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 1 answerLast reply by: Dr. William MurrayThu Jun 1, 2017 5:45 PMPost by Ernest Harris on May 30 at 06:36:20 PMHi Prof. Murray,Thank you, your lectures are helping me acquire a deeper understanding of the material.  For Example 1 part B:To be sure, the accumulation does not double count the argument 2 because only one set of values actually includes 2 where as the other set only approaches 2.  Is my thinking correct? 2 answersLast reply by: Dr. William MurraySat Apr 19, 2014 5:41 PMPost by Ali Momeni on April 18, 2014Hey Prof. Murray,Example 1 part B is super confusing. How did you know to add the INTEGRAL from 0 to 2 of (1/4 t) dt to the INTEGRAL from 2 to y of 1/4(4-t)dt? and why would you do that?Does the formula  -> INTEGRAL from a to b of f(y) dy = F(b) - F(a) apply here? If yes how?Thank you very very much for your work and the time you take to answer questions! - Ali

### Density & Cumulative 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
• Density Functions 0:43
• Density Functions
• Density Function to Calculate Probabilities
• Cumulative Distribution Functions 4:28
• Cumulative Distribution Functions
• Using F to Calculate Probabilities
• Properties of the CDF (Density & Cumulative Distribution Functions) 7:27
• F(-∞) = 0
• F(∞) = 1
• F is Increasing
• F'(y) = f(y)
• Example I: Density & Cumulative Distribution Functions, Part A 9:43
• Example I: Density & Cumulative Distribution Functions, Part B 14:16
• Example II: Density & Cumulative Distribution Functions, Part A 21:41
• Example II: Density & Cumulative Distribution Functions, Part B 26:16
• Example III: Density & Cumulative Distribution Functions, Part A 32:17
• Example III: Density & Cumulative Distribution Functions, Part B 37:08
• Example IV: Density & Cumulative Distribution Functions 43:34
• Example V: Density & Cumulative Distribution Functions, Part A 51:53
• Example V: Density & Cumulative Distribution Functions, Part B 54:19

### Transcription: Density & Cumulative Distribution Functions

Hello, and welcome to the probability lectures here on www.educator.com.0000

We are starting a chapter now on continuous distributions.0004

My name is Will Murray and I’m your guide today.0009

First section here is on density and cumulative distribution functions.0012

Anytime you have a continuous probability distribution, you will have a density function and a cumulative distribution function.0017

I’m going to explain what those are and make sure that you do not get those 2 mixed up.0026

We use F for both of them, that is always a little confusing but I will try to highlight the differences there,0032

and then we will do some example problems to practice them.0039

Let us jump in with density functions.0043

Y is a continuous random variable.0046

What that means is it contained values over a whole continuum of possibilities.0049

Instead of having discrete probability were Y would be a whole number or a certain number of possible values,0056

the values now for Y can be a whole range of things.0066

Things like the normal distribution would be a typical example of a continuous random variable.0070

The values of Y can be anything.0077

It can be any real number at this point.0081

It has a density function that, we are going to use f for density functions.0085

Be very careful here to distinguish between the density function and what we are going to learn next,0089

which is the cumulative distribution function.0095

We will use f for the density function and we will use F for the cumulative distribution function.0097

If you have sloppy handwriting, now is the time to be very careful to be clear about the difference between0106

f of Y which is the density function, and F of Y which is the cumulative distribution function.0113

We will learn about that in the next slide.0121

Be very careful about the difference between those two.0123

The density function, the properties it has to satisfy, it is always positive.0127

Essentially that means you cannot have negative probabilities, the probability is always positive.0132

The total amount of area under this graph from -infinity to infinity is always exactly equal to 1,0140

that is because it is a probability function.0151

The total probability of something happening has to be equal to 1.0155

Something has to happen and it happens with probability 1.0159

The way you use the density function to calculate probabilities is, you always talk in terms of ranges.0163

You never talk about the probability of Y being equal to a specific value.0170

Back when we are talking about discreet distributions, like the Poisson distribution,0176

the binomial distribution, geometric distribution, we would say what is the probability that Y is equal to 3?0180

What is the probability that there will be exactly 3 forest fires next year in California?0187

What is the probability that the coin will come up heads up 3 ×?0191

We would ask what is the probability that Y is equal to a particular value?0196

We will never ask the probability that Y is equal to 3, that would just be 0 because we have so many possible values.0203

Instead, we will ask about what is the probability that Y is between one number and another?0212

For example, we will find a probability that Y is between A and B.0220

The way you find that probability is, you calculate the area under the density function.0225

