WEBVTT mathematics/probability/murray
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Hi and welcome back to the probability lectures here on www.educator.com. My name is Will Murray
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We are starting a new chapter today which is on finding the distributions of functions of random variables.
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I need to explain what that is all about.
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We are going to use three different techniques to find distributions of functions of random variables.
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I have a lecture on each one of these three techniques.
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The first one is distribution functions.
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The second one is going to be transformations.
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The third one is the method of moment generating functions.
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In this lecture, I want to give an introduction to the whole idea.
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And then, we are going to spend some time studying the first method which is distribution functions.
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The premise here is that, we have one or more random variables.
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We might just have a single variable Y, we might have several Y1 and Y2, and so on.
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We want to study functions of these random variables.
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For example, we might want to study something like U can be defined to be Y1 + Y2.
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That is a function of two random variables.
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We want to find the distribution of U.
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Let me emphasize that, in the previous chapter, we had some theorems that helped us to find the mean and the variance of U.
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But, that is not necessarily enough to determine the whole distribution of U.
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The goal in the next couple of lectures is to actually find the full distribution for U.
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I need to show you, how we are going to do that.
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As I said, we want to find the full distribution function.
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I have called it F sub U of u.
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By definition, that means the probability that U will be less than some value of u.
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If we can find that distribution function which is F, then we can find the density function which is f.
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Remember, the density function is always the derivative of the distribution function.
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What we will do is, we will find the distribution for a function first F.
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And then, we will just take one derivative to get the density function.
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Assuming we can do that, then we can calculate probabilities on U.
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The probability that U will be in some range, you can calculate it as an integral of the density function.
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We will just use the distribution function F sub U of B - F sub U of A.
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That is the reason why we are interested in doing this is, is we want to be able to calculate probabilities on U.
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I’m not going to spend a lot of time on that particular element of it.
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The goal really for us today, is going to be to find this distribution function F.
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If we can do that, it will be just a quick step to find the density function, by taking the derivative.
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As I mentioned already, we are going to use three methods to find this F.
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Today, we are going to talk about distribution functions.
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That is what this lecture is focused on, distribution functions.
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We are going to be using a lot of geometry and we will have to do some double integral.
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I hope you are ready to go with your multivariable integrals.
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I hope your calculus 3 is up to snuff.
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If it is not, do not worry, we have some lectures here on www.educator.com of multivariable calculus.
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My colleague Raffi is just excellent.
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You can check out his lectures and practice your double integrals.
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The second technique is called transformations and we will cover that in the next lecture.
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That has got a lot of the methods from calculus 1, you will be up to speed on that.
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The final technique is on moment generating functions,
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and that is something that I taught you about in an earlier lecture here in the probability lectures.
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Today, we are to talk about distribution functions and let us jump into that.
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There is not a whole lot to say about distribution functions.
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The distribution function, by definition, is the probability that U is less than or equal to u.
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The way we want to approach that is, figure out what values of your variable Y or Y1 and Y2,
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correspond to U being less than or equal to u.
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In practice, you start out by saying F of U is equal to the probability that u is less than or equal to u.
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And then, you convert this U into a Y or Y1 and Y2.
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I do not mean you just change it to Y1 and Y2, use the definition of U as a function of Y1 and Y2.
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Convert into a function of Y1 and Y2.
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You convert that into Y1 and Y2, or maybe there is just one Y in there.
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We want to find the probability that Y1 and Y2 will be in that particular range.
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I do not have a good general way to tell you to do that, all I can say is it is a probability calculation.
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For single variable, since you are finding a probability, this usually means solving a single variable integral.
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For two variables, it means solving a double integral or sometimes just using some geometry
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because it often turns out to be a process of looking at a geometrical region,
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figuring out which combinations of Y1 and Y2 will give you that function less than or equal to u.
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It is very hard to give you general guidance, I think the strategy for learning about distribution functions is really to immerse yourself in examples.
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Practice a whole bunch of examples and it will start to make sense.
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All the examples are going to start with this probability of U less than or equal to u.
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And then, we will convert this into U of Y and possibly U of Y1 Y2.
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They will all start like that but after that, they sort of all go in different directions.
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Some will go geometrically, some turn into a single integral, some turn into a double integral.
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Just work through the examples with me and I think you will start to get the hang of it.
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In example 1, it is a single variable problem.
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We have a density function 3/2 Y² for the variable Y.
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We want to find the density function for U which is 3 -2Y.
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The first thing that I want to note here is the possible range of values for U.
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If I plugged in the range of values for Y, when Y is equal to -1 then my U will be 3 -2 × -1, which will be 3 + 2 is 5.
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When Y =1, that gives me a U of 3 -2 × 1 which is 1.
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That tells me a range for U, it is between 1 and 5, that is good progress there.
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We want to find the density function ultimately for U.
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The method here is to use distribution functions.
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We want to find the distribution function first.
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F sub U of u is by definition, the probability that U will be,
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let me erase that tail to make that look like a U, that is subscript there.
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The probability that U is less than or equal to any value of u.
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I want to convert that into a probability on Y.
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This is the probability, my U is defined to be 3 -2Y less than or equal to u.
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And now, I want to solve that for Y.
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Let me just say, if 3 -2Y is less than or equal to u,
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if I pull the Y over the other side and pull the u over to the left, that is 3 - u so that is equal to 2Y.
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That saying Y is bigger than or equal to 3 - u/2.
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This is the same as the probability that Y is bigger than or equal to 3 - u/2.
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Remember, I can solve that with an integral.
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This is the integral of the density function is 3/2 Y² DY.
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I want to integrate that as Y goes from 3 - u/2.
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Since, I want it to be bigger than that, up to the top of the range which is Y = 1.
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I’m going to do an integral there, it is not a bad integral.
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The integral is 3/2 Y² is ½ Y³.
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Because, the integral of 3Y² is Y³, so ½ Y³.
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I have to evaluate that from Y = 3 - u/2 up to Y = 1.
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I get ½ (1 - 3 - u/2³).
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What I just found here was the distribution function, that is F of U.
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That is the distribution function which is not quite what I'm looking for.
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I’m asked to find the density function distribution function.
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To get the density which is what I'm really looking for, I’m going to take the derivative of that.
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Density function is f of U and that is always the derivative of the distribution function.
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The derivative of what I just found above.
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Let us see, that is ½.
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If I take the derivative of 1, that is a constant so that goes away.
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The derivative of -3, I’m going to use the power rule, 3 -u/2².
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By the chain rule, I have to multiply by the derivative of 3 -u/2 which is - ½.
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I can collect all my constants, ½ × -3 × - ½ is ¾.
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I'm just going to leave 3 - u/2² and that is my density function.
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I should give a range on U, here it is, U goes from 1 to 5.
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That is my density function, my f on the variable U.
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Let me recap the steps there.
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The first thing I did was, I was given this density function for Y and I was given U in terms of Y.
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I wanted to first translate the range for Y into a range for U.
