Recitation: Uniform Probabilities on a Triangle

What this lecture covers

This is a recitation problem, not a lecture, that gives practice manipulating joint, marginal, and conditional densities for two continuous random variables. It picks a concrete case, X and Y uniformly distributed on the triangle with vertices (0,0), (1,0), and (0,1), and works through the full sequence of calculations you would need for any joint density.

It follows the lecture material on multiple continuous random variables and is meant to reinforce it with hands-on computation, including a symmetry trick for finishing off a calculation without extra integrals.

Key ideas

• Uniform joint density on a region: for a uniform distribution, the density is a constant equal to 1 divided by the area of the region, and 0 outside it.
• Marginal by integrating out a variable: the marginal density of Y comes from integrating the joint density over all valid x for each y, keeping careful track of the limits of integration.
• Conditional PDF as joint over marginal: dividing the joint density by the marginal of Y gives the conditional density of X given Y, which here turns out to be uniform on a shrinking interval.
• Conditional expectation: the expected value of X given Y = y is computed the same way as an ordinary expectation, but weighted by the conditional density.
• Total expectation theorem: integrating a conditional expectation against the marginal density recovers the unconditional expectation.
• Symmetry shortcut: when a joint density is unchanged by swapping X and Y, their expectations must be equal, which can close out an equation without further integration.

Walkthrough

Joint PDF and marginal of Y (0:00)

Because X and Y are uniform on the triangle bounded by x + y <= 1, x >= 0, y >= 0, the joint density is a constant equal to 2 inside the triangle (the reciprocal of its area, 1/2) and 0 elsewhere. Integrating this joint density over x from 0 to 1 - y gives the marginal density of Y as 2(1 - y) for y between 0 and 1, a triangular shape that is higher for small y because more x values pair with small y.

Conditional PDF of X given Y (7:10)

Dividing the joint density by the marginal of Y gives a conditional density of 1/(1-y) for x between 0 and 1 - y, which does not depend on x at all. That means, for a fixed y, X is uniformly distributed on [0, 1-y] — a direct consequence of slicing a uniform (flat) joint density, which always leaves a uniform slice.

Conditional and unconditional expectations (12:19)

Since X given Y = y is uniform on [0, 1-y], its conditional expectation is the midpoint, (1-y)/2. Applying the total expectation theorem and splitting the resulting integral expresses the unconditional expectation of X as 1/2 minus half the expectation of Y.

Finishing with symmetry (19:33)

Because the triangle and the joint density are symmetric under swapping x and y, E[X] and E[Y] must be equal. Substituting this into the equation from the total expectation theorem gives E[X] = E[Y] = 1/3.

Before you watch

• Know how to find a joint density for a uniform distribution over a region and how to integrate it to get a marginal density.
• Review the definition of conditional density as joint density divided by marginal density.
• Recall the total expectation theorem from the discrete case, since it is applied here with an integral in place of a sum.

Check your understanding

1. Why is the joint density in this problem equal to 2 rather than 1?
2. Why does slicing a uniform joint density always give a uniform conditional density?
3. How does the total expectation theorem turn a conditional expectation into an unconditional one?
4. Why does the symmetry between X and Y let you conclude E[X] = E[Y] without further calculation?