In order to calculate that area, what you do is you take an integral.0232

The probability that Y is between constant values A and B is the integral of the density function F of Y DY from Y = a to Y = b.0240

That is how you calculate probabilities from now on.0255

Remember, that is the density function, that is not to be confused with0258

the cumulative distribution function which is the next thing we are going to learn about.0262

The cumulative distribution function that is closely related to the density function but it is not the same thing.0270

The cumulative distribution function we use F of Y.0277

Remember, the density function was f.0281

The cumulative distribution function is the probability that Y is less than or equal to a particular cutoff value of Y.0283

Let me show you how you find that.0295

If this is the density function that I'm graphing right now, this is f of Y.0297

You have a particular cutoff value of Y and you want to find the probability of being less than that value.0303

The way you do it is you calculate the area to the left of that cutoff.0310

You can do that as an integral.0317

We already used y as the cutoff, I cannot use my Y as my variable of integration.0319

I cannot use of Y, I’m using T here, F of T DT.0325

By the way, that is a very common mistake that I see students make in doing probability problems is,0330

they will mix up their variables.0336

They will have a Y in here and then they will also have that Y over there in the limit.0338

That is very bad practice, you can get yourself in lots of problems that way.0342

Do not do that, use T when you are calculating the cumulative distribution function.0346

Use T as your variable of integration, and then use Y as your limit.0352

We can also use F, the cumulative distribution function to calculate probabilities.0359

The probability that you are within one range between A and B.0365

Let me graphically illustrate this.0373

The probability that you are between A and B.0378

That is the area in between A and B.0386

One way of calculating that is to calculate all the area less than B, and then to subtract off all the area that is less than A.0389

To subtract off all this area less than A.0402

What you are left with is exactly the area that you want, which is the area between A and B.0406

The way you calculate the area that is less than B is to use the cumulative distribution function F of B.0414

The area less than A is F of A.0421

Once you worked out what F is, you do not have to do any more integration.0424

You just plug in your limits B and A.0429

Essentially, this is the fundamental theorem of calculus coming through in a probability setting.0432

It means once you have done this integral, you just plug in the 2 limits F of B and F of A.0437

We are going to study some properties of the cumulative distribution function,0446

that is what CDF stand for cumulative distribution function.0451

What today's functions look like, remember that if the density function often looks like something like this.0455

This is the density function, f of Y.0465

The cumulative distribution function represents the area underneath or to the left of any given cutoff.0468

The cumulative distribution function therefore, as you start in the left hand side of the universe, at –infinity,0478

there is no area to the left.0488

It always has to start at 0, let me draw that in black here.0490

It always has to start at 0, that is why F of -infinity is always 0.0495

What I’m doing here is I’m going to graph F of Y.0504

As you start to increase Y here, you get more and more area until you get to the right hand edge of the universe at infinity.0511

You got all the area under the density function.0523

We said that total area is 1.0526

It always has to increase and it always has to end up at 1.0529

The cumulative distribution function always has the same general shape.0539

It always starts at 0, at –infinity, and it always increases and it finishes up at 1, at infinity.0544

That is why we have these properties F of infinity is 1, F is always increasing.0556

Its derivative, this is the fundamental theorem of calculus, get you back to the density function which is f of Y.0562

We will be using this property in particular, as we solve some problems.0571

Let us jump in and solve some problems with density functions and cumulative distribution functions.0579

The first one we are given that Y has density function f of Y, be careful, that is the distinction between f and F.0589

C × Y, when Y is between 0 and 2.0598

C × 4 – Y, when Y is between 2 and 4, and 0 elsewhere.0602

I will draw a little graph of that.0607

We do not know what C is.0611

In fact, the first part of the problem here is to figure out what C should be.0612

When Y goes from 0 to 2, we do know that it is something increasing.0619

It is linear and it is increasing.0623

4 - Y it will be decreasing again, as we go down to 4.0626

It is something like that but we want to find the exact value of C.0630

What we are going to use for that is, we are going to use the property of density functions which is that the integral of F of Y DY,0635

the total area always has to be equal to 1.0645

That was the first property of density function, I guess it was the second property of density functions.0650

It always has to be equal 1.0655

That represents the fact that this does represent probability,0657

the total probability of something happening always has to be 1.0661

Let us find that integral based on the information we are given.0666

I'm only going really look at the range between 0 and 4.0670

The integral from 0 to 4 of f of Y DY, I’m solving part A here.0676

It is supposed to be equal to 1.0683

Whatever that turns out to be, I'm going to set it equal to 1.0686

We have got 2 different parts of our density function.0691

I’m going to split up into integral from 0 to 2.0695

We also have an integral from 2 to 4.0700

Each one of those is multiplied by C, I will go ahead and factor the C out, put a C out here.0704

