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 1 answerLast reply by: Dr. William MurrayMon Mar 9, 2015 9:32 PMPost by Nick Nick on March 6, 2015Why are we assuming theta2 = 1 and theta1 = 0 ?

### Beta Distribution

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
• Beta Function 0:29
• Fixed parameters
• Defining the Beta Function
• Relationship between the Gamma & Beta Functions
• Beta Distribution 3:31
• Density Function for the Beta Distribution
• Key Properties of the Beta Distribution 6:56
• Mean
• Variance
• Standard Deviation
• Example I: Calculate B(3,4) 8:10
• Example II: Graphing the Density Functions for the Beta Distribution 12:25
• Example III: Show that the Uniform Distribution is a Special Case of the Beta Distribution 24:57
• Example IV: Show that this Triangular Distribution is a Special Case of the Beta Distribution 31:20
• Example V: Morning Commute 37:39
• Example V: Identify the Density Function
• Example V: Morning Commute, Part A
• Example V: Morning Commute, Part B
• Example V: Summary

### Transcription: Beta Distribution

Hello, you are watching the probability lectures here on www.educator.com, my name is Will Murray.0000

Currently, we are working our way through the continuous distributions.0005

We have already had videos on the uniform distribution and the Gamma distribution0009

which included the exponential distribution, and the Chi square distribution.0014

Today, we are going to talk about the last of the continuous distributions which is the β distribution.0018

It is similar to the Gamma distribution, you will see some of the same elements but of course, it is also different.0024

Let us check it out.0030

Before we jump into the actual β distribution, we have to learn what the β function is.0032

The first thing that you have to keep straight here is that, there is a β function and then there is β distribution.0037

They are not exactly the same thing.0044

We are going to use the β function in the process of defining the β distribution.0046

In that sense, it is like the Gamma distribution where we had a Gamma function and then we also had a Gamma distribution.0051

The Gamma function was just one part of the Gamma distribution.0057

Let us see what the β function is.0061

We had two fixed parameters, there is always an Α and there is a β.0063

The β distribution and the β function, it is the whole family of things.0069

Because you could pick any different value for α you want and any different value for β you want.0075

Let us define the β function.0080

What we do is, you want to think of Α and β as being numbers.0083

You take these numbers α and β, and you plug them into this integral.0088

We have the integral of Y ⁺α – 1 × 1 – Y ⁺β – 1 DY, that is just some integral.0092

Remember, α and β are constants.0101

We plug in constant values and then, you have an integral that0105

you can solve using calculus 2 methods and you will get some constant number.0109

The idea is that you plug in Α and β, the β function will spit out a particular number, a constant value.0114

There is a relationship between the Γ and the β functions,0125

which often makes it quite easy to evaluate the β function, which is that B of Α × β is equal to Γ of Α × Γ β ÷ Γ of Α + β.0129

That is very convenient if you have whole numbers because this relationship between Γ of N and the factorial function.0145

Γ of a whole number is exactly equal to N -1!.0156

If you know the numbers for Α and β here, you can just drop them into these 3 Gamma functions,0162

and evaluate those using factorials, and it turns out to be something fairly easy to evaluate.0169

You can find B of α and β fairly quickly as a number, just by calculating out some factorials.0176

If they are not whole numbers then it is a much more difficult proposition.0183

We are not going to get into that.0186

Remember, this is just the β function, I have not talked yet about the β distribution.0188

β function is just something we plug in two numbers, Α and β, and it spits out a number as an answer.0193

Next to a step is to see how that is incorporated into the β density function,0201

which is the density function for the β distribution.0208

Now, we are going to talk about the β distribution.0213

Remember, we already talked about the β function, that is different from the β distribution but0215

it is part of the β distribution.0220

The idea here is that, we want to define the double polynomial distribution.0222

It is always on the interval from 0 to 1, we always have Y going from 0 to 1.0229

We want to define this double polynomial distribution using essentially the function Y ⁺α -1 and 1 - Y ⁺β -1.0232

The point there is that, it is symmetric between 0 and 1.0247

Whatever the function does at 0, the 1 - Y behaves similarly at 1, depending on what the values of α and β are.0253

That is the basic function that we want to look at.0263

The problem is that, if we integrate that, we might not necessarily get it exactly equal to 1.0265

The probability density function must always satisfy the property that, when you integrate it,0272

we take the integral over the whole domain with the answer has to come out to be 1.0279

