WEBVTT mathematics/probability/murray
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Hello, welcome back to the probability lectures here on www.educator.com, my name is Will Murray.
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We are starting a chapter on probability distribution functions with two variables.
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From now on, we are going to have a Y1 and Y2.
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Today, we are going to talk about Bivariate density and Bivariate distribution functions.
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That is a lot to swallow, let us jump right into it.
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Bivariate density functions, the idea now is that we have two variables, Y1 and Y2.
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For example, you might be a student taking a certain number of units at college.
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Y1 is the number of math units you have taken and Y2 is the number of computer science units that a student has taken.
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All the different students at this university, each one has taken a certain number of math units
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and a certain number of computer science units.
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There is a density function which reflects the number of students who had taken
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any particular number of computer science units or math units.
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For example, we can say how many students have taken 10 or more math units, and 15 or more computer science units?
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There is a certain proportion of the population that has taken more than 10 math units
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and more than 15 computer science units.
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These are things that we will graphs on 2 axis.
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From now, all our graphs are going to be on 2 axis.
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We will always put Y1 on the horizontal axis and Y2 on the vertical axis.
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And then, the density will be distributed all over this plane of 2 axis.
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Lets us see what we do with these density functions.
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First of all, the density function always has to be positive.
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You cannot have a negative number of students who have taken a certain number of units.
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The smallest you can have would be, if I picked a particular combination of units,
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there might be 0 students that have had that combination of units.
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But, there would never be a negative number of students that have taken a certain number of units.
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Second, if we look at the total density, that means the density over the entire plane.
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All the possible combinations of units the students could have taken.
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If we integrate over the entire plane, it has to come out to be 1 that is because it is a probability function.
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The total density of students has to be 1, no matter how many students we have at this college,
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everybody is factored in there somewhere.
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We can calculate probabilities, when we graph this, as I explained it to you.
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If we want to find the probability of any particular region, I will graph a rectangular region
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because it makes it easy for me to describe.
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The rectangular region where Y1 goes from A to B and Y2 goes from C to D, there is Y2.
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We want to find the probability of landing within that region.
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For example, we want to find the probability, maybe proportion of students that had taken
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between 10 and 15 math units, and have taken between 20 and 30 computer science units.
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What proportion is the total student body of this college has taken between 10 and 15 math units,
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and between 20 and 30 computer science units.
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The way we do that is, we take a double integral, integral Y1 from A to B, Y2 from C to D.
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And then, we integrate the density function over that range.
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What this means is that you really have to remember calculus 3.
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If you do not remember how to do double integrals, what you want to do is
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you want to review your calculus 3, multivariable calculus.
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We do have a whole set of lectures devoted to multivariable calculus here on www.educator.com.
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My student colleague Raffi, teaches those lectures, he is amazing.
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If you cannot remember how to do a double integral, go after his lectures and
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you will be good to go for the rest of this chapter in probability.
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I have been talking about continuous distributions here.
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If you have a discreet distribution, it is basically the same idea.
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Instead of an F here, you will just change that to a P.
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Instead of integrals, you will have summation signs.
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In fact, you will have a double summation, instead of a double integral, if you have a discreet distribution.
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It is not as common though, usually Bivariate functions in probability classes,
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it just turns out that you usually study continuous ones.
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You usually end up studying double integrals.
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That is why you really have to know your multivariable calculus.
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If you are a little rusty on that, you want to brush up on your multivariable calculus.
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Another thing that we need to learn is the Bivariate distribution function,
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it is kind of the two variable analogue of the distribution functions we had before.
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The idea of the Bivariate distribution function is you have some cutoff values of Y1 and Y2.
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Here is Y1 and here is Y2, we have some cut off values.
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There is Y1 and there is Y2.
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What we are interested in, is the probability of being less than both of those cutoff values.
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You are interested in calculating the probability that Y1 is less than or equal to the cutoff value y1.
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Y2 is less than the cut off value y2.
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In other words, you want to find all the stuff in this region, right here.
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All the stuff where Y1 is less than y1 and Y2 is less than y2.
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If you think about that, that is just the double integral over that region.
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We are going to call that function F of Y1 Y2.
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It means you integrate from negative infinity to y1 and negative infinity to y2.
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I cannot call it Y1 and Y2 anymore for the variables because I'm using them for the cutoff values.
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I’m going to use T1 and T2, and then I’m going to integrate the density function.
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Just like in some sense, the distribution function was the integral of the density function.
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Back in single variable probability and Bivariate probability, the distribution function is
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the double integral of the density function, from negative infinity up to the cutoff values that you are interested in.
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Some properties of the distribution function satisfies, if E¹ Y1 Y2 is negative infinity,
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it means you are not looking in any area at all.
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Your value is going to be 0, no matter what the other variable is.
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If you plug in infinity for both of them, that means you are really looking at this entire plane.
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You are looking at all the possible density, it would have to be 1.
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That is a property that that has to satisfy.
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I think that is all the preliminaries now, we are ready to jump into the examples.
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We are going to be doing a lot of integrals for these examples.
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You really want to be ready to do some double integrals, even integrating over some non rectangular regions.
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We will work them out together.
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In the first example, we are going to consider the joint density function F of Y1 Y2 is defined to be K × Y2.
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Your K means it is a constant.
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On the triangle with coordinates at 0-0, 0-1, and 1-1.
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What we want to do with this is, we are going to find the value of K.
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We are going to keep using this same formula for examples 2 and 3.
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You really want to make sure you are up to speed on this.
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Let me first graph this region because we are going to be seeing this over and over again.
