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Lecture Comments (2)

1 answer

Last reply by: Dr. William Murray
Thu Mar 5, 2015 5:43 PM

Post by Shant Suvalian on March 3, 2015

The transformation method doesn't only work for single variable situations. Least according to my teacher it doesn't. For example, the following question:

Let Y1 and Y2 be independent and uniformly distributed over the interval (0; 1) Find
the probability density function of U1 = Y1/Y2 by the following methods
(a) The method of Transformation
(b) The Multivariable Transformations Using the Jacobians.

Food for thought. It would come in handy to have an example like that. Or better yet, a full lecture on multivariable transformations using the Jacobians.

But otherwise? The following 3 series lectures saved me. My textbook isn't exactly the best. To say the least.



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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
  • Premise 0:32
    • Premise
  • Goal 1:37
    • Goal Number 1: Find the Full Distribution Function
    • Goal Number 2: Find the Density Function
    • Goal Number 3: Calculate Probabilities
  • Three Methods 2:34
    • Method 1: Distribution Functions
    • Method 2: Transformations
    • Method 3: Moment-generating Functions
  • Requirements for Transformation Method 3:22
    • The Transformation Method Only Works for Single-variable Situations
    • Must be a Strictly Monotonic Function
    • Example: Strictly Monotonic Function
    • If the Function is Monotonic, Then It is Invertible
  • Formula for Transformations 7:09
    • Formula for Transformations
  • Example I: Determine whether the Function is Monotonic, and if so, Find Its Inverse 8:26
  • Example II: Find the Density Function 12:07
  • Example III: Determine whether the Function is Monotonic, and if so, Find Its Inverse 17:12
  • Example IV: Find the Density Function for the Magnitude of the Next Earthquake 21:30
  • Example V: Find the Expected Magnitude of the Next Earthquake 33:20
  • Example VI: Find the Density Function, Including the Range of Possible Values for u 47:42

Transcription: Transformations

Hi, welcome back to the probability lectures here on, my name is Will Murray.0000

Today, we are going to be talking about the method of transformations.0006

That is one of 3 methods that we are studying to find the distributions of functions of random variables.0009

This is actually the second of the 3 part lecture series.0016

You have already learned the method of distribution functions.0022

Today, we are going to talk about transformations.0026

In the next lecture, we will talk about the method of moment generating functions.0028

The premise is the same as in the previous lecture.0033

If you just watched the previous lecture or maybe, if you just watch the next one for moment generating functions,0037

the first couple slides here are going to be exactly the same.0042

If you are up to speed on them, you can really skip over the first couple of slides here.0045

It is nothing new, I’m just kind of introducing the same premise.0049

The idea is that we have several random variables Y1, Y2, and so on.0052

We want to study functions of them.0058

The function variable that I’m going to use is U of Y1 Y2 up through YN.0061

What we calculated in previous lectures, several lectures ago was methods to find the mean of U, 0068

and also the variance of U.0077

But, I did not find the entire distribution of U.0079

The purpose of this lecture, the one before it and the one after it, 0083

is to find several methods to calculate the formal distribution and density functions of this new variable U.0087

That is the idea, we want to find the full distribution function.0096

I’m calling it F sub U of u which represents the probability that U will be less than any cut off value, which I’m calling u.0101

If we have the distribution function, you can always find the density function as the derivative of the distribution function.0110

The density function is f, f is just the derivative of F.0118

Assuming, we can figure out F and f, we can actually calculate the probability that U will be in any given range, 0126

because the probability that U will be from between A and B, one way to calculate it is to just integrate the density function.0134

But of course, if you already know the distribution function that is just F of B - F of A.0141

We would be able to calculate probabilities, if we know the density and distribution function.0148

As I mentioned in the introductory text here, we have 3 methods to study the distribution of functions of random variables.0155

The first method was called distribution functions, and that is what we studied in the previous lecture.0169

If you are looking for the method distribution functions, just check out the previous lecture.0174

In this lecture, we are going to study what is called the method of transformations.0178

I will show you what that is all about, starting on the next slide.0182

In the next lecture, we will talk about moment generating functions.0185

We got three different methods to calculate the same general goal, 0190

but each method works better on different kinds of problems.0196

That is why we have to learn all three here.0200

The transformation method, that is the one we are studying this method.0204

First of all, it only works for single variable situation.0207

You cannot have a Y1, Y2, or YN.0210

You can only have U as a function of a single variable Y.0214

If you see Y1 and Y2, forget the transformation method, you are going to have to use one of the other two methods.0218

Maybe the method of distribution functions or maybe the method of moment generating functions.0225

The second requirement for the transformation method is that, the function U is a function of Y.0230

We are going to call our function H, that must be a strictly monotonic function.0237

A monotonic function means that it is always increasing or always decreasing.0241

For example, here is a function that is always increasing and here is a function that is always decreasing.0248

Either one of those kinds of functions for H, would be fair game for the transformation method.0258