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I looked at my endpoints for Y, -1 and 1.
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I plugged both of those into the definition of U and I got N points for U.
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U is between 1 and 5.
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And then, I took the key step here of saying that the distribution function for U, the F,
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it is always the probability that U is less than or equal to u.
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Then, I translated that U into a Y, into a function of Y.
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And then, I want to solve that out for Y, in terms of U.
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That is what I was doing over here, is solving that into Y bigger than 3 - u/2.
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The point of doing that is, I can then convert that into an integral.
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Since it is bigger than 3 - u/2, I made 3-u/2 my lower limit.
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I go up to the top of the range Y = 1, that comes from that one right there.
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I integrate my density function, do that integral, plug in the limits, and what I get is the distribution function for U.
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Which is not quite what I wanted, I was asked to find the density function.
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Remember, density and distribution are two different things.
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But you get from the distribution to the density, by taking the derivatives.
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You will get f, by taking the derivative of F.
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I took that derivative, in terms of U.
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The one dropped out, it was a constant, that 1 went away.
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And then, I took the derivative of 3 - u/2³ using the power rule.
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That was the power rule, right there.
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3 × 3 – u/2², and then, that was the chain rule.
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That is why I got that - ½, it came from the derivative of the inside 3 - u/2.
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Then, I just collected all my constants, simplified a little bit, and reminded myself of what the range on U is.
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Now, I have a nice density function for U and I know what range it is defined on.
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We are going to use this same starting premise of the 3/2Y² for example 2.
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But, we are going to use it for a different function U.
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Make sure you understand this, before you move on to example 2.
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Make sure everything makes sense and then, we are going to practice this again for a different function U.
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You can try it on your own, if you want, before you listen to my solution for example 2.
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In example 2, we have the same density function and the same range for Y that we had for example 1.
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The density function is 3/2Y² and the Y goes from -1 to 1.
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We want to find the density function for a new definition of U, it is going to be Y² now.
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The premises are the same, let me find the range on U.
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If Y goes from -1 to 1, let me just plug those endpoints in.
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If Y is -1 then U will be 1.
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If Y is 1, that will also make U equal to 1.
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I notice that in between there, we will have Y = 0.
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You could be as low as 0, Y = 0 gives me U = 0.
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My range on U will be between 0 and 1, that is the range that I will be looking for there.
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I have to find the density function for U, the way I’m going to do that is to first find the distribution function.
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I will use the distribution function initially, and then assuming that works out,
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I will take the derivative to find the density function.
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The distribution function is F sub U of u.
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By definition, that is the probability that U is less than or equal to some value of u.
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I'm going to plug in what U is, it is defined to be Y².
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This is the probability that Y² is less than or equal to u.
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Now, I want to solve that into a range on Y.
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If Y² is less than or equal to U, that really means that Y by itself, will be less than √ u and bigger than – √ u.
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Those are my ranges for Y, it is nice because I can now convert that into an integral.
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That is the integral as y goes from –√U to √U.
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What was my density function, 3/2 Y² DY.
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That is an easy integral, the integral of 3Y square is just Y³.
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We have ½ Y³ here and we want to evaluate that from Y = v-U to Y = √U, that is ½.
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Y³ at √U, that is the √ U³ will be U³/2 and – Y³, -√U³/2.
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√U³ will be U³/2, get some more parentheses in there and make it honest.
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What we have is two copies of U³/2, but we are also multiplying by ½.
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I would just get U³/2, that is kind of pleasant.
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That was the distribution function, that was my F of U.
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Let me figure out my density function which is my f.
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I have done all the hard work to get the density function, once you know the distribution function,
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what you do is just take the directive of the distribution.
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The density function is just F sub U of u derivative.
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Let me put in f of U.
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I just take the derivative of that which is 3/2 U ^½ or 3/2 √U.
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My range on that is, U goes from 0 to 1, 0 less than or equal to U less than or equal to 1.
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That is the density function because that is what we are asked to find.
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Let me go back over that, make sure that everybody is up to speed on this.
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We start out by finding the range on U.
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If Y varies from -1 to 1 and I’m keeping track of Y², Y² will always be positive.
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It will go from 0 to 1.
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The distribution function is the probability that U is less than or equal to some cutoff value of u.
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I just translated U into Y², I got that from the stem of the problem that U is Y².
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If Y² is less than or equal to u, that means Y by itself must be between –√U and √U.
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I set those up as my limits of integration and I integrated my density function for Y.
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I plug in my limits and it simplified down to U³/2, that is the distribution function F of U.
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To get the density function f, I took the derivative of the distribution.
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That is where I got the 3/2 U ^½.
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Then, I just reminded myself of the range, that came from here U goes from 0 to 1.
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In example 3, we have a multivariable example.
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F of Y1 Y2 is defined to be a joint density function E ⁻Y2.
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Our region there is Y1 and Y2 both go from 0 to infinity but Y2 is always bigger than Y1.
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Let me start out by graphing that because that is not the most obvious region.
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There is Y2, I always seem to switch my variables Y1 and Y2.
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Here is Y1 and there is Y2 on the vertical axis.
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Since, we are going interested in when Y2 is bigger than Y1 that is like saying Y is bigger than X.
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Let me graph line 1 = X.
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There it is, Y = X, or in terms of our variables, that is saying Y1 = Y2.
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I’m interested in the region where Y2 is bigger than Y1, that is the region above that line.
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That is my region right there.
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I want to find the cumulative distribution and density functions for U = Y1 + Y2.
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I’m going to find distribution function first.
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F is the distribution function, and by definition that is always the probability that U is less than some cutoff value of u.
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I want to convert that into on our range for Y1 and Y2.
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That is the probability that U is the same as Y1 + Y2 is less than or equal to u.
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Somehow, I want to describe that range on my graph.
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I'm going to try to draw that in red here, Y1 + Y2, I’m going to graph the line Y1 + Y2 is equal to U.
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That will intersect both axis at U.
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There is U on both axis and let me draw that line.
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There is the line at U, that is the line Y1 + Y2 is equal to U.
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The range that is less than that and is still in my region is this red region, right here.
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That is the range that I want to describe.
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need to set up a double integral on that range.
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I need to describe that, let me see if I can describe that range.
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I’m running out of colors here.
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I think I will describe it in that direction.
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First, I will describe what Y1 is, it looks like that will be U/2.
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My Y1 will go from 0 to U/2 and my Y2 will go from, that first diagonal line is just Y2 = Y1.
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That second diagonal line is the line, that red line Y1 + Y2 = U.
00:24:03.100 --> 00:24:08.500
In terms of Y2, that is Y2 is equal to U - Y1.
00:24:08.500 --> 00:24:12.500
I have got a description of that region that I'm interested in.
00:24:12.500 --> 00:24:16.600
I can use that to set up a double integral.
00:24:16.600 --> 00:24:19.100
Let me set up my double integral here.
00:24:19.100 --> 00:24:27.000
First variable is Y1 goes from 0 to U/2.