In order to fix that, what we do is we take this Y ⁺α -1 and 1 – Y ⁺β-1.0285

We just divide by the value of the integral, in order to make the integral come out to be 1.0292

That is how we create the density function for the β distribution.0299

We start out with Y ⁺α -1 × 1 – Y ⁺β -1.0303

And then, we divide it by the integral of that function, in order to not make the total integral come out to be 1.0309

You really want to think of this B of Α β in the denominator here,0316

it is just a correction term that we put in there to make the total integral come out to be 1.0321

It is just a constant and it is just sort of a fudge factor really,0326

that is probably the best way to think of it as a fudge factor to make the integral come out to be 1,0336

the integral of F of Y DY equal to 1.0344

It is not really the most important part of the function here.0349

The most important part are these two polynomial terms, the Y ⁺α -1 and 1 – Y ⁺β -1.0354

You want to think of those as the really important part of the function.0362

Those are the ones that give its shape and we will explore some of the graphs later.0365

This denominator is just a constant, it is a fudge factor the gets thrown in there,0370

in order to make the total integral come out to be 1.0374

The denominator is exactly the β function that we learn on the previous slide.0379

It is read up exactly to be the integral of Y ⁺α -1, 1 –Y ⁺β -1.0385

We divide by that constant, the whole integral now comes out to be 1.0393

That is the density function for the β distribution.0397

Remember, it is a kind of a polynomial thing and also remember that,0401

the density function and the β function are two different things, let us try to keep those straight.0405

We have the β distribution, we should figure out the key properties, I have listed them here.0414

Remember, the mean is always the same as expected value.0419

Expected value and mean are synonymous, they mean the same thing.0423

The mean for the β distribution is Α/Α + β.0427

That is a fairly easy formula to keep track of.0433

The variance is much more complicated and much harder to remember.0437

The variance turns out to be Α × β ÷ α + β² × Α + β + 1.0440

Much messier formula for the variance, we will see an example of that in the problems later on.0450

You will see how that gets used.0456

The standard deviation is usually not worth remembering, because you can always figure it out,0458

if you remember the variance.0463

The standard deviation is always the square root of the variance.0465

That is true for any distribution not just the β distribution.0469

The standard deviation, what we do is we just take that complicated formula for the variance0472

and we slap a square root around everything.0477

It is really not that enlightening by itself, it is more useful to remember the mean and variance of the β distribution.0481

In example 1, we are going to calculate B of 3, 4.0492

Just little practice with the β function first, before we jump into actually solving any probability problems.0498

The solution here is to, there are two ways you can do it.0505

One way would be to solve this as an integral.0510

Here is α, and here is β.0516

You could solve this out as an integral, the integral from 0 to 1 of Y ⁺α – 1 1 – Y ⁺β-1 DY.0518

Because that is the definition of B of Α β, that is the definition there.0533

You can solve out this integral, it would not be too bad because you plug in Α = 3 and β = 4.0538

You just do a little calculus and it would come out to be some number.0547

I'm not going to do it that way because I do not want to do that much calculus.0552

Let me show you another method to do it, instead.0555

The other method is, to remember this relationship between the β function and the Gamma function.0559

We learn that several slides ago, you can go back and check that if you did not pick up on that and register it the first time.0567

But, the relationship between the β function and the Gamma function is that, B of α and β is Γ of Α × Γ of β ÷ Γ of Α + β.0574

That is where it translates everything here into Gamma function.0591

I will go ahead and fill in α = 3 and β = 4.0595

Our Gamma of α + β will be Γ of 3 + 4 is 7.0602

I have to solve some Gamma functions but remember that Gamma function is,0609

it can be thought of as a generalization of the factorial function.0615

Γ of N is just N -1!.0619

This is, Γ of 3 would be 2!, Γ of 4 would be 3!, and Γ of 7 would be 6!.0624

Now, it is very easy just to calculate those factorials.0635

2! Is just 2, 3! Is 6, 6! is 9 × 2 × 3 × 4 × 5 × 6.0638

I'm writing it like that because now, I can cancel some things, I will not have to multiply by big numbers.0654

I will cancel the 6 there and I will cancel that 2 with that 2.0659

And now, I have got 1/3 × 4 × 5 which is 3 × 4 is 12 × 5 is 60, 1/60.0664

To remind you how that worked.0681

There are two ways I could have solve this.0683

I could use the original definition of the β function which is an integral formula.0685