Here is a function that is not always increasing.0264

That function is decreasing for Y, and then increasing, that is not monotonic.0268

It will not be available for the transformation function method.0274

Similarly, if you have a function like this, where it is increasing for a while and then decreasing, 0280

that is not something that you can use the transformation function method on.0285

An example of the function that is always going to be monotonic is linear functions.0291

U = AY + B, the special case there is that, if a 0 then you get a horizontal line, that does not count.0298

But, the any kind of linear function, except for a horizontal line is monotonic0313

because it will always be increasing like that or it will always be decreasing like that.0320

Any linear function is fair game for the transformation method.0326

Assuming that you do have a monotonic function, here is how you use it.0331

What you do is, you try to solve for Y in terms of U.0336

You have been given U, in terms of Y, U = H of Y.0339

What you want to do is try to find Y, in terms of U.0344

Let me write that down, you are trying to solve for Y in terms of U.0349

You are reverse engineering the Y, in terms of U.0360

You got Y to be a function of U, that function will be H inverse.0363

I want to now clarify here that, this notation is not an exponent.0369

This is not the same, let me write this in red so it will really look clear that, that is not what we are talking about.0373

This is not the same as 1/H of U.0381

That exponent there is little misleading, it does not mean H of U⁻¹.0385

What it really means is the inverse function.0390

That H inverse is an inverse function not 1/H of U.0397

We will get some practice with that, the first couple of examples are just practice finding inverse functions.0404

Make sure that you know the difference between an inverse function and just taking the reciprocal.0410

I still have not shown you how the formula for transformations works.0418

These were all sort of the requirements that you have to check, before you can even use the transformation method.0422

Let me show you how to use the transformation method.0429

Remember that we said, we get H inverse of U by solving, you have U = H of Y.0433

You solve that for Y, in terms of U.0449

You solve that to get Y is equal to some function H inverse of U.0453

Now, you have this H inverse and here is how you use it.0458

You get the density function for U, by taking the density function for Y.0462

Plugging in H inverse of U which is what you got here, when you solve for Y in terms of U.0467

This is really Y, in terms of U.0474

You also have to multiply on this correction term which is the absolute value of the derivative of H inverse of U.0482

That is quite a complicated formula, I think it will make sense after we do some examples.0492

I would say just hang onto the formula for now, let us practice some examples and I think it will start to be more clear.0497

Let us jump into the examples and practice it.0504

The first example, this is actually the same example that we studied in the previous lecture, 0507

using distribution functions.0514

We are going to have a different solution for this one.0516

If you want, you can compare this example to the solution we found using distribution functions.0518

Of course, we are going to get the same answer but we are going to use very different methods to find it.0525

We are given that Y has density function F sub Y is 3/2Y².0530

And then, the range for Y is -1 to 1.0536

We want to determine whether the function U = 3 -2Y is monotonic.0540

If so, we want to find its inverse.0545

Monotonic means increasing or decreasing.0549

If you graph U = 3 -2Y, it is a line and it is got slope -2 and it is got a Y vertical intercept 3.0552

It would look like that, and that is certainly a monotonic function, it is decreasing the entire way.0564

Another way to notice it is that, H of Y, U is H of Y is 3 -2Y is linear.0572

Any linear function is monotonic, as long as it is not a horizontal line.0585

It is linear, hence, monotonic, which means we can find its inverse and0591

we can use the method of transformations on this problem.0595

We are going to actually finish the problem in example 2.0600

But right now, we are just going to find the inverse function of H inverse of U.0603

If U = 3 -2Y, I’m going to use the lowercase letters now.0608

Then, I’m going to solve for Y, I will use the uppercase letters for this.0614

U = 3 -2Y, I’m going to solve for Y.0626

U + 2Y is equal to 3 and so 2Y is equal to 3 – U.0631

Y is equal to 3 – U/2, what that means is that this is H inverse of U.0642

It is the function you would use to get back from U to Y.0652

What we have done here is found the inverse function.0658

We have not found the density and distribution functions for U, we will do that over example 2.0661

I just wanted to make sure that you are comfortable with finding an inverse function, 0668

before we use it for anything more complicated.0672

Let me recap the steps here.0675

We had to determine whether the function U = 3 - 2Y is monotonic.0677

One way to do that is to graph it, and notice that it is decreasing the entire way.0681

That is what I did here, I graphed U = 3 -2Y.0687

Another way to notice that is to just observe that it is a linear function.0692

Linear functions are always monotonic, as long as they are horizontal lines.0697

This is monotonic and then to find its inverse, we take that equation U = 3 - 2Y, and try to solve it for Y in terms of U.0702

