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Lecture Comments (2)

1 answer

Last reply by: Dr. William Murray
Mon Dec 15, 2014 1:11 PM

Post by kunle fawole on December 14, 2014

How did you determine how to draw the y1 + y2 line? (red)

Distribution Functions

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Distribution Functions

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  • Intro 0:00
  • Premise 0:44
    • Premise
  • Goal 1:38
    • Goal Number 1: Find the Full Distribution Function
    • Goal Number 2: Find the Density Function
    • Goal Number 3: Calculate Probabilities
  • Three Methods 3:05
    • Method 1: Distribution Functions
    • Method 2: Transformations
    • Method 3: Moment-generating Functions
  • Distribution Functions 4:03
    • Distribution Functions
  • Example I: Find the Density Function 6:41
    • Step 1: Find the Distribution Function
    • Step 2: Find the Density Function
    • Summary
  • Example II: Find the Density Function 14:36
    • Step 1: Find the Distribution Function
    • Step 2: Find the Density Function
    • Summary
  • Example III: Find the Cumulative Distribution & Density Functions 20:39
    • Step 1: Find the Cumulative Distribution
    • Step 2: Find the Density Function
    • Summary
  • Example IV: Find the Density Function 33:01
    • Step 1: Setting Up the Equation & Graph
    • Step 2: If u ≤ 1
    • Step 3: If u ≥ 1
    • Step 4: Find the Distribution Function
    • Step 5: Find the Density Function
    • Summary
  • Example V: Find the Density Function 48:32
    • Step 1: Exponential
    • Step 2: Independence
    • Step 2: Find the Distribution Function
    • Step 3: Find the Density Function
    • Summary

Transcription: Distribution Functions

Hi and welcome back to the probability lectures here on www.educator.com. My name is Will Murray0000

We are starting a new chapter today which is on finding the distributions of functions of random variables.0006

I need to explain what that is all about.0014

We are going to use three different techniques to find distributions of functions of random variables.0016

I have a lecture on each one of these three techniques.0022

The first one is distribution functions.0025

The second one is going to be transformations.0028

The third one is the method of moment generating functions.0031

In this lecture, I want to give an introduction to the whole idea.0035

And then, we are going to spend some time studying the first method which is distribution functions.0038

The premise here is that, we have one or more random variables.0046

We might just have a single variable Y, we might have several Y1 and Y2, and so on.0050

We want to study functions of these random variables.0055

For example, we might want to study something like U can be defined to be Y1 + Y2.0059

That is a function of two random variables.0066

We want to find the distribution of U.0070

Let me emphasize that, in the previous chapter, we had some theorems that helped us to find the mean and the variance of U.0074

But, that is not necessarily enough to determine the whole distribution of U.0082

The goal in the next couple of lectures is to actually find the full distribution for U.0087

I need to show you, how we are going to do that.0096

As I said, we want to find the full distribution function.0099

I have called it F sub U of u.0103

By definition, that means the probability that U will be less than some value of u.0108

If we can find that distribution function which is F, then we can find the density function which is f.0116

Remember, the density function is always the derivative of the distribution function.0124

What we will do is, we will find the distribution for a function first F.0130

And then, we will just take one derivative to get the density function.0135

Assuming we can do that, then we can calculate probabilities on U.0138

The probability that U will be in some range, you can calculate it as an integral of the density function.0145

We will just use the distribution function F sub U of B - F sub U of A.0152

That is the reason why we are interested in doing this is, is we want to be able to calculate probabilities on U.0161

I’m not going to spend a lot of time on that particular element of it.0167

The goal really for us today, is going to be to find this distribution function F.0172

If we can do that, it will be just a quick step to find the density function, by taking the derivative.0178

As I mentioned already, we are going to use three methods to find this F.0186

Today, we are going to talk about distribution functions.0192

That is what this lecture is focused on, distribution functions.0195

We are going to be using a lot of geometry and we will have to do some double integral.0198

I hope you are ready to go with your multivariable integrals.0202

I hope your calculus 3 is up to snuff.0205

If it is not, do not worry, we have some lectures here on www.educator.com of multivariable calculus.0207

My colleague Raffi is just excellent.0213

You can check out his lectures and practice your double integrals.0216

The second technique is called transformations and we will cover that in the next lecture.0219

That has got a lot of the methods from calculus 1, you will be up to speed on that.0224

The final technique is on moment generating functions, 0228

and that is something that I taught you about in an earlier lecture here in the probability lectures.0232

Today, we are to talk about distribution functions and let us jump into that.0238

There is not a whole lot to say about distribution functions.0243

The distribution function, by definition, is the probability that U is less than or equal to u.0249

The way we want to approach that is, figure out what values of your variable Y or Y1 and Y2,0257

correspond to U being less than or equal to u.0266

In practice, you start out by saying F of U is equal to the probability that u is less than or equal to u.0270

And then, you convert this U into a Y or Y1 and Y2.0284

I do not mean you just change it to Y1 and Y2, use the definition of U as a function of Y1 and Y2.0290

Convert into a function of Y1 and Y2.0303

You convert that into Y1 and Y2, or maybe there is just one Y in there.0312

We want to find the probability that Y1 and Y2 will be in that particular range.0317

