For more information, please see full course syllabus of Probability

For more information, please see full course syllabus of Probability

### Expected Value of a Function of Random Variables

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Review of Single Variable Case 0:29
- Expected Value of a Single Variable
- Expected Value of a Function g(Y)
- Bivariate Case 2:11
- Expected Value of a Function g(Y₁, Y₂)
- Linearity of Expectation 3:24
- Linearity of Expectation 1
- Linearity of Expectation 2
- Linearity of Expectation 3: Additivity
- Example I: Calculate E (Y₁ + Y₂) 4:39
- Example II: Calculate E (Y₁Y₂) 14:47
- Example III: Calculate E (U₁) and E(U₂) 19:33
- Example IV: Calculate E (Y₁) and E(Y₂) 22:50
- Example V: Calculate E (2Y₁ + 3Y₂) 33:05

### Introduction to Probability Online Course

### Transcription: Expected Value of a Function of Random Variables

*Hello, welcome again to the probability lectures here on www.educator.com, my name is Will Murray.*0000

*We are working through a chapter on Bivariate density and distribution functions.*0006

*Everything in this chapter will have two variables, Y1 and Y2.*0011

*Today, we are going to talk about the expected value of the function of random variables.*0015

*That is something that we have seen in the single variable case.*0021

*We will see how the same ideas extend to the multivariable case.*0025

*I do want to start by reviewing the single variable case because*0031

*I think it will make the definitions make more sense for the multivariable case.*0034

*This kind of a review of stuff you seen before.*0038

*The expected value of a variable, if it is a discreet variable, what you do is you a sum up over all the possible values of the variable,*0041

*and then you just have Y by itself × the probability of that particular value of Y.*0050

*The analogous definition for the continuous case is that, instead of summing, you take an integral.*0057

*Instead of P of Y, you have F of Y which is the density function.*0064

*You still have this Y multiplied on it.*0070

*And then, we can also talk about the expected value.*0073

*Instead of Y itself, the expected value of G of Y which is some function.*0076

*For example, you might have G of Y to be Y².*0081

*The way you define that, if it is discreet is you sum up over all the values of Y,*0086

*you still have the probability of each value of Y there.*0090

*The difference is that, instead of the Y that we had before, we replace that with G of Y.*0094

*Then, the same kind of thing in a continuous case.*0101

*We take an integral, we still have the density function F of Y, and then*0104

*we replace the Y that we had before with the G of Y, and we have to solve that integral to find the expected value.*0108

*The multi variable cases, the new stuff that we are going to learn in this lecture.*0116

*The definitions are very similar, except that instead of single sums, we will have double sums.*0120

*Instead of single integrals, we will have double integrals.*0126

*Let us take a look at that.*0131

*In the Bivariate case, we will have a function of two variables.*0133

*G is now a function of Y1 and Y2.*0136

*The single sum, we have a double sum because there are two variables here.*0140

*There is a Y1 and there is a Y2.*0144

*We saw the probability function except it has two variables now.*0147

*P of Y1 Y2, and then the expected value of G of Y1 Y2, you will just multiply it on.*0151

*Instead of G of Y, we have G of Y1 Y2.*0157

*It is very similar to the single variable case, except that we have a Y1 and Y2.*0161

*In the Bivariate case, we still have that join density function F of Y1 Y2.*0167

*Again, we are integrating over both Y2 and Y1.*0173

*Instead of G of Y, we have G of Y1 Y2.*0179

*We will be doing a lot of double integrals in this lecture, I hope you are really fresh on your calculus 3,*0185

*your multivariable calculus because you will need to be able to do a lot of double integrals to solve the examples in this lecture.*0191

*Before we jump into the examples, there is a few more facts that I want to introduce you to.*0200

*In particular, linearity of expectation is a very useful principle to invoke, when you are calculating expected values.*0204

*The expected value of a constant is just a constant by itself.*0216

*Expectation is linear, in the sense that, if you have a constant multiplied by a function,*0221

*what you can do is just factor that constant on all of the expected value there.*0228