Then, I will have Y DY here on this range and 4 - Y on this range.0712

I'm just going to do the calculus there.0720

This is C × Y²/2 evaluated from 0 to 2 + 4 – Y, that is 4Y - Y²/2, evaluate that from 2 to 4,0722

that is supposed to be equal to 1.0740

This is all C, always multiply by C.0742

Y²/2 evaluated at 2 is 4/2 is just 2 evaluated at 0, nothing happens.0746

If I plug in 4 into Y for the second part, I will get 16 – 4² that is 8.0755

Plug in 2, I get -8, - and - is +, 2²/2 is + 4/2.0763

That is still supposed to equal 1.0776

I get C × 16 - 8 – 8, those cancel each other out.0778

C × 4 = 1.0783

C = ¼.0786

Let us see, that tells me the answer to part A there.0794

The key part there is that the total integral was supposed to come out to be 1, that is the property of a density function.0799

The way I worked that out was, I set the total of the integral is equal to 1.0808

I factored the C out of everything because that was a common factor on both parts there.0812

Then I worked out this integral on their respective ranges.0823

I had to split it up into 2 parts because the function was sort of defined piece wise.0826

Two different definitions on 2 different ranges.0832

We worked out the integral I got 2 × 4 is still equal to 1, it has to be equal to ¼.0836

I got my part C, I have found my value of C for part A.0842

Now, I have to find the cumulative distribution function of Y and that is going to be a little more work.0848

Let me go ahead and do that on the next slide.0853

With density function again, and the cumulative distribution function of Y, remember that F of Y,0860

the way we figure that out is, it is the integral from -infinity to Y of F of T DT.0869

That was our original definition of F, you can go back and you can find that.0880

Now, in this case, there is nothing going on between -infinity and 0.0885

I’m just going to start this integral at 0, the integral from 0 to Y.0894

I have 2 different ranges of Y here.0900

I’m going to split up 2 cases here.0903

The first case I’m going to do is Y is between 0 and 2.0905

What do I get there?0913

Then, on the previous slide, we figured out that C is ¼.0914

That is by the work we did on the previous slide.0923

If you do not remember that, just go back and check it, you will see where we got the C is ¼.0925

This is ¼, replacing my Y with T, I did the same mistake that I said that probability students often make, T DT.0929

This is T²/2T²/2 × 4 is 2²/8.0942

From 0 to Y which is just Y²/8, that is what my range of Y between 0 and 2.0949

For the range between 2 and 4, it is more complicated and this is very easy for students to mix up.0961

I hope you will follow me carefully here.0967

F of Y, it is still the integral from 0 to Y of F of T DT.0970

We have to split up into 2 parts because there is 2 different parts of this range.0977

It is the integral from 0 to 2 of ¼ of T DT + the integral from 2 to Y of the different definition which is C, C is × 4 – T DT.0981

I ‘m reading that off from this part of the definition here.1005

It is a little more complicated but if we be careful with that, we can keep it straight.1009

The integral of 1/4 of T DT, I can factor out a ¼ out of everything there.1013

I think that will make my life a little simpler here.1021

I factor the 1/4 out everything.1023

The integral of T DT is T²/2, we will be dividing that from 0 to 2 +, now that 1/4 is gone, it is 4T - T²/2,1026

evaluate that from T = 2 to T = Y.1040

This is 1/4 × 2²/2 is just 2, + I will plug in Y for T .1048

4Y - Y²/2, I will in 2 for T so -8.1057

And - is +, + 2²/2 is just 2, this is ¼.1064

Now what do I have here, I have 4Y - Y²/2 + 2 + 2 - 8 that is -4.1074

If I distribute that 4, I get Y - Y²/8 -1.1087

It is kind of a messy formula there but that is what we are stuck with.1098

Now, I found two different function for F of Y, depending on the different ranges we are in.1101

I’m going to summarize that.1108

My F of Y is equal to Y²/8 for 0 is less than Y, less than or equal to 2.1110

It is equal to Y - Y²/8 – 1 for 2 less than Y less than or equal to 4, that is my F of Y.1125

I should also mention what it does on the outsides of those ranges.1143

It will be 0/Y is less than 0 because, remember, the cumulative distribution function all starts out a 01146

and it always goes up to 1 for bigger values of Y.1154

For Y greater than 4, it will be 1.1160

It is worthwhile somtimes to check that the values match on the endpoints.1165

If you plug in 0 to Y²/8, you do indeed get 0.1170

If you plug in 4 to Y = 4, we will get to the second part of the function 4 - 4²/8 -1 will give us 4 -16/8 is 2 -1, gives us 1.1175

