we got here is adding up some fractions, they are all 1/36.*0602

*I think there is going to be 2, 3, 4, 5, 6, 7, that was 6 total.*0607

*It is 6/36, that is 1/6.*0612

*That agrees with what we had earlier, when we figure out the probability of getting 1 on the red dice.*0618

*We showed two different ways to calculate this.*0625

*In this case, if you look around, clearly one of them is much easier.*0628

*It was much easier to calculate just looking at the probability that Y1 is 1,*0632

*then by breaking it down over all the possible combinations of Y2.*0637

*I have to calculate the same kinds of things for all the other possible values of Y1.*0643

*But, I definitely going to use the first technique because it seems much easier.*0648

*P1 of 2 is the probability that Y1 is equal to 2.*0652

*Again, that is the probability that I'm getting a 2 on the red dice and that is also 1/6.*0660

*I could break it up using this long method that I did before, but I think it is clear now that,*0666

*that is not the most efficient method.*0675

*I could write it as probability of 2,2 + the probability of 2,3 all the way up to the probability of 2,12.*0678

*I could figure out each one of those probabilities but it is clear that that would take much longer.*0689

*I’m going to ignore that.*0694

*I will go ahead and calculate the rest of my probability function.*0696

*P1 of 3 it is the probability that Y1 is 3.*0700

*Again, it is getting 3 on the red dice, your probability is 1/6.*0706

*P1 of 4 is 1/6, P1 of 5 is 1/6, and P1 of 6 is 1/6.*0712

*All of those come out to be 1/6.*0728

*My marginal probability function P1 of Y1 is just given by P1 of 1 is 1/6, P1 of 2 is 1/6, P1 of 3 is 1/6, and so on,*0732

*for all the other possible values of Y1.*0746

*That gives us the probability function, that finishes example 1.*0750

*Let me recap that.*0754

*There are sort of 2 ways we could have calculated that.*0755

*We are going to look at each possible value of Y1.*0758

*And then, for each one, we can either calculate the probability that Y1 is that value,*0762

*or we can expand it out over all the possible values of Y2 and add up all the individual combinations.*0767

*What we discovered is that although it would be possible to do it by expanding it out,*0774

*we actually did it for the case of Y1 = 1 here.*0778

*It is much easier just to find the probability that Y1 = Y1.*0782

*For each one of the values, 1, 2, 3, 4, 5, and 6, we got probabilities of 1/6.*0788

*That would really answer the question.*0799

*If we want to do it the long way, we would have to look at all the possible values of Y2 going from 2 to 12.*0801

*Add up the probability of 1 and those values for each one.*0808

*For some of them we get 1/3 because those are the probabilities of rolling a 1 on the red dice and*0812

*you would have to be 1 on the blue dice.*0820

*Or 1 on the red dice and you do not have to be 2 on the blue dice.*0822

*Those will give you 1/36.*0825

*If we look at the combinations that have 1 on the red dice and anything bigger than 8 as a total, that cannot happen.*0827

*All those probabilities were 0, we will end up with 6/36 which gives us that same 1/6*0835

*that we got before, but it took much more work to find it that way.*0840

*Let us avoid that way, if we can.*0844

*In example 2, we have a following up on example 1.*0849

*We are rolling two dice, there is a red dice and a blue dice, same two variables as before.*0852

*Y1 is what shows on the red dice and Y2 is the total.*0858

*We are going to calculate the marginal probability function P2 of Y2.*0864

*Let me remind you that Y1 can take values from 1 to 6 here, because the red dice can show anything from 1 to 6.*0870

*Y2 is the total, that could be anywhere from 2 to 12.*0877

*We want to find the probabilities of each one of those individual values of Y2.*0883

*We have sort of two ways we can think about this.*0892

*P2 of Y2, one way to think about it is the probability that Y2 is going to be that particular value of y2.*0895

*The way I want to think about it is, to add up the probabilities of all the combinations of P1 of Y1 Y2 over all the possible values of Y1.*0903

