For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Bivariate Density & Distribution Functions

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
- Bivariate Density Functions
- Properties of the Density Function
- Properties of the Density Function 1
- Properties of the Density Function 2
- We Can Calculate Probabilities
- If You Have a Discrete Distribution
- Bivariate Distribution Functions
- Bivariate Distribution Functions
- Properties of the Bivariate Distribution Functions 1
- Properties of the Bivariate Distribution Functions 2
- Example I: Bivariate Density & Distribution Functions
- Example II: Bivariate Density & Distribution Functions
- Example III: Bivariate Density & Distribution Functions
- Example IV: Bivariate Density & Distribution Functions
- Example V: Bivariate Density & Distribution Functions

- Intro 0:00
- Bivariate Density Functions 0:21
- Two Variables
- Bivariate Density Function
- Properties of the Density Function 1:57
- Properties of the Density Function 1
- Properties of the Density Function 2
- We Can Calculate Probabilities
- If You Have a Discrete Distribution
- Bivariate Distribution Functions 5:25
- Bivariate Distribution Functions
- Properties of the Bivariate Distribution Functions 1
- Properties of the Bivariate Distribution Functions 2
- Example I: Bivariate Density & Distribution Functions 8:08
- Example II: Bivariate Density & Distribution Functions 14:40
- Example III: Bivariate Density & Distribution Functions 24:33
- Example IV: Bivariate Density & Distribution Functions 32:04
- Example V: Bivariate Density & Distribution Functions 40:26

### Introduction to Probability Online Course

### Transcription: Bivariate Density & Distribution Functions

*Hello, welcome back to the probability lectures here on www.educator.com, my name is Will Murray.*0000

*We are starting a chapter on probability distribution functions with two variables.*0006

*From now on, we are going to have a Y1 and Y2.*0012

*Today, we are going to talk about Bivariate density and Bivariate distribution functions.*0014

*That is a lot to swallow, let us jump right into it.*0020

*Bivariate density functions, the idea now is that we have two variables, Y1 and Y2.*0024

*For example, you might be a student taking a certain number of units at college.*0029

*Y1 is the number of math units you have taken and Y2 is the number of computer science units that a student has taken.*0036

*All the different students at this university, each one has taken a certain number of math units*0053

*and a certain number of computer science units.*0059

*There is a density function which reflects the number of students who had taken *0063

*any particular number of computer science units or math units.*0070

*For example, we can say how many students have taken 10 or more math units, and 15 or more computer science units?*0076

*There is a certain proportion of the population that has taken more than 10 math units *0084

*and more than 15 computer science units.*0090

*These are things that we will graphs on 2 axis.*0093

*From now, all our graphs are going to be on 2 axis.*0096

*We will always put Y1 on the horizontal axis and Y2 on the vertical axis.*0100

*And then, the density will be distributed all over this plane of 2 axis.*0107

*Lets us see what we do with these density functions.*0116

*First of all, the density function always has to be positive.*0119

*You cannot have a negative number of students who have taken a certain number of units.*0124

*The smallest you can have would be, if I picked a particular combination of units, *0129

*there might be 0 students that have had that combination of units.*0134

*But, there would never be a negative number of students that have taken a certain number of units.*0137

*Second, if we look at the total density, that means the density over the entire plane.*0141

*All the possible combinations of units the students could have taken.*0149

*If we integrate over the entire plane, it has to come out to be 1 that is because it is a probability function.*0156

*The total density of students has to be 1, no matter how many students we have at this college,*0161

*everybody is factored in there somewhere.*0170

*We can calculate probabilities, when we graph this, as I explained it to you.*0174

*If we want to find the probability of any particular region, I will graph a rectangular region*0181

*because it makes it easy for me to describe.*0187

*The rectangular region where Y1 goes from A to B and Y2 goes from C to D, there is Y2.*0191

*We want to find the probability of landing within that region.*0202

*For example, we want to find the probability, maybe proportion of students that had taken*0206

*between 10 and 15 math units, and have taken between 20 and 30 computer science units.*0211

*What proportion is the total student body of this college has taken between 10 and 15 math units,*0219

*and between 20 and 30 computer science units.*0226

*The way we do that is, we take a double integral, integral Y1 from A to B, Y2 from C to D.*0229

*And then, we integrate the density function over that range.*0236

*What this means is that you really have to remember calculus 3.*0241

*If you do not remember how to do double integrals, what you want to do is *0246

*you want to review your calculus 3, multivariable calculus.*0251

*We do have a whole set of lectures devoted to multivariable calculus here on www.educator.com.*0256

*My student colleague Raffi, teaches those lectures, he is amazing.*0264

*If you cannot remember how to do a double integral, go after his lectures and *0268