I do not have a good general way to tell you to do that, all I can say is it is a probability calculation.0324

For single variable, since you are finding a probability, this usually means solving a single variable integral.0331

For two variables, it means solving a double integral or sometimes just using some geometry0337

because it often turns out to be a process of looking at a geometrical region,0342

figuring out which combinations of Y1 and Y2 will give you that function less than or equal to u.0348

It is very hard to give you general guidance, I think the strategy for learning about distribution functions is really to immerse yourself in examples.0358

Practice a whole bunch of examples and it will start to make sense.0366

All the examples are going to start with this probability of U less than or equal to u.0373

And then, we will convert this into U of Y and possibly U of Y1 Y2.0380

They will all start like that but after that, they sort of all go in different directions.0387

Some will go geometrically, some turn into a single integral, some turn into a double integral.0391

Just work through the examples with me and I think you will start to get the hang of it.0396

In example 1, it is a single variable problem.0402

We have a density function 3/2 Y² for the variable Y.0405

We want to find the density function for U which is 3 -2Y.0411

The first thing that I want to note here is the possible range of values for U.0417

If I plugged in the range of values for Y, when Y is equal to -1 then my U will be 3 -2 × -1, which will be 3 + 2 is 5.0422

When Y =1, that gives me a U of 3 -2 × 1 which is 1.0436

That tells me a range for U, it is between 1 and 5, that is good progress there.0445

We want to find the density function ultimately for U. 0451

The method here is to use distribution functions.0455

We want to find the distribution function first.0458

F sub U of u is by definition, the probability that U will be,0463

let me erase that tail to make that look like a U, that is subscript there.0472

The probability that U is less than or equal to any value of u.0477

I want to convert that into a probability on Y.0481

This is the probability, my U is defined to be 3 -2Y less than or equal to u.0487

And now, I want to solve that for Y.0497

Let me just say, if 3 -2Y is less than or equal to u, 0500

if I pull the Y over the other side and pull the u over to the left, that is 3 - u so that is equal to 2Y.0509

That saying Y is bigger than or equal to 3 - u/2.0518

This is the same as the probability that Y is bigger than or equal to 3 - u/2.0525

Remember, I can solve that with an integral.0534

This is the integral of the density function is 3/2 Y² DY.0538

I want to integrate that as Y goes from 3 - u/2.0549

Since, I want it to be bigger than that, up to the top of the range which is Y = 1.0555

I’m going to do an integral there, it is not a bad integral.0562

The integral is 3/2 Y² is ½ Y³.0566

Because, the integral of 3Y² is Y³, so ½ Y³.0574

I have to evaluate that from Y = 3 - u/2 up to Y = 1.0582

I get ½ (1 - 3 - u/2³).0591

What I just found here was the distribution function, that is F of U.0603

That is the distribution function which is not quite what I'm looking for.0612

I’m asked to find the density function distribution function.0616

To get the density which is what I'm really looking for, I’m going to take the derivative of that.0623

Density function is f of U and that is always the derivative of the distribution function.0630

The derivative of what I just found above.0642

Let us see, that is ½.0645

If I take the derivative of 1, that is a constant so that goes away.0647

The derivative of -3, I’m going to use the power rule, 3 -u/2².0651

By the chain rule, I have to multiply by the derivative of 3 -u/2 which is - ½.0662

I can collect all my constants, ½ × -3 × - ½ is ¾.0673

I'm just going to leave 3 - u/2² and that is my density function.0682

I should give a range on U, here it is, U goes from 1 to 5.0691

That is my density function, my f on the variable U.0703

Let me recap the steps there.0712

The first thing I did was, I was given this density function for Y and I was given U in terms of Y.0714

I wanted to first translate the range for Y into a range for U.0723

I looked at my endpoints for Y, -1 and 1.0729

I plugged both of those into the definition of U and I got N points for U.0732

U is between 1 and 5.0736

And then, I took the key step here of saying that the distribution function for U, the F, 0738

it is always the probability that U is less than or equal to u.0747

Then, I translated that U into a Y, into a function of Y.0753

And then, I want to solve that out for Y, in terms of U.0760

That is what I was doing over here, is solving that into Y bigger than 3 - u/2.0763

The point of doing that is, I can then convert that into an integral.0770

Since it is bigger than 3 - u/2, I made 3-u/2 my lower limit.0775

I go up to the top of the range Y = 1, that comes from that one right there.0780

I integrate my density function, do that integral, plug in the limits, and what I get is the distribution function for U.0785

Which is not quite what I wanted, I was asked to find the density function.0796

Remember, density and distribution are two different things.0800

But you get from the distribution to the density, by taking the derivatives.0802

You will get f, by taking the derivative of F.0807

I took that derivative, in terms of U.0811

The one dropped out, it was a constant, that 1 went away.0814

And then, I took the derivative of 3 - u/2³ using the power rule.0818

That was the power rule, right there.0824

3 × 3 – u/2², and then, that was the chain rule.0827

That is why I got that - ½, it came from the derivative of the inside 3 - u/2.0833

Then, I just collected all my constants, simplified a little bit, and reminded myself of what the range on U is.0838