*We will the try a couple those out and we will see, maybe, which one is easier in each case.*0916

*We want to look at this for each possible value of Y2.*0924

*The first possible value of Y2 is 2, P2 of 2.*0928

*I will do it the long way first, I will add up over all the Y1.*0935

*That could be 1, 2, 3, all the way up to 6.*0941

*Let us figure out what the probabilities are for each one of those.*0955

*The probability of 1,2, that means you are getting a 1 on the red dice and a 2 total, which mean the blue dice would have to be a 1.*0957

*That probability of getting 1 on both dice is 1/36.*0967

*The probability of 2,2 means you get a 1 on the red dice and we have to get a 0 on the blue dice, to add up to 2.*0973

*That cannot happen, there is a 0 there.*0981

*The probability of getting a 3 on the red dice and a 2 total.*0984

*Certainly, it is not going to happen, and nor or any of these other possibilities, that is just 1/36.*0987

*The probability of, let us try calculating that directly because I think it might be a little faster.*0999

*P2 of 2 is also equal to the probability that Y2 is equal 2, which means what is the probability of getting a total of 2.*1008

*In order to get a total of 2, you have to get a 1 on the red dice and 1 on the blue dice, that is exactly 1/36.*1021

*Look, that is really much faster to calculate it that way.*1030

*Why do not we calculate it that way, from now on.*1034

*P2 of 3, I will just show a long way for one more but I also show the short way, and we will see how it goes.*1039

*That is the probability of 1, 3 + the probability of 2, 3 + the probability of 3, 3, up to the probability of 6, 3.*1048

*The probability of 1, 3 that means 1 on the red dice and 2 total.*1062

*I’m sorry, 3 total which means 2 on the blue dice.*1067

*That would mean that you have to get to 1 on the red dice, 2 on the blue dice, there is a 1/36 chance of that.*1072

*2 on the red dice and 1 on the blue dice, it would also be 1/36.*1080

*And then, 3 on the red dice would have to be 0 on the blue dice, that cannot happen.*1086

*All these others are 0, I will just write it as 2/36.*1092

*I'm not going to simplify that, of course, you could simplify it to 1/18.*1097

*I think it will be easier to spot a pattern, if I leave it unsimplified.*1101

*I will just leave it as 2/36.*1105

*Another way to calculate that would be, the probability that Y2 is equal to 3.*1108

*What is the probability of getting a total of 3, when you roll two dice?*1114

*The way you can get a total of 3 is red dice 1 blue dice 2 or blue dice 1 red dice 2.*1122

*There are two ways to get that, out of 36 possible rolls, that is 2 out of 36.*1132

*I think that is really much shorter.*1139

*I’m just going to calculate the others using the short way because otherwise, I will run out of space here.*1140

*P2 of 4, how many different ways are there to get a 4, when you roll two dice?*1148

*We could do 1-3, 2-2, or 3-1, that is 3 out of 36.*1154

*Again, I think I’m not going to simplify that, even though it is obvious that it could simplify.*1160

*But, it is going to be easy to spot a pattern.*1166

*In fact, maybe you already can spot a pattern because we got 1/36, 2/36, 3/36.*1168

*P2 of 5, how many ways are there to get a 5?*1176

*It could go 1-4, 2-3, 3-2, or 4-1, that is 4 out of 36.*1180

*P2 of 6, how many ways are there to get a 6?*1189

*There are 5 different ways because you can go 5-1, 4-2, 3-3, 2-4, or 1-5.*1194

*P2 of 7 is 6 out of 36 because there are 6 different ways that you can get a total of 7.*1205

*You can go 6-1, 5-2, 4-3, 3-4, 2-5, 1-6.*1215

*P2 of 8, these patterns changed its course here.*1228

*We had 1 out of 36, 2 out of 36, 4 out of 36, 5 out of 36, 6 out of 36.*1237

*You might think logically that we are going to go 7 out of 36.*1243

*It actually peaks at P2 of 7, it trails back to 5 out of 36.*1246

*The reason for that is that, there are only five ways to get an 8, when you roll two dice.*1252