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We got the triangle with coordinates at 0-0, and 0-1, and 1-1.
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I will graph that triangle and that is the region that we are interested in.
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I will color that in.
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There is a Y1 is equal to 1, Y2 is equal 1, there is 0.
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It looks like it is defined by the line Y = X, but since we are using Y1 and Y2 as our variables here, there is Y1 Y2,
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that is the line Y1 = Y2, like the line Y = X there.
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We want to find the value of K, the way we want to do that is, we want to remember that the total density has to be 1.
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We want to integrate over that region and our answer will have a K in it somehow.
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I will set the whole thing equal to be 1.
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First, I like to describe that region.
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The best way to describe that region, I think, is to describe it with Y2 first.
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I’m going to describe it, listing Y2 first, using constants for Y2.
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Y2 goes from 0 to 1 and Y1 goes from 0 to Y2.
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This is a hardcore multivariable calculus, if you do not remember how to set up these integrals
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on these triangular regions, you really got to review it right now.
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Go back and watch these lectures on multivariable calculus and you will get some practice with these triangular regions.
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Let me set up the integral on this region.
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The integral from Y2 = 0 to Y2 = 1.
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The integral from Y1 = 0 to Y1 = Y2.
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My density function K Y2, let me pull my K outside, it is just a constant, Y2.
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D Y1, I got to do first, and then D Y2.
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I'm going to integrate the inside one first, the integral with respect to Y1 is just Y2 Y1.
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Remember, you keep the Y2 constant when you are integrating with respect to Y1.
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We want to integrate that from Y1 = 0 to Y1 = Y2.
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If I plug those in, I will get Y2² -0.
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I have to do the integral there, from Y2 = 0 to Y2 = 1.
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I still have a K on the outside, I still have a D Y2.
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The integral of Y2² is just Y2³/3.
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I still have that K and I’m evaluating that from Y2 = 0 to Y2 = 1.
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Plug those in and I get K ×, it looks like 1/3.
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Remember that, the total density has to be equal to 1.
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K × 1/3 is equal to 1, that tells me then that K is equal to 3.
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I have just solved this problem.
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Let me show you again all the steps there.
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First thing was definitely to graph the region.
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That is kind of an excellent rule for any kind of multivariable calculus type problem.
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You always want to graph the region, it is very helpful to graph the region.
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I graphed the region, I graphed that triangle with those 3 points.
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I wrote down a little equation for the line of the boundary which is just Y = X or Y2 = Y1.
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To find the value of K, what I want to do is to use the fact that the total density has to be 1.
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The integral over this region of this density function, it has to come out to be 1.
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I describe that region, I chose Y2 to list first because that makes it a little bit simpler to set up the integral.
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You could reverse the order of the variables there, listing Y1 first,
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but I think you are going to get a slightly nastier integral.
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You would not get as many 0 in the bounds there.
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That is why I picked Y2 first to describe that region.
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And then, I set up my double integral, I integrated Y2.
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First, I integrated Y2 with respect to Y1.
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By the way, I use this as an example in my classes.
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A lot of students say the integral of Y2 should be Y2²/2.
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Not so, because we are not integrating with respect to Y2.
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We are integrating with respect to Y1 which means the integral is just Y2 Y1.
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And then, we evaluate for our boundaries on Y1 and we get Y2².
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Now, we integrate that with respect to Y2 that is a fairly easy integral.
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We come up with K × 1/3.
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Since, that is supposed to be equal 1, that is kind of our rule.
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Our rule is always that the double integral of DY1 DY2 is always equal to 1.
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That tells me that the K has to be 3.
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I hope this one made sense because we are going to keep using this example for problems 2 and 3.
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Make sure you understand this one, we are going to take the answer of this one
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and use it to answer some more complicated questions in example 2 and 3.
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In example 2, we are going to keep going with the same setup from example 1,
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except I will fill in the answer, the K was equal to 3.
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If you are a little foggy on what was going on in example 1, go back and watch example 1, it will make more sense.
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Let me go ahead and draw the region that we are interested in.
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It is the same one as before 0-0, 0-1, and 1-1.
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There is the region there, triangular region 0- 0, 0-1, and 1-1.
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There is 1, there is 1, this is Y2.
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That is Y1 there, my goodness what am I thinking.
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There is Y1, there is Y2.
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I want to find F, that is the distribution function of 1/3, ½.
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The first thing to do is to remember what that distribution notation means.
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F of 1/3, ½, the fractions are going to get nasty in this one.
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I’m just going to warn you in advance.
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That is the probability that Y1 is less than or equal to 1/3 and Y2 is less than or equal to ½.
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That was the definition of F.
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If you do not remember that definition of F, just click back a couple slides ago and
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you will see the definition of Bivariate distribution function.
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Remember, keep track of the difference between F, the distribution function, and f the density function.
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They do not mean the same thing.
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We want to find, the probability of Y1 being less than 1/3 and Y2 being less than ½.
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Let me graph that.
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There is 2/3, 1/3, and there is ½.
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I'm going to draw my region here.
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I got this region that sort of shape like a backwards state of Nevada.
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There it is right there, there is my backwards state of Nevada.
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What I want to do is to integrate over that region.
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To integrate over that region, I need to describe that region.
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It looks like, if I want to do it in one piece, I have to describe my Y1 first.
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That goes to 1/3, that goes to ½, and of course, these are both going to start at 0.
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Y1 is going to go, if I have constants for that, it is going to go from 0 to 1/3, that is Y1 1/3.
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Y2 is not going to go from 0 to ½, otherwise, I would have a rectangle, and Nevada is not a rectangle.