That matches up of what we said Y should be, when we get bigger than 4.1195

That checks my work there.1201

Let me box this whole solution here because this is all part of our solution.1203

Finally, we have found a cumulative distribution function Y there.1214

Let me remind you of the steps there.1218

We used the definition F of Y is the integral from -infinity to Y of F of T DT.1220

Now, that is pretty simple when Y is between 0 and 2.1229

You just take the first part of the definition and you work out the integral, and you get Y²/8.1235

That is where we got this part of the answer.1241

But when Y is between 2 and 4, it is much more complicated because you have1243

to take into account both parts of this definition.1248

You have to use both parts of this definition and split up the integral into two parts,1252

and evaluate both of those using T as your variable.1258

You do one from 0 to 2, one from 2 to Y, and then simplify that down into a much more complicated function.1263

That is how we got this more complicated function on the range between 2 and 4.1271

I want you to hang onto the answers that we got here for example 1,1277

because we are going to use the same density function and therefore, the same cumulative distribution function for example 2.1281

I want to make sure that you understand these answers for example 1.1293

Make sure you understand this example very well, before you move on to example 2.1296

In example 2 here, we are taking the same density function from example 1.1304

Remember, we figured out that the constant had to be 1/4 there.1309

I went ahead and write that in on example 2.1312

It is ¼ Y for the range between 0 and 2, and 1/4 of 4 – Y from the range between 2 and 4.1315

You got a density function that looks kind of like this.1325

What we want to find here is the probability that Y is between 1 and 3,1331

and the probability that Y is less than or equal to 2, given that Y is greater than or equal to 1.1335

Those are some conditional probability involved in there.1342

The useful thing to use at this point is not the density function that is given,1345

but the cumulative distribution function that we worked out in example 1.1351

If you have not just watched the video for example 1, you should go back and work out example 11356

because we are going to use that answer from example 1, to calculate the answers for example 2.21361

Let me remind you right now what the answer was from example 1.1368

F of Y, the cumulative distribution function was, I broke it down into two important parts there.1372

It was Y²/8, when Y is between 0 and 2.1380

It was more complicated Y - Y²/8 -1, when 2 is less than Y less than 4.1387

That was the cumulative distribution function, we did figure that out in example 1.1400

Quite a lot of integration we went into that, we are not going to redo it.1404

If you do not know where that is coming from, it is worth going back and working through example 1 because it will make sense.1408

For part A here, to find the probability that Y is between 1 and 3.1416

What we can do know is, we can use the cumulative distribution function F of 3 – F of 1.1421

We can also use an integral of the density function but then, we just end up redoing the work we did from example 1.1430

It is much easier to use F, if you already down that work.1437

F of 3, now, 3 is between 2 and 4.1443

Let me use this second version of the formula.1447

It is 3 -, 3²/8 is 9/8 – 1.1453

-F of 1, I have to use the first part of the formula because 1 is in the range between 0 and 2.1460

Y square/8 is 1/8 and I will just simplify those fractions.1468

3 -1 is 2 – 9/8 - 1/8 is 10/8.1474

10/8 is 2 - 5/4 and 2 is 8/4, that is just ¾.1480

That is my probability that Y is between 1 and 3.1487

3/4 probability that Y is between 1 and 3.1492

Let me show you that on the graph because I think it will makes sense there.1495

There is 1, 2, 3, 4, being between 1 and 3.1499

If you figure out how much of that area is between 1 and 3, if you do a little triangle geometry there,1505

you will figure out that, that ¾ of the total area is between 1 and 3.1512

We also checked that using our arithmetic here, using our integration.1520

I’m going to jump over onto the next line to do part B.1526

It will be a little more complicated, I need more space.1529

Just remind you how we did part A here.1532

I have recalled from example 1, the cumulative distribution function, the F of Y.1535

That is what we worked out in example 1.1541

And then, I just had to plug in F of 3 - F of 1.1543

The wrinkle in that was those are two different ranges so I have to use two different formulas.1549

One for F of 1, there is the F of 1 using that formula right there.1553

There is F of 3, using that formula right there.1559

Once I take those two values and plugged them in, I got some easy fractions to simplify.1567

It ended up with ¾ there.1573

We are still working on example 2.1578

We still have to do the second part of the problem here.1579

It is going to be really helpful to use the cumulative distribution function that we figured out in part 1.1583

Let me remind you what our cumulative distribution function was.1592

We figure this out in example 1.1597

There was two parts to this function, Y²/8 for Y between 0 and 2.1599

Y -1²/8 -1 for Y between 2 and 4.1611

We got to find conditional probability, it has been a long time since we did condition probability.1621