Now, I have a nice density function for U and I know what range it is defined on.0844

We are going to use this same starting premise of the 3/2Y² for example 2.0850

But, we are going to use it for a different function U.0858

Make sure you understand this, before you move on to example 2.0862

Make sure everything makes sense and then, we are going to practice this again for a different function U.0865

You can try it on your own, if you want, before you listen to my solution for example 2.0870

In example 2, we have the same density function and the same range for Y that we had for example 1.0878

The density function is 3/2Y² and the Y goes from -1 to 1.0885

We want to find the density function for a new definition of U, it is going to be Y² now.0890

The premises are the same, let me find the range on U.0898

If Y goes from -1 to 1, let me just plug those endpoints in.0906

If Y is -1 then U will be 1.0912

If Y is 1, that will also make U equal to 1.0915

I notice that in between there, we will have Y = 0.0919

You could be as low as 0, Y = 0 gives me U = 0.0924

My range on U will be between 0 and 1, that is the range that I will be looking for there.0931

I have to find the density function for U, the way I’m going to do that is to first find the distribution function.0939

I will use the distribution function initially, and then assuming that works out, 0948

I will take the derivative to find the density function.0952

The distribution function is F sub U of u.0955

By definition, that is the probability that U is less than or equal to some value of u.0968

I'm going to plug in what U is, it is defined to be Y².0979

This is the probability that Y² is less than or equal to u.0984

Now, I want to solve that into a range on Y.0992

If Y² is less than or equal to U, that really means that Y by itself, will be less than √ u and bigger than – √ u.0996

Those are my ranges for Y, it is nice because I can now convert that into an integral.1010

That is the integral as y goes from –√U to √U. 1018

What was my density function, 3/2 Y² DY.1026

That is an easy integral, the integral of 3Y square is just Y³.1033

We have ½ Y³ here and we want to evaluate that from Y = v-U to Y = √U, that is ½.1037

Y³ at √U, that is the √ U³ will be U³/2 and – Y³, -√U³/2.1058

√U³ will be U³/2, get some more parentheses in there and make it honest.1075

What we have is two copies of U³/2, but we are also multiplying by ½.1084

I would just get U³/2, that is kind of pleasant.1090

That was the distribution function, that was my F of U.1095

Let me figure out my density function which is my f.1101

I have done all the hard work to get the density function, once you know the distribution function,1107

what you do is just take the directive of the distribution.1112

The density function is just F sub U of u derivative.1118

Let me put in f of U.1129

I just take the derivative of that which is 3/2 U ^½ or 3/2 √U.1135

My range on that is, U goes from 0 to 1, 0 less than or equal to U less than or equal to 1.1146

That is the density function because that is what we are asked to find.1155

Let me go back over that, make sure that everybody is up to speed on this.1163

We start out by finding the range on U.1167

If Y varies from -1 to 1 and I’m keeping track of Y², Y² will always be positive.1171

It will go from 0 to 1.1178

The distribution function is the probability that U is less than or equal to some cutoff value of u.1181

I just translated U into Y², I got that from the stem of the problem that U is Y².1189

If Y² is less than or equal to u, that means Y by itself must be between –√U and √U.1196

I set those up as my limits of integration and I integrated my density function for Y.1204

I plug in my limits and it simplified down to U³/2, that is the distribution function F of U.1213

To get the density function f, I took the derivative of the distribution.1221

That is where I got the 3/2 U ^½.1226

Then, I just reminded myself of the range, that came from here U goes from 0 to 1.1233

In example 3, we have a multivariable example.1241

F of Y1 Y2 is defined to be a joint density function E ⁻Y2.1245

Our region there is Y1 and Y2 both go from 0 to infinity but Y2 is always bigger than Y1.1251

Let me start out by graphing that because that is not the most obvious region.1259

There is Y2, I always seem to switch my variables Y1 and Y2.1264

Here is Y1 and there is Y2 on the vertical axis.1270

Since, we are going interested in when Y2 is bigger than Y1 that is like saying Y is bigger than X.1274

Let me graph line 1 = X.1282

There it is, Y = X, or in terms of our variables, that is saying Y1 = Y2.1284

I’m interested in the region where Y2 is bigger than Y1, that is the region above that line.1293

That is my region right there.1301

I want to find the cumulative distribution and density functions for U = Y1 + Y2.1303

I’m going to find distribution function first.1312

F is the distribution function, and by definition that is always the probability that U is less than some cutoff value of u.1315

I want to convert that into on our range for Y1 and Y2.1330

That is the probability that U is the same as Y1 + Y2 is less than or equal to u.1336

Somehow, I want to describe that range on my graph.1346

I'm going to try to draw that in red here, Y1 + Y2, I’m going to graph the line Y1 + Y2 is equal to U.1350

That will intersect both axis at U.1362

There is U on both axis and let me draw that line.1368

There is the line at U, that is the line Y1 + Y2 is equal to U.1378

The range that is less than that and is still in my region is this red region, right here.1385

That is the range that I want to describe.1393

need to set up a double integral on that range.1397

I need to describe that, let me see if I can describe that range.1401

I’m running out of colors here.1406

I think I will describe it in that direction.1409

First, I will describe what Y1 is, it looks like that will be U/2.1413

My Y1 will go from 0 to U/2 and my Y2 will go from, that first diagonal line is just Y2 = Y1.1419