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It is going to go from, that line right there was the line Y2 is equal to Y1.
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Y2 goes from Y1 on up to Y2 goes from Y1.
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Maybe I should make that in black to make it a little more visible there.
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Y2 goes from Y1 on up to ½.
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Now, I have some boundaries, I can set up my integral.
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This probability, I'm going to integrate Y1 goes from 0 to 1/3.
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Let me go ahead and write the variables in there.
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Y1 is equal to 1/3 and Y2 goes from Y1 up to ½ there.
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I have a density function, there it is 3Y2.
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DY2 and DY1, I have a double integral to solve.
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Probably, the hardest part is setting up the double integral.
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Usually, solving the double integral is not bad.
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If you are lucky and your teacher is a nice person, and you can even use a calculator
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or a software to solve these double integrals.
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I'm going to go ahead and work it out by hand, just to prove that I'm an honest upstanding human being.
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We have to integrate 3Y2, DY2 the integral of Y2 is Y2²/2 Y2².
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We will pull out the 3/2.
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Y2², we are integrating that from Y2 = Y1 to Y2 = ½.
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I will keep the 3/2 here and Y2² from ½ to Y1 will give me, ½² is ¼ - Y1²,
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We are supposed to integrate that from Y1 = 0 to Y1 = 1/3.
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This is now DY1.
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I have to do calculus 1 integral, I get 3/2, ¼ Y1 - Y1² integrates to 1/3 Y1³.
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That is going to be a bit nasty to deal with.
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All of these evaluated from Y1 = 0 to Y1 = 1/3, I get 3/2.
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¼ × Y1 = 1/3 is ¼ × 1/3 is 1/12 -, 1/3 × Y1³.
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1/3³ is 1/27.
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Another 1/3 multiplied by that gives me 1/81, all the horrors here.
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I'm not going to write anything for Y1 = 0 because both of those terms will drop out.
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That is a small mercy there.
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These fractions simplify a bit because 3/2 × 1/12 is 3/24 is 1/8.
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3/2 × 1/81, the 3 will cancel with the 81 give me a 27 × 2 is 54.
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Not too bad, I think I'm going to have a common denominator there of 216.
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216, because 216 is 8 × 27, it is 54 × 4, that simplifies down to 23/216.
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I did plug that into a calculator, in case you are fond of decimals, 0.1064.
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If you are fond of percentages that is 10.64%.
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That is my probability, the probability that you will end up in that small Nevada State region.
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Or another way to think about that is what we just calculated is F of 1/3 and ½.
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That is what we calculated right here.
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Let me recap how I did that.
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First of all, use the definition of the Bivariate distribution function.
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It just means the probability that Y1 is less than 1/3 and Y2 is less than ½.
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Then, I went to try to draw that region on my full graph.
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I converted that description into a drawing there.
00:22:51.000 --> 00:22:55.900
In turn, I took that drawing which gave me a sort of Nevada shaped region.
00:22:55.900 --> 00:23:01.500
This rectangle with its lower coordinate cutoff.
00:23:01.500 --> 00:23:07.400
Then, I converted that, it is actually a backwards Nevada is not it.
00:23:07.400 --> 00:23:09.800
But I have been calling it a Nevada shaped region.
00:23:09.800 --> 00:23:13.200
If you look at Nevada in a mirror, this is what it looks like.
00:23:13.200 --> 00:23:17.000
Then, I tried to describe that in terms of variables.
00:23:17.000 --> 00:23:21.600
Prepare a tree to set it up and a double integral.
00:23:21.600 --> 00:23:29.600
I found this Y1 goes from 0 to 1/3, Y 2 goes from y1 to ½.
00:23:29.600 --> 00:23:38.100
Notice, I would like to say Y2 goes from 0 to ½ but that would be wrong because
00:23:38.100 --> 00:23:43.100
that would give me a rectangular region, that is not what I want.
00:23:43.100 --> 00:23:47.800
I have to say, Y2 goes from Y1 to ½.
00:23:47.800 --> 00:23:53.900
I took those limits and I set up a double integral here.
00:23:53.900 --> 00:24:02.500
The 3 Y2 comes from the density function up here, and then it is just a matter of cranking through the double integral.
00:24:02.500 --> 00:24:06.700
Not very hard, a little bit tedious, easy to make mistakes.
00:24:06.700 --> 00:24:10.100
But, you first integrate with respect to Y2.
00:24:10.100 --> 00:24:16.800
I factored out some constants, plug in your bounds which gives you everything in terms of Y1.
00:24:16.800 --> 00:24:25.200
Do another integral, get some nasty fractions, simplify them to a slightly less nasty fraction.
00:24:25.200 --> 00:24:27.600
If you like, you can leave your answer as a fraction.
00:24:27.600 --> 00:24:35.500
I converted it into a decimal and a percentage.
00:24:35.500 --> 00:24:43.100
In example 3, we are going to keep going with the same region and the same density function from example 1.
00:24:43.100 --> 00:24:49.200
Let me go ahead and draw that.
00:24:49.200 --> 00:24:59.100
We have got the triangular region from 0-0 to 1-1, and 0-1.
00:24:59.100 --> 00:25:06.500
There is 1, there is the Y2 axis, and there is the Y1 axis.
00:25:06.500 --> 00:25:13.300
We got this triangular region and we got a density function defined on that region.
00:25:13.300 --> 00:25:25.400
We want to find the probability that 2Y1 is less than Y2 or Y2 is bigger than 2Y1.