If you look back into some of the early videos in the series, you will find one that covers condition probability.1626

I will remind you of the formula that we have for condition probability.1633

The probability of A given B is the probability of A intersect B divided by the probability of B.1636

Remember, intersection is like saying N, you want both of those things to be true.1651

That formula is way back in the early videos for this course.1657

You can find it, just scroll back to the videos here on www.educator.com.1660

Our probability of Y being less than 2 given that it is greater than 1 is the probability of Y1666

less than or equal to 2 and is greater than or equal to 1 divided by the probability that Y is greater than or equal to 1.1675

I’m just filling in my formula for conditional probability.1685

Now, another way of saying that it is less than 2 and greater than 1 is to say 1 is less than or equal Y less than or equal to 2 divided by,1688

Now, the probability that Y is greater than or equal to 1, that is hard to compute directly but it is easy to compute1700

if write is a 1- the probability that Y is less than or equal to 1.1707

That is the easy way to calculate it because that sets it up to be something that we can answer1712

using the cumulative distribution function.1721

Let me write that on a new line here.1724

The probability that Y is less than or equal to 1 is just F of 1.1727

The probability that Y is between 1 and 2 is F of 2 – F of 1.1734

Now, I can just input all these values into my cumulative distribution function, F.1742

F of 2, it looks like everything here is on the first range, the Y²/8 which is nice, because that is the easier formula.1749

F of 2 is 2²/8 that is 4/8.1757

F of 1 is 1²/8, 1/8, 1 – F of 1 is 1 – 1/8.1763

This is 3/8 /7/8, and if you do the flip on the fractions, the 8’s will cancel.1770

I will just multiply top and bottom by 8, I will get 3/7.1780

That is my probability that Y is less than or equal to 2 given that Y is greater than or equal to 1.1784

You can also check that geometrically, if you like.1793

The graph we have on f of Y looked like an elongated triangle, f of Y.1797

There is 2, there is 1, and there is 3.1805

The probability that Y is greater than or equal to 1 is all that range there.1811

The probability that Y is less than or equal and 2, let me draw that in another color.1820

Less than or equal to 2 is that range right there.1827

If you break this up into little triangles, you can see that there is 1, 2, 3, 4, 5, 6, 7 triangles total.1833

3 of these 7 little triangles are in that region.1844

It does check graphically that we get this 3/7 answer.1853

But I probably, do not want to rely on that, I do like using the formulas.1858

Just to remind you how we did use the formulas there.1862

This formula for the cumulative distribution function, this F came from example 1, we work that out in example 1.1865

You can go back and check it, rewatch the video from example 1.1871

This break down here, we are using conditional probability.1878

This came from a very old video but it is on the series for conditional probability.1882

I'm using that formula for conditional probability here.1888

The probability that Y is less than or equal to 2 and greater than or equal 1, that just means it is between 1 and 2.1893

The probability that Y is greater than or equal 1 is hard to evaluate directly.1899

I flipped it around and that is because the probability that Y is less than or equal to 11904

is something we can answer easily using our cumulative distribution function, our F.1910

This is F of 2 - F of 1, 1 – F of 1.1918

I just dropped those values of Y into this F formula, since they are all in the first range there,1923

and simplify the fractions down to 3/7.1930

In examples 3, we are given a new density function for Y, F of Y, f of Y, f is the density function, is some C.1939

It does not tell us what C is, I guess we have to figure that out, on the range between 0 and 1.1951

2C is on the range between 1 and 2, and everywhere else it is going to be 0.1956

To the first task here is to find what c should be.1961

The second task is to find a cumulative distribution function of Y.1966

Let me go ahead and graph this.1971

We got between 0 and 1, the value is c.1975

Between 1 and 2, it jumps up to 2c.1984

That is our density function right there, it is 0 everywhere else.1994

We have a density function that looks like that, it is a step function there.2000

You can answer this question pretty easily graphically, if you know what you are doing.2005

Let me go ahead and show you the arithmetic here.2010

Just to make sure that it makes sense using either method.2013

The key point here is that the total area under the density function is always equal to 1.2018

In this case, the area is the integral from, in this case 0 to 2 of F of Y DY, should be equal to 1.2024

I want to figure out what the value of c should be, to make that equal to 1.2039

Let us evaluate that integral, that is the integral from, since we got the function defined differently2044

on the 2 different ranges, I will write it as the interval from 0 to 1.2050

We also have an integral from 1 to 2.2056

It looks like they are both going to be multiplied by c.2061

I will go ahead and factor that c out.2063

That will leave me with 1 on the first range, 1 DY.2067

2 on the second arrange, 2 DY there.2071

That is supposed to come out to be equal to 1.2079

Now, that is c × the integral of I DY is just Y from 0 to 1 + 2Y from 1 to 2 is equal to 1.2084