*Just factor out to the outside there.*0233

*That sometimes makes things a lot easier just because you can pull the constant outside,*0236

*not worry about them when you are doing your integral.*0241

*Even more useful is additivity.*0244

*If you have a sum of 2 functions and you want to find your expected value,*0246

*what you can do is calculate their expected value separately.*0252

*And then, just add them together.*0255

*That is extremely useful, if you try to find the expected value of something + something else,*0257

*just find your expected value separately and then, add them together.*0262

*We will see some examples, when we get to the problems of how that can be really useful.*0265

*It can save you doing a lot of very complicated integrals and sometimes reduce them to much simpler integrals.*0270

*Let us go ahead and start in on the examples.*0276

*First example here, we have a joint density function of E ^- Y2.*0280

*Remember that, := just means it is defined to be.*0285

*Our joint density function is defined to be E ⁻Y2.*0289

*Y1 and Y2 are both between 0 and infinity, but Y2 is always bigger than Y1.*0294

*We want to calculate the expected value of Y1 + Y2.*0300

*Right away, what I'm going to do is draw a graph of this region because it is not a perfectly square region,*0304

*it is not totally obvious what the region is.*0311

*I think the easiest thing is to start out with a graph.*0313

*There is Y1 on my horizontal axis, there is Y2 on my vertical axis.*0317

*I'm going to graph the line Y1 = Y2.*0321

*If your X and Y will be the line Y = X.*0326

*That diagonal line right there, that is the line Y1 = Y2.*0329

*What we want to look at, is the region where Y2 is bigger than Y1.*0338

*That is the region above that line, let me go ahead and color that in.*0343

*There is that region right there.*0348

*I want to describe that region because I’m going to be setting up a double integral to calculate this expected value.*0352

*It looks to me like, let me draw this in red.*0360

*It is easiest to describe it listing my Y2 first and then my Y1, in terms of Y2.*0367

*The point of that is that, I will have one more 0 to deal with, when I set up my limits.*0374

*That will make things a little bit easier.*0378

*If I list Y2 first, my Y2 will go from 0 to infinity.*0380

*My Y1 goes from Y1 = 0, that is the vertical axis, up to that diagonal line, that is the line Y1 is equal to Y2.*0388

*That is a good way to describe the region, we will use that to set up a double integral to find this expected value.*0400

*This expected value of Y1 + Y2 is equal to the double integral,*0409

*I will fill in the limits in a second, of Y1 + Y2 × the joint density function, the F of Y1 Y2 which is E ⁻Y2.*0414

*Let me fill in the limits on the integral, Y2 goes from 0 to infinity, Y2 = 0 to Y2, I will take the limit as it goes to infinity.*0427

*Y1 goes from 0 to Y2.*0444

*That means, I’m going to integrate Y1 first because it is on the inside.*0453

*I will put DY1 on the inside and DY2 on the outside.*0457

*Let me go ahead and do that integral.*0463

*Of course, if you are fortunate enough that you can use software for your integrals, at this point,*0465

*you just drop the whole thing right into a computer algebra system,*0472

*something like mathlab or mathematica, or something like that.*0477

*It will just spit out an answer for you.*0481

*If you are fortunate to have access to that, then go ahead and do that.*0483

*I want to go ahead work out the integral, just to show you that it is not that bad.*0487

*It can be done by hand.*0490

*When we do that first integral, notice that the inside integral, the variable is Y1 so that means Y2 is a constant.*0493

*In particular, the E ⁻Y2 was a constant.*0501

*I’m going to separate out that E ⁻Y2.*0504

*Then, the integral of Y1 + Y2 with respect to Y1 is Y1²/2 + Y2 Y1.*0508

*I just did that inside integral with respect to Y1.*0520

*I have to evaluate that from Y1 = 0 to Y1 = Y2.*0523

*I get E ⁻Y2 and if I plug in Y1 = Y2, I will get Y2²/2 + Y2².*0531

*I see I can combine those into 3/2 E ⁻Y2 and Y2².*0545

*3/2 Y2² E - Y2.*0554

*And that was just doing the inside integrals, I still need to evaluate the outside integral.*0559