*you will be good to go for the rest of this chapter in probability.*0272

*I have been talking about continuous distributions here.*0277

*If you have a discreet distribution, it is basically the same idea.*0281

*Instead of an F here, you will just change that to a P.*0286

*Instead of integrals, you will have summation signs.*0291

*In fact, you will have a double summation, instead of a double integral, if you have a discreet distribution.*0295

*It is not as common though, usually Bivariate functions in probability classes, *0301

*it just turns out that you usually study continuous ones.*0309

*You usually end up studying double integrals.*0313

*That is why you really have to know your multivariable calculus.*0315

*If you are a little rusty on that, you want to brush up on your multivariable calculus.*0319

*Another thing that we need to learn is the Bivariate distribution function, *0326

*it is kind of the two variable analogue of the distribution functions we had before.*0332

*The idea of the Bivariate distribution function is you have some cutoff values of Y1 and Y2.*0338

*Here is Y1 and here is Y2, we have some cut off values.*0348

*There is Y1 and there is Y2.*0352

*What we are interested in, is the probability of being less than both of those cutoff values.*0356

*You are interested in calculating the probability that Y1 is less than or equal to the cutoff value y1.*0365

*Y2 is less than the cut off value y2.*0374

*In other words, you want to find all the stuff in this region, right here.*0378

*All the stuff where Y1 is less than y1 and Y2 is less than y2.*0385

*If you think about that, that is just the double integral over that region.*0392

*We are going to call that function F of Y1 Y2.*0397

*It means you integrate from negative infinity to y1 and negative infinity to y2.*0401

*I cannot call it Y1 and Y2 anymore for the variables because I'm using them for the cutoff values.*0408

*I’m going to use T1 and T2, and then I’m going to integrate the density function.*0413

*Just like in some sense, the distribution function was the integral of the density function.*0421

*Back in single variable probability and Bivariate probability, the distribution function is*0426

*the double integral of the density function, from negative infinity up to the cutoff values that you are interested in.*0433

*Some properties of the distribution function satisfies, if E¹ Y1 Y2 is negative infinity,*0440

*it means you are not looking in any area at all.*0447

*Your value is going to be 0, no matter what the other variable is.*0452

*If you plug in infinity for both of them, that means you are really looking at this entire plane.*0457

*You are looking at all the possible density, it would have to be 1.*0464

*That is a property that that has to satisfy.*0469

*I think that is all the preliminaries now, we are ready to jump into the examples.*0472

*We are going to be doing a lot of integrals for these examples.*0476

*You really want to be ready to do some double integrals, even integrating over some non rectangular regions.*0479

*We will work them out together.*0487

*In the first example, we are going to consider the joint density function F of Y1 Y2 is defined to be K × Y2.*0490

*Your K means it is a constant.*0498

*On the triangle with coordinates at 0-0, 0-1, and 1-1.*0501

*What we want to do with this is, we are going to find the value of K.*0507

*We are going to keep using this same formula for examples 2 and 3.*0510

*You really want to make sure you are up to speed on this.*0514

*Let me first graph this region because we are going to be seeing this over and over again.*0517

*We got the triangle with coordinates at 0-0, and 0-1, and 1-1.*0524

*I will graph that triangle and that is the region that we are interested in.*0531

*I will color that in.*0537

*There is a Y1 is equal to 1, Y2 is equal 1, there is 0.*0540

*It looks like it is defined by the line Y = X, but since we are using Y1 and Y2 as our variables here, there is Y1 Y2, *0550

*that is the line Y1 = Y2, like the line Y = X there. *0562

*We want to find the value of K, the way we want to do that is, we want to remember that the total density has to be 1.*0571

*We want to integrate over that region and our answer will have a K in it somehow.*0579

*I will set the whole thing equal to be 1.*0585

*First, I like to describe that region.*0588

*The best way to describe that region, I think, is to describe it with Y2 first.*0591

*I’m going to describe it, listing Y2 first, using constants for Y2.*0596

*Y2 goes from 0 to 1 and Y1 goes from 0 to Y2.*0601

*This is a hardcore multivariable calculus, if you do not remember how to set up these integrals *0611

*on these triangular regions, you really got to review it right now.*0616

*Go back and watch these lectures on multivariable calculus and you will get some practice with these triangular regions.*0620

*Let me set up the integral on this region.*0628

*The integral from Y2 = 0 to Y2 = 1.*0633

*The integral from Y1 = 0 to Y1 = Y2.*0639

*My density function K Y2, let me pull my K outside, it is just a constant, Y2.*0647

*D Y1, I got to do first, and then D Y2.*0654

*I'm going to integrate the inside one first, the integral with respect to Y1 is just Y2 Y1.*0659

*Remember, you keep the Y2 constant when you are integrating with respect to Y1.*0666