00:25:25.400 --> 00:25:30.400
I'm going to go ahead and try to draw the region that we are interested in.
00:25:30.400 --> 00:25:33.700
I’m going to graph the region.
00:25:33.700 --> 00:25:42.300
If I say 2Y1 is equal to Y2, that is like saying Y2 is equal to 2Y1.
00:25:42.300 --> 00:25:49.900
To make that more familiar to people who graph things like this in algebra, it is like saying Y =2X.
00:25:49.900 --> 00:25:55.500
I’m going to graph line one = 2X.
00:25:55.500 --> 00:26:04.300
Let me put Y = 2X, that is going to be twice as steep.
00:26:04.300 --> 00:26:06.300
There it is right there.
00:26:06.300 --> 00:26:09.900
There is the line Y = 2X.
00:26:09.900 --> 00:26:17.100
We actually want to have Y2 greater than 2Y1, that greater than 2X.
00:26:17.100 --> 00:26:23.100
That means, we want the region of both that line.
00:26:23.100 --> 00:26:26.500
We want that blue region right there.
00:26:26.500 --> 00:26:31.700
We are going to find the probability of landing in that blue region.
00:26:31.700 --> 00:26:34.100
I have to describe that blue region.
00:26:34.100 --> 00:26:42.900
I think the best way to describe that blue region is to describe Y2 first.
00:26:42.900 --> 00:26:50.700
Y2, I can see it is going from 0 to 1.
00:26:50.700 --> 00:26:53.000
Let me show you how Y1 behaves.
00:26:53.000 --> 00:27:09.700
Y1 is going from 0 up to that line, that line was Y1 is equal to Y2/2.
00:27:09.700 --> 00:27:13.400
I will write that as ½ Y2.
00:27:13.400 --> 00:27:19.100
That line is Y1 is equal to ½ Y2.
00:27:19.100 --> 00:27:24.200
I have to make that my upper bound for Y1, ½ Y2.
00:27:24.200 --> 00:27:26.300
That is my description of the region.
00:27:26.300 --> 00:27:34.600
The reason I spent much time describing it that way is that, that sets me up for a nice double integral.
00:27:34.600 --> 00:27:41.100
My probability is equal to the double integral on that region.
00:27:41.100 --> 00:28:05.200
I can just use that description Y2 goes from 0 to 1 and Y1 goes from 0 to ½ Y2.
00:28:05.200 --> 00:28:09.000
I have my density function 3Y2.
00:28:09.000 --> 00:28:16.000
Once again, it is a multivariable calculus problem DY1 DY2.
00:28:16.000 --> 00:28:24.100
I'm just going to work that out as a multivariable calculus problem and integrating with respect to Y1 first.
00:28:24.100 --> 00:28:32.200
I will put a 3 on the outside, integral of Y2 with respect to Y1 is Y2 × Y1.
00:28:32.200 --> 00:28:37.300
It is not Y2²/2 because we are not integrating with respect to Y2.
00:28:37.300 --> 00:28:42.700
Be careful about that, that is a very common mistake that my own students make all the time.
00:28:42.700 --> 00:28:46.600
Even I, make that mistake, if I’m not being careful.
00:28:46.600 --> 00:28:54.500
Let me integrate that from Y1 = 0 to Y1 = ½ Y2.
00:28:54.500 --> 00:28:58.100
What I get there is, there is still a 3 on the outside.
00:28:58.100 --> 00:29:02.500
I’m just doing this first integral, not worrying about the second one yet.
00:29:02.500 --> 00:29:10.800
I get ½ Y2², when I plug in my Y1 = Y2.
00:29:10.800 --> 00:29:16.200
Y2², I will put the ½ on the outside.
00:29:16.200 --> 00:29:32.100
I have got the integral from Y2 = 0 to Y 2 = 1 of Y2² DY2, factoring the outside terms there.
00:29:32.100 --> 00:29:38.000
The integral of Y2² is Y2³/3.
00:29:38.000 --> 00:29:40.700
Personally, I have this 3 on the outside.
00:29:40.700 --> 00:29:45.300
I will just write that as ½ Y2³.
00:29:45.300 --> 00:29:52.100
Then, I will evaluate that from Y2 = 0 to Y2 = 1.
00:29:52.100 --> 00:29:59.000
I get ½ × 1 -0 which is just ½.
00:29:59.000 --> 00:30:04.000
That is nice and pleasant, much simpler answer than we have for the previous example.
00:30:04.000 --> 00:30:07.200
Let me walk you through that again.
00:30:07.200 --> 00:30:12.800
The key starting point here is we have that same region, that triangular region,
00:30:12.800 --> 00:30:18.000
with those coordinates of the triangle, just as before.
00:30:18.000 --> 00:30:23.100
We want to find the probability that 2Y1 is less than Y2.
00:30:23.100 --> 00:30:27.000
I wanted to graph that region, to see what part of the region that was.
00:30:27.000 --> 00:30:30.100
I graphed 2Y1 is equal to Y2.
00:30:30.100 --> 00:30:40.300
I got this line here, that is the line 2Y1 is equal to Y2, or you can write that as Y1 is ½ Y2.
00:30:40.300 --> 00:30:45.900
I want the region above the line because I want Y2 to be bigger than 2Y1.
00:30:45.900 --> 00:30:48.800
That is why I took the region above the line not below it.
00:30:48.800 --> 00:30:52.100
That is why I got this blue region colored in right here.
00:30:52.100 --> 00:31:00.000
I want to describe it and that would be more convenient to list Y2 first, so I can use constants for Y2.