*Y2 = 0 to Y2 goes to infinity, of this expression here, this is DY2.*0565

*Basically, I’m integrating X² –X, that is a classic case for integration by parts.*0573

*Let me go ahead and set up a little table to do my integration by parts.*0580

*I will pull the 3/2 out because that is really not relevant, it is not going to change anything in the integration by parts.*0588

*E ⁻Y2, this is my little shorthand trick for doing integration by parts.*0598

*If you do not know this trick or if you are rusty on integration by parts, we got some lectures,*0604

*it is in the level 2 college calculus section here on www.educator.com.*0610

*There is a whole lecture on integration by parts, you can just check it out and get all caught up to speed.*0617

*For the time being, I’m just going to use my short hand trick which is to take derivatives on the left,*0622

*that is 2Y2, and then 2, and then 0.*0627

*Integrals on the right, -E ⁻Y2 and then +E ⁻Y2, and then –E ⁻Y2.*0631

*You make these little diagonal hashes with a + - +.*0642

*You multiply down the diagonals and read off the answer.*0647

*This is 3/2 ×, it is negative because there is a negative there.*0651

*- Y2² E ^- Y2 - 2Y2 E ⁻Y2 - 2 E ⁻Y2.*0660

*All of these need to be evaluated from Y2 starting at 0 and then, we will take the limit as it goes to infinity.*0673

*Let us sort that out, we still have this 3/2.*0684

*When we plug in Y2 going to infinity, all of these terms are going to disappear.*0689

*There is as little bit of Patel’s rule in there but I'm not really showing you the details.*0695

*But, basically, the E⁻Y2 term dominates and it takes everything to 0.*0700

*Even though, it is being multiplied by infinity.*0708

*If you want to check the details of that, just go through Patel’s rule and check that out.*0710

*All the terms at infinity are going to 0, so 0 -0 -0.*0716

*I’m going to plug in Y2 = 0 and I will get, - and - so +, but it is 0 anyway, - and – so it is +2 E⁰ which is just 1, +2.*0721

*All the 0 disappear, I got 3/2 × 2.*0740

*My answer, my expected value of Y1 + Y2 is 3, that finishes that problem.*0744

*Let me go back over the steps and make sure that everybody is clear on everything.*0753

*The first thing to do here is to graph the region.*0757

*I looked at those limits there, Y1 and Y2 both go to infinity, but Y1 is always less than Y2.*0762

*That is how I got this region right here, this triangular region that goes on infinitely far.*0770

*And then, I want to describe that region in terms of Y1 and Y2.*0776

*I thought it would be easier to list Y2 first, because then I can get a 0 for the lower bound on Y1.*0781

*That makes my life slightly easier to have that 0 there.*0789

*That is why I picked Y2 to list first.*0792

*You can also done it with Y1 first, but it will made it a little more messy.*0795

*I’m going to use those limits to set up this integral, right here.*0800

*Because it is Y1 + Y2, that is why I multiplied everything by Y1 + Y2.*0805

*And then, I use the density function to get this E ⁻Y2.*0811

*Now, it is just a calculus 3 problem.*0816

*The only thing you have to be careful about is which variable you are integrating with respect to.*0819

*At first, I’m integrating with respect to Y1 which means Y2 is a constant.*0825

*That is why I got these different answers for Y1 and Y2 because the variable of integration was Y1.*0829

*Plugged in my values for Y1 and now it simplifies down to an integral in terms of Y2,*0835

*which is something that I needed parts for.*0845

*That is where I kind of outsourced to this tabular integration technique to do integration by parts.*0847

*That came back and gave me the answer, this long answer.*0857

*But, I plugged in my bounds, 0 and infinity.*0861

*The infinity terms all dropped out, that is really showing some Patel’s rule there.*0864

*I did not show the Patel’s rule but that is kind of what was going on in the background there.*0870