That is C × evaluation of Y from 0 to 1 is just 1.2101

2Y is just 2 × 2 -2 × 1, that is + 2 is equal to 1.2108

It looks like c is going to be 1/3, in order to make this be a valid density functions.2116

That tells us the answer to part A.2125

That is really not surprising if you go back and look at your graph here.2127

If you go back here, the area of that first block is definitely c, because height × width is I 2C.2133

A of that second block is 2C, the total area is 3C which should be 1.2141

Definitely, I want my C to be 1/3, in order for that to be a valid density function.2149

I got my C to be 1/3, and I still need to find F of Y.2158

That will be a little more work but let me recap the work here because that throws away the slide.2164

I use the fact that the total area under the density function is always equal to 1.2172

That is true for any density function, that is a requirement.2177

It is essentially saying that the total probability in any experiment has to be 1.2180

Because it was defined on two different ranges here, I just split up that integral into two parts.2187

It is 0 to 1, and 1 to 2.2192

Those were both very easy integrals, I factored out the C already.2195

That gave me 3C was equal to 1, I figure out that C is equal to 1/3.2205

I could have figured that out from the graph because I know that the total area should be 1.2210

If I just look at the blocks there, I get 3C is equal to 1, C is equal to 1/3.2217

We are going to hang onto this and make the jump to the next slide, where we will figure out part B here.2223

For part B of examples 3, we figured out in part A that C was equal to 1/3.2232

That is already done on the previous slide.2238

But we have to now find F of Y, we want to find the cumulative distribution function.2242

Let us remember the definition of F of Y.2252

It is the integral from -infinity to Y of f of T DT.2255

In this case, there is no density below 0.2264

I can just cut this off at 0, this is the integral from 0 to Y of F of T DT.2269

I need to separate two ranges here.2280

I have a range from Y going from 0 to 1, and then I will have another range from Y going from 1 to 2.2283

I need to separate this problem into two parts.2292

The second part will be a little more difficult.2295

The first part is pretty easy because then, I'm just looking at the definition of Y between 0 and 1.2297

This is we get that from, 0 to Y is just 1/3, that was my value of C DT.2305

That is 1/3 T evaluated from 0 to Y which is just 1/3Y, that part is fairly easy.2316

The second part is more complicated, you want to be careful about that.2328

On the second range, Y goes from, I said 0 to 2, and I should have said 1 to 2.2333

You do not just want to use this formula from 1 to 2 because we really need to find the integral from 0 to Y of F of T DT.2340

That breaks that up into two parts, the integral from 0 to 1 of F of T DT and the integral from 1 to Y of F of T DT.2350

We need to do separate integrals for each of those.2364

I have to do that because I know that Y is bigger than 1.2367

Y is somewhere in this range between 1 and 2.2370

The first part is not too bad, 0 to 1 of 1/3 DT +, 1 to Y between 1 and 2, the density function is 2 × C, that is 2/3 DT.2374

I get 1/3 T from the 0 to 1 + 2/3 T from 1 to Y, from T = 1 to T = Y.2395

I get 1/3 × one – 0 so 1/3 × 1 + 2/3 Y - 2/3 × 1.2410

I guess that simplifies a bits into 2/3 Y + 1/3 - 2/3 is -1/3.2421

I got two different ranges in two different functions.2431

I will put those together to give my answer, my F of Y is equal to 1/3Y, when 0 is less than Y less than 1.2436

It is equal to 2/3 Y - 1/3, when one is less than or equal to Y less than or equal to 2.2452

Those are the important parts of my cumulative distribution function.2467

We just tacked on the parts of the N.2471

If Y is less than 0, it will just be 0 because the left hand side is always 0, the right hand side is always 1.2473

If Y is greater than 2 then we will be looking at a function of 1.2484

You can always plug that in and check, I will plug in Y = 2 and I get 2/3 × 2.2491

4/3 - 1/3, I’m plugging in Y = 2 in here and it does come out to be 1.2498

That reassures me that I have probably been doing my arithmetic correctly.2503

Let me box this up.2511

I got my cumulative distribution function.2518

We will be using this again in the next examples.2521

I want to make sure that you understand this very well, before you move on to example 4.2524

Let me remind you of how we got that.2530

Our cumulative distribution function, the F, we always get that by integrating the f, the density function.2532

You integrate it, you change the variable from Y to T, and then you integrate from 0 to Y.2539