*We want to integrate that from Y1 = 0 to Y1 = Y2.*0671

*If I plug those in, I will get Y2² -0.*0679

*I have to do the integral there, from Y2 = 0 to Y2 = 1.*0685

*I still have a K on the outside, I still have a D Y2.*0692

*The integral of Y2² is just Y2³/3.*0696

*I still have that K and I’m evaluating that from Y2 = 0 to Y2 = 1.*0702

*Plug those in and I get K ×, it looks like 1/3.*0712

*Remember that, the total density has to be equal to 1.*0719

*K × 1/3 is equal to 1, that tells me then that K is equal to 3.*0723

*I have just solved this problem.*0731

*Let me show you again all the steps there.*0734

*First thing was definitely to graph the region.*0737

*That is kind of an excellent rule for any kind of multivariable calculus type problem.*0740

*You always want to graph the region, it is very helpful to graph the region.*0745

*I graphed the region, I graphed that triangle with those 3 points.*0752

*I wrote down a little equation for the line of the boundary which is just Y = X or Y2 = Y1.*0757

*To find the value of K, what I want to do is to use the fact that the total density has to be 1.*0766

*The integral over this region of this density function, it has to come out to be 1.*0773

*I describe that region, I chose Y2 to list first because that makes it a little bit simpler to set up the integral.*0781

*You could reverse the order of the variables there, listing Y1 first, *0790

*but I think you are going to get a slightly nastier integral.*0795

*You would not get as many 0 in the bounds there.*0798

*That is why I picked Y2 first to describe that region.*0802

*And then, I set up my double integral, I integrated Y2.*0806

*First, I integrated Y2 with respect to Y1.*0810

*By the way, I use this as an example in my classes.*0813

*A lot of students say the integral of Y2 should be Y2²/2.*0816

*Not so, because we are not integrating with respect to Y2.*0821

*We are integrating with respect to Y1 which means the integral is just Y2 Y1.*0824

*And then, we evaluate for our boundaries on Y1 and we get Y2².*0829

*Now, we integrate that with respect to Y2 that is a fairly easy integral.*0836

*We come up with K × 1/3.*0840

*Since, that is supposed to be equal 1, that is kind of our rule.*0843

*Our rule is always that the double integral of DY1 DY2 is always equal to 1.*0847

*That tells me that the K has to be 3.*0859

*I hope this one made sense because we are going to keep using this example for problems 2 and 3.*0862

*Make sure you understand this one, we are going to take the answer of this one*0867

*and use it to answer some more complicated questions in example 2 and 3.*0871

*In example 2, we are going to keep going with the same setup from example 1, *0881

*except I will fill in the answer, the K was equal to 3.*0888

*If you are a little foggy on what was going on in example 1, go back and watch example 1, it will make more sense.*0892

*Let me go ahead and draw the region that we are interested in.*0897

*It is the same one as before 0-0, 0-1, and 1-1.*0902

*There is the region there, triangular region 0- 0, 0-1, and 1-1.*0909

*There is 1, there is 1, this is Y2.*0914

*That is Y1 there, my goodness what am I thinking.*0920

*There is Y1, there is Y2.*0924

*I want to find F, that is the distribution function of 1/3, ½.*0928

*The first thing to do is to remember what that distribution notation means.*0933

*F of 1/3, ½, the fractions are going to get nasty in this one.*0940

*I’m just going to warn you in advance.*0947

*That is the probability that Y1 is less than or equal to 1/3 and Y2 is less than or equal to ½.*0949

*That was the definition of F.*0962

*If you do not remember that definition of F, just click back a couple slides ago and *0965

*you will see the definition of Bivariate distribution function.*0969

*Remember, keep track of the difference between F, the distribution function, and f the density function.*0973

*They do not mean the same thing.*0981

*We want to find, the probability of Y1 being less than 1/3 and Y2 being less than ½.*0986

*Let me graph that.*0994

*There is 2/3, 1/3, and there is ½.*0996

*I'm going to draw my region here.*1001

*I got this region that sort of shape like a backwards state of Nevada.*1007

*There it is right there, there is my backwards state of Nevada.*1012

*What I want to do is to integrate over that region.*1017

*To integrate over that region, I need to describe that region.*1021

*It looks like, if I want to do it in one piece, I have to describe my Y1 first.*1026

*That goes to 1/3, that goes to ½, and of course, these are both going to start at 0.*1032

*Y1 is going to go, if I have constants for that, it is going to go from 0 to 1/3, that is Y1 1/3.*1039

*Y2 is not going to go from 0 to ½, otherwise, I would have a rectangle, and Nevada is not a rectangle.*1050

*It is going to go from, that line right there was the line Y2 is equal to Y1.*1060