*The infinity terms all dropped out, and most of the 0 term dropout but this 0 term gave me a value of 2.*0874

*When I multiply that by 3/2, that is how I got my expected value of 3.*0881

*In example 2 here, we have got a joint density function of 2 × 1 - Y2 and Y1 and Y2 are both bigger than 0 and less than 1.*0889

*Let me graph that, before we go any farther, that is a simpler region than we had in example 1,*0901

*because that is a square region.*0907

*Here is Y1 and here is Y2, and there is 0 and there is 1, and there is 1.*0910

*We are just looking at a square region here.*0918

*Let me color than in and we will need to integrate that.*0922

*We will need to describe that region.*0926

*It is very easy to describe, as Y1 goes from 0 to 1 and Y2 goes from 0 to 1.*0932

*Let us go ahead and figure out what we are calculating.*0941

*E of Y1 Y2, the expected value of Y1 × Y2.*0944

*Remember, the way you calculate that is you set up a double integral on your region, Y1 = 0 to 1.*0951

*Maybe, I will list the Y2 on the outside, they are both constant so it will work the same way, either way.*0964

*Y2 = 0 to Y2 = 1, Y1 = 0 to Y1 = 1.*0972

*The function that I'm trying to find the expected value of, is Y1 Y2.*0982

*I will multiply Y1 Y2 in there and then I will put in the density function that we are given, which is 2 × 1 - Y2.*0987

*While, I got to setup my first integral as DY1 and my outside integral is DY2.*0996

*I’m going to go ahead and solve that integral.*1004

*The first one on the inside, I’m integrating with respect to Y1.*1007

*I see I have a 2Y1, I’m going to use those together and just get Y1².*1013

*And then, everything else is a constant because I’m integrating with respect to Y1.*1019

*Y2 × 1 - Y2, and I evaluate that from Y1 = 0 to Y1 = 1.*1023

*If I plug in Y1 = 1, I just get Y2 × 1 - Y2 and Y1 = 0 drops out.*1033

*This is what I’m integrating, I’m integrating this DY2.*1041

*That is Y2 - Y2², the integral of Y2 is Y2²/2.*1048

*The integral of Y2² is Y2³/3.*1055

*I need to evaluate that from Y2 = 0 to Y2 = 1.*1062

*If I plug in Y2 = 1, I get ½ -1/3.*1071

*Plug in Y2 = 0, they are both 0.*1075

*½ - 1/3, if you put that over a common denominator is 3/6 -2/6.*1080

*My expected value of Y1 × Y2 is exactly 1/6.*1087

*That finishes that problem, let me recap the steps.*1093

*First thing I did was, I looked at the region there and I drew a graph there Y1 and Y2*1097

*are both between 0 and 1 so I got a square.*1104

*And then, I describe that region in terms of Y1 and Y2, they are both constants because it is a square region.*1107

*And then, I use that to set up a double integral.*1114

*The Y1 Y2 here, that came from the fact that we are trying to find the expected value of Y1 Y2.*1121

*The 2 × 1 - Y2 came from the density function that we are given.*1128

*At that point, it is just a calculus 3 problem, multivariable calculus, solving a double integral.*1134

*The important thing to notice is the first variable of integration, the inside one is DY1.*1140

*That is why all the Y2 were just constants.*1146

*I integrated 2Y1 to get Y1², dropped in my values for Y1 and it simplified down to something, in terms of Y2.*1149

*I integrated that with respect to Y2, dropped in my values for Y2,*1161

*and got a number that represents the expected value of Y1 × Y2.*1167

*In example 3, we have two variables Y1 and Y2, we have not been told the joint density function,*1175

*but what we have been told is that the mean of Y1 is 7 and the mean of Y2 is 5.*1183

*We have a couple of functions defined here, U1 is Y1 + 2Y2 and U2 is Y1 - Y2.*1192

*We have to calculate the expected values of the U.*1201

*The point of this problem is to link linearity of expectation.*1206

*If you do not remember that, go back to the introductory slides for this lecture and*1211

