# Uniform Probabilities on a Triangle

Hi. In this problem, we’re going

to get a bunch of practice working with multiple random

variables together. And so we’ll look at joint

PDFs, marginal PDFs, conditional PDFs, and also get

some practice calculating expectations as well. So the problem gives us a pair

of random variables– x and y. And we’re told that the joint

distribution is uniformly distributed on this triangle

here, with the vertices being 0, 0 1, 0, and 0, 1. So it’s uniform in

this triangle. And the first part of the

problem is just to figure out what exactly is disjoint PDF of

the two random variables. So in this case, it’s pretty

easy to calculate, because we have a uniform distribution. And remember, when you have a

uniform distribution, you can just imagine it being

a sort of plateau coming out of the board. And it’s flat. And so the height of the

plateau, in order to calculate it, you just need to figure

out what the area of this thing is, of this triangle is. So remember, when you had single

random variables, what we had to do was calculate, for

uniform distribution, we had to integrate to 1. So you took the length, and you

took 1 over the length was the correct scaling factor. Here, you take the area. And the height has to make it so

that the entire volume here integrates to 1. So the joint PDF is just

going to be 1 over whatever this area is. And the area is pretty

simple to calculate. It’s 1/2 base times height. So it’s 1/2. And so what we have is

that the area is 1/2. And so the joint PDF of x and

y is going to equal 2. But remember, you always have

to be careful when writing these things to remember

the ranges when these things are valid. So it’s only 2 within

this triangle. And outside of the

triangle, it’s 0. So what exactly does inside

the triangle mean? Well, we can write it

more mathematically. So this diagonal line, it’s

given by x plus y equals 1. So everything in the triangle

is really x plus y is less than or equal to 1. It means everything under

this triangle. And so we need x plus y to be

less then or equal to 1 and also x to be non-negative and

y to be non-negative. So with these inequalities,

that captures everything within this triangle. And otherwise, the joint

PDF is going to be 0. The next part asks us to find,

using this joint PDF, the marginal of y. And remember, when you have

a joint PDF of two random variables, you essentially have

everything that you need, because from this joint PDF, you

can calculate marginals, you can calculate from the

margins, you can calculate conditionals. The joint PDF captures

everything that there is to know about this pair of

random variables. Now, to calculate a marginal PDF

of y, remember a marginal really just means collapsing

the other random variable down. And so you can just imagine

taking this thing and collapsing it down

onto the y-axis. And mathematically, that is just

saying that we integrate out the other random variable. So the other random variable

in this case will be x. We take x and we get rid of

it by integrating out from negative infinity to infinity. Of course, this joint PDF

is 0 in a lot of places. And so a lot of these

will be 0. And only for a certain range

of x’s will this integral actually be non-zero. And so again, the other time

when we have to be careful is when we have these limits of

integration, we need to make sure that we have the

right limits. And so we know that the

joint PDF is 2. It’s nonzero only within

this triangle. And so it’s only 2 within

this triangle, which means what for x? Well, depending on what

x and y are, this will be either 2 or 0. So let’s just fix

some value of y. Pretend that we’ve picked some

value y, let’s say here. We want this value of y. Well, what are the values of x

such that the joint PDF for that value y is actually

nonzero, it’s actually 2? Well, it’s everything from

x equals 0 to whatever x value this is. But this x value, actually, if

you think about it, is just 1 minus y, because this line

is x plus y equals 1. So whatever y is, x is going

to be 1 minus that. And so the correct limits

would actually be from 0 to 1 minus y. And then the rest of that

is pretty simple. You integrate this. This is a pretty simple

integral. And you get that it’s actually

two times 1 minus y. That’s a y. But of course, again, we need to

make sure that we have the right regions. So this is not always true

for y, of course. This is only true for

y between 0 and 1. And otherwise, it’s actually 0,

because when you take a y down here, well, there’s no

values of x that will give you a nonzero joint PDF. And if you take a value of y

higher than this, the same thing happens. So we can actually draw

this out and see what it looks like. So let’s actually draw

a small picture here. Here’s y. Here’s the marginal PDF of y. And here’s 2. And it actually looks