That is what I was doing here, just a little algebra to get Y in terms of U.0713

And then, the answer I got 3 –U/2 is H inverse of U.0717

Hang onto this result, we are going to use it again, in example 2.0723

In example 2, this is a follow-up to example 1.0729

It is the exact same setting from example 1.0733

We have Y has density function F sub Y is 3/2Y², Y goes from -1 to 1.0737

We want to find the density function for U = 3 -2Y.0744

We did start out studying this problem in example 1.0748

Let me remind you what we figure out there, in example 1.0751

We found the inverse function, H inverse of U, example 1.0757

We solved that equation for Y, in terms of U.0762

We got Y is equal to H inverse of U which was 3 – U/2.0766

I’m going to try to find the density function for U.0775

I’m going to use my generic formula for the transformation method.0778

Let me remind you what that formula is for the transformation method.0783

Just checking it here and make sure I do not write it down wrong.0790

F sub U is F sub Y of H inverse of U × the absolute value of the derivative DY DU of H inverse of U.0793

That should have been a closing absolute value sign there.0810

In this case, my H inverse of U is what we found out above, that was from example 1, that is 3 –U/2.0813

What I’m going to do is I’m going to plug that in to F sub Y.0830

This is F sub Y, here is 3/2Y².0835

It is 3/2 × 3 –U/2².0840

Now, I need to find the derivative of that.0846

The derivative of that DY DU of H inverse of U is, 3/2 goes nowhere.0848

-U/2 has derivative - ½.0856

The absolute value of that is + ½, that term gives me a +½.0861

I can simplify this down to ¾, if I move that ½ to the front.0867

3 – (U/2)², that is my density function.0873

Let me figure out the range for U, really quick.0880

If Y is equal to -1 then U is 3 -2Y, that is 3 + 2 which would be 5.0882

If Y is equal to 1 and U is 3 - 2Y would be 1.0892

This is my U goes from 1 to 5, that is my density function, the f of U.0899

I have done the problem now.0910

This problem is one we did in the previous lecture, the lecture on distribution functions.0912

We used a very different method to solve it, involved doing an integral.0918

This one requires doing derivatives, the method we used using distribution functions involved taking an integral.0922

You might go back and check the example from the previous lecture, when we used the distribution functions.0931

We do get the same answer at the end of it, but using very different methods.0937

I think this method is a little bit quicker, for this particular problem.0942

Let me remind you how we got this answer.0945

First, I was invoking what I learned from example 1, the one just previous to this one,0949

where we solve for H inverse of U to be 3 – U/2.0954

That is just solving this equation for Y, in terms of U, nothing difficult there.0959

Then, I used my generic formula for the transformation method,0966

F sub U of U is FY of H inverse of U × the derivative of H inverse of U.0969

I took my H inverse of U and I just plug it in into F sub Y.0975

There is my F sub Y right there.0980

I plugged H inverse of U in for Y there, that is why I got 3 – U/2².0983

My derivative, I worked that out over here.0990

The derivative of 3 – U/2 is - ½, then I take the absolute value so I got a +½ here, and then the 3/2, and ½ gave me that ¾.0992

I also want to give a range for U and I got that by plugging in the boundaries of the range for Y,1005

back into that function and then that gave me the boundaries for U, going from 1 to 5.1011

I end up in the same answer that I got in the previous lecture for this problem, using completely different method.1017

It is really worth comparing one method to the other and seeing which one think is more efficient, 1023

and making sure that both methods give you the same answer.1028

In example 3, Y has density function FY is 3/2Y².1035

This is actually the same density function that we had back in examples 1 and 2.1040

Y goes from -1 and 1, we want to determine whether the function U = Y² is monotonic.1046

If so, we want to find its inverse.1052

Remember, the whole point of checking whether a function is monotonic is that 1054

we can only use the transformation method, when this function, in terms of Y is monotonic.1059

Let me go ahead and graph U = Y².1067

Let me set up YU axis, here is Y on the horizontal axis, and here is U on the vertical axis.1071

Y goes from -1 to 1, there is -1 and 1.1080

U = Y², of course that is the familiar parabola.1084

We are looking at just that part of it.1089

But eve though just that part of it, I can see that it is decreasing on the first segment and1092

increasing on the second segment, it is not monotonic.1098

A monotonic function has to be consistently decreasing or consistently increasing.1102

U = Y² is decreasing on the interval from -1 to 0 and increasing on the interval from 0 to 1.1112

It is not monotonic.1138

Because it is not monotonic, we cannot find an inverse function.1148

No inverse function and that also means, we cannot use the method of transformations to solve this problem.1156

We cannot use transformations to solve this problem.1173

That is kind of the end of the road, as far as this lecture is concerned on this problem.1180

Because the method that I'm trying to teach you right now, does not work on this problem.1185

We checked the requirements and it just does not work.1189

We cannot solve this problem using the method of transformations.1193

We did solve this exact problem in the previous lecture, using distribution functions.1197

See previous lecture for distribution function solution.1205

We still can solve this problem, if you are willing to watch the video for the previous lecture.1225