00:24:27.000 --> 00:24:37.800
And then, my second variable is Y2 = Y1 to Y2 is equal to U - Y1.
00:24:37.800 --> 00:24:45.400
I have to put in my joint density function E ⁻Y2.
00:24:45.400 --> 00:24:50.800
It looks like my inner variable is Y2, my outer variable is Y1.
00:24:50.800 --> 00:24:54.100
Now, it is just a calculus 3 problem.
00:24:54.100 --> 00:24:58.200
I just got to do this double integral and just grind through it.
00:24:58.200 --> 00:25:03.800
If you are one of the fortunate people who is allowed to do double integrals on computers,
00:25:03.800 --> 00:25:07.800
if your probability teacher allows that, go ahead and throw this in your computer.
00:25:07.800 --> 00:25:11.200
I do not mind, see if your answer checks with mine.
00:25:11.200 --> 00:25:14.400
I will show you that it is possible to do it by hand.
00:25:14.400 --> 00:25:16.600
I'm going to do this integration by hand.
00:25:16.600 --> 00:25:24.700
First variable is Y2, integral of E ⁻Y2 is –E ⁻Y2.
00:25:24.700 --> 00:25:36.600
Just do the inside integral first, E ⁻Y2 goes from Y2 = Y1 to Y2 = U - Y1.
00:25:36.600 --> 00:25:47.800
The top part will give me –E, if Y2 is U - Y1, then -Y2 is Y1 – U.
00:25:47.800 --> 00:26:03.300
A – and - is +E ^- Y1, and we want to integrate that with respect to Y1 from 0 to U2.
00:26:03.300 --> 00:26:05.300
Let me go ahead and integrate that.
00:26:05.300 --> 00:26:13.400
The first term E ⁻Y 1 will be just, let me write the second term it –E ⁻Y1.
00:26:13.400 --> 00:26:23.500
That first term, if you think about it, this is like saying E ⁺Y1 × E ⁻U.
00:26:23.500 --> 00:26:35.700
The E ⁻U is a constant, this is –E ⁻U × E ⁺Y1.
00:26:35.700 --> 00:26:43.100
That is not too bad there, I’m a little worried here.
00:26:43.100 --> 00:26:47.600
I think I’m good, I think I have done everything right.
00:26:47.600 --> 00:26:56.800
But, I need to a plug in my limits which is Y1 = 0 to Y1 = U/2.
00:26:56.800 --> 00:27:13.700
Let me plug those in, this is -E ⁻U × E ⁻Y1 which is E ⁺U/2 – E ^- Y1 E ⁻U/2.
00:27:13.700 --> 00:27:34.200
If I plug in Y1 = 0, then for my first term I get - and - which is +E ⁻U, a - and - which is +E⁰ which is just 1.
00:27:34.200 --> 00:27:38.900
Let me see if I can simplify this a bit because it looks a bit messy.
00:27:38.900 --> 00:27:45.200
This right here is –E ⁻U and E ⁺U/2.
00:27:45.200 --> 00:27:47.200
Let me add the exponents.
00:27:47.200 --> 00:27:55.500
That is U/2 – U, that is E ⁻U/2.
00:27:55.500 --> 00:28:03.700
I can combine that with the other one, I will get 1 + E ⁻U.
00:28:03.700 --> 00:28:12.000
I got two copies of E ⁻U/2.
00:28:12.000 --> 00:28:21.200
I forgot to include my 2 there, -2 copies of E ⁻U/2.
00:28:21.200 --> 00:28:32.500
My range there is, here is Y1 + Y2 that could go between 0 and infinity.
00:28:32.500 --> 00:28:41.900
U goes from 0 to infinity there, that was my distribution function, my F of U.
00:28:41.900 --> 00:28:49.800
That is certainly an accomplishment to have found that.
00:28:49.800 --> 00:28:58.600
But, I also need to find the density function, that was the distribution function that I just found.
00:28:58.600 --> 00:29:06.000
The density function, remember you take the derivative of the distribution function.
00:29:06.000 --> 00:29:09.600
I do not think the density has an i there.
00:29:09.600 --> 00:29:21.600
The derivative of the distribution function f of U is just F prime of U.
00:29:21.600 --> 00:29:32.600
I need to take that derivative, the derivative of 1 is 0, the derivative of E ⁻U is –E ⁻U.
00:29:32.600 --> 00:29:46.500
The derivative of -2 E ⁻U/2 -2 × E ⁻U/2 ×, by the chain rule, another - ½ there.
00:29:46.500 --> 00:29:49.500
Maybe, I can write that a little more nicely.
00:29:49.500 --> 00:29:59.400
The 2 and ½ cancel, I get E ⁻U/2 – E ⁻U.
00:29:59.400 --> 00:30:12.200
That is my density function, same range for U, U goes from 0 to infinity there.
00:30:12.200 --> 00:30:14.900
Let me box that up and then let me go back over the steps,
00:30:14.900 --> 00:30:22.000
just in case anybody is still wondering where some of those points came from.
00:30:22.000 --> 00:30:24.200
This is our first multivariable example.
00:30:24.200 --> 00:30:28.700
The first thing I did was to graph the range here.
00:30:28.700 --> 00:30:34.100
Y1 and Y2 both go from 0 to infinity.
00:30:34.100 --> 00:30:37.800
We also know that Y2 is bigger than Y1.
00:30:37.800 --> 00:30:44.700
I graphed the Y1 = Y2, and the general range here is all of these blue shaded region.
00:30:44.700 --> 00:30:48.100
Do not think about the red yet because that comes in later.
00:30:48.100 --> 00:30:51.700
It is all of these blue shaded triangular regions.
00:30:51.700 --> 00:30:58.200
I’m going to find the distribution function first, the U.
00:30:58.200 --> 00:30:59.800
That is what I’m finding right now.
00:30:59.800 --> 00:31:06.300
It is the probability that F is the probability that U is less than u.
00:31:06.300 --> 00:31:11.300
I plugged in my definition of U, Y1 + Y2.
00:31:11.300 --> 00:31:14.900
I got to find the probability that that is less than u.
00:31:14.900 --> 00:31:19.800
But, I want to describe that in terms of a region.
00:31:19.800 --> 00:31:24.800
I graphed Y1 + Y2 is equal to U, that is this red line here.
00:31:24.800 --> 00:31:31.500
I took all the regions less than that line, that is why I got this sort of a doubly shaded red and blue area.
00:31:31.500 --> 00:31:37.600
I want to describe that area, it is useful to describe it with Y1 first.
00:31:37.600 --> 00:31:41.700
Otherwise, you have to break it up into two pieces and that would make it even worse.
00:31:41.700 --> 00:31:48.600
I graphed Y1 goes from 0 to U2, that is the biggest possible value of Y1.
00:31:48.600 --> 00:31:56.000
Y2 goes from this lower diagonal line up to this upper diagonal line.
