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Lecture Comments (2)

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.

Bivariate Density & Distribution Functions

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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 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

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 units0053

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 region0181

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 taken0206

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 is0426

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 one0867

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 limit2336

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