Introduction
A joint-distribution problem can present the same probability model in several forms: a table of point probabilities, a density over a region, or a joint cumulative distribution function. The main task is to choose the representation that makes the requested quantity direct to calculate. A table is often best for discrete outcomes, a density for continuous regions, and a CDF for cumulative probabilities or rectangles.
This tutorial brings those tools together. The workflow is to identify the support, compute the marginals, evaluate the requested event using an appropriate representation, and test independence using the full joint law. The calculations are aligned with the SOA VEE Mathematical Statistics syllabus.
A Reliable Problem-Solving Sequence
Start by identifying whether the variables are discrete, continuous, or mixed, and write down where probability can occur. Then match the calculation to the representation:
- For a discrete joint PMF, sum the relevant cells.
- For a continuous joint density, integrate over the event and respect the support boundaries.
- For a joint CDF, use boundary limits for marginals and inclusion-exclusion for rectangles.
To test independence, compare the joint law with the product of its marginals. For discrete variables this means checking whether \(p_{X,Y}(x,y)=p_X(x)p_Y(y)\) for every possible pair. For continuous variables, the equality \(f_{X,Y}(x,y)=f_X(x)f_Y(y)\) must hold almost everywhere, meaning except possibly on a set of area zero. Equivalently, independence implies \(F_{X,Y}(x,y)=F_X(x)F_Y(y)\) for every \(x,y\).
Identify the possible values or region before summing or integrating. A formula outside the support contributes no probability.
Sum a joint PMF over the other variable, integrate a joint density over the other coordinate, or take a joint-CDF boundary limit.
Translate its inequalities into cells, integration limits, or CDF differences.
Compare a joint probability or the joint law with the product of the marginals; use a region of positive probability where needed.
Worked Examples
Worked Example: A PMF Table, Marginals, and a Region Event
Let \(X\) be the number of sensor alerts in a test interval, with possible values \(0,1,2\), and let \(Y\) indicate whether a separate warning light is on, with values \(0,1\). The joint PMF is given below. Find the marginal PMFs, calculate \(P(X+Y\geq 2)\), and determine whether \(X\) and \(Y\) are independent.
| \(X\backslash Y\) | 0 | 1 | Row total |
|---|---|---|---|
| 0 | 0.10 | 0.20 | 0.30 |
| 1 | 0.20 | 0.10 | 0.30 |
| 2 | 0.10 | 0.30 | 0.40 |
| Column total | 0.40 | 0.60 | 1.00 |
Method: sum across rows or columns. The marginal PMF of \(X\) is given by the row totals: \(p_X(0)=0.30\), \(p_X(1)=0.30\), and \(p_X(2)=0.40\). The marginal PMF of \(Y\) is given by the column totals: \(p_Y(0)=0.40\) and \(p_Y(1)=0.60\). The table is normalized because its entries sum to \(1.00\).
Region probability. The pairs satisfying \(X+Y\geq 2\) are \((1,1)\), \((2,0)\), and \((2,1)\). Therefore,
Independence test. If the variables were independent, the joint probability at \((0,0)\) would equal \(p_X(0)p_Y(0)=0.30(0.40)=0.12\). The table gives \(p_{X,Y}(0,0)=0.10\), which is not \(0.12\). Thus \(X\) and \(Y\) are dependent.
Common mistake avoided: Checking just that the table sums to one does not establish independence. Normalization checks that the entries define a PMF; independence requires the joint probabilities to factor into the corresponding marginal probabilities.
Worked Example: Integrating Over a Triangular Support
Suppose continuous variables \(X\) and \(Y\) have joint density \(f_{X,Y}(x,y)=2\) on the triangular region \(0<x<y<1\), and density zero elsewhere. Find both marginal densities, calculate \(P(X<1/2,Y<1/2)\), and test independence.
Method: integrate over the support. The density is normalized: the triangle has area \(1/2\), so its integral is \(2(1/2)=1\). For a fixed \(x\), the support allows \(x<y<1\). For a fixed \(y\), it allows \(0<x<y\). Hence
Both marginal densities are zero outside the stated intervals. As a check, \(\int_0^1 2(1-x)\,dx=1\) and \(\int_0^1 2y\,dy=1\).
For \(X<1/2\) and \(Y<1/2\), the support requires \(0<y<1/2\) and \(0<x<y\). Integrate over that triangular part of the support:
For independence, compare this with the product of the marginal probabilities. They are
Their product is \((3/4)(1/4)=3/16\), not \(1/4\), so \(X\) and \(Y\) are dependent.