*You can go 6-2, 5-3, 4-4, 3-5, or 2-6.*1259

*Your pattern changed it course and it starts to drop off again.*1268

*P2 of 9, there are four ways to get a 9 because it can go 6-3, 5-4, 4-5, 3-6, that is 4 out of 36.*1273

*P2 of 10 is 3 out of 36, there are three ways to get a 10.*1283

*P2 of 11 is 2 out of 36, and finally, P2 of 12, how many ways are there to get a 12?*1291

*There is just one way you have to get double 6, 1 out of 36.*1301

*All of these, you could calculate using the expanded form.*1306

*But the expanded form was obviously taking much too long.*1311

*I got my answer more easily, just by thinking directly about Y2, about the total of the dice.*1315

*That gives my answers, I found the marginal probably function P2 of Y2.*1323

*1/36 from 2, 2/36 for 3, and so on.*1329

*Going down for all the possible values of Y2, I gave you a fraction representing the probability.*1334

*To recap there, there are two ways you can calculate this.*1343

*You can calculate directly the probability that Y2 will be equal to any given value y2.*1346

*Or you can sum it up over all the possible Y1.*1353

*Summing it up over all the possible Y1, we did that for Y2 = 2.*1357

*We took Y1 = 1, 2, 3, 4, 5, 6.*1362

*Discovered that only one of those gives us any probability because for all the other possible values of Y1,*1368

*there is no way to make the given total.*1373

*We just got 1/36.*1376

*But then, a much easier way we discover to calculate it was that, just to find the probability that the total will be 2.*1378

*The only way to get that is by getting a 1-1, when you roll, it is 1/36.*1385

*P2 of 3 means we sum up over all the possible values of Y1, 1, 2, 3, 4, 5, 6.*1391

*Two of them gives us positive probabilities, the rest are all 0.*1399

*It gives us 2 out of 36.*1403

*But then, the probability directly that Y2 is 3 just means you are looking at two different possible rolls,*1406

*a 1 and 2 or 2 and 1.*1414

*You will get 2 out of 36 that way.*1416

*It really looks like that is the shorter way.*1418

*For all of these other values, I just thought about all the possible ways to get the given total and then counted them up.*1420

*That is how I got these different numbers, that seems to be a better way to solve that.*1427

*At least, in the discreet case.*1432

*We will see in the next few examples, how we are going to use an integral to solve the same kind of thing in the continuous case.*1434

*In examples 3, we have a continuous problem with the marginal probability.*1443

*We are given the joint density function F of Y1 Y2 is 6 × (1 - Y2).*1450

*It is important to pay attention to the domain that we are given here which is that Y1 and Y2 are both between 0 and 1.*1458

*Let me go ahead and draw this out, as I talk about it.*1467

*Y1 and Y2, there is Y1, goes from 0 to 1 and Y2 also goes from 0 to 1.*1470

*But we are told that Y2 has to be bigger than or equal to Y1.*1479

*If you think about that, in terms of X and Y, we got the same values bigger than X.*1486

*Let me go ahead and draw the line Y = X.*1491

*There it is, right there.*1496

*We want the region that is above that line because we want Y2 to be bigger than or equal to Y1.*1498

*It is that triangular region above that diagonal line there.*1506

*Let me colored that in blue.*1510

*That is the region that we are looking at.*1513

*We have been asked now to find the marginal density function F1 of Y1.*1517

*Let me remind you of the definition of marginal density function.*1522

*F1 of Y1, remember, there is this variable change where, if you are trying to find F1 of Y1,*1527

*what you do is you integrate over Y2.*1534

*And then, you find the density function F of Y1 Y2, the joint density function.*1538

*You integrate that with respect to Y2.*1543

*That means we have to describe this region, in terms of Y2.*1546

*Let me draw what the region looks like from the perspective of Y2.*1552

*Y2 rows up from that diagonal line up to, let me draw that a little bit longer, up to the line Y2 = 1.*1559