That is sometimes a bit subtle, especially, when you have these functions that are defined piece wise.2545

Because the integral from 0 to Y, Y is between 0 and 1, no problem, you just use the definition of the f of Y between 0 and 1.2550

You integrate that and you get 1/3Y.2561

If Y is between 1 and 2, then you get a more complicated integral because2564

you have to break it up between 0 to 1, and 1 to Y.2570

And then, you have to use the different definitions, the 1/3 the 1C, the 2/3 the 2C, on the different ranges.2574

You plug in those two different definitions, the C and the 2 C, and then you do the integral and it simplifies fairly nicely.2583

You want to assemble these two answers into a definition of F of Y defined on the different ranges.2593

You always have the left hand then being 0 and the right hand then being 1.2600

That is our cumulative distribution function.2606

You do want to remember this because we will be using it again for example 4.2609

Let us go ahead and take a look at that.2613

In example 4, we are using the same density function from example 3.2616

F of Y is 1/3 on the range from 0 to 1, 2/3 on the range from 1 to 2.2621

Here, we want to find a condition probability, the probability that Y is less than 3/2 given that it is bigger than ½.2630

I did do a lot of work on this density function back in examples 3.2640

If you did not just watch the video on example 3, you should go back and work through example 3,2645

in order to understand example 4.2652

I’m going to use the results right now.2654

Let me remind you what our answer was from example 3.2657

Let us see, where do I have space for that, I will put that over here.2666

In example 3, this was our answer from examples 3 and it did take us a fair amount of work to get to this point.2668

We found the cumulative distribution function F of Y not f, you got to be very careful to keep those clear.2676

I will just put the two important ranges here.2684

The range from 0 to 1, and then we had a different answer when Y was between 1 and 2.2688

Our two answers there, where Y was 3, where do I have that in my notes there.2696

That was 1/3 Y on the first range, and then on the second range it was 2/3 Y -1/3.2703

We work those out in example 3, I’m not the showing you the work right now to do those.2713

You have to go back and watch example 3.2717

For the probability of what we are asking for here, this is the conditional probability,2720

let me remind you of the formula for conditional probability.2726

That was done in one of our earlier probability videos here on www.educator.com.2729

Let me remind you that the probability of A given B, the conditional probability is2735

the probability of A intersect B divided by the probability of B.2741

Remember, A intersect B is the same as A and B, mathematical, when events are happening at the same time,2750

divided by the probability of B.2760

That is what we are going to apply to this conditional probability that we are asked to calculate here.2762

This is the probability that Y is less than or equal to 3/2 and Y is greater than or equal to ½ divided by the probability of B.2767

This is our event A and this is B right here.2780

B is the probability that Y is greater than or equal to ½.2783

Being less than 3/2 and bigger than ½, that just means you are in the range between ½ and 3/2.2791

Being greater than ½, the easier way to think about that is to write that as 1 - the probability that Y is less than ½.2803

The reason I put it that way is because that is making it in a form where2817

we can easily use our cumulative distribution function.2823

Remember, F of Y represents by definition, it represents the probability that Y is less than or equal to some value y.2826

This is equal to, in the denominator that would be 1 – F of ½.2841

In the numerator, we want the probability of being between ½ and 3/2, that is F of 3/2 – F of ½.2851

Now that I have an F function, I work that out in example 3, I can just drop in all the values.2865

3/2 is bigger than 1, I have to use the second range there F of 3/2.2874

2/3 × 3/2 - 1/3 -1/2, now that ½ is less than 1, I will use the first range - 1/3 × ½ all divided by 1 - F of ½ is just 1/3 × ½.2881

Now, I just have some fractions to simplify.2901

2/3 and 3/2 that is 1 -1/3 - 1/3 × ½ is 1/6, 1 -1/6.2903

1/3 + 1/6 is ½, I get 1 -1/2 is just ½ in the numerator.2915

1 -1/6 is 5/6 in the denominator, that is 6/5 × ½ is 3/5 as my answer.2922

I think this is one we can also check graphically.2938

Let me draw that function.2942

We graph this out in examples 3 but that was the density function that I'm graphing right now.2943

It had a big jump at 1 and it really went from 0 to 2.2954

The probability that you are less than 3/2, let me draw 3/2 there.2958

There is 3/2, given that you are bigger than ½.2964

There is ½ and ½, we are given that we are bigger than ½, let me go ahead and color in that area there.2970

There is the probability that we are bigger than ½ and being less than 3/2 would mean that we are in that area right there.2980