And then, I would just plug in the values of the Α and β and done some calculus to work out the integral formula.0691

I see that with the benefit of having gotten the other way, that my integral better has worked out to be 1/60.0698

The way I actually calculated it, was to use this relationship between0708

the β function and Gamma function which I gave you a couple slides ago.0713

Translates a β function into three computations using Gamma functions.0719

I dropped in the values for Α and β, and converted those into factorials.0725

Remember, the Gamma function is just like the factorial function except offset by 1,0730

when you have a whole number in there.0735

I plugged in those factorials, calculated it out, and simplified it down to a relatively nice fraction there.0738

In example 2, we are asked to graph several density functions for the β distribution,0747

using several combinations of Α and β.0753

Let me first remind you what the density function is for the β distribution.0756

The density function for the β distribution is F of Y is equal to Y ⁺α – 1 1- Y ⁺β -1.0763

Those are really the important factors in the density function for the β distribution.0777

There is one other factor but it is just a constant, and it is the sort of correction term.0782

This fudge factor that gets introduced, in order to make the total integral be 1.0787

I'm just going to write that as a constant here.0793

I’m not going to even bother to work that out for the different combinations here because0795

that is not so important to determining the shape.0800

What is really important to determining the shape is the values of Α and β,0803

and those factors of Y ⁺α -1 and 1 - Y ⁺β -1 in the numerator.0808

Let me set up some axis here and we will take a look at these different combinations of Α and β.0817

Here are my axes.0826

Remember, the β distribution is always defined from Y equal 0 to Y = 1.0828

Up at Y here, it is a little confusing on the X axis.0835

But here is Y = 0 and here is Y = 1.0839

The important thing to look at here, are the different values of Α and β.0845

In particular, whether they are greater than 1 or less than 1.0850

Here is why, let us look at the key part of the density function here is Y ⁺α – 1.0856

Let us think about what that does for different values of Α.0868

At Y = 0, let us think about what that does.0876

If Α is bigger than 1, that means our exponent Y ⁺α -1 is positive.0882

Y ⁺α-1 will go to 0.0890

If Α is equal to 1 then Y ⁺Α -1 would just be Y at 0 go to 1.0900

If Α is less than 1 then Y ⁺α -1 will be Y to a negative number.0910

0 to negative power, it is like trying to divide by 0, that will go to positive infinity.0917

That helps me characterize the behavior at Y = 0 depending on the different values of Α.0927

The same kind of thing happens, if we look at 1 - Y ⁺β -1.0934

This similar kinds of phenomenon will occur at Y = 1, because if Y goes to 1 then 1 - Y goes to 0.0944

If β is bigger than 1 then 1 - Y ⁺β -1 goes to 0.0956

If β is equal to 1 then 1 - Y ⁺β -1 goes to 1.0964

If β is less than 1 then 1 - Y ⁺β -1 goes to infinity, because we are trying to take 0 to a negative exponent there.0972

Now, we will look at the different combinations of α and β.0985

In the first combination here, I see I got an Α of ½ and a β of 2.0988

Α of ½ is less than 1 that means it is going to go to infinity at Y = 0.0994

A β of 2 is bigger than 1, that means that Y = 1, it is going to go to 0.1002

I got something that goes to infinity as Y goes to 0.1012

And then, it comes down, let me make it a little steeper and in it is declined1021

because we are only allowed at the total area 1 here.1027

We are allowed to have total area 1.1033

It has to come down and hit 0 at Y = 1.1037

What we got right there is the graph of Α = ½ and β is equal to 2.1043

Let me do the next one in a different color.1052

I will do 1 and 4 in green, that Α is 1 and β is 4.1055

I see when α is equal to 1 and Y is equal to 0, it goes to 1.1064

Now, that is a little bit misleading.1070

I cannot say it is exactly equal to 1 because I'm kind of ignoring the effect of this constant here.1072

It is going to be 1 ÷ some constant.1079

Let me just show it at some finite value.1084

The important thing there is, it is not going 0 and it is not going to infinity.1088

I see that, as it goes to Y = 1, the β value is still bigger than 1.1093

It still going to go down to 0 there.1100

Let me make that a little more rounded, more curved, it is not a completely straight line.1116