*If we describe that region, in terms of Y2, we would say, the diagonal line was Y1 = Y2 or Y2 = Y1.*1572

*We would say Y2, the limits on Y2, if you want to describe that region, the Y2 goes from Y1, the diagonal line, up to 1.*1583

*That is what we use as the limits on our integral.*1596

*This is the integral from Y2 = Y1 to Y2 = 1.*1600

*The density function, I’m just going fill it in, I will expand it out.*1606

*6 -6 Y2, and we are going to integrate this with respect to Y2.*1611

*That is a pretty easy integral, the integral of 6 is just 6Y2.*1618

*The integral of 6Y2 is 3Y2².*1624

*I’m supposed to evaluate that from Y2 = Y1 to Y2 = 1.*1629

*I said Y2 = 1, I should have said Y2 = Y1 on the lower limit there and Y2 = 1 on the upper limit.*1637

*If we plug those in, we get 6.*1646

*I said 3Y2, it should have been 3Y2² because we are integrating 6Y2 there.*1649

*If we plug in Y2 = 1, then we get 6 -3.*1657

*Plugging in Y2 = Y1 for the lower limit, I get -6Y1 + 3Y1².*1662

*I can simplify that a little bit and rearrange a bit.*1676

*F1 of Y1 is 3Y1² -6 Y1 + 6 -3 is + 3, that is my marginal density function for Y1.*1680

*Let me recap the steps there.*1696

*I want to find the marginal density function F1 of Y1.*1699

*There is this variable switch where you integrate the joint density function over Y2.*1702

*You have to describe your region, in terms of Y2.*1711

*Our region, in terms of Y2, goes from the lower line, the diagonal line Y2 = Y1 to that upper bound, is Y2 = 1.*1715

*That is where I got these limits from, it comes from that graph right there.*1726

*That is where I got those limits.*1731

*And then, I plug those limits into the integral.*1733

*I also plugged in the density function that we are given.*1736

*I integrate that with respect to Y2 and that is a pretty easy integral.*1739

*I plug in my limits and what I end up with is a function, in terms of Y1 which is what is supposed to happen.*1743

*You always want your marginal density function of Y1 to be a function of Y1.*1751

*In the next example, in example 4 of this lecture, what we are going to do is*1756

*we are going to come back and look at this same scenario.*1760

*We are going to find the marginal density function of Y2.*1764

*If we do it right then we will get a function of Y2.*1767

*This is important to notice that we get a function of Y1 at the end here.*1772

*If we had not, if we had some Y2 in our answer, we know we would have made some kind of mistake.*1779

*By the way, this example is kind of a setup from example that we are going to use in the next lecture*1784

*on conditional probability and conditional expectation.*1792

*We would use this answer in the next lecture.*1796

*You want to make sure that you understand this.*1800

*We are going to use it for, I think it is going to be example 3 in the next lecture*1802

*which is on conditional probability and conditional expectation.*1816

*If you are wondering, what do we just calculate here, probability.*1824

*What is it good for, what we can use this for, if you want to take a sneak peek ahead,*1830

*just get forward to the next lecture and look at example 3.*1836

*You will see the same setup and you will see the same function.*1839

*We are actually going to use it to calculate some probability.*1841

*In example 4, we are looking at the same setup that we had in example 3.*1848

*But, instead of calculating the marginal density function in terms of Y1,*1852

*that is what we did back in example 3, we are going to calculate the marginal density function in terms of Y2.*1856

*Let me just redraw that graph, it is the same graph that we had in example 3.*1864

*We have Y1 and Y2 here, they both go from 0 to 1 but we are given that Y2 is bigger than Y1,*1870

*which means we are only going to stay above the line, the Y =X line, the Y2 = Y1 line.*1879

*We are looking at this triangular region above the line Y=X.*1888

*That is the domain of definition for our joint density function.*1893

*Now, we are trying to find the marginal density function F2 of Y2.*1900

*Remember, there is always this variable switch.*1904

*When you try to find the marginal density function for Y2, you integrate over Y1 and then*1906