like this. It’s a triangle and a 0

outside this range. So does that make sense? Well, first of all, you see

that actually does in fact integrates to 1,

which is good. And the other thing we notice

is that there is a higher density for smaller

values of y. So why is that? Why are smaller values

of y more likely than larger values of y? Well, because when you have

smaller values of y, you’re down here. And it’s more likely because

there are more values of x that go along with it that

make that value of y more likely to appear. Say you have a large

value of y. Then you’re up here

at the tip. Well, there aren’t very many

combinations of x and y that give you that large

a value of y. And so that large value of

y becomes less likely. Another way to think about it

is, when you collapse this down, there’s a lot more stuff

to collapse down its base. There’s a lot of x’s

to collapse down. But up here, there’s only a

very little bit of x to collapse down. And the PDF of y becomes

more skewed towards smaller values of y. So now, the next thing that we

want to do is calculate the conditional PDF of x, given y. Well, let’s just recall

what that means. This is what we’re looking for–

the conditional PDF of x, given y. And remember, this is calculated

by taking the joint and dividing by the

marginal of y. So we actually have the

top and the bottom. We have to joint PDF from part

A. And from part B, we calculated the marginal

PDF of y. So we have both pieces. So let’s actually

plug them in. Again, the thing that you have

to be careful here is about the ranges of x and y where

these things are valid, because this is only non-zero

when x and y fall within this triangle. And this is only non-zero when

y is between 0 and 1. So we need to be careful. So the top, when it’s

non-zero, it’s 2. And the bottom, when it’s

non-zero, it’s 2 times 1 minus y. So we can simplify that to

be 1 over 1 minus y. And when is this true? Well, it’s true when x and y are

in the triangle and y is between 0 and 1. So put another way, that means

that this is valid when y is between 0 and 1 and x is between

0 and 1 minus y, because whatever x has to be,

it has to be such that they actually still fall within

this triangle. And outside of this, it’s 0. So let’s see what this

actually looks like. So this is x, and this is the

conditional PDF of x, given y. Let’s say this is

1 right here. Then what it’s saying is, let’s

say we’re given that y is some little y. Let’s say it’s somewhere here. Then it’s saying that the

conditional PDF of x given y is this thing. But notice that this value,

1 over 1 minus y, does not depend on x. So in fact, it actually

is uniform. So it’s uniform between

0 and 1 minus y. And the height is something

like 1 over 1 minus y. And this is so that the scaling

makes it so that actually is a valid PDF, because

the integral is to 1. So why is the case? Why is that when you condition

on y being some value, you get that the PDF of x is

actually uniform? Well, when you look over here,

let’s again just pretend that you’re taking this value of y. Well, when you’re conditioning

on y being this value, you’re basically taking a slice of this

joint PDF at this point. But remember, the original

joint PDF was uniform. So when you take a slice of a

uniform distribution, joint uniform distribution,

you still get something that is uniform. Just imagine that you have

a cake that is flat. Now, you take a slice

at this level. Then whatever slice you have

is also going to be imagine being a flat rectangle. So it’s still going

to be uniform. And that’s why the conditional

PDF of x given y is also uniform. Part D now asks us to find a

conditional expectation of x. So we want to find the

expectation of x, given that y is some little y. And for this, we can

use the definition. Remember, expectations are

really just weighted sums. Or in the [? continuous ?]

case, it’s an integral. So you take the value. And then you weight

it by the density. And in this case, because we’re

taking conditional a expectation, what we weight it

by is the conditional density. So it’s the conditional

density of x given that y is little y. We integrate with

respect to x. And fortunately, we know what

this conditional PDF is, because we calculated it earlier

in part C. And we know that it’s this– 1 over 1 minus y. But again, we have to be

careful, because this formula, 1 over 1 minus y, is only

valid certain cases. So let’s think about

this first. Let’s think about some

extreme cases. What if y, little

y, is negative? If little y is negative,

we’re conditioning on something over here. And so there is no density for

y being negative or for y, say, in other cases when

y is greater than 1. And so in those cases, this

expectation is just undefined, because conditioning on that

doesn’t really make sense, because there’s no density

for those values of y. Now, let’s consider the case

that actually makes, sense where y is between 0 and 1. Now, we’re in business, because

that is the range where this formula is valid. So this formula is valid,

and we can plug it in. So it’s 1 over 1 minus y dx. And then the final thing that we

again need to check is what the limits of this

integration is. So we’re integrating

with respect to x. So we need to write down what

values of x, what ranges of x is this conditional PDF valid. Well, luckily, we specified

that here. x has to be between

0 and 1 minus y. So let’s actually calculate

this integral. This 1 over 1 minus y is a

constant with respect to x. You can just pull that out. And then now, you’re really