That is my combination for Α is equal to 1 and β is equal to 4.1128

And that means that, at 0 it is going to go to a finite limit and at 1 it is going to go down to 0.1140

I will do the next one in red, α is 10 and β is 2.1151

What is that mean, Α is 10, that is way bigger than 1 which means at Y = 0, it is definitely going to go to 0.1159

It is so much bigger than 1 that, that term is going to sort of drag it down for a long time.1168

Here it is at 0, at Y = 0 and it is going to drag along near 0 for quite a long time.1174

Since β is 2, that means when it gets to the right hand and point here, it is still going down to 0.1184

It does have to have area equal to 1, the area underneath the curve has to be equal 1 because all of these functions do.1193

At some point, it is going to get some area.1200

Let me show it getting some area right at, as it approaches 1 there.1203

It is skewed to the right hand side, that Α = 10 drags it down on the left.1213

The β = 2 does drag it down at Y = 1 but we have to have an area of 1 in there, somewhere.1220

That is my 10, 2, α = 10 and β = 2.1228

I’m going to try out my new purple marker for that.1237

Here is α and that looks a lot like a blue to me.1243

Α is 1.1 and β is 2.1247

What that means, α is 1.1 and it is going to behave very similarly to the Α = 1,1251

which means it is going to trying to go into a finite limit.1262

Β = 2 means it is also going to be tied down at 0.1267

Let me show this graph that sort of trying to go a finite limit, as it approaches Y is equal to 0.1272

There it is, as it approaches Y = 0, it is trying to go to a positive limit.1281

What happens is, when it gets right down to Y = 0, that 1.1 is still bigger than 1 which means it is forced to go to 0.1287

Here is this graph that sort of trying to go to a finite limit and then at the last moment,1298

it has to turn sharply downwards and go to 0.1303

This purple graph, if you can tell the difference between the purple and the blue,1310

this purple graph is Α = 1.1 and β is equal to 2.1313

Let me recap all the different of things we are exploring here.1323

I have to highlight them in yellow, as I go along.1327

The important thing is to remember the general form of the β density function.1329

Here it is right here, Y ⁺α-1 1 - Y ⁺β -1.1335

And then, that constant in the denominator is really not important.1341

I’m not going to worry about it, it does not affect the shape of the graph.1345

There are two sub components there, the Y ⁺Α -1 determines what happens at Y = 0.1351

You want to look at the value of Α to determine what happens at Y= 0.1358

If Α is bigger than 1, then you got a positive exponent on Y, it is going to go to 0.1363

If α is equal to 1, that terms drops out and it goes to a finite limit.1369

If α is less than 1, then you got 0 to a negative number which means you are trying to divide by 0,1374

which means you are going to go to infinity.1381

If you look at the first graph here, there is an Α less than 1 and that is why it is going to infinity.1383

The second graph, α is equal to 1, goes to a finite limit.1391

The third graph, Α is much bigger than 1 which is why it starts off at 0 and stays at 0 for a long time.1396

Finally, the 4th graph, Α is bigger than 1 but very close to 1, that is why it is sort of1408

trying to go to a finite limit and then at the last minute, it has to drop down to 0.1414

The second component of this function is the 1 - Y ⁺β -1.1420

That determines what the graph does at Y = 1, in very much asymmetric fashion to the Y = 0,1426

except that it is looking at the value of β, instead.1434

If β is bigger than 1, you got something going to 0.1437

If β is equal to 1, you got something going to 1, and or at least a constant.1441

If β is less than 1, then you are trying to divide by 0, you are going to go to infinity again.1447

In this case all, of the β that we looked at were all bigger than 1, which means all of these graphs sort of tied down to 0 at Y = 1.1455

That is why all these graphs tie down here, but they all exhibit different behavior on their way to getting to that point,1466

which is why we get this sort of interesting of differences in all these graphs up to that point.1475

It is worth trying some of these out in your calculator, if you want to try graphing some of these in your calculator,1481

just throw these values of α and β into the density function, and put the graphs on your calculator,1486

and see what kinds of shapes you get.1492

You get quite a lot of riots, it is kind of fun.1494

In example 3 here, we are going to make a connection to a previous distribution that1498

we learn which was the uniform distribution.1504

It turns out that it is a special case of the β distribution.1507

I did a whole video on the uniform distribution.1511

If you do not remember the definition of the uniform distribution, or if you do not know what is it all,1515

you can go back and look at the previous video covering the uniform distribution.1521