You will see that we solved this problem using distribution functions.1230

But in this one, the method of transformations does not work because we are not working with a monotonic function.1233

Just to remind you how I realize that, I graphed U = Y², this is the graph of U = Y².1240

In particular, I looked to the interval from -1 to 1.1246

I looked at this interval right here which is this part of the curve, 1250

and I noticed that it was decreasing for the first half of the interval, increasing on the second half.1254

Monotonic means it is always increasing or always decreasing.1260

Since, it is sometimes decreasing and sometimes increasing, 1264

It is not monotonic meaning there is no inverse function, meaning we cannot use the method of transformations here.1267

That is sort of the end of the line for method of transformations, if you want solve this problem,1275

go look at the previous lecture, we actually did this exact example in the previous lecture1280

using distribution functions, it did work but you cannot use transformations.1284

In example 4, we have got a word problem here.1292

Major earthquakes in California occur once every 2 decades on average, according to an exponential distribution.1295

The magnitude of an earthquake is 1 + Y² where Y is the time since the last earthquake.1303

We want to find density function for the magnitude of the next earthquake.1308

This is quite complicated.1313

First, we need some equations to get started and I see the words exponential distribution.1315

That is where I'm going to get started here.1321

I’m going to write down my density function for an exponential distribution.1323

The exponential distribution, that is one we studied back in an earlier lecture here in the probability lectures.1328

If you scroll back, I think the lecture was actually on the γ distribution.1338

And then, the exponential distribution came in as a special case of the γ distribution.1342

Scroll back, look for the lecture on the γ distribution, and you will see as part of that, something on the exponential distribution.1348

I will remind you now what the density function for the exponential distribution was.1356

It was F of Y is equal to 1/β E ⁻Y/β.1362

And then, the range on Y was from 0 to infinity.1372

One other thing that was really useful for the exponential distribution was to know the mean.1378

The mean of the exponential distribution is that parameter β.1383

Every exponential distribution has a parameter β, that is a fixed constant.1388

You plug in that β and then you get the density function here, and the mean is exactly that β.1393

In this case, we have an exponential distribution and it says that the mean is every 2 decades.1399

That means that on average, you wait 2 decades to see a major earthquake in California.1408

It is actually have been about 2 decades, since the last major earthquake in California.1413

The Northridge quake in 1994, that is 20 years since then.1418

This is perhaps, we are overdue for another earthquake.1425

What that means is the β is equal to 2, because we are going to measure time in decades here.1429

Our density function for Y, I will write FY of Y.1435

I'm just going to plug in β = 2 to my exponential distribution here.1441

1/2 × E ⁻Y/2, and that range on Y goes from 0 to infinity.1445

We are going to take a function, I will call it U.1457

U is going to be 1 + Y², I will write this as Y² + 1.1460

Notice that, if Y goes from 0 to infinity then U, if Y is 0, Y² + 1 will be 1.1466

As Y goes to infinity, U will also go to infinity.1474

The range for U is from 1 to infinity.1477

What I notice if I graph that, if Y goes from 0 to infinity, here is Y and here is U, Y² + 1 looks like that.1482

It is symmetric, Y² + 1 looks like that, U = Y² + 1.1500

On the range from 0 to infinity, it is increasing the entire way.1505

The fact that it is decreasing on the range below 0, it does not matter because we are only interested in the range above 0.1511

U is increasing on the range where Y goes from 0 to infinity.1518

Increasing means that it is monotonic.1530

Monotonic which means it does have an inverse, we can use the method of transformations here.1534

That means, I can solve for Y in terms of U and get an H inverse, this was H of Y. 1543

Now, I’m going to solve for Y in terms of U.1553

What I get is Y² is equal to U -1.1556

Y is equal to √U -1 and that is my inverse function, H inverse of U.1562

I’m going to use that in my generic formula for the method of transformations.1575

My generic formula for the method of transformations is the density function for U is equal to the density function for Y.1581

But then, you plug in H inverse of U.1592

You plug in that expression that we just figured out.1596

You also have to multiply by the absolute value of the derivative of H inverse of U.1598

Maybe, I keep trying to twist my last absolute value sign, I noticed.1606

It might be helpful to find the derivative separately.1613

The derivative D by DU of H inverse U is the derivative of √U -1.1616

That is like U-1 ^ ½, its derivative is U -1 ^- ½.1626

That is 1/√U -1, ½ of that because of the power rule.1632

This is U -1 ^ ½, the derivative is ½ U -1 ^- ½ × the derivative of U -1, by the chain rule.1640

The derivative of U -1 is just 1, that is all I get there.1651

That is positive, I do not have to worry about my absolute values in the next step.1656

Let me go ahead and say, F sub Y of H inverse of U.1660

F sub Y is given right here, I'm going to go ½ E ^-, in place of Y, I’m going to put H inverse of U which is this √U -1.1665

E ^- √U -1/2, and then I have to multiply on by D by DU of H inverse of U, which I calculated over here.1679