*you integrate the joint density function F of Y1 Y2, your variable is DY1.*1912

*That means, we have to describe this region in terms of Y1.*1919

*Let us think about what Y1 does here.*1925

*Y1 grows up from the left hand side is the vertical line Y1 = 0.*1928

*The right hand side is the line Y1 = Y2, the diagonal line there.*1936

*My limits here on Y1 are from 0 to Y2, that is what we are going to use to set up the integral.*1942

*We had the integral from Y1 = 0 to Y1 = Y2.*1953

*Our joint density function is still 6 × 1 - Y2.*1959

*We have DY1.*1965

*There is a very seductive mistake to be made here and my students often make it.*1969

*It is very easy to make.*1975

*You see a Y2 in your integral and you want to integrate it with respect to Y2.*1976

*You see that Y2 and think that the integral is Y2²/2.*1985

*Not so, because we are integrating with respect to Y1.*1991

*Y1 is our variable of integration, when we see the Y2 that is just a big old constant.*1996

*This integral 6 × 1 - Y2, integrate that just as a constant, we just get all that × Y1.*2002

*We are going to evaluate that from Y1 = 0 to Y1 = Y2.*2014

*We get 6 × 1 - Y2 × Y2, when Y1 is 0 we get nothing, I just get -0.*2024

*If I simplify this down a little bit, I will get 6Y2.*2034

*I’m just going to distribute everything across.*2039

*6Y2 -6Y2², that is my joint density function in terms of Y2.*2041

*Notice, as I mentioned before in example 3, we do want to get a function in terms of Y2.*2051

*That is a function of Y2 which is a good thing, that kind of confirms that I have been doing at least some of my work correctly.*2060

*Because, I do not want to get a function that involves Y1 in any way, at the end of this*2070

*because the joint density function is always a function of Y2.*2074

*That finishes off example 4, but let me recap the steps here.*2082

*This is the same graph that we drew for example 3.*2086

*It comes from looking at this definition of the region here.*2090

*We draw the graph, the same graph we have for example 3.*2094

*But, the difference from example 3 now is that, we are finding the marginal density function of Y2, instead of Y1.*2097

*There is this variable switch that means we are integrating over Y1.*2104

*I need to describe my region, in terms of bounds for Y1.*2112

*That is why I drew a horizontal line here, instead of the vertical line that we had before.*2117

*I describe this region as going from Y1 = 0 to the diagonal line Y1 = Y2.*2122

*That gave me my limits that I use on the integral.*2128

*It is relatively easy integral, as long as you do not fall into the trap of thinking, look I have a Y2, I will just integrate that Y2.*2133

*Remember, we are integrating with respect to Y1.*2145

*The 6 × 1 - Y2 is a constant.*2148

*You just get that constant × Y1, plug in our values for Y1, and we end up with a function of Y2.*2151

*That represents our marginal density function, in terms of Y2.*2158

*In our last example here, example 5, we have a joint density function F of Y1 Y2 = E ⁻Y2.*2167

*Our region here, both Y1 and Y2 go from 0 to infinity.*2176

*Let me set up my axis and we will try to draw that.*2183

*There is Y1, always on the horizontal axis and Y2 is always on the vertical axis.*2185

*They both start at 0.*2191

*The catch here is that Y2 was always bigger than Y1.*2193

*Let me draw the line Y1 = Y2 or Y = X.*2198

*We want Y2 bigger than Y1 which means we want the region above this line.*2203

*Let me color in that region above this line.*2209

*Of course, it goes on to infinity.*2212

*I’m not going to draw all of it but that is the general shape.*2215

*It is this triangular region.*2218

*What we want to do is, find the marginal density function F1 of Y1.*2220

*Let me remind you of the definition of marginal density.*2225

*F1 of Y1, I gave you the formula for this back on the third slide to this lecture.*2228

*There is this variable switch where you always integrate over the other variable.*2236