You will see lots of information about the uniform distribution.1525

You just need a quick refresher on the uniform distribution.1528

I will remind you that the uniform distribution, the density function is just F of Y1531

is always equal to 1/θ2 - θ1, where θ1 and θ2 are constants here.1539

Those tripe lines there, the triple = means it is always equal to something.1548

It is not varying at all, it is equal to this constant the whole way.1554

That is where Y varies between θ1 and θ2.1560

What we are going to do now is show how the β distribution, if you choose the right parameters1569

turns into the uniform distribution.1575

That was the uniform distribution that I showed you up above.1578

Now, let me show you the formula for the β distribution which is much more complicated.1581

F of Y is equal to Y ⁺Α -1 × 1 - Y ⁺β -1 ÷ B of Α β and that goes from Y between 0 and 1.1589

I want to show you how you can choose the right parameters and turn the β distribution into the uniform distribution.1615

I'm going to choose my Α is equal to 1 and my β is also equal to 1.1621

Let us see how that worked out.1630

First of all, let us find the constant value B of Α β.1631

I'm going to use the relationship with the Gamma function, in order to figure that out.1637

That is Γ of 1 × Γ of 1 ÷ Γ of 1 + 1.1643

This is the relationship between the β function and the Gamma function,1653

that I mentioned back in one of the earlier slides in the lecture.1656

If you do not remember this, just go back and check your earlier slide and you will see the β and Γ relationship.1659

Also, remember here that Γ of N is N -1!, when N is a whole number.1667

This is 0! × 0! ÷ Γ of 2 would be 1!.1676

Of course, all those factorials, 0 and 1 are just 1, this is all just 1.1684

The β and density function is F of Y, that denominator is now going to be 1, we just work that out.1690

Y ⁺α -1, α is 1 so α-1 is 0 × 1 - Y ⁺β -1 is also 0.1698

And, that just comes out to be 1.1710

It looks like it is going to be constant.1714

Notice, by the way that, if it is constant then I can put three lines there,1717

that is equal to 1/1 -0 that is 1/θ2 – θ1, where θ1 is 0 and θ2 is 1.1723

What I found is that, I got the same density function for the β distribution as I would have gotten for the uniform distribution.1738

This is the same as the uniform distribution.1757

I discovered the uniform distribution as a special distinguished member of the β family.1769

If you choose your α and β right, the β distribution just turns into the uniform distribution.1775

Let me recap that, first of all, I recalled the uniform distribution and we did have a whole lecture on the uniform distribution.1783

You can check the video on that, if you do not really remember how that worked out.1791

The idea is you take Y between two values, θ1 and θ2.1794

F of Y is just 1 ÷ θ2 – θ1.1800

It is a constant distribution, there is no Y term appearing in there.1804

And then, I reminded myself of the density function for the β distribution.1809

That is our density function for the β distribution.1814

I picked good values in the parameters, I pick α to be 1 and β to be 1.1817

Then, I just plug those in.1823

First, I had to calculate B of Α β and I turn that into an expression using Gamma functions.1825

In turn, the Gamma functions turn into factorial.1832

It just reduced down to 1 which why my denominator here was 1, that is where that one came from.1835

And then, I plugged in my values of Α and β into the exponent, I got 0 is for the exponents,1843

which why everything just disintegrated into big old 1 here.1848

We just got that constant distribution F of Y is equal to 1, all the way across,1854

which is the same as the uniform distribution on the interval 0-1.1859

If you take θ1 = 0 and θ2 = 1, then we get the uniform distribution of 1 on that interval.1866

The uniform distribution, it turns out, is a special case of the β distribution.1875

In example 4, we are going to look at the triangular distribution F of Y = 2Y from 0 to 1,1882

and show that that is also a special case of the β distribution.1889

Let me just draw a quick graph of that triangular distribution.1893

It is obvious why we are calling it the triangle distribution.1896

Let me make my axis in black, I think that will show things a little better.1901

We are going from Y = 0 to Y = 1 here.1909

You always do that with the β distribution, it always goes from 0 to 1.1912

F of Y = 2Y, that is just a straight line is not it.1917

Probably, a little bit steeper than that, let me make that a little bit steeper.1923

F of Y = 2Y, and then, the challenge here is to recognize that as a special case of the β distribution.1932

