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Expected Value of a Function of Random Variables

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Expected Value of a Function of Random Variables

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  • 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

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 because0031

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 then0104

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 Y21097

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 and1211

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 looking1436

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 and1807

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