For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### 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
- Premise
- Goal
- Goal Number 1: Find the Full Distribution Function
- Goal Number 2: Find the Density Function
- Goal Number 3: Calculate Probabilities
- Three Methods
- Distribution Functions
- Example I: Find the Density Function
- Example II: Find the Density Function
- Example III: Find the Cumulative Distribution & Density Functions
- Example IV: Find the Density Function
- 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

- 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

### Introduction to Probability Online Course

### Transcription: Distribution Functions

*Hi and welcome back to the probability lectures here on www.educator.com. My name is Will Murray*0000

*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 geometry*0337

*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

1 answer

Last reply by: Dr. William Murray

Wed Feb 1, 2017 5:29 PM

Post by renia sari on February 1 at 06:39:15 AM

Hi Professor Murray,

In example III, why you set the upper limit of Y1 to be u/2, not Y2?

Thanks.

1 answer

Last reply by: Dr. William Murray

Mon Jan 16, 2017 9:47 AM

Post by Bharathi Viswanathan on January 14 at 03:07:24 PM

Unit VI

Problem 4: U>1 scenario; Small triangle

Should that be 1/2 * 2-u * 1-u (since the ht of big triangle is 1)?

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)