00:31:00.000 --> 00:31:07.000
And then, I want my bounds for Y1 would be 0 and the other bound is ½ Y2, I got that from the line.
00:31:07.000 --> 00:31:14.900
That ½ Y2, that comes from right here, that is where that comes from.
00:31:14.900 --> 00:31:20.900
I took these bounds and I set them up as the limits on my integral.
00:31:20.900 --> 00:31:24.100
The function I’m integrating is the density function.
00:31:24.100 --> 00:31:28.800
That comes from the stem of the problem, the 3Y2.
00:31:28.800 --> 00:31:31.000
Now, it is just a matter of working through a double integral.
00:31:31.000 --> 00:31:34.900
But be careful always which variable you are integrating.
00:31:34.900 --> 00:31:41.400
The first variable I’m integrating is DY1, that is why the integral is Y2 × Y1.
00:31:41.400 --> 00:31:43.600
I'm holding Y2 constant there.
00:31:43.600 --> 00:31:47.400
It is not Y2²/2.
00:31:47.400 --> 00:31:52.600
Run that through the limits, get Y2², integrate that with respect to Y2.
00:31:52.600 --> 00:32:06.600
Now, it just simplifies down into the very friendly fraction of ½.
00:32:06.600 --> 00:32:09.900
In example 4, we have a new joint density function here.
00:32:09.900 --> 00:32:16.000
F of Y1 Y2 is defined to be E ⁻Y1 + Y2.
00:32:16.000 --> 00:32:21.200
The region we are looking at is Y1 bigger than 0, Y2 bigger than 0.
00:32:21.200 --> 00:32:27.700
Note that, there is no upper bounds given on that, that means Y1 and Y2 can go all the way to infinity.
00:32:27.700 --> 00:32:32.000
Let me graph that region.
00:32:32.000 --> 00:32:43.200
There is Y1 and there is Y2, we want to find the probability that Y1 is less than 2 and Y2 is bigger than 3.
00:32:43.200 --> 00:32:48.400
Y1 should be less than 2 here.
00:32:48.400 --> 00:32:53.800
Y2 should be bigger than 3.
00:32:53.800 --> 00:32:58.500
We want to find, let me go ahead and draw those lines there.
00:32:58.500 --> 00:33:04.400
Y1 should be less than 2, we want to go to the left of that vertical line.
00:33:04.400 --> 00:33:09.100
Y2 should be bigger than 3, I want to go above that horizontal line.
00:33:09.100 --> 00:33:14.500
Somewhat this region right here, that region right there.
00:33:14.500 --> 00:33:20.600
And that is the region that we are going to integrate over, in order to find this probability.
00:33:20.600 --> 00:33:25.300
I'm going to set up a double integral on that region.
00:33:25.300 --> 00:33:29.000
The integral, I think I can list Y1 first safely.
00:33:29.000 --> 00:33:39.300
Y1 goes from 0 to 2 because 2 is the upper bound for Y1.
00:33:39.300 --> 00:33:43.600
Y2, that is where I bring in the 3, Y2 is 3.
00:33:43.600 --> 00:33:49.000
I have to run that to infinity because that one just goes on forever.
00:33:49.000 --> 00:33:52.400
Maybe, you are uncomfortable saying Y2 is equal to infinity.
00:33:52.400 --> 00:33:58.900
Maybe, I will say Y2 approaches infinity but it does not really affect the calculations that we will be doing.
00:33:58.900 --> 00:34:03.600
It will be fairly easy to plug in infinity, after we do the calculations.
00:34:03.600 --> 00:34:11.900
The density function is E ⁻Y1 + Y2, that is a quantity there.
00:34:11.900 --> 00:34:17.000
I have DY2 first and then DY1.
00:34:17.000 --> 00:34:25.800
I think the best way to approach this is to factor the density function into E ⁻Y1 × E ⁻Y2.
00:34:25.800 --> 00:34:33.700
The point of doing that, is that in the first integral, we are integrating with respect to Y2.
00:34:33.700 --> 00:34:38.000
We can take the E ⁻Y1, that is just a big old constant now.
00:34:38.000 --> 00:34:41.300
We can pull it all the way out of the integral, let me go ahead and do that.
00:34:41.300 --> 00:34:57.800
We have the integral of E ⁻Y1, and now the integral from Y2 = 3 to Y2 goes to infinity of E ⁻Y2 DY2.
00:34:57.800 --> 00:35:03.400
There is a DY1 on the outside but let me just handle that first integral inside.
00:35:03.400 --> 00:35:08.100
The integral of E ⁻Y2 is just –E ⁻Y2.
00:35:08.100 --> 00:35:12.600
That is a little substitution there, a little old calculus 1 trick.
00:35:12.600 --> 00:35:18.900
I'm evaluating that from Y2 = 3 to Y2 goes to infinity.
00:35:18.900 --> 00:35:20.900
Technically, I should be writing limits in here.
00:35:20.900 --> 00:35:24.800
I should be introducing a T and take the limit as T goes to infinity.
00:35:24.800 --> 00:35:32.400
I’m being a little sloppy about that, that is kind of the privilege of having been through so many calculus classes.
00:35:32.400 --> 00:35:39.200
When Y2 goes to infinity, we get E ⁻infinity here, that is 1/E ⁺infinity.
00:35:39.200 --> 00:35:51.500
That is just 0- E⁻³, that all simplifies down into E⁻³.
00:35:51.500 --> 00:35:55.800
I still have that E ⁻Y1, I’m going to bring that back in here.
00:35:55.800 --> 00:36:00.800
I’m going to integrate that DY1.