Let me remind you of the density function for the β distribution.1944

And then, we will take a look at it and see if we can make it match what we have here.1948

For β distribution, F of Y is equal to Y ⁺Α -1 × 1 – Y ⁺β -1 ÷ B of Α, β.1954

I want that to match 2Y, and I think what I want to do there is I want to pick Α equal to 2.1967

That will make the exponent on the Y match.1978

That 1 - Y does not really seem to match.1981

In order to make that dropout, I'm going to take my β equal to 1.1983

Let us plug those in and see how it works out.1989

B of Α β, I could use the original integral definition to calculate that but I'm fond of the relationship1992

between the β function and the Gamma function.2000

I’m going to use that.2003

It is Γ of Α × Γ of β ÷ Γ of α + β.2004

Those are the relationship that we had back on one of the earlier slides in this lecture.2013

You can look that up, if you do not remember it.2017

Γ of Α is Γ of 2 × Γ of 1, β is 1 here.2020

Γ of 2 + 1 is Γ of 3.2028

Now, let us remember that the Gamma function is just a sort of2031

a generalized version of the factorial function, except it shifted over by 1.2036

This is 1! × 0! ÷ 2!.2041

I’m shifting everything back by 1 that is because Γ of N is equal to N -1!, for whole numbers there.2047

This is easy to solve, 1! Is 1, 0! Is 1, and 2! Is 2, that is ½.2057

F of Y, now I know what my constant is, it is ½, is Y ⁺Α – 1.2066

That is Y¹, 1 - Y ⁺β – 1, that is 1 - Y = 0 ÷ ½.2074

That simplifies, the 1 - Y = 0 is just 1, it goes away.2083

What we get here is Y by itself, but then, ÷ ½ is the same as multiplying by 2.2089

It is 2Y, and of course Y is trapped between 0 and 1 here.2101

That is very encouraging because that is the distribution, that is the density function that we are looking at.2107

We started out with F of Y is 2Y, we found that to occur, if we pick the right parameters in the β distribution.2119

Let me recap here, we started out wanting F of Y is equal 2Y.2130

I wrote down the density function for the β distribution, Y ⁺α -1 1- Y ⁺β – 1/B of α, β.2136

I’m trying to make it match to Y, I kind of looked at this exponent α -1.2145

And I said, how can I make that be equal to 1, that will work if α is equal to 2.2150

That 1 - Y really is not represented over here.2156

The 1 - Y has a power of 0 here, I will take β equal 1 to make that work out.2160

And then, I had to calculate the constant B of Α, β.2167

I did that, by converting that into Gamma functions, according to the formula2171

that we had on the earlier slides for this lecture.2175

It is Γ of α × Γ of β ÷ Γ of α + β.2178

If you drop the values of α and β data in there, then you will get something that we can simplify into factorials.2183

Remember, there is a relationship between the Gamma function and the factorial function.2190

Once, you simplified it into factorials, it simplifies quite nicely down into a fraction.2197

The F of Y, if I fill in my α is 2 then I get Y¹.2204

Β is 1 so I get 1 – Y = 0.2211

That Y ÷ /2 which is 2Y.2217

2Y from Y going from 0 to 1 is exactly the density function that we started with.2221

It does match what we are given, we do achieve this triangular distribution as a special case of the β distribution.2229

Let me show you quickly why it is called the triangular distribution.2241

If you look at all the area there, the area is a triangle.2244

The density is sort of spread out over a triangle there, which is why we call it the triangular distribution.2251

In example 5, we got morning commute, maybe this is you driving to work in the morning or driving to school in the morning.2261

It is a random variable, apparently, that has a β distribution with α and β both being 2.2269

This is measuring your commune in hour.2277

Remember, Y is always between 0 and 1 in the β distribution.2280

I guess that means that your commute could be 0 or it could be up to an hour.2286

In part A, we are going to find the chance that it will take longer than 30 minutes.2293

What we will actually be looking for there is the probability that Y is bigger than or equal2298

to 30 minutes is ½ an hour, it is bigger than or equal to ½.2304

In part B, apparently, you have a rage level which is a function of how long your commute is.2310

You want to find the expected value of rage that you will arrive at work tomorrow.2316

Let us work this out.2324

The first thing I'm going to do is try to identify the density function for this distribution.2326

We got F of Y, the generic density function for the β distribution is Y ⁺α – 1 1- Y ⁺β – 1 ÷ B of Α, β.2333