That second diagonal line is the line, that red line Y1 + Y2 = U.1437

In terms of Y2, that is Y2 is equal to U - Y1.1443

I have got a description of that region that I'm interested in.1448

I can use that to set up a double integral.1452

Let me set up my double integral here.1456

First variable is Y1 goes from 0 to U/2.1459

And then, my second variable is Y2 = Y1 to Y2 is equal to U - Y1.1467

I have to put in my joint density function E ⁻Y2.1478

It looks like my inner variable is Y2, my outer variable is Y1.1485

Now, it is just a calculus 3 problem.1491

I just got to do this double integral and just grind through it.1494

If you are one of the fortunate people who is allowed to do double integrals on computers, 1498

if your probability teacher allows that, go ahead and throw this in your computer.1504

I do not mind, see if your answer checks with mine.1508

I will show you that it is possible to do it by hand.1511

I'm going to do this integration by hand.1514

First variable is Y2, integral of E ⁻Y2 is –E ⁻Y2.1516

Just do the inside integral first, E ⁻Y2 goes from Y2 = Y1 to Y2 = U - Y1.1525

The top part will give me –E, if Y2 is U - Y1, then -Y2 is Y1 – U.1536

A – and - is +E ^- Y1, and we want to integrate that with respect to Y1 from 0 to U2.1548

Let me go ahead and integrate that.1563

The first term E ⁻Y 1 will be just, let me write the second term it –E ⁻Y1.1565

That first term, if you think about it, this is like saying E ⁺Y1 × E ⁻U.1573

The E ⁻U is a constant, this is –E ⁻U × E ⁺Y1.1583

That is not too bad there, I’m a little worried here.1596

I think I’m good, I think I have done everything right.1603

But, I need to a plug in my limits which is Y1 = 0 to Y1 = U/2.1607

Let me plug those in, this is -E ⁻U × E ⁻Y1 which is E ⁺U/2 – E ^- Y1 E ⁻U/2.1617

If I plug in Y1 = 0, then for my first term I get - and - which is +E ⁻U, a - and - which is +E⁰ which is just 1.1634

Let me see if I can simplify this a bit because it looks a bit messy.1654

This right here is –E ⁻U and E ⁺U/2.1659

Let me add the exponents.1665

That is U/2 – U, that is E ⁻U/2.1667

I can combine that with the other one, I will get 1 + E ⁻U.1675

I got two copies of E ⁻U/2.1684

I forgot to include my 2 there, -2 copies of E ⁻U/2.1692

My range there is, here is Y1 + Y2 that could go between 0 and infinity.1701

U goes from 0 to infinity there, that was my distribution function, my F of U.1712

That is certainly an accomplishment to have found that.1722

But, I also need to find the density function, that was the distribution function that I just found.1730

The density function, remember you take the derivative of the distribution function.1738

I do not think the density has an i there.1746

The derivative of the distribution function f of U is just F prime of U.1749

I need to take that derivative, the derivative of 1 is 0, the derivative of E ⁻U is –E ⁻U.1761

The derivative of -2 E ⁻U/2 -2 × E ⁻U/2 ×, by the chain rule, another - ½ there.1772

Maybe, I can write that a little more nicely.1786

The 2 and ½ cancel, I get E ⁻U/2 – E ⁻U.1789

That is my density function, same range for U, U goes from 0 to infinity there.1799

Let me box that up and then let me go back over the steps,1812

just in case anybody is still wondering where some of those points came from.1815

This is our first multivariable example.1822

The first thing I did was to graph the range here.1824

Y1 and Y2 both go from 0 to infinity.1829

We also know that Y2 is bigger than Y1.1834

I graphed the Y1 = Y2, and the general range here is all of these blue shaded region.1838

Do not think about the red yet because that comes in later.1845

It is all of these blue shaded triangular regions.1848

I’m going to find the distribution function first, the U.1852

That is what I’m finding right now.1858

It is the probability that F is the probability that U is less than u.1860

I plugged in my definition of U, Y1 + Y2.1866

I got to find the probability that that is less than u.1871

But, I want to describe that in terms of a region.1875

I graphed Y1 + Y2 is equal to U, that is this red line here.1880

I took all the regions less than that line, that is why I got this sort of a doubly shaded red and blue area.1885

I want to describe that area, it is useful to describe it with Y1 first.1891

Otherwise, you have to break it up into two pieces and that would make it even worse.1897

I graphed Y1 goes from 0 to U2, that is the biggest possible value of Y1.1902

Y2 goes from this lower diagonal line up to this upper diagonal line.1908

I found the equations for each one of those, Y2 = Y1 and Y2 = U - Y1.1916

That gave me the limits of this double integral, that is where I got those from, that came from over there.1923

And then, I integrated the density function that I was given, that came from right here.1930

It is just solving a double integral and this was a little bit messy to solve.1935

But, I do not think I even had to do integration by parts or anything like that.1939

It is just a little messy, I solved the first one with respect to Y2.1944

I plug in my bounds, got some things in terms of Y1, integrated that.1948

What I finally found here was the distribution function, the F that we are looking for.1953