00:36:00.800 --> 00:36:04.500
E⁻³ is just a constant, I will write that separately.
00:36:04.500 --> 00:36:10.000
E⁻³ × -E ⁻Y1.
00:36:10.000 --> 00:36:11.700
What are my limits on Y1?
00:36:11.700 --> 00:36:19.200
My Y1 went from 0 to 2, Y1 = 0 to Y1 = 2.
00:36:19.200 --> 00:36:23.000
I'm going to need some more space on this.
00:36:23.000 --> 00:36:30.400
This is E⁻³, -E⁻², plugging in 2.
00:36:30.400 --> 00:36:41.200
Plugging in Y1 = 0, I get E⁰ which is just 1, it is - -1.
00:36:41.200 --> 00:36:46.200
If I simplify this, I get E⁻³, that one becomes positive.
00:36:46.200 --> 00:36:50.600
It is 1 – E⁻², and I could factor that through.
00:36:50.600 --> 00:37:03.200
I get E⁻³ – E⁻³ × E⁻², you add the exponents, it is E⁻⁵.
00:37:03.200 --> 00:37:07.900
That is really my exact answer but it is not very illuminating.
00:37:07.900 --> 00:37:18.900
I did find the decimal for that, and my calculator told me that that is approximately equal to 0.043.
00:37:18.900 --> 00:37:31.100
if you want to convert that into a percentage, then it is 4.3%.
00:37:31.100 --> 00:37:35.100
We have an answer for that one. Let me show you again the steps involved to finding it.
00:37:35.100 --> 00:37:40.100
First of all, I graphed the whole region Y1 greater than 0 and Y2 greater than 0.
00:37:40.100 --> 00:37:51.300
That is a whole planar region, quarter plane because it is just where Y1 and Y2 are positive, and that is my whole region.
00:37:51.300 --> 00:37:59.000
What I'm really interested in, is the probability of Y1 being less than 2 and Y2 being bigger than 3.
00:37:59.000 --> 00:38:03.900
I chopped that up and I found that region was this blue region colored in here.
00:38:03.900 --> 00:38:09.100
That is where Y1 is less than 2 and Y2 is bigger than 3.
00:38:09.100 --> 00:38:13.300
In order to integrate that, I had to describe that limits.
00:38:13.300 --> 00:38:18.700
Y1 goes from 0 to 2, Y2 goes from 3 to infinity.
00:38:18.700 --> 00:38:24.100
I plugged in my density function, there is my density function right there, and I plug it in here.
00:38:24.100 --> 00:38:25.600
I had to do that double integral.
00:38:25.600 --> 00:38:33.100
The nice thing about that is I can factor that density function into E ⁻Y1 and E ⁻Y2.
00:38:33.100 --> 00:38:39.400
Since, my first integral, the inside integral is going to be a Y2, I can pull out the E ⁻Y1.
00:38:39.400 --> 00:38:45.200
That is just a constant, and that got pullout as a constant, outside the first integral.
00:38:45.200 --> 00:38:50.600
I’m just left with E ⁻Y2 which integrates to –E ⁻Y2.
00:38:50.600 --> 00:38:56.300
Do a u substitution in there, u= -Y2.
00:38:56.300 --> 00:39:01.400
Evaluate that from 3 to infinity, when we plug in infinity or take the limit
00:39:01.400 --> 00:39:06.800
as the variable approaches infinity, we will get 1/E ⁺infinity.
00:39:06.800 --> 00:39:09.500
That is 1/infinity is just 0.
00:39:09.500 --> 00:39:13.300
That is what that infinity term gives you, is the 0.
00:39:13.300 --> 00:39:18.900
And then, it turn out to be a +E⁻³, that is just a constant in the next step because
00:39:18.900 --> 00:39:22.500
we are integrating with respect to Y1 now.
00:39:22.500 --> 00:39:29.500
I get -E ⁻Y1, plug in the values, do a little bit of algebra and simplifying.
00:39:29.500 --> 00:39:36.000
Get down to E⁻³ - E⁻⁵.
00:39:36.000 --> 00:39:40.300
I plugged that into a calculator, just to see what kind of decimal we are talking about.
00:39:40.300 --> 00:39:43.000
It should always come out to be positive, when we are finding these probabilities.
00:39:43.000 --> 00:39:48.800
If you do not get a positive probability, in fact, if you do not get something between 0 and 1, you know you screwed up.
00:39:48.800 --> 00:39:52.600
I like to plug things in and just get a number.
00:39:52.600 --> 00:39:56.900
In this case, I got 0.043 which is 4.3%.
00:39:56.900 --> 00:40:01.400
Yes, that is between 0 and 1, it is not too surprising.
00:40:01.400 --> 00:40:06.600
That is my probability of landing in that region with that density function.
00:40:06.600 --> 00:40:11.300
We will use the same region and density function for example 5.
00:40:11.300 --> 00:40:16.700
Make sure you understand this very well, before you move on to example 5.
00:40:16.700 --> 00:40:27.700
It is it is the same density function, we will be integrating a different corner of the region, let me put it that way.
00:40:27.700 --> 00:40:35.200
In example 5, we are going to look at the joint density function, the same one that we had in example 4.
00:40:35.200 --> 00:40:38.300
Let me go ahead and remind you of what that looked like there.
00:40:38.300 --> 00:40:43.900
We have this graph and we are looking at the entire positive quarter plane there.
00:40:43.900 --> 00:40:50.900
There is Y1 and there is Y2, and everything is going from 0 to infinity.
