Imagine a pair (X,Y) uniformly distributed over the triangle 0<x<y<2. Uniform here means constant density per unit area, not that both coordinates have uniform marginal distributions. Predict which Y values admit more horizontal source length and therefore carry more marginal probability.
Predict
For fixed y, x ranges from zero to y. For fixed x, y ranges from x to two. Predict why replacing either interval by (0,2) would integrate over points outside the triangle. The geometry of the support belongs inside every marginal calculation.
Experiment
The triangle has area two, so joint density is 1/2. Integrating over x gives f_Y(y)=y/2 for 0<y<2. Integrating over y gives f_X(x)=(2−x)/2 for 0<x<2. Each integrates to one.
Fixed coordinate
Other-coordinate range
Marginal density
Y=y
0<x<y
y/2
X=x
x<y<2
(2−x)/2
Draw one horizontal and one vertical slice. Then compute P(Y<1)=∫₀¹y/2 dy=1/4. The lower half of the vertical coordinate range contains only a quarter of the total triangular area.
Understand
A joint mass function assigns probabilities to pairs of values; a joint density assigns probability per unit area and integrates over a support region. Both must be nonnegative and normalize to one. A joint CDF P(X≤x,Y≤y) provides another description that also accommodates discrete and mixed structures.
Marginalization sums or integrates over the unreported coordinate. For a density, f_X(x)=∫f_(X,Y)(x,y)dy and similarly for Y. Writing the density as zero outside its support is equivalent to using explicit slice bounds, but the support cannot simply be ignored. Nonnegative integrands permit Tonelli’s interchange; signed expectation integrals need suitable integrability.
Joint probabilities require describing the event region. For X<Y in this triangular model the probability is one, while for an independent pair with the same marginals the relationship would generally differ. The support itself encodes dependence. A constant joint density over a nonrectangular region does not imply independent coordinates.
Marginals do not determine the joint law. Two fair bits can be independent, with all four pair probabilities one quarter, or equal almost surely, with only 00 and 11 having probability one half each. Both constructions have the same one-variable marginals but different probabilities of agreement, sums and conditional events.
An empirical scatterplot can suggest support or dependence, but a finite cloud does not establish a continuous joint density or prove independence. The explicit triangle here is an original mathematical model chosen to expose the geometry. Keep its assumed support separate from evidence about a real population.
When a variable pair has a deterministic relationship, such as Y=X², its joint law lies on a curve and generally has no ordinary two-dimensional density with respect to area. One-dimensional marginals may still have densities. Do not force every joint law into an area-density formula.
Apply
Compute the practice marginals using two different slice directions, then check both normalizations. For the fair-bit example, write complete joint tables to demonstrate nonuniqueness. The purpose is to show exactly which probabilities are lost when a report presents only separate summaries.
Continue
Next normalize a slice to obtain a conditional law. This differs from marginalizing: conditioning preserves one coordinate as information rather than averaging it away.
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Practice and reasoning
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A density equals c on 0<x<y<1 and zero elsewhere. Find c.
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For that density, find P(Y<1/2).
Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.
For density 2 on 0<x<y<1, f_X(x) on (0,1) is:
A joint law is specified by its two marginals alone:
For density 6x on 0<x<y<1, derive both marginals and check total mass.
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Construct two different joint laws for fair bits X,Y with the same marginals, one independent and one with X=Y almost surely.
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Reflect and continue
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