00:31:56.000 --> 00:32:02.800
I found the equations for each one of those, Y2 = Y1 and Y2 = U - Y1.
00:32:02.800 --> 00:32:10.000
That gave me the limits of this double integral, that is where I got those from, that came from over there.
00:32:10.000 --> 00:32:15.000
And then, I integrated the density function that I was given, that came from right here.
00:32:15.000 --> 00:32:19.000
It is just solving a double integral and this was a little bit messy to solve.
00:32:19.000 --> 00:32:23.700
But, I do not think I even had to do integration by parts or anything like that.
00:32:23.700 --> 00:32:28.000
It is just a little messy, I solved the first one with respect to Y2.
00:32:28.000 --> 00:32:33.400
I plug in my bounds, got some things in terms of Y1, integrated that.
00:32:33.400 --> 00:32:40.100
What I finally found here was the distribution function, the F that we are looking for.
00:32:40.100 --> 00:32:45.300
To get the density function, I took the derivative of what I had just found.
00:32:45.300 --> 00:32:57.400
I just did D by DU to get to the density function and that simplify down to this the density function.
00:32:57.400 --> 00:33:03.200
That gives me both of the things that I was looking for, and that wraps up that example.
00:33:03.200 --> 00:33:09.100
In example 4, we are given that Y1 and Y2 had joint density function.
00:33:09.100 --> 00:33:11.400
Let me remind you what this notation means.
00:33:11.400 --> 00:33:15.900
That colon means it is defined to be.
00:33:15.900 --> 00:33:24.600
The joint density function is defined to be, that = means always =.
00:33:24.600 --> 00:33:31.600
That just means we have a constant density function, it is always equal to 1.
00:33:31.600 --> 00:33:37.000
We have got a region here, the triangle bounded by 0,0.
00:33:37.000 --> 00:33:43.500
I need to graph this, 0,0 and 2,0 and 2,1.
00:33:43.500 --> 00:34:01.100
There is Y2 on the horizontal axis, there is 2,0, 2,1 I will put up here, and 0,0.
00:34:01.100 --> 00:34:07.200
There is the Y2 axis, I keep writing Y2 on the horizontal axis.
00:34:07.200 --> 00:34:10.300
I do not know why I always do that.
00:34:10.300 --> 00:34:12.800
I have done that about 4 × in a row now.
00:34:12.800 --> 00:34:16.000
But, there is the point 0,0.
00:34:16.000 --> 00:34:21.600
The triangle we are interested in is that triangle right there.
00:34:21.600 --> 00:34:27.700
We are told that we have a uniform density, constant density 1 on the triangle.
00:34:27.700 --> 00:34:32.100
I’m not going to color in that triangle because I have to make a couple other pictures on here.
00:34:32.100 --> 00:34:38.400
That will get a little crowded but the region here is really this region inside that triangle.
00:34:38.400 --> 00:34:43.600
And then, we are given a function U is Y1 - Y2.
00:34:43.600 --> 00:34:47.000
We want to find the density function for that.
00:34:47.000 --> 00:34:51.500
But of course, we are going to find the distribution function first, the F.
00:34:51.500 --> 00:34:54.900
Then, we will take the derivative to find the density function, the f.
00:34:54.900 --> 00:34:58.300
Let me go ahead and try to figure out what the F should be.
00:34:58.300 --> 00:35:16.700
F of u, by definition it is the probability that U is less than u.
00:35:16.700 --> 00:35:29.500
In turn, I’m going to translate that into Y, the probability that Y1 - Y2 is less than or equal to this cutoff value of u.
00:35:29.500 --> 00:35:37.200
I want to describe that geometrically and see if I can describe the region.
00:35:37.200 --> 00:35:42.600
Let me think about the line Y1 - Y2 is equal to U.
00:35:42.600 --> 00:35:50.400
If I solve that then, let us see, that is saying, let me say less than or equal to U.
00:35:50.400 --> 00:35:53.800
If I solve that for Y2, I get Y2 on the other side.
00:35:53.800 --> 00:36:00.500
Y2 is greater than or equal to Y1 –U.
00:36:00.500 --> 00:36:08.800
It looks like I got line with slope 1, if I just graph this as a line, Y2 would be equal to Y1 – U.
00:36:08.800 --> 00:36:13.100
I got a line with a slope 1 and Y intercept –U.
00:36:13.100 --> 00:36:18.900
Let me try and draw that on my axis here.
00:36:18.900 --> 00:36:26.400
Let us say that is U, my line has slope 1 here.
00:36:26.400 --> 00:36:35.100
I’m going to draw a line with slope 1, and there it is, right there.
00:36:35.100 --> 00:36:43.700
There is my line with slope 1, and I want all the region above that line because I want Y2 to be bigger than Y1 –U.
00:36:43.700 --> 00:36:46.800
I’m looking at that region right here.
00:36:46.800 --> 00:36:55.600
What you notice is that we sort of have two cases here, because it really depends on where you are drawing that line.
00:36:55.600 --> 00:37:03.100
If you draw that line to the left of this right hand corner, then you are going to get this little triangle.
00:37:03.100 --> 00:37:09.600
Let me draw another version of this.
00:37:09.600 --> 00:37:17.700
If you draw that line, it goes to the right of that top right corner, you are going to get a slightly different shape.
00:37:17.700 --> 00:37:24.700
That is going to make this thing kind of awkward to deal with.
00:37:24.700 --> 00:37:32.100
If we draw the line to the right, it goes to the right then we get a different and more awkward shape there.
00:37:32.100 --> 00:37:47.400
If U is bigger than 1 then we will get this awkward shape that it will be a little hard to evaluate.
00:37:47.400 --> 00:37:50.000
We are going to have to do two different cases here.
00:37:50.000 --> 00:38:00.800
I noticed that, the place where that blue line intersects the top horizontal axis is also at U.
00:38:00.800 --> 00:38:05.100
Let me make the blue line a little more prominent.
00:38:05.100 --> 00:38:07.800
The place where it intersects the axis is at U.
00:38:07.800 --> 00:38:16.600
It looks like we are going to have two different cases, depending on whether U is less than or bigger than 1.
00:38:16.600 --> 00:38:19.500
I'm going to have to find those areas.
00:38:19.500 --> 00:38:29.000
Technically, I should integrate over those areas but the one saving grace of this problem is that, we have a uniform density of 1.
00:38:29.000 --> 00:38:34.600
Instead of integrating, I can just calculate those areas.
00:38:34.600 --> 00:38:39.100
Let me make some cases now.
00:38:39.100 --> 00:38:51.700
If U is less than or equal to 1, then we are looking at this little triangle sort of above the line.
00:38:51.700 --> 00:38:54.400
Let me just notice somewhere up here.
00:38:54.400 --> 00:38:58.900
The double integral of 1 is just equal to the area.
00:38:58.900 --> 00:39:03.700
Instead of doing a double integral, I just have to find the area of these regions here.