To get the density function, I took the derivative of what I had just found.1960

I just did D by DU to get to the density function and that simplify down to this the density function.1965

That gives me both of the things that I was looking for, and that wraps up that example.1977

In example 4, we are given that Y1 and Y2 had joint density function.1983

Let me remind you what this notation means.1989

That colon means it is defined to be.1991

The joint density function is defined to be, that = means always =.1996

That just means we have a constant density function, it is always equal to 1.2004

We have got a region here, the triangle bounded by 0,0.2011

I need to graph this, 0,0 and 2,0 and 2,1.2017

There is Y2 on the horizontal axis, there is 2,0, 2,1 I will put up here, and 0,0.2023

There is the Y2 axis, I keep writing Y2 on the horizontal axis.2041

I do not know why I always do that.2047

I have done that about 4 × in a row now.2050

But, there is the point 0,0.2053

The triangle we are interested in is that triangle right there.2056

We are told that we have a uniform density, constant density 1 on the triangle.2061

I’m not going to color in that triangle because I have to make a couple other pictures on here.2068

That will get a little crowded but the region here is really this region inside that triangle.2072

And then, we are given a function U is Y1 - Y2.2078

We want to find the density function for that.2083

But of course, we are going to find the distribution function first, the F.2087

Then, we will take the derivative to find the density function, the f.2091

Let me go ahead and try to figure out what the F should be.2095

F of u, by definition it is the probability that U is less than u.2098

In turn, I’m going to translate that into Y, the probability that Y1 - Y2 is less than or equal to this cutoff value of u.2117

I want to describe that geometrically and see if I can describe the region.2129

Let me think about the line Y1 - Y2 is equal to U.2137

If I solve that then, let us see, that is saying, let me say less than or equal to U.2142

If I solve that for Y2, I get Y2 on the other side.2150

Y2 is greater than or equal to Y1 –U.2154

It looks like I got line with slope 1, if I just graph this as a line, Y2 would be equal to Y1 – U.2160

I got a line with a slope 1 and Y intercept –U.2169

Let me try and draw that on my axis here.2173

Let us say that is U, my line has slope 1 here.2179

I’m going to draw a line with slope 1, and there it is, right there.2186

There is my line with slope 1, and I want all the region above that line because I want Y2 to be bigger than Y1 –U.2195

I’m looking at that region right here.2204

What you notice is that we sort of have two cases here, because it really depends on where you are drawing that line.2207

If you draw that line to the left of this right hand corner, then you are going to get this little triangle.2215

Let me draw another version of this.2223

If you draw that line, it goes to the right of that top right corner, you are going to get a slightly different shape.2229

That is going to make this thing kind of awkward to deal with.2238

If we draw the line to the right, it goes to the right then we get a different and more awkward shape there.2245

If U is bigger than 1 then we will get this awkward shape that it will be a little hard to evaluate.2252

We are going to have to do two different cases here.2267

I noticed that, the place where that blue line intersects the top horizontal axis is also at U.2270

Let me make the blue line a little more prominent.2281

The place where it intersects the axis is at U.2285

It looks like we are going to have two different cases, depending on whether U is less than or bigger than 1.2288

I'm going to have to find those areas.2296

Technically, I should integrate over those areas but the one saving grace of this problem is that, we have a uniform density of 1.2299

Instead of integrating, I can just calculate those areas.2309

Let me make some cases now.2314

If U is less than or equal to 1, then we are looking at this little triangle sort of above the line.2319

Let me just notice somewhere up here.2332

The double integral of 1 is just equal to the area.2334

Instead of doing a double integral, I just have to find the area of these regions here.2339

If U is less than or equal to 1, I need to find the area of the triangle which of course is ½ × base × height.2344

That is the area of the triangle, base × height.2357

The base of my triangle is exactly U, that is the easy part.2362

The height is a little harder to find.2373

How high is that triangle?2375

I need to find out how high that triangle is.2380

The way I'm going to figure that out is by finding the intersection point of those two lines.2383

That blue line right there is, let us see, that was the line Y2 = Y1 – U.2389

The black line is the line Y2 is equal to ½ Y1.2398

I can solve those together, the height will be the vertical coordinate. 2407

The height will be the Y2, let me try to solve for Y2.2412

Y2 is equal to Y1, let us see, ½ Y1.2416

If I write that as 2Y2 is equal to Y1, and if I plug that into the first line, I get Y2 is equal to 2Y2 – U.2422

I’m solving this line and this line together, right now.2436

If I switch things around, I will get U = Y2.2441

That is actually quite nice, that means the height is also U.2447

This is ½ U × U, I get ½ U² as the area of the triangle, that is really quite nice.2454

Now, I have to find this other awkward area.2463

If U is bigger than 1, that was if U is less than 1.2465

If U is bigger than 1, I think the easiest way to find the shaded area there 2470

is to subtract off the area of this little triangle right here.2483

The area is equal to, the total area - the area of the small triangle.2488

Of course, I mean the area of the small triangle.2502

Let us figure out what the total area is.2505

The total area is ½ × the base of the big triangle × the height of the big triangle which is 1.2506