Let me go ahead and find the value of that constant, B of Α, β.2348

I think the easiest way to calculate these, if you have whole numbers, is to convert it into Γ.2355

Let me convert that into Γ of Α × Γ of β ÷ Γ of α + β.2364

That is a formula that we learn back on one of the first slides in this lecture.2374

In this case, we got Γ of 2 × Γ of 2 ÷ Γ of 4.2378

Because α and β are both 2, that was given in the problem.2385

This is 1! × 1!.2389

Γ of 4 would be 3!.2395

Remember, a part of the Gamma function is, it sort of the generalization of the factorial function but it is offset by 1.2398

Γ of 4 is 3!, 1! Is 1, 3! Is 6, this is 1/6 here.2406

This F of Y is Y ⁺α – 1, that is Y¹, 1 - Y ⁺β -1 is also 1.2415

And now, we know that we are dividing it by 1/6.2425

This simplifies down into 6Y × 1 - Y is Y - Y².2429

I can go ahead and distribute that, 6 Y - 6 Y².2439

That is my density function for this distribution.2445

We are going to go ahead and use that to solve these two problems.2450

I'm going to jump over onto next slide and use that density function to solve these two problems.2453

But, let me recap where these came from.2461

First, I was writing out the generic density function for the β distribution.2463

There is the formula there.2468

I have to figure out what B of Α, β was, I did that down below.2470

I converted that into a bunch of Γ because I know easily how to solve the Gamma function,2473

when you have a whole number in there.2481

It is just the corresponding factorial, except you have to shift it down by 1 to plug the number into factorial.2483

That is why B of Α, β turns out to be 1/6.2494

I plugged that into my density function.2498

I also plugged in the values of Α and β.2510

2 -1 is 1, and I got 6 × Y - Y², that is the density function that I'm going to work with,2513

in order to solve these two problems.2521

On the next page, I’m not going to copy the problems because I need the space.2526

We are going to solve the probability that Y is greater than ½.2529

And then, we are going to find the expected value of R.2533

Those are the two things that we are going to solve on the next page there.2537

We are still working on example 5, we have the setup on the previous side.2545

What we figured out was the density function is equal to 6 × Y - Y².2549

For part A, we want to find the probability that your commute will be longer than 30 minutes.2556

The probability that Y will be bigger than or equal to, we are measuring things in terms of hours.2562

I converted 30 minutes into an hour, it is ½, that is the integral.2570

Our range is from 0 to 1, ½ to 1 because you want it bigger than ½ of 6Y -6Y² DY.2576

We have a little calculus problem to solve and it is not a hard one at all.2589

Let us see, integral of 6Y is 3Y².2594

The integral of 6Y² is 2Y³.2602

We want to integrate that, to evaluate that from Y = ½ to Y = 1.2607

That is 3 -2 -3 × ½² is ¾.2614

+ 2 × ½³, ½³ is 1/8.2626

2 × 1/8 is ¼.2631

We get ¼ there.2635

3 -2 is 1 - ¾ is ¼.2641

This is ¼ + ¼, and that is ½, that is the probability that you are going to spend more than 30 minutes in traffic tomorrow morning.2647

For part B, we had this rage level, your rage level was R is equal to Y² + 2Y + 1.2661

What we want to calculate there was your expected rage level.2672

How angry you expected to be, when you get to work?2675

What we will use heavily here is the linearity of expectation.2679

The expected value of R is the expected value Y² + 2Y + 1.2683

But that is equal to the expected value of Y², this is linearity now, + 2 × the expected value of Y +,2693

We can say that the expected value of 1 is just going to be 1.2704

We need to find expected value of Y and Y².2710

In order to figure out those, I'm going to remember what we learned on one of the very early slide.2714

You can flip back and I think the slide was called key properties of β distribution.2721