00:39:03.700 --> 00:39:16.700
If U is less than or equal to 1, I need to find the area of the triangle which of course is ½ × base × height.
00:39:16.700 --> 00:39:22.300
That is the area of the triangle, base × height.
00:39:22.300 --> 00:39:32.900
The base of my triangle is exactly U, that is the easy part.
00:39:32.900 --> 00:39:34.700
The height is a little harder to find.
00:39:34.700 --> 00:39:40.100
How high is that triangle?
00:39:40.100 --> 00:39:43.000
I need to find out how high that triangle is.
00:39:43.000 --> 00:39:49.100
The way I'm going to figure that out is by finding the intersection point of those two lines.
00:39:49.100 --> 00:39:58.100
That blue line right there is, let us see, that was the line Y2 = Y1 – U.
00:39:58.100 --> 00:40:07.100
The black line is the line Y2 is equal to ½ Y1.
00:40:07.100 --> 00:40:12.200
I can solve those together, the height will be the vertical coordinate.
00:40:12.200 --> 00:40:15.900
The height will be the Y2, let me try to solve for Y2.
00:40:15.900 --> 00:40:22.400
Y2 is equal to Y1, let us see, ½ Y1.
00:40:22.400 --> 00:40:35.900
If I write that as 2Y2 is equal to Y1, and if I plug that into the first line, I get Y2 is equal to 2Y2 – U.
00:40:35.900 --> 00:40:41.500
I’m solving this line and this line together, right now.
00:40:41.500 --> 00:40:47.600
If I switch things around, I will get U = Y2.
00:40:47.600 --> 00:40:54.400
That is actually quite nice, that means the height is also U.
00:40:54.400 --> 00:41:03.100
This is ½ U × U, I get ½ U² as the area of the triangle, that is really quite nice.
00:41:03.100 --> 00:41:05.600
Now, I have to find this other awkward area.
00:41:05.600 --> 00:41:10.600
If U is bigger than 1, that was if U is less than 1.
00:41:10.600 --> 00:41:22.700
If U is bigger than 1, I think the easiest way to find the shaded area there
00:41:22.700 --> 00:41:28.600
is to subtract off the area of this little triangle right here.
00:41:28.600 --> 00:41:42.100
The area is equal to, the total area - the area of the small triangle.
00:41:42.100 --> 00:41:44.800
Of course, I mean the area of the small triangle.
00:41:44.800 --> 00:41:46.400
Let us figure out what the total area is.
00:41:46.400 --> 00:41:56.500
The total area is ½ × the base of the big triangle × the height of the big triangle which is 1.
00:41:56.500 --> 00:41:59.400
The small triangle, let us think about that.
00:41:59.400 --> 00:42:02.800
The base of the small triangle, it is not U.
00:42:02.800 --> 00:42:08.900
Let me move that U over a little bit to make it a little more obvious.
00:42:08.900 --> 00:42:14.000
That base is not U, but that base distance is 2 – U.
00:42:14.000 --> 00:42:19.700
Since, we are looking at a 45° line, that height is also 2 – U.
00:42:19.700 --> 00:42:27.800
It is ½ × 2 - U², but that is a little bit more messy.
00:42:27.800 --> 00:42:36.300
½ × 2 × 1 is just 1 - (2 - U)²/2.
00:42:36.300 --> 00:42:44.000
What I just figured out here was my distribution function.
00:42:44.000 --> 00:43:07.600
My F of U is equal to ½ of U², if U is less than 1 and 1 - (2 - U)²/2, if U is bigger than or equal to 1.
00:43:07.600 --> 00:43:11.600
That was my distribution function, that is what I figure out, right there is the distribution function.
00:43:11.600 --> 00:43:17.900
That is not what the problem was asking for, I have to take it one more step and find the density function.
00:43:17.900 --> 00:43:23.600
Remember, you find the density function just by taking the derivative of the distribution function.
00:43:23.600 --> 00:43:26.000
That would not be much more work.
00:43:26.000 --> 00:43:40.900
The density is f of U, but you get that by just taking the derivative of F, the distribution function, take the derivative of that.
00:43:40.900 --> 00:43:49.400
If I take the derivative of each of those parts, the derivative of ½ U² is just U.
00:43:49.400 --> 00:43:56.600
The derivative of 1 -2 - U²/2, the derivative of 1 is 0.
00:43:56.600 --> 00:44:06.100
The derivative of 2 - U²/2 is - ½ × 2 × 2 – U.
00:44:06.100 --> 00:44:10.100
By the chain rule, we have to multiply a -1, the inside.
00:44:10.100 --> 00:44:15.800
The derivative of 2 – U is -1, that cancels nicely.
00:44:15.800 --> 00:44:27.900
What we get is, we get U for, when 0 is less than or equal to U less than or equal to 1.
00:44:27.900 --> 00:44:34.700
I see that my value of U here could go anywhere from 0 to 2.
00:44:34.700 --> 00:44:36.700
That is a possible range for U.
00:44:36.700 --> 00:44:50.200
And then, this part simplifies down to, the ½ canceled, the negatives cancel, we get 2 – U for U being between 1 and 2.
00:44:50.200 --> 00:44:55.400
That is my density function and that is what I needed to find.
00:44:55.400 --> 00:44:58.800
A lot of geometry in there, we did do some calculus.
00:44:58.800 --> 00:45:01.200
I was going to say that we have not done any integration, and that is true.
00:45:01.200 --> 00:45:04.400
But, we did take some derivatives in there.
00:45:04.400 --> 00:45:07.100
Let me recap that problem, kind of some tricky geometry here.
00:45:07.100 --> 00:45:17.200
The first thing I did was graph the triangle, graph these three points 0,0 and 2,0 and 2,1.
00:45:17.200 --> 00:45:21.900
I know I'm looking at this lopsided triangle here.
00:45:21.900 --> 00:45:26.400
I’m trying to find the density function for Y1 - Y2.
00:45:26.400 --> 00:45:34.100
I’m going to start out by finding the distribution function F, which is the probability that U is less than u.
00:45:34.100 --> 00:45:40.600
I plugged in my definition, Y1 - Y2 into here.
00:45:40.600 --> 00:45:43.500
That turns into a geometric constraint.
00:45:43.500 --> 00:45:53.800
If I graph Y1 - Y2 less than U, that graphed out into Y2 bigger than Y1 – U.
00:45:53.800 --> 00:46:00.800
We get all the regions above this line, this is the line Y2 = Y1 – U.
00:46:00.800 --> 00:46:02.900
It is all the regions above that.
00:46:02.900 --> 00:46:08.300
But I see that depending on where I draw that line, I can be looking at two different possible shapes.
00:46:08.300 --> 00:46:11.900
If U is over here less than 1, I got this little triangle.
00:46:11.900 --> 00:46:21.100
If U is bigger than 1, then I got this sort of strange quadrilateral.
00:46:21.100 --> 00:46:24.300
The cases here, this is the case U bigger than or equal 1.
00:46:24.300 --> 00:46:27.400
This is the case that U less than or equal to 1.