*look for the one called linearity of expectation.*1215

*That is what we are going to use very heavily to solve this problem.*1219

*Linearity of expectation is the key to solving this problem.*1223

*The expected value of U1 is equal to the expected value, just by definition of U1 of Y1 + 2Y2.*1234

*But then, expectation is linear, we can separate this out into the expected value of Y1 + 2 × the expected value of Y2.*1244

*We can just plug in the means that we have been given.*1256

*The mean of Y1 was 7 and the mean of Y2 is 5/7 +, 7 + 2 × 5 is 7 + 10 that is 17.*1260

*I almost got tripped up on my arithmetic, at the end there.*1276

*U2 behaves exactly the same way.*1279

*The expected value of U2 is the expected value of Y1 and - Y2.*1282

*Again, it splits up using linearity there.*1293

*That is the expected value of Y1 - the expected value of Y2 which is 7 -5.*1296

*Our expected value of Y1 - Y2 is 2.*1305

*We have an answer for both of those.*1312

*The key to finding those answers was really the fact about linearity of expectation.*1314

*But, if you have the expected value, you can split it up and take expected value separately.*1320

*You can also pull out constants.*1325

*That is true for expectation, by the way, that is not true for variance.*1328

*You do not want to be monkeying around the linearity of variance because there is no such thing.*1331

*But for expectation, that is true.*1336

*What we did here was, I just plug in what U1 was, Y1 + 2Y2.*1340

*Use linearity to split up into expected values of Y1 and Y2, then I just dropped in those expected values.*1345

*Remember, expected value and mean are the same thing.*1352

*I just dropped in the values of 7 and 5, and I got 17 there.*1355

*And then, the same thing for U2, I plug in Y1 - Y2, split up, and dropped in my expected values of 7 and 5,*1359

*and that is where that 2 came from.*1368

*In example 4, we have got joint density function F of Y1 Y2 defined to be 6 × 1 - Y2,*1372

*where Y1 and Y2 are both between 0 and 1, but Y2 is bigger than Y1.*1383

*Let me graph out this region, before we go any farther with that.*1389

*There is Y2, here is Y1, and we are looking between 0 and 1 on both axis.*1394

*But, we only want to take the region where Y1 is bigger than Y2.*1405

*That is the area above the line Y = X.*1410

*Let me draw that region, color that a bit.*1414

*That blue region is what we are going to be looking at.*1420

*We are going to end up doing a double integral for this one, actually, two double integrals.*1423

*I think it is going to be useful to describe the limits of this region.*1428

*I switched my Y1 and Y2 on the axis, not sure why I seemed to like to do that by looking*1436

*but I know I have done that several times.*1442

*Y1 is always the horizontal axis and Y2 is always the vertical axis.*1444

*This should be if I done it correctly.*1449

*I think the best way to describe this is in terms of Y2 first.*1452

*Y2 has to be constant, goes from 0 to 1.*1456

*And then Y1, Y1 we cannot use constants because we will get the whole square.*1460

*We just want the region from Y1 = 0 up to Y1, that line there is the line Y1 = Y2.*1466

*Y1 stays less than or equal to Y2, that describes that triangular region.*1480

*We use that to set up a double integral or a couple of double integral, since we have two expected values to solve here.*1486

*The expected value of Y1 is the double integral and I will follow those limits there.*1494

*Y2 goes from 0 to 1 and Y1 goes from 0 up to Y2.*1500

*Since, we are finding the expected value of Y1, I'm going to multiply in a Y1 × the density function, 6 × 1 - Y2.*1511

*And then, the inside integral is DY1, the outside integral is DY2.*1522

*Now, it is just a double integral problem, we can solve it using what we learned in calculus 3.*1526

*The first variable, the inside one is DY1 which means that 1 - Y2 is just a big old constant.*1534

*The integral of 6 × Y1 is 3Y1².*1544

*We want to integrate that or evaluate that from Y1 = 0 to Y1 = Y2.*1554

*If I plug in my limits for Y1, 3Y1² will give me 3Y2².*1563

*3Y2² × 1 - Y2, the Y1 = 0 just drops out.*1572

*That one does not play a role.*1578

*That is finishing the first integral, I still need to integrate that with respect to Y2.*1581