*Y2 goes from Y1 on up to Y2 goes from Y1.*1068

*Maybe I should make that in black to make it a little more visible there.*1077

*Y2 goes from Y1 on up to ½.*1081

*Now, I have some boundaries, I can set up my integral.*1085

*This probability, I'm going to integrate Y1 goes from 0 to 1/3.*1089

*Let me go ahead and write the variables in there.*1100

*Y1 is equal to 1/3 and Y2 goes from Y1 up to ½ there.*1102

*I have a density function, there it is 3Y2.*1116

*DY2 and DY1, I have a double integral to solve.*1122

*Probably, the hardest part is setting up the double integral.*1131

*Usually, solving the double integral is not bad.*1133

*If you are lucky and your teacher is a nice person, and you can even use a calculator *1136

*or a software to solve these double integrals.*1141

*I'm going to go ahead and work it out by hand, just to prove that I'm an honest upstanding human being.*1144

*We have to integrate 3Y2, DY2 the integral of Y2 is Y2²/2 Y2².*1152

*We will pull out the 3/2.*1161

*Y2², we are integrating that from Y2 = Y1 to Y2 = ½.*1165

*I will keep the 3/2 here and Y2² from ½ to Y1 will give me, ½² is ¼ - Y1²,*1175

*We are supposed to integrate that from Y1 = 0 to Y1 = 1/3.*1189

*This is now DY1.*1198

*I have to do calculus 1 integral, I get 3/2, ¼ Y1 - Y1² integrates to 1/3 Y1³.*1201

*That is going to be a bit nasty to deal with.*1219

*All of these evaluated from Y1 = 0 to Y1 = 1/3, I get 3/2.*1222

*¼ × Y1 = 1/3 is ¼ × 1/3 is 1/12 -, 1/3 × Y1³.*1234

*1/3³ is 1/27.*1243

*Another 1/3 multiplied by that gives me 1/81, all the horrors here.*1247

*I'm not going to write anything for Y1 = 0 because both of those terms will drop out.*1253

*That is a small mercy there.*1258

*These fractions simplify a bit because 3/2 × 1/12 is 3/24 is 1/8.*1261

*3/2 × 1/81, the 3 will cancel with the 81 give me a 27 × 2 is 54.*1270

*Not too bad, I think I'm going to have a common denominator there of 216.*1280

*216, because 216 is 8 × 27, it is 54 × 4, that simplifies down to 23/216.*1288

*I did plug that into a calculator, in case you are fond of decimals, 0.1064.*1303

*If you are fond of percentages that is 10.64%.*1315

*That is my probability, the probability that you will end up in that small Nevada State region.*1323

*Or another way to think about that is what we just calculated is F of 1/3 and ½.*1332

*That is what we calculated right here.*1340

*Let me recap how I did that.*1344

*First of all, use the definition of the Bivariate distribution function.*1346

*It just means the probability that Y1 is less than 1/3 and Y2 is less than ½.*1351

*Then, I went to try to draw that region on my full graph.*1358

*I converted that description into a drawing there.*1364

*In turn, I took that drawing which gave me a sort of Nevada shaped region.*1371

*This rectangle with its lower coordinate cutoff.*1376

*Then, I converted that, it is actually a backwards Nevada is not it.*1381

*But I have been calling it a Nevada shaped region.*1387

*If you look at Nevada in a mirror, this is what it looks like.*1390

*Then, I tried to describe that in terms of variables.*1393

*Prepare a tree to set it up and a double integral.*1397

*I found this Y1 goes from 0 to 1/3, Y 2 goes from y1 to ½.*1401

*Notice, I would like to say Y2 goes from 0 to ½ but that would be wrong because *1409

*that would give me a rectangular region, that is not what I want.*1418

*I have to say, Y2 goes from Y1 to ½.*1423

*I took those limits and I set up a double integral here.*1428

*The 3 Y2 comes from the density function up here, and then it is just a matter of cranking through the double integral.*1434

*Not very hard, a little bit tedious, easy to make mistakes.*1442

*But, you first integrate with respect to Y2.*1447

*I factored out some constants, plug in your bounds which gives you everything in terms of Y1.*1450

*Do another integral, get some nasty fractions, simplify them to a slightly less nasty fraction.*1457

*If you like, you can leave your answer as a fraction.*1465

*I converted it into a decimal and a percentage.*1467

*In example 3, we are going to keep going with the same region and the same density function from example 1.*1475

*Let me go ahead and draw that.*1483

*We have got the triangular region from 0-0 to 1-1, and 0-1.*1489

*There is 1, there is the Y2 axis, and there is the Y1 axis.*1499

*We got this triangular region and we got a density function defined on that region.*1506

*We want to find the probability that 2Y1 is less than Y2 or Y2 is bigger than 2Y1.*1513

*I'm going to go ahead and try to draw the region that we are interested in.*1525

*I’m going to graph the region.*1530

*If I say 2Y1 is equal to Y2, that is like saying Y2 is equal to 2Y1.*1534

*To make that more familiar to people who graph things like this in algebra, it is like saying Y =2X.*1542