00:46:27.400 --> 00:46:31.900
I really have to evaluate two different areas here.
00:46:31.900 --> 00:46:36.900
The good news is that, my density function is consistently 1 which means
00:46:36.900 --> 00:46:40.700
that instead of doing a double integral, I can just find the areas.
00:46:40.700 --> 00:46:45.400
That is only because the density is equal to 1.
00:46:45.400 --> 00:46:52.600
For the first case, when U is less than or equal to 1, I had to find the area of this triangle right here.
00:46:52.600 --> 00:46:58.200
It is ½ base × height, and I figure out that both my base and my height were U.
00:46:58.200 --> 00:47:00.300
The tricky part is finding the height there.
00:47:00.300 --> 00:47:09.700
The way I found that height of U is by solving this line and this line together.
00:47:09.700 --> 00:47:14.900
That was the algebra that I was doing right here, was by solving those two lines together,
00:47:14.900 --> 00:47:17.400
in order to find that intersection point.
00:47:17.400 --> 00:47:21.400
I found that intersection point was Y2 = U.
00:47:21.400 --> 00:47:24.300
My height was U, that is where that U came from.
00:47:24.300 --> 00:47:28.400
That simplified nicely into ½ U².
00:47:28.400 --> 00:47:32.700
If U is bigger than 1, I’m looking at this sort of lopsided region.
00:47:32.700 --> 00:47:34.700
I really did not want to calculate that area.
00:47:34.700 --> 00:47:42.400
The easy way to calculate that area is to take the total area and then subtract off this small triangle, right here.
00:47:42.400 --> 00:47:49.100
I figure out that, that small triangle has base and height 2 - U for both them.
00:47:49.100 --> 00:47:53.800
That is really because this is a 45° line.
00:47:53.800 --> 00:47:58.100
The area of the small triangle is 2 - U²/2.
00:47:58.100 --> 00:48:01.900
The area of the big triangle is just 1.
00:48:01.900 --> 00:48:06.400
That was my distribution function, if U is bigger than 1.
00:48:06.400 --> 00:48:10.300
Both of my distribution and my density functions, we have common cases here,
00:48:10.300 --> 00:48:13.600
depending on whether U is less than 1 or bigger than 1.
00:48:13.600 --> 00:48:20.300
To get to the density function, what I did was I took the derivative of those two pieces.
00:48:20.300 --> 00:48:34.100
The derivative simplified it quite a bit, I just got this nice U and 2 - U for my two cases of my density function.
00:48:34.100 --> 00:48:41.500
In the last example here, we have Y1 Y2 are independent exponential variables with mean 1 and U is their average.
00:48:41.500 --> 00:48:44.900
I want to find the density function for U.
00:48:44.900 --> 00:48:47.300
There are a lot of words here, not a lot of equations.
00:48:47.300 --> 00:48:53.200
The first thing is try and figure out how we translate these words into equations.
00:48:53.200 --> 00:48:57.200
The first thing is to notice that we have exponential variables.
00:48:57.200 --> 00:49:03.600
Let me remind you of the generic of formula for an exponential random variable.
00:49:03.600 --> 00:49:11.000
An exponential random variable has the following density function,
00:49:11.000 --> 00:49:23.400
F of Y is equal to 1/β E - Y/β and that is as Y ranges between 0 and infinity.
00:49:23.400 --> 00:49:27.900
If that sounds like totally gobbledygook to you, if you have never heard that before in your life,
00:49:27.900 --> 00:49:32.800
that means you did not watch my earlier lecture on the different distributions.
00:49:32.800 --> 00:49:37.500
Just go back, it is the same series of probability videos here on www.educator.com.
00:49:37.500 --> 00:49:42.500
Just go back and scroll back, and you will see a lecture on the exponential distribution, that is what I'm quoting right now.
00:49:42.500 --> 00:49:48.600
That was the density function that we learned back in the earlier lecture.
00:49:48.600 --> 00:49:56.000
The other thing we learned is that the mean there or the expected value was just β.
00:49:56.000 --> 00:50:00.100
That is expected value of the exponential distribution.
00:50:00.100 --> 00:50:06.100
In this case, we are given that we have exponential variables with mean 1.
00:50:06.100 --> 00:50:13.600
That means my density function for Y1 is just, it is the function I just listed.
00:50:13.600 --> 00:50:25.700
Except, I plug in β = 1, I get something fairly nice, E ⁻Y1 where Y1 goes from 0 to infinity.
00:50:25.700 --> 00:50:37.400
The same thing for Y2, F2 of Y2 is E ⁻Y2 where Y2 goes from 0 to infinity.
00:50:37.400 --> 00:50:41.900
What we are told here is that these variables are independent.
00:50:41.900 --> 00:50:49.600
What that means is that the joint density function, you get just by multiplying the two individual density functions.
00:50:49.600 --> 00:51:05.500
Using independence, F of Y1 Y2, the joint density function, you get by multiplying the two marginal density functions.
00:51:05.500 --> 00:51:19.900
F of Y1 of × F2 of Y2, that is E ⁻Y1 × E ⁻Y2.
00:51:19.900 --> 00:51:24.200
Let me draw the range that we are looking at there.
00:51:24.200 --> 00:51:30.700
Because, that will be something we will need to look at.
00:51:30.700 --> 00:51:33.000
Look, I got the variable right this time.
00:51:33.000 --> 00:51:40.600
Y1 and Y2 both go from 0 to infinity, the range that we are generically looking at is,
00:51:40.600 --> 00:51:50.300
all of these upper quarter plane here, that is my range.
00:51:50.300 --> 00:51:55.500
I think I'm about ready to start looking at what U is.
00:51:55.500 --> 00:52:03.600
U, in this case, by definition, we are given that U is their average.
00:52:03.600 --> 00:52:09.200
What is the average, that means Y1 + Y2/2.
00:52:09.200 --> 00:52:16.800
I want to find the density function for U, but the whole point of this lecture is to use the method of distribution functions.
00:52:16.800 --> 00:52:25.400
The method of distribution functions says, you first find the distribution function F of U.
00:52:25.400 --> 00:52:35.800
That is by definition, the probability that U, I’m trying to avoid writing tails of my U so that will look less like u.
00:52:35.800 --> 00:52:41.200
Because, we have u coming right up here.
00:52:41.200 --> 00:52:45.500
U is less than or equal to some cutoff value of u.
00:52:45.500 --> 00:52:56.500
Now, I'm going to plug in what U is, that is the probability of Y1 + Y2 being less than or equal to u.
00:52:56.500 --> 00:53:03.000
I said, the way you deal with this is, you try to convert that into a region.
00:53:03.000 --> 00:53:08.000
And then, you try to find the probability of being in that region.
00:53:08.000 --> 00:53:16.100
I want to look at the region where Y1 + Y2 is less than or equal to u,
00:53:16.100 --> 00:53:21.000
which means I want to graph the line where Y1 + Y2 is equal to U.
