Domain

The domain is the set of allowed input points.

Example:

$$f(x,y)=\sqrt{1-x^2-y^2}$$

For this to be real:

$$1-x^2-y^2\ge 0$$

So:

$$x^2+y^2\le 1$$

That is the unit disk.

So this function only exists above the disk:

$$x^2+y^2\le 1$$

This matters later because joint densities also often live only on some region.

Support

Support means the region where the function is nonzero.

Example:

$$f(x,y)=\{\begin{array}{ll}2,& 0<y<x<1\\ 0,& \text{otherwise}\end{array}$$

The function is nonzero only where:

$$0<y<x<1$$

This is a triangle.

So before doing any integral, you must identify the region.

This is the first major skill:

$$\text{formula}\to \text{region in the }xy\text{-plane}$$

Level Curves

A level curve is where the function has constant value.

For:

$$f(x,y)=x+y$$

level curves are:

$$x+y=c$$

These are straight lines.

For:

$$f(x,y)=x^2+y^2$$

level curves are:

$$x^2+y^2=c$$

These are circles.

For:

$$f(x,y)=xy$$

level curves are:

$$xy=c$$

Lesson 2: Double Integrals from First Principles

In single-variable calculus:

$${\int }_a^{b}f(x)dx$$

means: split the line from $a$ to $b$ into tiny intervals, take height $f(x)$, multiply by tiny width $dx$, and add.

So each tiny piece is:

$$f(x)dx$$

For two variables:

$${\iint }_{R}f(x,y)dA$$

means: split a 2D region $R$ into tiny rectangles, take height $f(x,y)$, multiply by tiny area $dA$, and add.

Each tiny piece is:

$$f(x,y)dA$$

If the tiny rectangle has width $dx$ and height $dy$, then:

$$dA=dxdy$$

So:

$${\iint }_{R}f(x,y)dA={\iint }_{R}f(x,y)dxdy$$

Order Matters in Notation

This:

$${\int }_a^b{\int }_c^{d}f(x,y)dydx$$

means integrate $y$ first, then $x$.

This:

$${\int }_c^d{\int }_a^{b}f(x,y)dxdy$$

means integrate $x$ first, then $y$.

For rectangular regions, both usually give the same result:

$${\int }_a^b{\int }_c^{d}f(x,y)dydx={\int }_c^d{\int }_a^{b}f(x,y)dxdy$$

This is a version of Fubini's theorem.

Lesson 3: Non-Rectangular Regions and Building the Limits of Integration

So far we've integrated over rectangles:

$$a\le x\le b,c\le y\le d$$

Life is easy because the limits are constants.

But probability almost never gives you rectangles. It gives you things like

$$0<y<x<1$$

or

$$x+y<1$$

The question is:

How do I convert a geometric region into integral limits?

Example 1

Consider

$$0<y<x<1$$

Let's not jump to the integral.

First decode the inequalities.

The three conditions are

$$y>0,$$$$y<x,$$$$x<1.$$

Now imagine moving a vertical line across the region.

Why vertical?

Because we're going to integrate with respect to $y$ first.

For every fixed $x$,

the lowest point is

$$y=0,$$

the highest point is

$$y=x.$$

Now ask:

What values can $x$ itself take?

Obviously

$$0\le x\le 1.$$

So the integral becomes

$${\int }_0^1{\int }_0^{x}f(x,y)dydx$$

Notice what happened.

The outer variable always has constant limits.

The inner variable usually depends on the outer variable.

Why Would We Ever Change the Order?

Because one order may be impossible.

Example:

$${\int }_0^1{\int }_0^x{e}^{y^2}dydx$$

The inside integral

$$\int e^{y^2}dy$$

has no elementary antiderivative.

You're stuck.

Now reverse the order.

The region becomes

$$0\le y\le 1,$$$$y\le x\le 1.$$

So

$${\int }_0^1{\int }_y^1{e}^{y^2}dxdy$$

Notice something beautiful.

The integrand

$$e^{y^2}$$

doesn't depend on

$$x.$$

So integrating over

$$x$$

is trivial.

You get

$${\int }_y^1{e}^{y^2}dx=(1-y)e^{y^2}.$$

Lesson 5: Change of Variables

Until now, we changed only the order:

$$dydx\leftrightarrow dxdy$$

But the variables stayed the same:

$$x,y$$

Now we ask something bigger.

What if we define new variables?

$$u=x+y$$$$v=x-y$$

Now instead of describing points using $(x,y)$, we describe the same points using $(u,v)$.

This is called change of variables.

Why do this?

Because sometimes the original variables are awkward.

Example:

$$x^2+y^2$$

is annoying in $x,y$, but simple in polar coordinates:

$$x^2+y^2=r^2$$

So changing variables can make the region or the function simpler.

Same point. Different coordinate system. Humanity invents notation, then spends centuries escaping it.

Simple 1D Version First

In single-variable calculus, suppose:

$$u=2x$$

Then:

$$du=2dx$$

So:

$$dx=\frac{1}{2}du$$

Therefore:

$$\int f(2x)dx=\int f(u)\frac{1}{2}du$$

That extra factor $\frac{1}{2}$ appears because the new variable $u$ stretches distances.

If:

$$u=2x$$

then a tiny change $dx$ becomes:

$$du=2dx$$

So $u\text{-space}$ is stretched by factor $2$. Therefore $dx$ is only half of $du$.

That is the whole idea behind Jacobians too.

In 2D, Area Changes

In two variables, we care about tiny areas.

Originally:

$$dA=dxdy$$

After changing variables:

$$dA=dudv$$

But usually:

$$dxdy\ne dudv$$

There is an area-scaling factor.

That factor is the Jacobian determinant.

Example: Simple Scaling

Let:

$$u=2x$$$$v=3y$$

Then:

$$x=\frac{u}{2}$$$$y=\frac{v}{3}$$

A tiny rectangle in $x,y\text{-space}$ has sides:

$$dx,dy$$

Area:

$$dxdy$$

In $u,v\text{-space}$:

$$du=2dx$$$$dv=3dy$$

So:

$$dudv=6dxdy$$

Therefore:

$$dxdy=\frac{1}{6}dudv$$

So if you transform from $(x,y)$ to $(u,v)$, you must multiply by:

$$\frac{1}{6}$$

otherwise your areas are wrong.

And if your areas are wrong, your probabilities are wrong, which is bad because the whole point of probability is already pretending uncertainty is manageable.

General Formula

Suppose:

$$x=x(u,v)$$$$y=y(u,v)$$

Then:

dxdy=​∂(u,v)∂(x,y)​​dudv

where

∂(u,v)∂(x,y)​=​∂u∂x​∂u∂y​​∂v∂x​∂v∂y​​​

That determinant tells you how much tiny area changes.

Why Determinant?

A tiny rectangle in the $uv\text{-plane}$ is spanned by two small movements.

Move in the $u\text{-direction}$:

$$du$$

This changes $(x,y)$ by approximately:

$$\left(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u}\right)du$$

Move in the $v\text{-direction}$:

$$dv$$

This changes $(x,y)$ by approximately:

$$\left(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v}\right)dv$$

These two vectors form a tiny parallelogram in the $xy\text{-plane}$.

The area of that parallelogram is:

​​∂u∂x​∂u∂y​​∂v∂x​∂v∂y​​​​dudv

That is the proof idea.

The Jacobian determinant is not a random punishment. It is the area of the transformed tiny square.

Core Formula

When changing from $(x,y)$ to $(u,v)$:

∬R​f(x,y)dxdy=∬S​f(x(u,v),y(u,v))​∂(u,v)∂(x,y)​​dudv