*It is the integral over Y2 of the joint density function F of Y1 Y2 DY2.*2240

*I need to describe that region, in terms of Y2 so that I can set up my limits on the integral.*2251

*I want to think about what the limits would be on that region, in terms of Y2.*2262

*Y2, that means take a vertical arrow here.*2268

*Y2 grows up from that line, it goes on to infinity, does not it.*2272

*There is infinity and there is the line Y2 = Y1.*2278

*My region can be described as Y2 goes from the diagonal line Y1 up to infinity.*2284

*Those are the bounds on my integral, Y2 = Y1 to Y2 goes to infinity.*2296

*My joint density function is just E ^- Y2.*2306

*I’m integrating with respect to Y2.*2310

*The integral of E ⁻Y2 with respect to Y2 is - E ^- Y2.*2315

*I did a u substitution in my head to get that integral.*2322

*I want to evaluate that from Y2 = Y1 = Y1 to Y2, take the limit as it goes to infinity.*2328

*If I plug in infinity, that is E ^- infinity.*2339

*Which means, 1 divided by E ⁺infinity.*2346

*1/ infinity is just 0 - - E ⁻Y1 because I plugged in Y2 = Y1.*2350

*That simplifies down to E ⁻Y1.*2361

*Let me remind you what we are calculating here.*2367

*F1 of Y1, the marginal density function Y1.*2370

*It is kind of reassuring here that, when we calculate that, we get a function of Y1.*2374

*That is what we should get, when we calculate a marginal density function.*2380

*That is our answer, let me recap the steps on this.*2386

*First, I tried to graph this region.*2392

*Y1 and Y2 both go from 0 to infinity but we are looking at the region where Y2 is bigger than Y1.*2395

*That is why we have this upper triangular region here, all the region above the line Y=X there.*2402

*Then, to find the marginal density function for Y1, we have to integrate over Y2.*2410

*There is always this variable switch, that is always very confusing.*2416

*In order to find the region of integration, I tried to describe that region in terms of Y2.*2422

*The lower bound for Y2 is this diagonal line, that is where I got this Y2 = Y1.*2429

*Upper bound, it goes on forever, that is why there is an infinity there.*2434

*I plugged those in and I get this region of integration.*2438

*The density function just comes from the stem of the problem E ⁻Y2.*2443

*Integrate that with respect to Y2, did a u substitution in my head, u = -Y2.*2448

*It came up with –E ⁻Y2, and then I plugged in Y2.*2454

*The limit is, as it goes to infinity that just takes you to 0 and Y2 = Y1 gives me my Y1 in the exponent.*2459

*It simplifies down to this nice function of Y1 which is appropriate, because we should get a function of Y1,*2467

*when we are looking for the marginal density function of Y1.*2475

*I really want you to understand this example, because we are going to use this example again*2479

*for calculating some conditional and expected probability, in the next lecture.*2485

*I think this example is set to be example 5, in the next lecture.*2492

*We are going to use the answer to this example in the next lecture which is on conditional probability and conditional expectation.*2499

*You will see this coming back, I want to make sure you understand this now so that when we dropped the answer in,*2514

*you will not be confused by it, in the next lecture probability.*2520

*Make sure that everything is good with this example.*2528

*In the meantime, that wraps up this lecture on marginal probability.*2532

*As I mentioned in the beginning, marginal probability is really a tool in the service of conditional probability.*2536

*You will see a lot of the stuff used in the next lecture on condition probability and conditional expectation.*2541

*I hope you will stick around for this.*2547

*These are a part of a larger series of lectures on probability, here on www.educator.com.*2549

*My name is Will Murray, thank you for joining us today, bye.*2556

1 answer

Last reply by: Dr. William Murray

Mon Dec 14, 2015 10:12 AM

Post by Alexander Karakosta on December 12, 2015

using Y1 and Y2 makes it more difficult than it already is to follow some of these notes. I don't understand why you would want to use this method instead of a traditional X and Y.