00:40:50.900 --> 00:40:57.000
We want to find the probability that Y1 + Y 2 will be less than or equal to 2.
00:40:57.000 --> 00:41:02.400
Let me draw the line Y1 + Y 2 = 2.
00:41:02.400 --> 00:41:07.600
There it is, it is a diagonal line and it got a slope of -1.
00:41:07.600 --> 00:41:15.000
That is the line Y1 + Y2, Y1 + Y2 is equal to 2.
00:41:15.000 --> 00:41:18.100
It is just X + Y = 2 and you can solve that out.
00:41:18.100 --> 00:41:20.100
You can do a little algebra to find that.
00:41:20.100 --> 00:41:26.000
It does have intercept 2 on each axis.
00:41:26.000 --> 00:41:35.400
I want it to be less than or equal 2, which means I need to look at the region underneath that line.
00:41:35.400 --> 00:41:44.400
I want to describe that region and then do a double integral, in order to find the probability of landing in that region.
00:41:44.400 --> 00:41:48.900
The first thing I’m going to do is try to describe that region.
00:41:48.900 --> 00:41:56.100
It does not really matter which variable you list first here, but I listed Y1 first.
00:41:56.100 --> 00:41:59.700
I can use constants for Y1, I’m going to go from 0 to 2.
00:41:59.700 --> 00:42:05.600
And then, I listed Y2 but I cannot use constants for that because otherwise, I will get a rectangle.
00:42:05.600 --> 00:42:16.900
Y2 goes from 0 to, if you solve that line out, you get Y1 + Y2 = 2.
00:42:16.900 --> 00:42:22.900
If you solve for Y2, you will get Y2 is equal to 2 - Y1.
00:42:22.900 --> 00:42:28.100
I’m going to use that as my upper bound, that is going to make for some nasty integration but there is no way around it.
00:42:28.100 --> 00:42:30.400
We just have to go through it.
00:42:30.400 --> 00:42:36.200
My probability is, it will be the double integral on that region.
00:42:36.200 --> 00:42:40.900
I already set up the limits here, I have done the hard part.
00:42:40.900 --> 00:42:43.900
The rest of it is just tedious integration.
00:42:43.900 --> 00:42:56.500
Y1 = 0 to Y1 = 2 and Y2 = 0 to Y2 = 2 -Y1.
00:42:56.500 --> 00:43:03.400
I have that same density function E ⁻Y1 + Y2.
00:43:03.400 --> 00:43:08.800
I have DY2 and then DY1.
00:43:08.800 --> 00:43:19.600
Remember, the old trick that we use back in example 4 works again, is to write that density function as E ⁻Y1 × E ⁻Y2.
00:43:19.600 --> 00:43:25.900
The utility of that is that is we are integrating Y2 first.
00:43:25.900 --> 00:43:31.800
And that means, E ⁻Y1 is a constant and I can pull it out of the integral.
00:43:31.800 --> 00:43:36.100
I'm going to pull that out of the first integral, of the inside integral.
00:43:36.100 --> 00:43:39.500
E ^- Y1 just sits there on the outside.
00:43:39.500 --> 00:43:46.200
I have the integral of E ⁻Y2 DY2.
00:43:46.200 --> 00:43:51.600
And then later on, I will do the integral with respect to Y1.
00:43:51.600 --> 00:43:58.600
E ⁻Y1, the integral of E ⁻Y2 is –E ⁻Y2.
00:43:58.600 --> 00:44:07.100
I need to evaluate this from Y2 = 0 to Y2 is equal to 2 -Y1.
00:44:07.100 --> 00:44:10.700
There is still DY1 here.
00:44:10.700 --> 00:44:16.400
This is a little bit nasty, I got –E ⁻Y2.
00:44:16.400 --> 00:44:33.300
If Y2 is 2 -Y1, -Y2 will be Y1 -2, --E⁻⁰, --1.
00:44:33.300 --> 00:44:40.900
I will get +1 - E ⁺Y1 -2 at the next step.
00:44:40.900 --> 00:44:45.800
I’m going to bring this E ⁻Y1 from over on the left.
00:44:45.800 --> 00:44:50.100
I think that is going to be useful because I’m going to go ahead and multiply that through.
00:44:50.100 --> 00:44:51.900
I think I will simplify things a bit.
00:44:51.900 --> 00:44:58.400
I get E ⁻Y1 -, E ^- Y1 + Y1 – 2.
00:44:58.400 --> 00:45:06.100
E⁻², that one is a little sideways there.
00:45:06.100 --> 00:45:08.800
That is not too bad, what am I supposed to do this.
00:45:08.800 --> 00:45:16.900
I’m supposed to integrate it with respect to Y1 DY1.
00:45:16.900 --> 00:45:22.000
Let me go ahead and keep going on the next column here.
00:45:22.000 --> 00:45:28.800
The integral of E ⁻Y1, do a little substitution is -E –Y1.
00:45:28.800 --> 00:45:36.400
E⁻² was just a constant, it is –E⁻² × Y1.
00:45:36.400 --> 00:45:42.700
This whole expression is supposed to be evaluated from, where are my limits, right there at the beginning.
00:45:42.700 --> 00:45:47.200
Y1 = 0 to Y1 = 2.
00:45:47.200 --> 00:45:57.800
I plug in Y1 = 2, I get -E⁻² - E⁻² × 2.
00:45:57.800 --> 00:46:07.400
If I plug in Y1 = 0, I get + 1 because the two negatives cancel, I’m subtracting a negative.
00:46:07.400 --> 00:46:15.100
And then, + 0 because we got Y1 = 0 in the last term there.