*I’m going to graph line one = 2X.*1550

*Let me put Y = 2X, that is going to be twice as steep.*1555

*There it is right there.*1564

*There is the line Y = 2X.*1566

*We actually want to have Y2 greater than 2Y1, that greater than 2X.*1570

*That means, we want the region of both that line.*1577

*We want that blue region right there.*1583

*We are going to find the probability of landing in that blue region.*1586

*I have to describe that blue region.*1592

*I think the best way to describe that blue region is to describe Y2 first.*1594

*Y2, I can see it is going from 0 to 1.*1603

*Let me show you how Y1 behaves.*1611

*Y1 is going from 0 up to that line, that line was Y1 is equal to Y2/2.*1613

*I will write that as ½ Y2.*1630

*That line is Y1 is equal to ½ Y2.*1633

*I have to make that my upper bound for Y1, ½ Y2.*1639

*That is my description of the region.*1644

*The reason I spent much time describing it that way is that, that sets me up for a nice double integral.*1646

*My probability is equal to the double integral on that region.*1654

*I can just use that description Y2 goes from 0 to 1 and Y1 goes from 0 to ½ Y2.*1661

*I have my density function 3Y2.*1685

*Once again, it is a multivariable calculus problem DY1 DY2.*1689

*I'm just going to work that out as a multivariable calculus problem and integrating with respect to Y1 first.*1696

*I will put a 3 on the outside, integral of Y2 with respect to Y1 is Y2 × Y1.*1704

*It is not Y2²/2 because we are not integrating with respect to Y2.*1712

*Be careful about that, that is a very common mistake that my own students make all the time.*1717

*Even I, make that mistake, if I’m not being careful.*1723

*Let me integrate that from Y1 = 0 to Y1 = ½ Y2.*1726

*What I get there is, there is still a 3 on the outside.*1734

*I’m just doing this first integral, not worrying about the second one yet.*1738

*I get ½ Y2², when I plug in my Y1 = Y2.*1742

*Y2², I will put the ½ on the outside.*1751

*I have got the integral from Y2 = 0 to Y 2 = 1 of Y2² DY2, factoring the outside terms there.*1756

*The integral of Y2² is Y2³/3.*1772

*Personally, I have this 3 on the outside.*1778

*I will just write that as ½ Y2³.*1781

*Then, I will evaluate that from Y2 = 0 to Y2 = 1.*1785

*I get ½ × 1 -0 which is just ½.*1792

*That is nice and pleasant, much simpler answer than we have for the previous example.*1799

*Let me walk you through that again.*1804

*The key starting point here is we have that same region, that triangular region, *1807

*with those coordinates of the triangle, just as before.*1813

*We want to find the probability that 2Y1 is less than Y2.*1818

*I wanted to graph that region, to see what part of the region that was.*1823

*I graphed 2Y1 is equal to Y2.*1827

*I got this line here, that is the line 2Y1 is equal to Y2, or you can write that as Y1 is ½ Y2.*1830

*I want the region above the line because I want Y2 to be bigger than 2Y1.*1840

*That is why I took the region above the line not below it.*1846

*That is why I got this blue region colored in right here.*1849

*I want to describe it and that would be more convenient to list Y2 first, so I can use constants for Y2.*1852

*And then, I want my bounds for Y1 would be 0 and the other bound is ½ Y2, I got that from the line.*1860

*That ½ Y2, that comes from right here, that is where that comes from.*1867

*I took these bounds and I set them up as the limits on my integral.*1875

*The function I’m integrating is the density function.*1881

*That comes from the stem of the problem, the 3Y2.*1884

*Now, it is just a matter of working through a double integral.*1889

*But be careful always which variable you are integrating.*1891

*The first variable I’m integrating is DY1, that is why the integral is Y2 × Y1.*1895

*I'm holding Y2 constant there.*1901

*It is not Y2²/2.*1903

*Run that through the limits, get Y2², integrate that with respect to Y2.*1907

*Now, it just simplifies down into the very friendly fraction of ½.*1912

*In example 4, we have a new joint density function here.*1926

*F of Y1 Y2 is defined to be E ⁻Y1 + Y2.*1930

*The region we are looking at is Y1 bigger than 0, Y2 bigger than 0.*1936

*Note that, there is no upper bounds given on that, that means Y1 and Y2 can go all the way to infinity.*1941

*Let me graph that region.*1948

*There is Y1 and there is Y2, we want to find the probability that Y1 is less than 2 and Y2 is bigger than 3.*1952

*Y1 should be less than 2 here.*1963

*Y2 should be bigger than 3.*1968

*We want to find, let me go ahead and draw those lines there.*1974

*Y1 should be less than 2, we want to go to the left of that vertical line.*1978

*Y2 should be bigger than 3, I want to go above that horizontal line.*1984

*Somewhat this region right here, that region right there.*1989

*And that is the region that we are going to integrate over, in order to find this probability.*1994