I’m going to cut off at 3/2.2991

If you just kind of work by blocks there, there is sort of 5 blocks in the color blue1, 2, 3, 4, 5,2994

and three of them are colored green.3004

That is where we get the 3/5 coming from, if you want to do it graphically, instead of doing the equations.3007

You will still get 3/5.3014

Let me recap how we did that.3018

We are using conditional probability here.3021

I went back to my old conditional probability formula.3024

The probability of A and B divided by the probability of B.3027

In this case, my A and B where Y being less than 3/2 and bigger than ½.3032

I get the probability that Y is between ½ and 3/2.3038

The probability of Y being bigger than ½, I flip that around and said it is 1 - the probability of Y being less than ½.3042

The reason I flipped it around was, I could easily convert that into a value of my cumulative distribution function, my F.3050

I can also find this probability of the range using F.3060

I recalled the F that we calculated back in examples 3.3064

I recalled this F here and I just dropped in a different values.3072

Of course, I have to use the different parts of the formula because 3/2 was in this second range and ½ was in this first range.3078

That is why I use the second formula for 3/2 and the first formula for ½.3088

Then, I just simplify down the fractions and it simplify down to 3/5.3096

I could have done all that just by looking at the graph and by measuring up the areas.3100

I get 3 blocks out of 5 blocks, that is why that checks my answer to be 3/5.3106

In example 5 here, we have a cumulative distribution function given to us, the F of Y.3116

It looks like the most important ranges here are Y being between 0 and 1, and Y being between 1 and 2.3123

We got two problems here, we want to find f of Y, the density function.3130

We also want to find the probability that Y is in a particular range.3135

It looks like I have really left myself no space to solve this problem.3143

Let me jump over to the next slide and solve the problem.3146

Here is my cumulative distribution function, I put off the less important parts of the definition there.3154

We are going to find f of Y and the probability that Y is in a certain range.3164

This is fairly easy relative to the earlier problems that we did.3170

You get a little bit of break this time.3175

What is this X doing here, we do not use X in probability.3177

F of Y is equal to, remember, it is the derivative of F of Y.3185

I can find this just by taking the derivative of F of Y.3191

The derivative of Y/4 is ¼ and the derivative of Y²/4 is 2Y/4, that is just Y/2.3196

That is for the particular ranges 0 less than Y less than or equal to 1, and 1 less than Y less than or equal 2.3206

Outside those ranges, F of Y, remember it was just the constant 0 and 1.3216

Its derivative will just be 0 for other values of y.3223

If Y is less that or equal to 0, f will be 0.3230

If Y is greater than 2, f is greater than 2, f will also be 0.3234

That is my answer for f of Y, we found the density function there by taking3242

the derivative of the cumulative distribution function.3250

Those are the answer to part A.3257

Part B, we want to find the probability that Y is between ½ and 3/2.3259

You can do that quickly just by using the cumulative distribution function F of 3/2 - F of ½.3267

You do not have to do any integral here because we already have the cumulative distribution function.3277

Let us just drop those in.3282

Now, 3/2 is between 1 and 2, I’m going to use the second formula there, 3/2²/4.3283

For ½, I use the first part of the definition that is because ½ is between 0 and 1.3294

½ /4, there should be a - there not =.3301

I just work out the fractions, that is 9/4 /4 - ½ /4.3308

½ /4 is 9/16 - 1/8, we can write 1/8 as 2/16.3315

9/16 - 2/16 is 7/16, that is my answer.3326

You do not have to do any integrals there, and that is because the cumulative distribution function was already given to us.3335

If it had been a density function, we would have been doing the integral to calculate that.3341

Let me recap the steps there.3345

We are given in this one, we are given the cumulative distribution function and we are trying to find the density functions.3348

We are given F, we are trying to find f.3354

To find f, what you will just do is, go and take the derivative.3357

You took the derivative of Y/4, that is where that 1/4 came from.3360

Took the derivative of Y²/4, that is where that Y/2 came from because it is 2Y/4.3366

The derivatives of the constants on either end are just 0.3373

We get those two definitions on those ranges.3378

That is how we find f of Y.3382

To find probabilities, if you know the cumulative distribution function, it is just a matter of plugging the endpoints into F.3384

The only issue there is you have to be careful which of the two definitions for F you use.3391

You figured that out because 3/2 is between 1 and 2.3398

1/2 is between 0 and 1.3403

That is why I used those two respected definitions for F.3406

Drop the numbers in, simplify it down, get a nice fraction as my answer.3412

That wraps up our lecture on density functions and cumulative distribution functions.3418

This is part of the chapter on continuous probability,3424

which in turn is part of the probability lecture series here on www.educator.com.3428

My name is Will Murray, thank you very much for joining us, bye.3434