Exam PMultivariate probabilityFree to read

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

$$\iint f(x,y)dxdy=1$$

Marginal

$$f_X(x)=\int f(x,y)dy$$

Conditional

$$f_{Y\mid X}(y\mid x)=\frac{f(x,y)}{f_X(x)}$$

Independence

$$f(x,y)=f_X(x)f_Y(y)\text{ on a rectangular support}$$

Probability of a region

$$P((X,Y)\in A)={\iint }_{A}f(x,y)dxdy$$

Where it comes from

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

Worked example

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

• Covariance, correlation and sums
• Conditional expectation and the variance decomposition
• Transformations and order statistics

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