*I'm going to set up a double integral on that region.*2000

*The integral, I think I can list Y1 first safely.*2005

*Y1 goes from 0 to 2 because 2 is the upper bound for Y1.*2009

*Y2, that is where I bring in the 3, Y2 is 3.*2019

*I have to run that to infinity because that one just goes on forever.*2023

*Maybe, you are uncomfortable saying Y2 is equal to infinity.*2029

*Maybe, I will say Y2 approaches infinity but it does not really affect the calculations that we will be doing.*2032

*It will be fairly easy to plug in infinity, after we do the calculations.*2039

*The density function is E ⁻Y1 + Y2, that is a quantity there.*2043

*I have DY2 first and then DY1.*2052

*I think the best way to approach this is to factor the density function into E ⁻Y1 × E ⁻Y2.*2057

*The point of doing that, is that in the first integral, we are integrating with respect to Y2.*2066

*We can take the E ⁻Y1, that is just a big old constant now.*2074

*We can pull it all the way out of the integral, let me go ahead and do that.*2078

*We have the integral of E ⁻Y1, and now the integral from Y2 = 3 to Y2 goes to infinity of E ⁻Y2 DY2.*2081

*There is a DY1 on the outside but let me just handle that first integral inside.*2098

*The integral of E ⁻Y2 is just –E ⁻Y2.*2103

*That is a little substitution there, a little old calculus 1 trick.*2108

*I'm evaluating that from Y2 = 3 to Y2 goes to infinity.*2112

*Technically, I should be writing limits in here.*2119

*I should be introducing a T and take the limit as T goes to infinity.*2121

*I’m being a little sloppy about that, that is kind of the privilege of having been through so many calculus classes. *2125

*When Y2 goes to infinity, we get E ⁻infinity here, that is 1/E ⁺infinity.*2132

*That is just 0- E⁻³, that all simplifies down into E⁻³.*2139

*I still have that E ⁻Y1, I’m going to bring that back in here.*2151

*I’m going to integrate that DY1.*2156

*E⁻³ is just a constant, I will write that separately.*2161

*E⁻³ × -E ⁻Y1.*2164

*What are my limits on Y1?*2170

*My Y1 went from 0 to 2, Y1 = 0 to Y1 = 2.*2172

*I'm going to need some more space on this.*2179

*This is E⁻³, -E⁻², plugging in 2.*2183

*Plugging in Y1 = 0, I get E⁰ which is just 1, it is - -1.*2190

*If I simplify this, I get E⁻³, that one becomes positive.*2201

*It is 1 – E⁻², and I could factor that through.*2206

*I get E⁻³ – E⁻³ × E⁻², you add the exponents, it is E⁻⁵.*2210

*That is really my exact answer but it is not very illuminating.*2223

*I did find the decimal for that, and my calculator told me that that is approximately equal to 0.043.*2228

*if you want to convert that into a percentage, then it is 4.3%.*2239

*We have an answer for that one. Let me show you again the steps involved to finding it.*2251

*First of all, I graphed the whole region Y1 greater than 0 and Y2 greater than 0.*2255

*That is a whole planar region, quarter plane because it is just where Y1 and Y2 are positive, and that is my whole region.*2260

*What I'm really interested in, is the probability of Y1 being less than 2 and Y2 being bigger than 3.*2271

*I chopped that up and I found that region was this blue region colored in here.*2279

*That is where Y1 is less than 2 and Y2 is bigger than 3.*2284

*In order to integrate that, I had to describe that limits.*2289

*Y1 goes from 0 to 2, Y2 goes from 3 to infinity.*2293

*I plugged in my density function, there is my density function right there, and I plug it in here.*2299

*I had to do that double integral.*2304

*The nice thing about that is I can factor that density function into E ⁻Y1 and E ⁻Y2.*2305

*Since, my first integral, the inside integral is going to be a Y2, I can pull out the E ⁻Y1.*2313

*That is just a constant, and that got pullout as a constant, outside the first integral.*2319

*I’m just left with E ⁻Y2 which integrates to –E ⁻Y2.*2325

*Do a u substitution in there, u= -Y2.*2330

*Evaluate that from 3 to infinity, when we plug in infinity or take the limit*2336

*as the variable approaches infinity, we will get 1/E ⁺infinity.*2341

*That is 1/infinity is just 0.*2347

*That is what that infinity term gives you, is the 0.*2349

*And then, it turn out to be a +E⁻³, that is just a constant in the next step because *2353

*we are integrating with respect to Y1 now.*2359

*I get -E ⁻Y1, plug in the values, do a little bit of algebra and simplifying.*2362

*Get down to E⁻³ - E⁻⁵.*2369

*I plugged that into a calculator, just to see what kind of decimal we are talking about.*2376