That is 1/2 × √U -1.1693

And I'm really done there except I'm going to combine some terms and simplify a little bit.1698

I'm going to bring all these terms out to the front, 1/½ × ½ is ¼.1704

I will put √U -1 over here and now I’m going to multiply my E term.1710

E ^-√U -1/2, and that is as simple as it is going to get, that is my density function for U.1716

The range of possible values for U, I already figured that out, that goes from 1 to infinity.1728

That is my density function for U.1736

We are going to use this again in the next example, in example 5.1738

I want to make sure that you understand this and I want to make sure that you remember this,1742

because I do not want to figure it out again from scratch in example 5.1746

Let me go over that and make sure that all the steps are clear.1752

First of all, it said that major earthquakes in California occur, according to an exponential distribution.1755

I went back and found my formula for exponential distribution.1762

That was in the earlier lecture on the γ distributions, just go back and watch that,1766

if you have completely forgotten what the exponential distribution is.1770

Watch the γ distribution because the exponential distribution is a special case of it.1773

What I learned in that from that earlier lecture is that the density function1779

for the exponential distribution is 1/β E ⁺Y/β, where β is the mean of the distribution.1784

In this case, the mean is 2 that comes from the fact that earthquakes occur every 2 decades, on average.1793

My β is 2, I plug that in here E ⁻Y/2 and ½.1800

Y runs from 0 to infinity that also comes from the definition of the exponential distribution.1807

And then, I defined the variable U to be the magnitude of the quake which is this Y² + 1.1813

I got the Y² + 1 from the stem of the problem.1820

If Y goes from 0 to infinity then Y² + 1 will go from 1 to infinity.1824

That is just plugging those values in for Y.1830

I wanted to graph that, this is the graph here of Y² + 1.1834

I want to check that it was monotonic.1840

It does not look monotonic because it is decreasing on the negative values and then increasing on the positive values.1843

What is going on there?1850

What is going on is that the range we are interested in, is just the range from 0 to infinity.1851

We are only really looking at this part of the graph.1857

That part of the graph, U is increasing.1864

It is increasing all the way, it is monotonic.1868

We can use the method of transformations.1871

The first part of the method of transformations is to find the inverse function.1875

You take your equation and you try to solve it for Y, in terms of U.1879

That is what I did here, I solved that to Y is equal to U -1 ^½.1884

That was not very clear, the way I wrote that.1889

Let me see if I can write that a little more clearly.1892

That is ½, much better.1895

I got the inverse function U -1 ^½.1898

I also found its derivative because I know that I’m going to need it in the formula coming up.1902

½ U ⁺E -1/2, there is no need to invoke the chain rule there, because the derivative of U -1 is just 1.1906

Then, I use my generic formula for transformations.1917

I got this off one of the early slides for this lecture, one of the intro slides for this lecture.1920

That is just the formula for transformations that we are using through out these exercises.1926

It is the same every time.1930

I plugged in H inverse of U in place of Y, into the density function for Y.1932

That was plugging in √U-1, I plugged that in as H inverse of U.1940

I plug that in for Y, there is also this ½, that is where that ½ came from.1948

And then, this term comes from the derivative here.1958

That is where that term comes from.1962

I just collected terms, ½ and ½ give me ¼.1964

I still have √U -1 in the denominator and we still have that exponential expression.1967

We already figure out the range for U was right here, that is where that came from.1974

I want to make sure you are very comfortable with this because in the next example, example 5.1979

We are going to use this density function to calculate the mean of the magnitude.1984

I want to make sure that this density function is okay with you, 1989

and you feel comfortable with it before you move on to example 5.1992

Example 5 here is a follow-up to example 4.2000

If you have not just watched example 4, go back and watch that first.2004

We are going to use the density function that we found in example 4, to find the expected magnitude of the next earthquake.2008

We want to check our answer using the properties of the exponential distribution.2016

Let me just remind you the short summary of example 4.2020

We had U is equal to Y² + 1.2026

Y followed an exponential distribution.2030

The density function for Y FY of Y was 1/β but β was 2 so E ^- Y/β.2036

Y ranges from 0 to infinity and U ranges from 1 to infinity.2051

We figured out the density function for U, we did this in example 4.2059

F sub U of U was 1/4 √U -1 × E ^ -√U -1/2.2063

That is as far as we got in example 4, but in this problem we have to find the mean of U 2078

because we want to find the expected magnitude.2084

That is the same as expected value, and expected value is the same as the mean.2086

Let me show you how that works out.2093

E of U, by definition of expected value, we know this way back into the early lectures.2095

It is just the integral on U × F of U DU.2101

Let me plug in everything I can do here.2108

The F sub U of U, that is this function we figure out in example 4, that came from example 4.2112