00:53:21.000 --> 00:53:28.000
If a graph that out, let me graph that in red.
00:53:28.000 --> 00:53:36.100
There is the line in red, where Y1 + Y2, let me not write it right there.
00:53:36.100 --> 00:53:45.800
There is the line where Y1 + Y2 is equal to U.
00:53:45.800 --> 00:53:51.400
The intercepts on both axis there are U.
00:53:51.400 --> 00:53:57.900
There is U and there is U, let me write Y2 a little further down here.
00:53:57.900 --> 00:54:01.800
And then, I can write U right by the intercept there.
00:54:01.800 --> 00:54:08.300
We are looking for the region where Y1 + Y2 is less than or equal to U.
00:54:08.300 --> 00:54:18.000
That is all the regions south of this line, that is all that region colored in red there.
00:54:18.000 --> 00:54:26.800
I hope you are not colorblind because it helps to see the color in understanding these examples.
00:54:26.800 --> 00:54:31.700
I would like to describe that region, in terms of values of Y1 and Y2.
00:54:31.700 --> 00:54:34.600
I think I will describe it like this.
00:54:34.600 --> 00:54:42.000
First, I will describe Y1 is going from 0, the biggest value I see there is U.
00:54:42.000 --> 00:54:57.100
Y2, I see that I have a small mistake here, my U is not Y1 + Y2, it is Y1 + Y2/2.
00:54:57.100 --> 00:55:00.700
I need to fix that mistake.
00:55:00.700 --> 00:55:08.600
If I solve this, this is the probability of Y1 + Y2 is less than or equal to 2U, not U.
00:55:08.600 --> 00:55:13.600
Let me see if I can change that on my picture.
00:55:13.600 --> 00:55:21.700
The general shape will be the same but let me just change all my intercepts there.
00:55:21.700 --> 00:55:29.500
This is Y1 + Y2 is 2U, and each of those intercepts will be 2U.
00:55:29.500 --> 00:55:34.500
My line right here is the line Y1 + Y2 less than or equal to 2U.
00:55:34.500 --> 00:55:41.300
If you are following along, you will get a little confuse by that, you are very right to be confused.
00:55:41.300 --> 00:55:49.800
I think it is right, Y2 is going from 0 to 2U.
00:55:49.800 --> 00:55:53.000
That should have been a 2U up above.
00:55:53.000 --> 00:55:58.600
This should be 2U - Y1, that describes my region now.
00:55:58.600 --> 00:56:06.500
I think I got that describe right and I'm ready to set up a double integral over that region.
00:56:06.500 --> 00:56:24.700
This is a double integral of Y1 goes from 0 to 2U and Y2 goes from 0 to 2U - Y1.
00:56:24.700 --> 00:56:34.600
And then, the function we are integrating, that joint density function, what we figure out is E ⁻Y1 × E ⁻Y2.
00:56:34.600 --> 00:56:39.300
It looks like I got a Y2 on the inside, DY2 DY1.
00:56:39.300 --> 00:56:43.800
I got to solve that integral, it is going to be a little bit messy.
00:56:43.800 --> 00:56:48.300
But, I notice that my first integral is with respect to Y2.
00:56:48.300 --> 00:56:57.800
The inside integral is with respect to Y2, that means E ⁻Y1 is a constant, that is one mercy there.
00:56:57.800 --> 00:57:05.700
Let me pull that one out of the first integral of the inside integral E ⁻Y1, because it is just a big constant.
00:57:05.700 --> 00:57:13.100
The interval of E ⁻Y2 is - E ⁻Y2, those are being multiplied.
00:57:13.100 --> 00:57:21.900
I have to evaluate that fromy2 = 0 to Y2 = 2U - Y1.
00:57:21.900 --> 00:57:39.400
I get E ^- Y1, if I plugged in –E ⁻Y2, Y2 is 2U - Y1, -Y2 will be Y1 -2U.
00:57:39.400 --> 00:57:48.000
E ⁺Y1 -2U -, if I plug in Y2 = 0, I will get 1.
00:57:48.000 --> 00:57:52.700
But it is – a negative so it is +1 there.
00:57:52.700 --> 00:57:59.900
I think I’m going to distribute my E ⁻Y1, let me keep going over here.
00:57:59.900 --> 00:58:08.000
If I distribute E ^- Y1, E ⁻Y1 × 1 is E ⁻Y1.
00:58:08.000 --> 00:58:13.600
E ⁻Y1 × E ⁺Y1 -2U, remember you add the exponents.
00:58:13.600 --> 00:58:17.000
Those Y1's will cancel, that is pretty nice.
00:58:17.000 --> 00:58:23.100
I will get E ⁻Y1 –E ⁻2U.
00:58:23.100 --> 00:58:34.100
Of course ,that was all the inside integral and I still need to integrate that with respect to Y1 DY1.
00:58:34.100 --> 00:58:42.900
The integral of E -Y1 is –E ⁻Y1.
00:58:42.900 --> 00:58:50.100
The integral of E ⁻U, U is a constant because our variables right now are Y1 and Y2.
00:58:50.100 --> 00:59:07.200
E ⁻2U, the integral of that is just –E ⁻2U × Y1, that is because E ⁻2U is a constant.
00:59:07.200 --> 00:59:21.600
If I integrate that or evaluate that from Y1 = 0 to Y1 = 2U.
00:59:21.600 --> 00:59:26.600
Let me plug those limits in.
00:59:26.600 --> 00:59:39.600
The first one, I get –E ⁻2U -, Y1 = 2U so -2U E ⁻2U.
00:59:39.600 --> 00:59:48.600
Y1 = 0, I have - or +, Y1 = 0 gives me just E⁰ which is 1.
00:59:48.600 --> 00:59:54.700
My last term is just 0 because of the Y1 term.
00:59:54.700 --> 01:00:04.200
I think I have finally figured out my distribution function, my F of U, let me rearrange the terms a little bit,
01:00:04.200 --> 01:00:06.200
get the positive one at the front.
01:00:06.200 --> 01:00:17.200
1 – E ⁻2U – 2U, that is my distribution function.
01:00:17.200 --> 01:00:21.000
The goal here was to find the density function, but fortunately, I have done the hard work.
01:00:21.000 --> 01:00:30.200
The density function, you just get by finding the derivative of the distribution function, F prime of U.
01:00:30.200 --> 01:00:46.500
The derivative of 1 is 0, the derivative of E ⁻2U is -2E ⁻2U but I got one -, it is +2 E ⁺2U.
01:00:46.500 --> 01:00:48.300
I got to use the product rule.
01:00:48.300 --> 01:01:08.100
-2 × U × the derivative of E ⁻2U, -2U E ⁻2U + E ⁻2U × the derivative of U + E ⁻2U.
01:01:08.100 --> 01:01:11.000
Surely, some of this simplifies.
01:01:11.000 --> 01:01:25.800
I see that I have got 4U E ⁻2U and I got 2E ⁺2U – 2E ⁻2U.