*It should always come out to be positive, when we are finding these probabilities.*2380

*If you do not get a positive probability, in fact, if you do not get something between 0 and 1, you know you screwed up.*2383

*I like to plug things in and just get a number.*2389

*In this case, I got 0.043 which is 4.3%.*2392

*Yes, that is between 0 and 1, it is not too surprising.*2397

*That is my probability of landing in that region with that density function.*2401

*We will use the same region and density function for example 5.*2406

*Make sure you understand this very well, before you move on to example 5.*2411

*It is it is the same density function, we will be integrating a different corner of the region, let me put it that way.*2417

*In example 5, we are going to look at the joint density function, the same one that we had in example 4.*2428

*Let me go ahead and remind you of what that looked like there.*2435

*We have this graph and we are looking at the entire positive quarter plane there.*2438

*There is Y1 and there is Y2, and everything is going from 0 to infinity.*2444

*We want to find the probability that Y1 + Y 2 will be less than or equal to 2.*2451

*Let me draw the line Y1 + Y 2 = 2.*2457

*There it is, it is a diagonal line and it got a slope of -1.*2462

*That is the line Y1 + Y2, Y1 + Y2 is equal to 2.*2467

*It is just X + Y = 2 and you can solve that out.*2475

*You can do a little algebra to find that.*2478

*It does have intercept 2 on each axis.*2480

*I want it to be less than or equal 2, which means I need to look at the region underneath that line.*2486

*I want to describe that region and then do a double integral, in order to find the probability of landing in that region.*2495

*The first thing I’m going to do is try to describe that region.*2504

*It does not really matter which variable you list first here, but I listed Y1 first.*2509

*I can use constants for Y1, I’m going to go from 0 to 2.*2516

*And then, I listed Y2 but I cannot use constants for that because otherwise, I will get a rectangle.*2520

*Y2 goes from 0 to, if you solve that line out, you get Y1 + Y2 = 2.*2525

*If you solve for Y2, you will get Y2 is equal to 2 - Y1.*2537

*I’m going to use that as my upper bound, that is going to make for some nasty integration but there is no way around it.*2543

*We just have to go through it.*2548

*My probability is, it will be the double integral on that region.*2550

*I already set up the limits here, I have done the hard part.*2556

*The rest of it is just tedious integration.*2561

*Y1 = 0 to Y1 = 2 and Y2 = 0 to Y2 = 2 -Y1.*2564

*I have that same density function E ⁻Y1 + Y2.*2576

*I have DY2 and then DY1.*2583

*Remember, the old trick that we use back in example 4 works again, is to write that density function as E ⁻Y1 × E ⁻Y2.*2589

*The utility of that is that is we are integrating Y2 first.*2599

*And that means, E ⁻Y1 is a constant and I can pull it out of the integral.*2606

*I'm going to pull that out of the first integral, of the inside integral.*2612

*E ^- Y1 just sits there on the outside.*2616

*I have the integral of E ⁻Y2 DY2.*2619

*And then later on, I will do the integral with respect to Y1.*2626

*E ⁻Y1, the integral of E ⁻Y2 is –E ⁻Y2.*2631

*I need to evaluate this from Y2 = 0 to Y2 is equal to 2 -Y1.*2638

*There is still DY1 here.*2647

*This is a little bit nasty, I got –E ⁻Y2.*2651

*If Y2 is 2 -Y1, -Y2 will be Y1 -2, --E⁰, --1.*2656

*I will get +1 - E ⁺Y1 -2 at the next step.*2673

*I’m going to bring this E ⁻Y1 from over on the left.*2681

*I think that is going to be useful because I’m going to go ahead and multiply that through.*2686

*I think I will simplify things a bit.*2690

*I get E ⁻Y1 -, E ^- Y1 + Y1 – 2.*2692

*E⁻², that one is a little sideways there.*2698

*That is not too bad, what am I supposed to do this.*2706

*I’m supposed to integrate it with respect to Y1 DY1.*2709

*Let me go ahead and keep going on the next column here.*2717

*The integral of E ⁻Y1, do a little substitution is -E –Y1.*2722

*E⁻² was just a constant, it is –E⁻² × Y1.*2729

*This whole expression is supposed to be evaluated from, where are my limits, right there at the beginning.*2736

*Y1 = 0 to Y1 = 2.*2743

*I plug in Y1 = 2, I get -E⁻² - E⁻² × 2.*2747

*If I plug in Y1 = 0, I get + 1 because the two negatives cancel, I’m subtracting a negative.*2758

*And then, + 0 because we got Y1 = 0 in the last term there.*2767

*This simplifies a bit, I got 1 – E⁻² – 2 E⁻².*2775

*1 -3 E⁻², that is as good as it is going to get.*2782

*I did plug that into a calculator to get a decimal approximation.*2788

*My calculator told me that that was 0.594.*2795

*Again, it is between 0 and 1, that is reassuring every probability answer should be between 0 and 1.*2801