The FS sub U of U is 1/4 √U -1 E ^-√U -1/2.2122

But there is one more power of U, that U right there.2133

I’m going to stick in a U right there and DU.2136

What is my range on U, there it is right there.2142

I want U to go from 1 to infinity.2144

And now, I have an integral that I need to solve.2150

You can solve this on your favorite software, if you like.2154

I do want to show you that it is possible to solve this by hand, using just what you learned in calculus 1 and 2.2157

I work it out using calculus 1 and 2, if you want to go ahead and cheat, 2164

throw it into your favorite software, that is totally okay with me.2168

Here is my method, I’m going to use a new variable S.2171

I will define my S to be that thing in the exponent of the E.2180

The √U -1/2 and then, if I make that substitution, I also have to find my DS, which will be ½ ×,2185

The derivative of √U -1 is ½ × U -1⁻¹/2, U -1.2198

That looks a little awkward the way I written it.2209

Let me write it as DU/4 √U-1.2211

That is actually quite encouraging, because we have something very close to that inside the integral.2220

I’m a little encouraged by that.2224

I’m also going to want to solve for U, in terms of S.2226

I got 2S is √U -1 and, if I square both sides there, I get 4S² is equal to U -1.2230

4s² + 1 is equal to U.2245

Am I finished here, No, I still I also have to convert my limits.2251

When U is equal 1, S is equal to, if I plug in U = 1 into S, it looks like S is going to be 0.2257

When U goes to infinity, √U-1/2 will also go to infinity.2267

If I convert this integral into terms of S, I got the integral of S goes from 0 to infinity.2274

This U right here converts into 4S² + 1.2284

Everything else there that 4 √U-1 in the denominator, I can use that as part of the DS here.2291

I'm just going to build that into the DS and then I have E ⁻S as E ^-√U-1/2.2302

That was rigged up to work, I defined S in order to make that work.2310

I got an integral which I can solve using calculus 2 methods.2315

In particular, I’m going to use parts here and integration by parts.2320

I’m going to use the kind of cheaters trick for integration by parts, this tabular integration trick and get it done quickly.2324

4S square + 1, if you do not remember this, check out the section on integration by parts in my calculus 2 lectures, 2330

here on because I showed you this trick there.2338

I got E ⁻S here, I’m going to take derivatives on the left.2342

The derivative of 4S² + 1 is 8S, the 1 goes away.2346

The derivative of that is 8 and then 0.2351

The integrals on the right, the integral of E ⁻S is –E ⁻S.2356

The integral of that is E ⁻S and the integral of that is –E ⁻S.2361

I will connect them up and go + - +, it got a little sloppy there, that was supposed to be a + there.2369

What I get for my integral is, it is like everything is going to be negative.2378

4S² + 1 × –E ⁻S -8S E ⁻S, + × -, -8 E ^- S.2385

I'm evaluating this from S =0 to S goes to infinity.2402

If I plug in my S going to infinity, this E ⁻S term dominates everything else possible.2412

An exponential term always beats a polynomial term.2421

E ⁻S is going to drag all of those terms to 0 because it is like saying 1/E ⁺infinity.2425

That 0 -0 -0 for the S going to infinity.2434

I plug in S =0, in the first term I have - and - so + E⁰ + E⁰, for the first term.2438

+ 0 and then + 8 E⁰.2449

E⁰ is exactly 1, E⁰ + 8E⁰ is 9 × 1 is just 9.2454

That is the expected magnitude of the next earthquake.2463

That is rather unsettling because that tells me that I’m looking forward to a magnitude 9 earthquake2467

on the Richter scale in California, that would be absolutely devastating.2475

I certainly hope that this example is wrong.2479

The math is correct but the geological premises are probably a little bit faulty there.2483

Let me check my answer using the properties of the exponential distribution, 2490

because that is what it asked me to do in the example.2495

What I know about the exponential distribution is that the mean is β and the σ², 2499

the variance of the exponential distribution is β².2510

Let me see how I can use that to find the expected value of U.2516

The expected value of U is the expected value of Y² + 1.2522

Let me split that up using linearity of expectation.2530

The expected value of Y² + the expected value of 1 which is just 1.2532

The expected value of Y², you can find that as σ² - the expected value of (Y)².2538

That is because you can find the variance as the expected value of Y².2551

There is a + there, the σ² is the expected value of Y² - the expected value of Y².2560

If you just move this term over to the other side then the expected value of Y² is just σ² + E of (Y)², I still have + 1 there.2572

If I plug in my variance for the exponential distribution, that is β² + E of Y is β is the mean, 2586

another β² + 1 which is 2 β² + 1.2598

I was told that the mean was 2, β is equal to 2.2603

That is 2 × 4 + 1, we get 9.2608

That is very reassuring in the mathematical sense, because it does agree with my previous answer.2614

It is a little worrying geologically because I happen to live in Southern California and2621

