Joint support, marginalization and conditional densities

What this concept is

A joint density describes two measurements together. For the simple regions here, compute probabilities by nested one-variable integrals: hold x fixed, integrate over allowed y, then integrate over x. This is an introductory iterated-integral bridge, not a full multivariable calculus course.

Why it matters here

Independence is a property of the whole joint model, including support. Separate histograms cannot establish it.

Concrete example

Let f(x,y)=2 on 0<x<y<1, zero elsewhere. For fixed x, y ranges from x to 1. Hence f_X(x)=∫ₓ¹2dy=2(1−x) for 0<x<1. For fixed y, x ranges from 0 to y, so f_Y(y)=2y. Normalization is ∫₀¹2(1−x)dx=1. They are dependent: positive marginal densities on the square would give a positive product even where y<x, but the joint is zero there.

Common confusion / misconception

Conditioning a continuous variable at X=x is not division by P(X=x), which is zero. A conditional density is f(x,y)/f_X(x) where the marginal is positive, defined up to the usual zero-probability exceptions.

How to use it

Draw the support before writing bounds. Here f_(Y|X=x)(y)=1/(1−x) for x<y<1, so E[Y|X=x]=(x+1)/2. Integrating this conditional mean against 2(1−x) gives E[Y]=2/3, agreeing with ∫₀¹y·2y dy. At x=1 the displayed quotient is undefined; do not insert that endpoint into the formula.

Check yourself

Why are integrations over the whole unit square with constant density 2 wrong? What variable is held fixed in the inner integral?

Mini drill or application

Compute P(Y>3/4) from f_Y; answer 7/16. Then compute P(Y>3/4|X=1/2); answer 1/2. Explain why different conditional and unconditional values are consistent.

Read this only if stuck

Primary source support. See the extension guide for practice, scope and evidence requirements. Local examples and lab code are original.