00:46:15.100 --> 00:46:21.900
This simplifies a bit, I got 1 – E⁻² – 2 E⁻².
00:46:21.900 --> 00:46:28.600
1 -3 E⁻², that is as good as it is going to get.
00:46:28.600 --> 00:46:35.100
I did plug that into a calculator to get a decimal approximation.
00:46:35.100 --> 00:46:40.800
My calculator told me that that was 0.594.
00:46:40.800 --> 00:46:49.800
Again, it is between 0 and 1, that is reassuring every probability answer should be between 0 and 1.
00:46:49.800 --> 00:46:57.900
What I get there, if I wanted to make that into a percent, that is 59.4%.
00:46:57.900 --> 00:47:03.900
That is my probability of landing in that sort of triangular region in the corner.
00:47:03.900 --> 00:47:10.200
A probability that Y1 + Y2 is less than or equal to 2.
00:47:10.200 --> 00:47:15.000
That is the probability that I have been asked to compute there.
00:47:15.000 --> 00:47:24.400
Let us recap that, first of all, I graphed the whole region which is the positive quarter plane here.
00:47:24.400 --> 00:47:30.600
Let me see if I can draw that without making things too messy on the graph.
00:47:30.600 --> 00:47:36.200
And that is the whole region but that is not the region we are interested in.
00:47:36.200 --> 00:47:42.100
We are interested in the region where Y1 + Y2 is less than or equal to 2.
00:47:42.100 --> 00:47:45.200
I have to graph that part of the region.
00:47:45.200 --> 00:47:51.100
That is what this diagonal line is, it is line Y1 + Y2 is equal to 2.
00:47:51.100 --> 00:47:53.600
We want all the regions less than that.
00:47:53.600 --> 00:47:57.600
That is why I colored in this blue triangular region here.
00:47:57.600 --> 00:48:01.000
I was trying to describe that region, in terms of variables.
00:48:01.000 --> 00:48:05.100
I did use constants for the first, Y1 goes from 0 to 2.
00:48:05.100 --> 00:48:08.900
But I cannot say Y2 goes from 0 to 2, otherwise, I will have a square.
00:48:08.900 --> 00:48:11.300
I do not want a square, I need a triangle.
00:48:11.300 --> 00:48:21.900
I said Y2 was less than 2 - Y1 and that came from solving out the equation of the line in terms of Y2.
00:48:21.900 --> 00:48:23.700
That is how I got that.
00:48:23.700 --> 00:48:26.600
Once I had that description, that was the hardest part of the problem.
00:48:26.600 --> 00:48:33.300
Then, I just dropped those in as my limit for the integral, I dropped my density function in.
00:48:33.300 --> 00:48:34.400
At this point, you could drop the entire thing into a calculator.
00:48:34.400 --> 00:48:41.000
If your calculator will do multivariable integrals, otherwise,
00:48:41.000 --> 00:48:47.400
you could throw into some kind of computer algebra system, or an online on integration system.
00:48:47.400 --> 00:48:51.500
I’m trying to be honest with you, I’m trying to do it by hand.
00:48:51.500 --> 00:48:58.300
I factored E ⁻Y1 + Y2 as E ⁻Y1 E ⁻Y2.
00:48:58.300 --> 00:49:03.000
The important part about that is that, in this first integral, this inner integral,
00:49:03.000 --> 00:49:08.800
we are integrating with respect to Y2, which means we can treat Y1 as a constant.
00:49:08.800 --> 00:49:18.500
That is why I pulled E ⁻Y1 all the way out of the integral, which gives me a nicer integral E ⁻Y2 on the inside.
00:49:18.500 --> 00:49:25.500
That integrates to E ⁻Y2, and then I plugged in my bounds here to get something little messy.
00:49:25.500 --> 00:49:30.200
I multiplied E ⁻Y1 back through and it simplified a bit.
00:49:30.200 --> 00:49:37.700
And then, I integrated that E ⁻Y1 integrates to –E ⁻Y1.
00:49:37.700 --> 00:49:45.300
E⁻² is a constant, when you integrate that, it is E⁻² × Y1.
00:49:45.300 --> 00:49:51.600
I plug in the bounds Y1 = 2 all the way through and Y1 = 0.
00:49:51.600 --> 00:49:58.600
Simplified it down a little bit to get this slightly mysterious number 1 -3 E⁻².
00:49:58.600 --> 00:50:02.800
When I converted that into a decimal, I got something that was between 0 and 1.
00:50:02.800 --> 00:50:05.800
That is a little reassuring that we are doing a probability problem.
00:50:05.800 --> 00:50:11.000
If it had not been between 0 and 1, I would have known that I was wrong.
00:50:11.000 --> 00:50:13.400
And then, I convert that into a percentage.
00:50:13.400 --> 00:50:17.900
Any one of those forms, if you gave it to me in my probability class, I will be happy.
00:50:17.900 --> 00:50:25.400
You do not have to convert it into a percentage, but if you like to know what it is as a percentage, there it is.
00:50:25.400 --> 00:50:30.500
That wraps up this lecture on Bivariate density and distribution functions.
00:50:30.500 --> 00:50:36.800
This is part of the chapter on multivariate probability density and distributions.
00:50:36.800 --> 00:50:42.500
We are going to move on to marginally conditional probability, in our next video.
00:50:42.500 --> 00:50:48.800
This is all part of the larger lecture series on probability, here on www.educator.com.
00:50:48.800 --> 00:50:52.000
I'm your host, Will Murray, thank you very much for watching today, bye now.