I certainly do not want to experience a magnitude 9 earthquake.2626

Let me go ahead and check the steps here.2631

What we did was, we took the density function that we found in example 4.2633

Here is the density function that we found in example 4.2638

It wants us to find the expected value of U.2641

I'm going to integrate that, multiplying it by U.2645

That is because we are trying to find the expected value of U.2648

That is where that extra factor of U came in.2651

To solve the integral, it was a bit of tedious calculus.2654

You can just drop it into some kind of integration program or calculator, or something online.2658

What I did was I need a little substitution, S is √U -1.2666

I solve for U in terms of S, when I plug in that U, that turned into the 4S² + 1.2671

I also had to calculate my DS, that is me calculating the DS in terms of U.2681

All of that part became the DS, that part became the 4S² + 1 and that part right there converted into S.2689

That is how I got this integral E ⁻S DS.2703

And then, I also had to convert my limit.2707

I converted U = 1 and infinity into S = 0 to infinity.2710

That sets up an integration by parts problem.2717

Since, I'm too lazy to do the full integration by parts formula, I used tabular integration right here,2719

in order to work out the integral of 4S² + 1 × E ⁺S.2726

The derivatives on the left, integrals on the right.2731

I get this large expression, it is not that bad though because when you plug in infinity for S, 2733

all of these terms disappear because E ⁻S goes to 0 faster than any polynomial can go.2741

That is why we get 0 for all the infinity terms.2749

And then, when I plugged in S =0 then, not all the terms disappeared, I got a couple of E⁰ terms which of course E⁰ is 1.2753

It simplified down to 9, that represents the expected magnitude of the next earthquake in California.2766

I can also check this using the properties of the exponential distribution.2773

What I know is that U was defined to be Y² + 1, there was my definition of U.2780

Since, it is expected value it is just the expected value of Y² + 1, that is linearity of expectation.2786

The expected value of Y², I can get this using the original formula for variance, 2792

that we learned long ago in a lecture from way back at the top of the list.2799

The formula for variance was E of Y² – E (Y)².2805

I can solve that for E of Y² to be σ² + E of (Y)².2811

I know what σ² is, for the exponential distribution.2818

These came from properties of the exponential distribution which is something we learned about, 2822

in the lecture on the γ distribution.2830

Scroll way back and you will see these properties for the exponential distribution.2833

I plugged in σ² is β², E of Y is β, and I know the β is 2, that something I got from example 4,2838

because we said that the major earthquakes occur every 2 years on average.2846

I plug that in, do a little arithmetic, and I come up with 9.2852

Since those two answers agree with each other, I'm pretty confident that I did everything right there.2857

In example 6, Y has a β distribution with α and β equal to 2.2864

U is Y² + 2Y + 1.2869

We want to find the density function for U.2873

The first thing here is to recognize what is a β distribution.2878

We had a lecture on β distributions, it is part of the series on continuous distributions.2882

If you scroll up, you will see a lecture on β distributions.2886

Let me remind you of the generic formula for density function for β distribution.2891

F of Y is equal to Y ⁺α -1, 1 – Y ⁺β – 1.2898

And then, we divide that by B of α β.2909

B of α β, in turn is Γ of α, Γ of β/Γ of α + β.2917

What we have to do is interpret this with α and β equal to 2.2932

Our F sub Y of Y is equal to, if I flip that denominator, I will have γ of 2 + 2 = γ 4 divided by γ of 2 × γ of 2.2938

Y ⁺α – 1 is Y¹ and 1- Y ⁺β – 1 is, β is 2, β -1 is 1.2954

What is γ of 4, remember there is this relationship between the γ function and factorials, at least for whole numbers.2967

Γ of N is equal to N -1!.2977

Γ of 4 is 3! which is 6, γ of 2 is 1! which is just 1.2980

The denominators cancel, we get 6 × Y × 1 – Y.2993

6 × Y - Y², that is my F sub Y of Y.2999

It actually let me now solve for, U is a function of Y.3009

I think I need to solve for my inverse function there.3019

U is H of Y which is Y² + 2Y + 1.3023

An easy way to think about that would be as Y + 1².3030

I think that is going to make it easier to deal with.3034

I want to solve for Y, in terms of U, that is how I get the inverse function.3037

I’m going to solve for Y, √U is equal to Y + 1.3044

Y is equal to √U -1, that right there is my H inverse of U.3057

Let me remind you of the generic formula for the density function, when you use the method of transformations.3071

It says that F sub U of variable U is equal to, we started with a density function for Y F sub Y.3078

Then, you plug in H inverse of U and then multiply on the derivative D by DU of H inverse of U.3087

In case that is negative, you want to take its absolute value.3098

We are going to need to know H inverse of U.3101

I already found it, √U -1.3104

We are also going to need to know H inverse of U, its derivative.3108

The derivative there, let us see.3115

I will go ahead and plug everything in here.3118

F sub Y, there is my F sub Y there.3121

This is 6 ×, it says Y but I have to convert that into U, convert that into H inverse of U.3123