What we learn was that the expected value of Y was equal to Α ÷ Α + β.2726

In this case, α and β are both 2.2735

This is 2 ÷ 2 + 2, that is 2 ÷ 4 is ½.2738

That was the easy one, the expected value of Y² is a little trickier.2745

What we are going to do is use the variance.2749

Σ², and that was kind of complicated formula, let me remind you what it was.2752

This is coming from, I think it was the third slide in this lecture,.2758

It was called key properties of the β distribution.2761

(α + β)² × α + β + 1.2763

We are going to work that out, α × β.2771

Α and β are both 2, that was given to us on the previous slide.2773

2 × 2 is 4 ÷ 2 + 2² is 16, and then 2 + 2 + 1 is 5.2779

This 4 ÷ 16 is ¼ × 1/5 is 1/20.2789

That is not the expected value of Y², that is the variance.2797

Let me remind you how we use to calculate the variance.2802

That is the expected value of Y² - the expected value of (Y)².2806

What we can do is, we can use this to solve for the expected value of Y².2814

This is a little old trick in probability is, if you know the variance, you can sort of reverse engineer for E of Y².2821

E of Y² is equal to 1/20 + (E of Y)².2828

We can write that as 1/20, we figure out what E of Y is, that is ½.2840

We figure that out up above, ½², this is 1/20 + ¼ which is my common denominator, there is 20, 1/20 + ¼ is 5/20.2845

I get here 6/20 and that reduces down to 3/10.2864

My rage level, I decomposed it into E of Y² + 2E of Y + E of 1.2870

Now, I can solve that, E of R is E of Y² + 2 E of Y + 1.2878

I can solve that E of Y² is 3/10.2890

2 E of Y is 2 × ½, I should have written E of 1 up above, here is E of 1.2896

E of 1 is just 1 because that is the expected value of a constant.2907

A constant is always going to be a constant, it is 1.2912

This is 3/10 + 1 + 1, 3/10 + 2 is 23/10.2916

All the Y is known as 2.3.2927

I do not know what the units are there, I’m just going to leave them.2930

I guess 2.3 range units is what you are going to carry into work,2934

or at least the expected value of your rage as you could work tomorrow morning.2939

2.3 is the expected value there.2945

That answers both things that we were looking for here in this problem.2950

Let me remind you where everything came from.2953

This F of Y, this density function is something we figure out on the previous side.2957

You can check back on the previous idea and see all the steps calculating that.2961

In part A, we had to find the probability that it will take longer than 30 minutes to drive to work.2965

30 minutes is ½ an hour and we want it to be longer than 30 minutes.2971

We are going to integrate from ½ to 1, the density function, and that turns out to be a fairly straightforward integral.2977

I’m going to keep the fractions straight.2985

We get that the probability there is exactly ½.2989

Half the time your commute will be longer than 30 minutes, half the time it would be shorter than 30 minutes.2993

In part B, we had to find the expected value of your rage level, where it is defined like this.3001

The expected value decomposes into these three terms.3008

That is using linearity of expectation, very useful for this kind of problem.3013

The expected value of Y, I am looking at the formulas for mean and variance on earlier slide from this talk.3019

If you go back, just scroll up and you will see a slide called key properties of β distribution,3027

and that is where I got these formulas, these complicated formulas using α and β.3032

But then, I just plug in Α = 2 and β = 2, that was given to me in the problem.3036

And, simplify those down into fractions ½ and 1/20.3042

However, this was the variance, it was not the expected value of Y² directly.3048

Remember, the old way to calculate variance is to find the expected value of Y² - the expected value of (Y)².3053

What we can do is reverse engineer this, in order to solve for the term we want.3062

Here is it, the term we want is the expected value of Y².3069

I bring this E of Y² over to the other side and that is where I get 1/20 + E of Y².3073

And then, this half right here is that half right there.3081

That is where that half comes from.3086

That is some more fractions, doing little simplification of the fractions.3088

We get that E of Y² is exactly 3/10.3092

I drop that into our expected value for R, and then this ½ is also that ½ from the E of Y.3096

The expected value of 1 is always 1 because 1 is constant.3109

You expect it to have its value.3113

And then, simplifying down 3/10 + 1 + 1 is 23/10 or 2.3.3116

That wraps up this lecture on the β distribution.3124

This was the last of the continuous distributions.3128

We worked through the uniform distribution, and the normal distribution, and the Gamma distribution.3131

The Γ, of course, includes the exponential and the Chi square distribution.3137

We had one big lecture on all of those.3142

Finally, we got the β distribution.3144

You are supposed to be an expert now on the continuous distributions.3147

I got one more lecture in this chapter, it is going to cover moment generating functions.3151

That is what you will see, if you stick around for the next lecture.3155

In the meantime, you have been watching the probability lectures here on www.educator.com.3158

My name is Will Murray, thank you for joining me today, bye.3163