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Joint, marginal and conditional distributions

Integrate out what you do not care about to get a marginal; divide by a marginal to get a conditional. Drawing the region is most of the work.

The formulas

Joint density
∬f(x,y) dx dy=1\iint f(x,y)\,dx\,dy = 1
Marginal
fX(x)=∫f(x,y) dyf_X(x) = \int f(x,y)\,dy
Conditional
fY∣X(y∣x)=f(x,y)fX(x)f_{Y\mid X}(y\mid x) = \dfrac{f(x,y)}{f_X(x)}
Independence
f(x,y)=fX(x)fY(y) on a rectangular supportf(x,y) = f_X(x)f_Y(y) \text{ on a rectangular support}
Probability of a region
P((X,Y)∈A)=∬Af(x,y) dx dyP((X,Y)\in A) = \iint_A f(x,y)\,dx\,dy

Where it comes from

  1. The marginal density of XX collects all the probability at each xx regardless of yy, which is exactly integrating yy out.
  2. The conditional density renormalises the slice at xx so that it integrates to 1 - the continuous version of dividing by P(B)P(B).
  3. Independence requires the density to factor AND the support to be a product region; a triangular support alone rules independence out.

Worked example

Two standardised losses have joint density $f(x,y) = x + y$ on the unit square. Find $P(X + Y < 1)$.

  1. The region is the triangle below the line y=1−xy = 1-x, so integrate yy from 0 to 1−x1-x and xx from 0 to 1.
  2. Inner integral: ∫01−x(x+y) dy=x(1−x)+(1−x)22\int_0^{1-x}(x+y)\,dy = x(1-x) + \tfrac{(1-x)^2}{2}.
  3. Outer integral: ∫01[x−x2+1−2x+x22]dx=12−13+16\int_0^1\left[x - x^2 + \tfrac{1-2x+x^2}{2}\right]dx = \tfrac12 - \tfrac13 + \tfrac16.
  4. P(X+Y<1)=13=0.3333P(X+Y<1) = \tfrac13 = 0.3333.

Answer: 1/3 ≈ 0.3333

This answer is recomputed from the site’s own interest-theory and probability functions every time the test suite runs, so the page and the mathematics cannot drift apart.

Memory hooks

  • Sketch the region first. The limits, not the integrand, are where marks are lost.
  • A non-rectangular support means the variables cannot be independent, whatever the density looks like.

Traps

  • Using constant limits on a triangular region.
  • Calling variables independent because the density factors, without checking the support.

Related

Drill this: the Exam P question bank has original questions on this topic, and today’s free round is open to everyone.