Joint Distributions: Bivariate Random Variables for Exam P

Joint PMF and PDF

For two discrete random variables X and Y, the joint PMF is p(x, y) = P(X = x, Y = y). It must satisfy p(x, y) >= 0 for all (x, y) and the double sum over all (x, y) equals 1. For continuous variables, the joint PDF f(x, y) satisfies P((X, Y) in A) = double integral over A of f(x, y) dx dy. Again, f(x, y) >= 0 everywhere and the total integral equals 1.

On Exam P, a common setup is f(x, y) = c * g(x, y) for (x, y) in some region, and you must find c by integrating. For example, if f(x, y) = c * (x + y) for 0 < x < 1, 0 < y < 1, then integrating over the unit square gives c * [integral from 0 to 1 of integral from 0 to 1 of (x + y) dy dx] = c * 1 = 1, so c = 1.

When the support region is not a simple rectangle, setting up the integration limits requires care. If f(x, y) = c * x for 0 < x < y < 1, the region is a triangle. Integrating: integral from 0 to 1 of integral from 0 to y of c * x dx dy = c * integral from 0 to 1 of y^2/2 dy = c/6 = 1, giving c = 6. Always sketch the support region before setting up your integral on Exam P.

Joint CDF

The joint CDF is F(x, y) = P(X <= x, Y <= y). For continuous variables, F(x, y) = integral from negative infinity to x of integral from negative infinity to y of f(s, t) dt ds. The joint PDF can be recovered by differentiating: f(x, y) = partial^2 F / (partial x * partial y).

Properties of the joint CDF include: F(negative infinity, y) = 0, F(x, negative infinity) = 0, and F(positive infinity, positive infinity) = 1. Also, F is non-decreasing in each argument. On Exam P, you may be given the joint CDF and asked to find probabilities for rectangular regions using the inclusion-exclusion formula: P(a < X <= b, c < Y <= d) = F(b, d) - F(a, d) - F(b, c) + F(a, c).

This rectangle probability formula is one of the most useful tools for joint CDF problems. It generalizes the one-dimensional result P(a < X <= b) = F(b) - F(a) to two dimensions. Getting the signs right is essential, and many students find it helpful to think of this as an inclusion-exclusion on the four corners of the rectangle.

Independence

X and Y are independent if and only if f(x, y) = f_X(x) * f_Y(y) for all x, y. Equivalently, the joint CDF factors: F(x, y) = F_X(x) * F_Y(y). A quick test for independence: if the support region is rectangular and f(x, y) can be written as a function of x alone times a function of y alone, then X and Y are independent.

Independence fails when the support region is not rectangular. For instance, if f(x, y) = 2 for 0 < x < y < 1, the triangular support means X and Y cannot be independent, regardless of the form of the density. The restriction that y > x creates a dependence between X and Y: knowing X constrains the possible values of Y.

To verify independence formally, you must find the marginals f_X(x) and f_Y(y) by integrating, then check whether their product equals the joint density everywhere. Even if the density factors into a product, the support must also factor. For example, f(x, y) = 4xy on 0 < x < 1, 0 < y < 1 factors as (2x)(2y) on a rectangular region, so X and Y are independent. But f(x, y) = 4xy on the triangular region 0 < x + y < 1, x > 0, y > 0 does not yield independence despite the same functional form.

Computing Probabilities from Joint Distributions

To find P(X + Y < 1), P(X > Y), or any event defined by a region in the (x, y) plane, set up the double integral of f(x, y) over that region. The challenge lies in correctly determining the limits of integration. Always draw the support region and the event region on the same plot, then integrate over their intersection.

For the density f(x, y) = 2 on 0 < x < y < 1, finding P(X + Y < 1) requires integrating over the intersection of the triangle 0 < x < y < 1 with the half-plane x + y < 1. The effective region is 0 < x < y and x + y < 1, which means x < y < 1 - x and x < 1/2. So P(X + Y < 1) = integral from 0 to 1/2 of integral from x to (1 - x) of 2 dy dx = integral from 0 to 1/2 of 2(1 - 2x) dx = 1/2.

Another common setup: find P(Y < 2X) for f(x, y) = 6x on 0 < x < 1, 0 < y < 1 - x. The region y < 2x intersected with the support gives different integration sub-regions depending on whether 2x exceeds 1 - x (i.e., x > 1/3). Breaking the integral at x = 1/3 is essential. These region-splitting problems frequently appear on Exam P.

Joint Distributions of Discrete Variables

For discrete bivariate distributions, the joint PMF is often given in a table. Each cell contains P(X = x_i, Y = y_j). The row sums give the marginal PMF of X, and the column sums give the marginal PMF of Y. To compute E[XY], sum x_i * y_j * p(x_i, y_j) over all cells.

Example: Suppose X takes values 1 and 2, and Y takes values 0 and 1, with joint PMF p(1, 0) = 0.2, p(1, 1) = 0.3, p(2, 0) = 0.1, p(2, 1) = 0.4. The marginal of X is p_X(1) = 0.5, p_X(2) = 0.5. The marginal of Y is p_Y(0) = 0.3, p_Y(1) = 0.7. Since p(1, 0) = 0.2 is not equal to p_X(1) * p_Y(0) = 0.15, X and Y are not independent.

For computing expectations, E[X] = 1(0.5) + 2(0.5) = 1.5 and E[Y] = 0(0.3) + 1(0.7) = 0.7. E[XY] = 1(0)(0.2) + 1(1)(0.3) + 2(0)(0.1) + 2(1)(0.4) = 0 + 0.3 + 0 + 0.8 = 1.1. Therefore Cov(X, Y) = 1.1 - 1.5(0.7) = 1.1 - 1.05 = 0.05.

Common Pitfalls

The most frequent error on joint distribution problems is setting up incorrect integration limits. When the support is not rectangular, the inner limits depend on the outer variable. Always draw the region and clearly label which variable is integrated first. If you integrate y first, the y-limits are functions of x (and vice versa).

Another common mistake is forgetting that the joint PDF can be zero outside the support region. When computing P((X, Y) in A), only the portion of A that overlaps with the support contributes to the integral. Ignoring the support leads to integrating over regions where f(x, y) = 0, which wastes time and can produce incorrect limits.

Students also sometimes confuse joint and marginal densities. Remember that f(x, y) is not the same as f_X(x) or f_Y(y). To get a marginal, you must integrate out the other variable. Plugging a specific value of y into f(x, y) gives the conditional density f(x|y) (up to a normalizing constant), not the marginal f_X(x).

Study Tips

For Exam P, practice drawing support regions and setting up double integrals. The most common regions are rectangles, triangles (0 < x < y < 1 or x + y < 1), and unbounded regions (x > 0, y > 0). Work through problems that require splitting the integration region, as these test your geometric reasoning. Memorize the joint CDF rectangle formula and the factorization test for independence. Aim to solve joint distribution problems in three to four minutes, with at least 30 seconds spent sketching the region before writing any integrals.