The small triangle, let us think about that.2516

The base of the small triangle, it is not U.2519

Let me move that U over a little bit to make it a little more obvious.2523

That base is not U, but that base distance is 2 – U.2529

Since, we are looking at a 45° line, that height is also 2 – U.2534

It is ½ × 2 - U², but that is a little bit more messy.2540

½ × 2 × 1 is just 1 - (2 - U)²/2.2548

What I just figured out here was my distribution function.2556

My F of U is equal to ½ of U², if U is less than 1 and 1 - (2 - U)²/2, if U is bigger than or equal to 1.2564

That was my distribution function, that is what I figure out, right there is the distribution function.2587

That is not what the problem was asking for, I have to take it one more step and find the density function.2591

Remember, you find the density function just by taking the derivative of the distribution function.2598

That would not be much more work.2603

The density is f of U, but you get that by just taking the derivative of F, the distribution function, take the derivative of that.2606

If I take the derivative of each of those parts, the derivative of ½ U² is just U.2621

The derivative of 1 -2 - U²/2, the derivative of 1 is 0.2629

The derivative of 2 - U²/2 is - ½ × 2 × 2 – U.2636

By the chain rule, we have to multiply a -1, the inside.2646

The derivative of 2 – U is -1, that cancels nicely.2650

What we get is, we get U for, when 0 is less than or equal to U less than or equal to 1.2656

I see that my value of U here could go anywhere from 0 to 2.2668

That is a possible range for U.2675

And then, this part simplifies down to, the ½ canceled, the negatives cancel, we get 2 – U for U being between 1 and 2.2677

That is my density function and that is what I needed to find.2690

A lot of geometry in there, we did do some calculus.2695

I was going to say that we have not done any integration, and that is true.2699

But, we did take some derivatives in there.2701

Let me recap that problem, kind of some tricky geometry here.2704

The first thing I did was graph the triangle, graph these three points 0,0 and 2,0 and 2,1.2707

I know I'm looking at this lopsided triangle here.2717

I’m trying to find the density function for Y1 - Y2.2722

I’m going to start out by finding the distribution function F, which is the probability that U is less than u.2726

I plugged in my definition, Y1 - Y2 into here.2734

That turns into a geometric constraint.2740

If I graph Y1 - Y2 less than U, that graphed out into Y2 bigger than Y1 – U.2743

We get all the regions above this line, this is the line Y2 = Y1 – U.2754

It is all the regions above that.2761

But I see that depending on where I draw that line, I can be looking at two different possible shapes.2763

If U is over here less than 1, I got this little triangle.2768

If U is bigger than 1, then I got this sort of strange quadrilateral.2772

The cases here, this is the case U bigger than or equal 1.2781

This is the case that U less than or equal to 1.2784

I really have to evaluate two different areas here.2787

The good news is that, my density function is consistently 1 which means 2792

that instead of doing a double integral, I can just find the areas.2797

That is only because the density is equal to 1.2801

For the first case, when U is less than or equal to 1, I had to find the area of this triangle right here.2805

It is ½ base × height, and I figure out that both my base and my height were U.2812

The tricky part is finding the height there.2818

The way I found that height of U is by solving this line and this line together.2820

That was the algebra that I was doing right here, was by solving those two lines together,2830

in order to find that intersection point.2835

I found that intersection point was Y2 = U.2837

My height was U, that is where that U came from.2841

That simplified nicely into ½ U².2844

If U is bigger than 1, I’m looking at this sort of lopsided region.2848

I really did not want to calculate that area.2853

The easy way to calculate that area is to take the total area and then subtract off this small triangle, right here.2855

I figure out that, that small triangle has base and height 2 - U for both them.2862

That is really because this is a 45° line.2869

The area of the small triangle is 2 - U²/2.2874

The area of the big triangle is just 1.2878

That was my distribution function, if U is bigger than 1.2882

Both of my distribution and my density functions, we have common cases here,2886

depending on whether U is less than 1 or bigger than 1.2890

To get to the density function, what I did was I took the derivative of those two pieces.2893

The derivative simplified it quite a bit, I just got this nice U and 2 - U for my two cases of my density function.2900

In the last example here, we have Y1 Y2 are independent exponential variables with mean 1 and U is their average.2914

I want to find the density function for U.2921

There are a lot of words here, not a lot of equations.2925

The first thing is try and figure out how we translate these words into equations.2927

The first thing is to notice that we have exponential variables.2933

Let me remind you of the generic of formula for an exponential random variable.2937

An exponential random variable has the following density function, 2943

F of Y is equal to 1/β E - Y/β and that is as Y ranges between 0 and infinity.2951

If that sounds like totally gobbledygook to you, if you have never heard that before in your life, 2963

that means you did not watch my earlier lecture on the different distributions.2968

Just go back, it is the same series of probability videos here on www.educator.com.2973

Just go back and scroll back, and you will see a lecture on the exponential distribution, that is what I'm quoting right now.2977

That was the density function that we learned back in the earlier lecture.2982

The other thing we learned is that the mean there or the expected value was just β.2988

That is expected value of the exponential distribution.2996

In this case, we are given that we have exponential variables with mean 1.3000

That means my density function for Y1 is just, it is the function I just listed.3006