*What I get there, if I wanted to make that into a percent, that is 59.4%.*2810

*That is my probability of landing in that sort of triangular region in the corner.*2818

*A probability that Y1 + Y2 is less than or equal to 2.*2824

*That is the probability that I have been asked to compute there.*2830

*Let us recap that, first of all, I graphed the whole region which is the positive quarter plane here.*2835

*Let me see if I can draw that without making things too messy on the graph.*2844

*And that is the whole region but that is not the region we are interested in.*2850

*We are interested in the region where Y1 + Y2 is less than or equal to 2.*2856

*I have to graph that part of the region.*2862

*That is what this diagonal line is, it is line Y1 + Y2 is equal to 2.*2865

*We want all the regions less than that.*2871

*That is why I colored in this blue triangular region here.*2873

*I was trying to describe that region, in terms of variables.*2877

*I did use constants for the first, Y1 goes from 0 to 2.*2881

*But I cannot say Y2 goes from 0 to 2, otherwise, I will have a square.*2885

*I do not want a square, I need a triangle.*2889

*I said Y2 was less than 2 - Y1 and that came from solving out the equation of the line in terms of Y2.*2891

*That is how I got that.*2902

*Once I had that description, that was the hardest part of the problem.*2904

*Then, I just dropped those in as my limit for the integral, I dropped my density function in.*2906

*At this point, you could drop the entire thing into a calculator.*2913

*If your calculator will do multivariable integrals, otherwise, *2914

*you could throw into some kind of computer algebra system, or an online on integration system.*2921

*I’m trying to be honest with you, I’m trying to do it by hand.*2927

*I factored E ⁻Y1 + Y2 as E ⁻Y1 E ⁻Y2.*2931

*The important part about that is that, in this first integral, this inner integral, *2938

*we are integrating with respect to Y2, which means we can treat Y1 as a constant.*2943

*That is why I pulled E ⁻Y1 all the way out of the integral, which gives me a nicer integral E ⁻Y2 on the inside.*2949

*That integrates to E ⁻Y2, and then I plugged in my bounds here to get something little messy.*2958

*I multiplied E ⁻Y1 back through and it simplified a bit.*2965

*And then, I integrated that E ⁻Y1 integrates to –E ⁻Y1.*2970

*E⁻² is a constant, when you integrate that, it is E⁻² × Y1.*2978

*I plug in the bounds Y1 = 2 all the way through and Y1 = 0.*2985

*Simplified it down a little bit to get this slightly mysterious number 1 -3 E⁻².*2991

*When I converted that into a decimal, I got something that was between 0 and 1.*2998

*That is a little reassuring that we are doing a probability problem.*3003

*If it had not been between 0 and 1, I would have known that I was wrong.*3006

*And then, I convert that into a percentage.*3011

*Any one of those forms, if you gave it to me in my probability class, I will be happy.*3013

*You do not have to convert it into a percentage, but if you like to know what it is as a percentage, there it is.*3018

*That wraps up this lecture on Bivariate density and distribution functions.*3025

*This is part of the chapter on multivariate probability density and distributions.*3030

*We are going to move on to marginally conditional probability, in our next video.*3037

*This is all part of the larger lecture series on probability, here on www.educator.com.*3042

*I'm your host, Will Murray, thank you very much for watching today, bye now.*3049

1 answer

Last reply by: Dr. William Murray

Wed Apr 8, 2015 6:06 PM

Post by Arash Mosharraf on April 6, 2015

Thanks for the great lecture. I have the below problem that has to do with the joint density function. I just wanted to make sure I did it right.

A firm manufactures electronic equipment. Total production time per unit is the sum of the time it takes to assemble the item (assembly time), and how long it takes to inspect the item and package it (packing time). Suppose that assembly time is a random variable (X) ranging from 20 minutes to 40 minutes, and packing time is a random variable (Y) ranging from 5 minutes to 15 minutes. Assume that assembly time and packing time are independent and jointly uniformly distributed.

a. State the joint probability density function of X and Y, fXY (x, y) .

b. The production line must pause whenever a unit takes more than 45 minutes to produce in total. What

is the probability this will occur? Show how you obtain your answer.

I stated the joint probability function as double integral of fXY(x,y) dydx=1 with bounds [20-40]and [5-15] on the first and second integral respectively.

For the second part I drew a rectangle with the width of 10-40 and 5-15 on the XY axis and then guessed in order to get to 45 min, if the assembly time is 40 the packing time should be 5 and if the packing time is 15(its max) the assembly time must be 35. Then I drew and line and calculated he area of that rectangle and multiplied by 1/200. I was wondering if I did it right and if yes whether there is another way I could do this. Thank you.