*I changed a Y1 to a Y2, that is very important.*1590

*3Y2² × 1 is just 3Y2², that integrates to Y2³ -3Y2³.*1594

*I have to integrate that, that would integrate to 3 × Y2⁴/4.*1607

*¾ Y2⁴, I have to evaluate that from Y2 = 0 to Y2 = 1.*1617

*If I plug in 1 to both of those, I get 1 - ¾.*1628

*If I plug in Y2 = 0 to both of those, they both drop out, -0.*1634

*That simplifies down to 1 - ¾ is ¼, that is my expected value for Y1.*1640

*I still have to calculate the expected value of Y2, because the problem asks for both of those.*1647

*I have already done some of the work, I already described the region.*1652

*It will be the same integral, Y2 = 0 to Y2 = 1.*1655

*Y1 = 0 to Y1 = Y2, the same integral except that, instead of multiplying by Y1,*1662

*I’m going to multiply by Y2, because that is what I'm finding the expected value of.*1671

*Y2 × 6 × 1 - Y2, DY1 and DY2.*1676

*Still doing the inside integral, since I’m integrating with respect to Y1, all the variables here are in terms of Y2.*1686

*I'm integrating just a huge constant.*1697

*It is 6Y2 × 1 - Y2, that is all a constant just × Y1.*1700

*I need to evaluate that from Y1 = 0 to Y1 = Y2.*1708

*Let us see, if I plug in Y1 = Y2, I will get 6Y2² × 1 - Y2.*1716

*There is a Y2 there and that Y1 becomes a Y2, when I evaluate it.*1728

*If I plug in Y1 = 0, it just drops out.*1734

*I have the integral from Y2 = 0 to 1 of 6Y2² DY2.*1738

*What I notice here is that, that is the same as the integral I had above, 3Y2² 1 – Y2 except that,*1748

*I have a 6 instead of 3, that is 2 × the same integral as above.*1757

*Maybe, I can plug that here with a star.*1763

*This is 2 × integral *.*1767

*If I evaluate that integral, what I will just get is 2 × the answer that I got above.*1769

*2 × ¼ and 2 × ¼ is just ½.*1774

*That gives me an answer for the expected value of Y2 of being ½.*1780

*If you do not like the way I did that, by sort of citing the integral above, just go ahead and work out this integral.*1787

*It is just a calculus 1 problem, it should not take you very long.*1792

*You should end up getting ½, it should work out.*1795

*That side gives us both the answers that we are looking for.*1801

*Let me remind you how we set that up.*1804

*The first thing I did, like in all of these problems is, I looked at the region that was given to me and*1807

*I graphed the regions.*1814

*In this case, Y1 and Y2 are both between 0 and 1, but Y2 is bigger than Y1.*1815

*I just looked at the region above that diagonal line, the Y = X line.*1821

*I decided that it was easier to describe that region, if we kind of march horizontally.*1827

*That means describing the Y2 first, in terms of constants.*1833

*And then, describing Y1 going from 0 to Y2.*1838

*If you just said Y1 goes from 0 to 1, all of the sudden you described a square, and that is not the region that we are looking at.*1843

*I set up those limits, by looking at the region.*1851

*And then, I pull those limits over and used them to set up a double integral.*1855

*I took the density function 6 × 1 - Y2, and I multiplied it by the thing we are finding the expected value of,*1862

*which in the first case is Y1, but in the second case is Y2.*1870

*That is because of the different requirements, the first one is Y1 and the second one is Y2.*1874

*In each case, I got a double integral to solve.*1881

*And then, I just did a double integral, integrated DY1 first which is why 6Y1 turn into 3Y1².*1884

*1 - Y 2 was just a big old constant.*1893

*Plugged in my values for Y1, got an integral in terms of Y2, pretty easy integral in the sense of,*1896