Except, I plug in β = 1, I get something fairly nice, E ⁻Y1 where Y1 goes from 0 to infinity.3013

The same thing for Y2, F2 of Y2 is E ⁻Y2 where Y2 goes from 0 to infinity.3026

What we are told here is that these variables are independent.3037

What that means is that the joint density function, you get just by multiplying the two individual density functions.3042

Using independence, F of Y1 Y2, the joint density function, you get by multiplying the two marginal density functions.3049

F of Y1 of × F2 of Y2, that is E ⁻Y1 × E ⁻Y2.3065

Let me draw the range that we are looking at there.3080

Because, that will be something we will need to look at.3084

Look, I got the variable right this time.3091

Y1 and Y2 both go from 0 to infinity, the range that we are generically looking at is, 3093

all of these upper quarter plane here, that is my range.3100

I think I'm about ready to start looking at what U is.3110

U, in this case, by definition, we are given that U is their average.3115

What is the average, that means Y1 + Y2/2.3123

I want to find the density function for U, but the whole point of this lecture is to use the method of distribution functions.3129

The method of distribution functions says, you first find the distribution function F of U.3137

That is by definition, the probability that U, I’m trying to avoid writing tails of my U so that will look less like u.3145

Because, we have u coming right up here.3156

U is less than or equal to some cutoff value of u.3161

Now, I'm going to plug in what U is, that is the probability of Y1 + Y2 being less than or equal to u.3165

I said, the way you deal with this is, you try to convert that into a region.3176

And then, you try to find the probability of being in that region.3183

I want to look at the region where Y1 + Y2 is less than or equal to u,3188

which means I want to graph the line where Y1 + Y2 is equal to U.3196

If a graph that out, let me graph that in red. 3201

There is the line in red, where Y1 + Y2, let me not write it right there.3208

There is the line where Y1 + Y2 is equal to U.3216

The intercepts on both axis there are U.3226

There is U and there is U, let me write Y2 a little further down here.3231

And then, I can write U right by the intercept there.3238

We are looking for the region where Y1 + Y2 is less than or equal to U.3242

That is all the regions south of this line, that is all that region colored in red there.3248

I hope you are not colorblind because it helps to see the color in understanding these examples.3258

I would like to describe that region, in terms of values of Y1 and Y2.3267

I think I will describe it like this.3272

First, I will describe Y1 is going from 0, the biggest value I see there is U.3274

Y2, I see that I have a small mistake here, my U is not Y1 + Y2, it is Y1 + Y2/2.3282

I need to fix that mistake.3297

If I solve this, this is the probability of Y1 + Y2 is less than or equal to 2U, not U.3301

Let me see if I can change that on my picture.3308

The general shape will be the same but let me just change all my intercepts there.3313

This is Y1 + Y2 is 2U, and each of those intercepts will be 2U.3322

My line right here is the line Y1 + Y2 less than or equal to 2U.3329

If you are following along, you will get a little confuse by that, you are very right to be confused.3334

I think it is right, Y2 is going from 0 to 2U.3341

That should have been a 2U up above.3350

This should be 2U - Y1, that describes my region now.3353

I think I got that describe right and I'm ready to set up a double integral over that region.3358

This is a double integral of Y1 goes from 0 to 2U and Y2 goes from 0 to 2U - Y1.3366

And then, the function we are integrating, that joint density function, what we figure out is E ⁻Y1 × E ⁻Y2.3385

It looks like I got a Y2 on the inside, DY2 DY1.3394

I got to solve that integral, it is going to be a little bit messy.3399

But, I notice that my first integral is with respect to Y2.3404

The inside integral is with respect to Y2, that means E ⁻Y1 is a constant, that is one mercy there.3408

Let me pull that one out of the first integral of the inside integral E ⁻Y1, because it is just a big constant.3418

The interval of E ⁻Y2 is - E ⁻Y2, those are being multiplied.3426

I have to evaluate that fromy2 = 0 to Y2 = 2U - Y1.3433

I get E ^- Y1, if I plugged in –E ⁻Y2, Y2 is 2U - Y1, -Y2 will be Y1 -2U.3442

E ⁺Y1 -2U -, if I plug in Y2 = 0, I will get 1.3459

But it is – a negative so it is +1 there.3468

I think I’m going to distribute my E ⁻Y1, let me keep going over here.3473

If I distribute E ^- Y1, E ⁻Y1 × 1 is E ⁻Y1.3480

E ⁻Y1 × E ⁺Y1 -2U, remember you add the exponents.3488

Those Y1's will cancel, that is pretty nice.3493

I will get E ⁻Y1 –E ⁻2U.3497

Of course ,that was all the inside integral and I still need to integrate that with respect to Y1 DY1.3503

The integral of E -Y1 is –E ⁻Y1.3514

The integral of E ⁻U, U is a constant because our variables right now are Y1 and Y2.3523

E ⁻2U, the integral of that is just –E ⁻2U × Y1, that is because E ⁻2U is a constant.3530