√U -1, I'm looking at this Y - Y² but I'm plug in √U-1 for Y.3132

-√U – 1², it is going to get a little messy, is not it.3144

The derivative of H inverse of U is the derivative of √U-1.3151

That is 1/2 √U, that is because you can think of that as U¹/2.3157

The derivative is ½ U ^½, that -1 there that just goes away when we take the derivative.3165

We are supposed to take the absolute value of that, but that is already positive, 1/2 √U.3171

I'm just going to do some algebra to work this out.3177

Hopefully, I will simplify a bit, I do not think it is going to get great.3180

But, let us go ahead and see what we can do with this.3183

The 6/2 simplifies a little bit, right away.3185

We get 3/√U, inside the parenthesis I have √U -1.3189

If I square out √U - 1², I will get U -2 √U + 1 there.3199

Maybe, I can simplify that a little bit more, 3 √U -1- U + 2 √U – 1 there.3213

I guess this will simplify a bit, = 3/√U, √U + 2 √U is 3 √U -1- 1 – 2 – √U.3231

I can distribute that √U from the outside into the inside, 3 × 3 – 2/√U.3250

√U divided by √U is √U.3264

That is what I get, I do not think it is going to get any simpler than that.3270

I think we found the density function, that is F sub U/U.3276

The problem reminds us that we should also find the range of possible values for U.3282

If Y, for β distribution function, Y will go from 0 to 1, that is always the range for β distribution function, 3287

whereas U in terms of Y, Y is √U -1.3303

U is Y + 1², U if it is Y + 1², 0 + 1² is 1.3311

1 + 1² is 4, my range of values for U is from 1 to 4.3321

That is my density function for U, kind of ugly there but I did not see any way in making it any nicer.3330

I think we are stuck with that.3338

In the end, it did simplify a little bit.3340

I guess we have to be happy with that.3342

Let me recap the steps there.3345

First of all, I saw that we had a β distribution.3346

I went back and looked at my formula for β distribution which we learned in a lecture on β distributions.3349

It is quite a long time ago in the chapter on continuous distributions.3355

Just go back and look, you will see a lecture on β distributions and you will see this formula Y ⁺α – 1 1- Y the β -1/B of AB.3359

You will also see an expansion for B of α β, I said B of AB, I mean B of α β, in terms of γ functions.3370

That is the expansion of B, since that is in the denominator, I flip that over when I pull it outside in the denominator here.3380

Here, my α and β were 2, I was given that in the problem.3391

I plug those in, I got exponents 1 on both of those.3394

To find those values of Γ, I just remembered that γ of a whole number N is equal to N -1!.3399

Γ of 4 is 3!, that is where that 6 came from.3409

Γ of 2 is just 1!, both of those dropped out.3413

If I multiply the Y through, get 1 - Y².3418

Now, I try to solve U = H of Y to find the inverse function.3426

The convenient way to write that was Y + 1².3431

And then, I want to solve it for Y, I took the square root of both sides.3434

√U = Y + 1, Y = √U -1.3438

I solved for Y in terms of U, that right there, let me see if I can write down a little more cleanly,3443

Is H inverse of U, and that is kind of the key ingredient to our transformation formula.3450

This is the generic transformation formula that I gave you on the introductory slides to this lecture.3458

That is the generic transformation formula.3465

I dropped the H inverse of U in here but that was in F of Y.3467

I plugged it in here, that is why I got my 6.3472

This right here is Y, that is coming from that Y.3477

This right here is Y² but I’m plugging in Y = H inverse of U to both of those.3482

And then, my derive of H inverse of U is how I got the 1/2 √U.3491

I got that by taking the derivative of this right here.3498

Remember, we take the derivative, the 1 drops out, we just get 1/2 √U.3503

This got a little messy but it was just algebra from here.3509

I expanded out my perfect square, A² + 2 AB + B².3513

I combine terms as best I could, I distributed the 3/√U.3518

I did distribute the 3, but I distributed the √U into the parentheses.3524

I got something a little bit nicer but still not that nice.3530

The way I got the range for U was, I remembered that the range for β distribution is always Y goes from 0 to 1.3534

I took those values of Y and I plugged those into my function for U.3545

U is Y + 1², that is how I got U goes from 1 to 4, when Y goes from 0 to 1.3553

That is where my range of values for U comes from.3560

That wraps up example 6, and that is the last example for this lecture on the method of transformations.3565

Remember, this is part of the three lectures series, these are all different methods 3571

for how to find the density and distribution functions for a function of random variables.3577

This was the middle one, this is a method of transformations.3586

In the previous lecture, we covered the method of distribution functions.3588

I hope you will stick around for the next lecture which is the method of moment generating functions.3592

It might be worthwhile to review moment generating functions, before you jump in to the next lecture.3598

This is all part of a larger chapter on functions of random variables.3604

In turn, that is part of the probability lectures here on, with your host Will Murray.3608

Thank you very much for joining me today, bye.3615