01:01:25.800 --> 01:01:34.600
Those cancel and that is my whole density function is just for 4U E ⁻2U.
01:01:34.600 --> 01:01:37.500
Let me of put a range on U.
01:01:37.500 --> 01:01:45.200
Y1 and Y2 are both going from 0 to infinity, which means their average will also go from 0 to infinity.
01:01:45.200 --> 01:01:51.700
I have found a density function, just 4U E ⁻2U.
01:01:51.700 --> 01:02:00.300
Let me box that up and present that as my solution.
01:02:00.300 --> 01:02:05.700
Let me go back and go over the steps there, lots and lots of steps of this problem.
01:02:05.700 --> 01:02:08.800
We start out with independent exponential variables.
01:02:08.800 --> 01:02:10.100
What is an exponential variable?
01:02:10.100 --> 01:02:17.300
Fortunately, there is a whole lecture on exponential variables earlier on in this series here,
01:02:17.300 --> 01:02:19.600
on the probability lectures on www.educator.com.
01:02:19.600 --> 01:02:25.400
Go back and check it out, the earlier lecture on exponential variables.
01:02:25.400 --> 01:02:36.000
You will see the definition of the density function was F of Y is 1/β E ⁻Y/β.
01:02:36.000 --> 01:02:38.000
The mean there is β.
01:02:38.000 --> 01:02:42.000
In this case, we have mean 1,that is why I plug in β = 1.
01:02:42.000 --> 01:02:48.100
I just got for my density function for Y1 is E ⁻Y1.
01:02:48.100 --> 01:02:52.600
Similarly, my density function for Y2 is E ⁻Y2.
01:02:52.600 --> 01:03:00.400
Now, independence means that the joint density function, you get just by multiplying the marginal density functions.
01:03:00.400 --> 01:03:03.800
I just multiply those together.
01:03:03.800 --> 01:03:11.200
And now, I want to find the density function for this U.
01:03:11.200 --> 01:03:19.300
It says that U is their average, the average of Y1 and Y2 is Y1 + Y2/2.
01:03:19.300 --> 01:03:25.100
To find the density function, I'm going to first find the distribution function.
01:03:25.100 --> 01:03:31.400
That means, the F is a probability that U is less than some cutoff value.
01:03:31.400 --> 01:03:38.000
And then, I plugged in the definition of U is Y1 + Y2/2.
01:03:38.000 --> 01:03:44.700
Then, I simplified that into, I just move the 2 over, Y1 + Y2 was less than 2U.
01:03:44.700 --> 01:03:48.300
I want to figure out, what region that is describing.
01:03:48.300 --> 01:03:51.900
I graphed the line Y1 + Y2 = 2U.
01:03:51.900 --> 01:03:56.200
There it is that red right there, that red line.
01:03:56.200 --> 01:04:02.500
I found the region less than that and I tried to describe it in terms of Y1 and Y2.
01:04:02.500 --> 01:04:10.400
First, Y1 goes from 0 to 2U and then, Y2 goes from 0 up to that diagonal line.
01:04:10.400 --> 01:04:17.800
If you write that in terms of Y2, you solve that and you get Y2 is equal to 2U - Y1.
01:04:17.800 --> 01:04:21.000
That is where those limits came from.
01:04:21.000 --> 01:04:27.000
And then, those limits turned into the limits that I use for the double integral.
01:04:27.000 --> 01:04:35.800
Those got used right there, and then I used my joint density function, that is what I'm integrating.
01:04:35.800 --> 01:04:39.900
Now, it turns into a somewhat messy calculus 3 problem.
01:04:39.900 --> 01:04:47.100
You do not like doing all the calculus 3 and you want to throw this into an integration program,
01:04:47.100 --> 01:04:48.400
that is totally fine with me.
01:04:48.400 --> 01:04:54.700
If you want to throw it into an integration program, you should end up getting the same thing I do for the distribution function.
01:04:54.700 --> 01:05:00.500
I integrated with respect to Y2 first, E ⁻Y1 is a constant at this point.
01:05:00.500 --> 01:05:10.600
Plug in my values for Y2, that is a little hard to read, let me see if I can write that a little more clearly because that was U.
01:05:10.600 --> 01:05:13.500
There is a U, a lot much better.
01:05:13.500 --> 01:05:18.900
Plug in the values for Y2 and simplified it down a bit.
01:05:18.900 --> 01:05:24.600
I distributed this E ⁻Y1 which is how I got this function right here.
01:05:24.600 --> 01:05:27.300
I still need to do the integral with respect to Y1.
01:05:27.300 --> 01:05:30.700
I did that integral, remember that U is a constant.
01:05:30.700 --> 01:05:35.400
The integral of E ⁻2U is just E ⁻2U × Y1.
01:05:35.400 --> 01:05:43.500
And then, I plugged in my values for Y1 and simplified it down and I got my distribution function.
01:05:43.500 --> 01:05:50.700
That is still not the density function, you get the density function by taking the derivative of the distribution function.
01:05:50.700 --> 01:05:56.100
I took that derivative and used a little product rule here, which made it get a little messy.
01:05:56.100 --> 01:06:04.000
But then, it turned out that some terms canceled and I got a fairly simple density function for U × E ⁻2U.
01:06:04.000 --> 01:06:08.700
Of course, my range on U, it is all the possible values of U.
01:06:08.700 --> 01:06:20.900
Since, U is Y1 + Y2, Y1 + Y2/2 that can be as small as 0 and unboundedly large.
01:06:20.900 --> 01:06:26.300
We have to say the range of U is from 0 to infinity.
01:06:26.300 --> 01:06:29.200
That wraps up this lecture on distribution functions.
01:06:29.200 --> 01:06:33.900
Remember that, distribution function is the first of three methods
01:06:33.900 --> 01:06:41.600
that we are going to use to find the distribution and densities of functions of random variables.
01:06:41.600 --> 01:06:45.600
Distribution functions is the first one, that is what we have just been talking about.
01:06:45.600 --> 01:06:48.500
The next one is the method of transformations.
01:06:48.500 --> 01:06:54.400
I hope you will stick around for the next video that will cover the method of transformations.
01:06:54.400 --> 01:06:59.800
The third method is moment generating functions.
01:06:59.800 --> 01:07:04.300
I got another video coming up after that, about moment generating functions.
01:07:04.300 --> 01:07:07.200
There are all sort of different techniques to solve the same problem
01:07:07.200 --> 01:07:15.100
but some of them work better in different circumstances, which is why we learn all three.
01:07:15.100 --> 01:07:23.400
This is part of the larger chapter on finding distributions of functions of random variables.
01:07:23.400 --> 01:07:29.500
That, in turn, is part of the whole probability lecture series here on www.educator.com.
01:07:29.500 --> 01:07:33.100
I have been working with you today, my name is Will Murray.
01:07:33.100 --> 01:07:35.000
I hope you will stick around, thank you very much, bye.