If I integrate that or evaluate that from Y1 = 0 to Y1 = 2U.3547

Let me plug those limits in.3561

The first one, I get –E ⁻2U -, Y1 = 2U so -2U E ⁻2U.3566

Y1 = 0, I have - or +, Y1 = 0 gives me just E⁰ which is 1.3579

My last term is just 0 because of the Y1 term.3588

I think I have finally figured out my distribution function, my F of U, let me rearrange the terms a little bit, 3595

get the positive one at the front.3604

1 – E ⁻2U – 2U, that is my distribution function.3606

The goal here was to find the density function, but fortunately, I have done the hard work.3617

The density function, you just get by finding the derivative of the distribution function, F prime of U.3621

The derivative of 1 is 0, the derivative of E ⁻2U is -2E ⁻2U but I got one -, it is +2 E ⁺2U.3630

I got to use the product rule.3646

-2 × U × the derivative of E ⁻2U, -2U E ⁻2U + E ⁻2U × the derivative of U + E ⁻2U.3648

Surely, some of this simplifies.3668

I see that I have got 4U E ⁻2U and I got 2E ⁺2U – 2E ⁻2U.3671

Those cancel and that is my whole density function is just for 4U E ⁻2U.3686

Let me of put a range on U.3694

Y1 and Y2 are both going from 0 to infinity, which means their average will also go from 0 to infinity.3697

I have found a density function, just 4U E ⁻2U.3705

Let me box that up and present that as my solution.3712

Let me go back and go over the steps there, lots and lots of steps of this problem.3720

We start out with independent exponential variables.3726

What is an exponential variable?3729

Fortunately, there is a whole lecture on exponential variables earlier on in this series here,3730

on the probability lectures on www.educator.com.3737

Go back and check it out, the earlier lecture on exponential variables.3739

You will see the definition of the density function was F of Y is 1/β E ⁻Y/β.3745

The mean there is β.3756

In this case, we have mean 1,that is why I plug in β = 1.3758

I just got for my density function for Y1 is E ⁻Y1.3762

Similarly, my density function for Y2 is E ⁻Y2.3768

Now, independence means that the joint density function, you get just by multiplying the marginal density functions.3772

I just multiply those together.3780

And now, I want to find the density function for this U.3784

It says that U is their average, the average of Y1 and Y2 is Y1 + Y2/2.3791

To find the density function, I'm going to first find the distribution function.3799

That means, the F is a probability that U is less than some cutoff value.3805

And then, I plugged in the definition of U is Y1 + Y2/2.3811

Then, I simplified that into, I just move the 2 over, Y1 + Y2 was less than 2U.3818

I want to figure out, what region that is describing.3825

I graphed the line Y1 + Y2 = 2U.3828

There it is that red right there, that red line.3832

I found the region less than that and I tried to describe it in terms of Y1 and Y2.3836

First, Y1 goes from 0 to 2U and then, Y2 goes from 0 up to that diagonal line.3842

If you write that in terms of Y2, you solve that and you get Y2 is equal to 2U - Y1.3850

That is where those limits came from.3858

And then, those limits turned into the limits that I use for the double integral.3861

Those got used right there, and then I used my joint density function, that is what I'm integrating.3867

Now, it turns into a somewhat messy calculus 3 problem.3876

You do not like doing all the calculus 3 and you want to throw this into an integration program, 3880

that is totally fine with me.3887

If you want to throw it into an integration program, you should end up getting the same thing I do for the distribution function.3888

I integrated with respect to Y2 first, E ⁻Y1 is a constant at this point.3895

Plug in my values for Y2, that is a little hard to read, let me see if I can write that a little more clearly because that was U.3900

There is a U, a lot much better.3910

Plug in the values for Y2 and simplified it down a bit.3913

I distributed this E ⁻Y1 which is how I got this function right here.3919

I still need to do the integral with respect to Y1.3924

I did that integral, remember that U is a constant.3927

The integral of E ⁻2U is just E ⁻2U × Y1.3931

And then, I plugged in my values for Y1 and simplified it down and I got my distribution function.3935

That is still not the density function, you get the density function by taking the derivative of the distribution function.3943

I took that derivative and used a little product rule here, which made it get a little messy.3951

But then, it turned out that some terms canceled and I got a fairly simple density function for U × E ⁻2U.3956

Of course, my range on U, it is all the possible values of U.3964

Since, U is Y1 + Y2, Y1 + Y2/2 that can be as small as 0 and unboundedly large.3969

We have to say the range of U is from 0 to infinity.3981

That wraps up this lecture on distribution functions.3986

Remember that, distribution function is the first of three methods 3989

that we are going to use to find the distribution and densities of functions of random variables.3994

Distribution functions is the first one, that is what we have just been talking about.4001

The next one is the method of transformations.4005

I hope you will stick around for the next video that will cover the method of transformations.4008

The third method is moment generating functions.4014

I got another video coming up after that, about moment generating functions.4020

There are all sort of different techniques to solve the same problem 4024

but some of them work better in different circumstances, which is why we learn all three.4027

This is part of the larger chapter on finding distributions of functions of random variables.4035

That, in turn, is part of the whole probability lecture series here on www.educator.com.4043

I have been working with you today, my name is Will Murray.4049

I hope you will stick around, thank you very much, bye.4053