*probably you can hand that off to first semester calculus student, and they can solve it for you.*1904

*The answer that they hopefully come up with would be ¼.*1910

*For Y2, same kind of thing happens except that, when you are doing that first integral, there are no Y1 at all,*1914

*which means you are integrating a constant.*1921

*You will just get that constant × Y1, plug in the value of Y1 = Y2 and we get 6Y2² 1 - Y2.*1924

*Because I'm lazy, I noticed that that was the same integral that we had back here,*1933

*except it is multiplied by 2 which means we can just take the old answer multiplied by 2.*1937

*2 × ¼ is ½.*1944

*If you had not noticed that, that integral was the same as the previous one, that is okay.*1946

*Just go ahead and work it out, and do one or more steps of calculus 1, and you will get an answer.*1953

*I want you to really understand these two answers.*1960

*I want you to remember them because the next example, example 5,*1964

*we are going to be using these answers from example 4.*1968

*It is the same setup as in example 4, I’m going to use these answers to take it, separate farther.*1972

*Make sure you understand these, before you go on to example 5.*1980

*In example 5, we have got the same setup that we had from example 4.*1987

*We are going to be using the answers that we derived in example 4, to solve example 5.*1993

*If you have not just watched example 4, go back and watch example 4, or are work it on your own.*1999

*Just make sure your answers agree with mine.*2006

*You want to have those answers fresh in your mind and ready to go for example 5.*2009

*Let me solve example 5, we are given the same setup F of Y1 Y2 is 6 × 1 - Y2 and then, they describe the region for us.*2015

*Let me just emphasize that, that is all the same as example 4.*2026

*Because of that, I’m not even going to try to graph it, like I did with all the other examples.*2034

*If you want to kind of work that out and you want my help, just go back and watch example 4,*2038

*you will see that that carefully worked out, I drew the graph and everything.*2046

*In this case, we do not have to do that again, because we are finding the expected value of 2Y1 + 3Y2.*2050

*The expected value of 2Y1 + 3Y2, the trick here is not to do an integral but to use linearity of expectation.*2060

*That was something that I introduced you to, on the third introductory slide to this lecture, linearity of expectation.*2070

*It is super useful for problems like this.*2079

*I should spell expectation right, since it is such a useful concept.*2084

*Linearity of expectation says, you can distribute and split up this expected value into 2 ×*2088

*the expected value of Y1 + 3 × the expected value of Y2.*2096

*Both of those were things that we calculated in example 4.*2103

*I'm not going to recalculate those now because we did no little job of work in calculating those.*2108

*The expected value for Y1, back when work it out in example 4 was ¼.*2117

*The expected value for Y2 was ½.*2125

*Go back and watch example 4, if you are unsure where those numbers are coming from.*2130

*This is 2 × ¼ is ½ + 3/2, that is just 4/2 or 2.*2134

*We got an answer right away, we did not have to do another double integral.*2143

*The reason we did not have to do another integral is because this is the same setup as example 4.*2149

*We already worked out the basic values in example 4.*2155

*We can just extend that using linearity of expectation.*2159

*If the phrase linearity of expectation does not quite trip off your tongue yet,*2163

*it is worth going back and watching that third slide again.*2168

*You will see how nicely it helps us in this problem because it allows us to take 2Y1 + 3Y2,*2171

*and split up and just find the expected values of Y1 and Y2, and then combine them back together.*2178

*Those expected values we calculated in example 4, to be ¼ and ½, just drop those numbers in and then simplify the fractions.*2184

*We get our expected value of 2 for 2Y1 + 3Y2.*2194

*That wraps up this lecture on expected value of a function of random variables.*2202

*Next up, I got a lecture on Covariance, it is a big topic, there is a lot of stuff in there.*2208

*I hope you will stick around for that.*2212

*In the meantime, this is the chapter on Bivariate density functions and distributions.*2214

*This is all part of the larger lecture series on probability, here on www.educator.com.*2219

*My name is Will Murray, thank you for joining me today, bye.*2225

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