just integrating x from 0 to 1 minus y. So the integral of x is

[? 1 ?], 1/2x squared. So you get a 1/2x squared, and

you integrate that from 0 to 1 minus y. And so when you plug in

the limits, you’ll get a 1 minus y squared. That will cancel out the

1 over 1 minus y. And what you’re left with is

just 1 minus y over 2. And again, we have to specify

that this is only true for y between 0 and 1. Now, we can again actually

verify that this makes sense. What we’re really looking for is

the conditional expectation of x given some value of y. And we already said that

condition on y being some value of x is uniformly

distributed between 0 and 1 minus y. And so remember for our uniform

distribution, the expectation is simple. It’s just the midpoint. So the midpoint of 0

and 1 minus y is exactly 1 minus y/2. So that’s a nice way of

verifying that this answer is actually correct. Now, the second part of

part D asks us to do a little bit more. We have to use the total

expectation theorem in order to somehow write the expectation

of x in terms of the expectation of y. So the first thing we’ll

do is use the total expectation theorem. So the total expectation theorem

is just saying, well, we can take these conditional

expectations. And now, we can integrate this

by the marginal density of y, then we’ll get the actual

expectation of x. You can think of it as just kind

of applying the law of iterated expectations as well. So this integral is going

to look like this. You take the conditional

expectation. So this is the expectation of x

if y were equal to little y. And now, what is that

probability? Well, now we just multiply that

by the density of y at that actual value of little y. And we integrate with

respect to y. Now, we’ve already calculated

what this conditional expectation is. It’s 1 minus y/2. So let’s plug that in. 1 minus y/2 times the

marginal of y. There’s a couple ways of

attacking this problem now. One way is, we can actually

just plug in that marginal of y. We’ve already calculated that

out in part B. And then we can do this integral and calculate

out the expectation. But maybe we don’t really want

to do so much calculus. So let’s do what the

problem says and try a different approach. So what the problem suggests is

to write this in terms of the expectation of y. And what is the expectation

of y? Well, the expectation of y is

going to look something like the integral of y times

the marginal of y. So let’s see if we can identify

something like that and pull it out. Well, yeah, we actually

do have that. We have y times the marginal

of y, integrated. So let’s isolate that. So besides that, we

also have this. We have the integral of the

first term, is 1/2 times the marginal of y. And then the second term is

minus 1/2 times the integral of y of dy. This is just me splitting

this integral up into two separate integrals. Now, we know what this is. The 1/2 we can pull out. And then the rest of it is

just the integral of a marginal of a density from minus

infinity to infinity. And by definition, that

has to be equal to 1. So this just gives us a 1/2. And now, what is this? We get a minus 1/2. And now this, we already said

that is the expectation of y. So what we have is the

expectation of y. So in the second part of this

part D, we’ve expressed the expectation of x in terms

of the expectation of y. Now, maybe that seems like

that’s not too helpful, because we don’t know what

either of those two are. But if we think about this

problem, and as part E suggests, we can see that

there’s symmetry in this problem, because x and y are

essentially symmetric. So imagine this is x equals y. There’s symmetry in this

problem, because if you were to swap the roles of x and y,

you would have exactly the same joint PDF. So what that suggests is that

by symmetry then, it must be that the expectation of x and

the expectation of y are exactly the same. And that is using the

symmetry argument. And that helps us now, because

we can plug that in and solve for expectation of x. So expectation of x is 1/2 minus

1/2 expectation of x. So we have 3/2 expectation

of x equals 1/2. So expectation of

x equals 1/3. And of course, expectation

of y is also 1/3. And so it turns out that the

expectation is around there. So this problem had

several parts. And it allowed us to start

out from just a raw joint distribution, calculate

marginals, calculate conditionals, and then from

there, calculate all kinds of conditional expectations

and expectations. And a couple of important points

to remember are, when you do these joint

distributions, it’s very important to consider where

values are valid. So you have to keep in mind

when you write out these conditional PDFs and joint PDFs

and marginal PDFs, what ranges the formulas you

calculated are valid for. And that also translates to

when you’re calculating expectations and such. When you have integrals, you

need to be very careful about the limits of your integration,

to make sure that they line up with the range

where the values are actually valid. And the last thing, which is

kind of unrelated, but it is actually a common tool that’s

used in a lot of problems is, when you see symmetry in these

problems, that can help a lot, because it will simplify things

and allow you to use facts like these to help

you calculate what the final answer is. Of course, this is also comes

along with practice. You may not immediately see that

there could be a symmetry argument that will help

with this problem. But with practice, when you do

more of these problems, you’ll eventually build up

that kind of–

## 3 Replies to “Uniform Probabilities on a Triangle”

I love math, especially when I understand something new. I knew this, though.

Jimmy li is my MANS oh my god why can't he teach me every class jimmy li for prom king 2018

He's the best TA of this course, at least I think so, explaining things very clear