Common mistake avoided: The double integral must follow the triangular support. Integrating over the whole square \(0<x<1,\ 0<y<1\) would include points where \(x\geq y\), where the density is zero, and would use incorrect limits for the event.
Using a Joint CDF
A joint CDF is particularly useful when the problem gives cumulative probabilities or asks about a rectangle. Its definition is \(F_{X,Y}(x,y)=P(X\leq x,Y\leq y)\). The marginal CDFs follow by taking the other threshold to infinity:
For any real numbers \(a<b\) and \(c<d\), the probability of the half-open rectangle \((a,b]\times(c,d]\) is
This formula works for discrete, continuous, and mixed distributions. The half-open boundaries matter when there may be point masses; the formula includes the upper endpoints and excludes the lower endpoints.
Worked Example: Recovering Probabilities From a Joint CDF
Let \(X\) and \(Y\) each take values in \(\{0,1\}\). Their joint CDF is zero when either argument is negative, and has these values:
| Range of \(x\) | Range of \(y\) | \(F_{X,Y}(x,y)\) |
|---|---|---|
| \(0\leq x<1\) | \(0\leq y<1\) | 0.20 |
| \(x\geq1\) | \(0\leq y<1\) | 0.50 |
| \(0\leq x<1\) | \(y\geq1\) | 0.60 |
| \(x\geq1\) | \(y\geq1\) | 1.00 |
Find the four point probabilities, the marginals, \(P(X+Y\geq1)\), and whether \(X\) and \(Y\) are independent.
Method: take the CDF increments. The probability at \((0,0)\) is \(F_{X,Y}(0,0)=0.20\). Increasing the \(x\) threshold from \(0\) to \(1\) while \(y=0\) adds the mass at \((1,0)\): \(0.50-0.20=0.30\). Increasing the \(y\) threshold while \(x=0\) adds the mass at \((0,1)\): \(0.60-0.20=0.40\). The remaining mass is at \((1,1)\):
The four masses sum to \(0.20+0.30+0.40+0.10=1.00\). Summing by \(X\), \(p_X(0)=0.20+0.40=0.60\) and \(p_X(1)=0.30+0.10=0.40\). Summing by \(Y\), \(p_Y(0)=0.20+0.30=0.50\) and \(p_Y(1)=0.40+0.10=0.50\).
The event \(X+Y\geq1\) includes \((1,0),(0,1),(1,1)\), so its probability is \(0.30+0.40+0.10=0.80\). For independence, the product of the marginal probabilities at \((0,0)\) is \(p_X(0)p_Y(0)=0.60(0.50)=0.30\), whereas the joint probability is \(0.20\). The variables are dependent.
Common mistake avoided: A CDF value is cumulative, not the probability at the point where it is evaluated. For example, \(F_{X,Y}(1,1)=1\) is the total probability of all four points, not the mass at \((1,1)\); obtaining a point mass requires CDF increments.
Independence Checks and Exam Traps
A single failed factorization is enough to show dependence. For example, if \(p_{X,Y}(x,y)\ne p_X(x)p_Y(y)\) for even one possible pair, the variables are not independent. For continuous variables, avoid relying on a mismatch at just one point: densities are only required to factor almost everywhere, and changing a density at one point does not change any probability. Instead compare probabilities of events, or establish a mismatch over a region with positive area.
Support geometry can make that comparison efficient. If a joint support excludes an entire rectangle, while each marginal assigns positive probability to the corresponding coordinate intervals, independence is impossible. Under independence, the probability of that rectangle would be the product of two positive marginal probabilities. A single excluded point, by contrast, is not enough for a continuous-density argument.
The following incorrect calculation illustrates a common trap. In the first worked example, the marginals at \(X=1\) and \(Y=1\) are \(0.30\) and \(0.60\). A wrong conclusion would be: “The joint probability is \(0.10\), and \(0.30(0.60)=0.18\), so the variables are independent.” The arithmetic actually contradicts independence: \(0.10\ne0.18\). The correction is to state that the variables are dependent. Always write the equality being tested and compare both sides before deciding.
Check Your Understanding
For each problem, identify the useful representation and work from the support or CDF values given.
- A joint PMF has rows \(X=0,1\) and columns \(Y=0,1\), with entries \(p(0,0)=0.15\), \(p(0,1)=0.25\), \(p(1,0)=0.35\), and \(p(1,1)=0.25\). Find \(P(X=1)\), \(P(Y=1)\), \(P(X=1\text{ or }Y=1)\), and decide whether the variables are independent. (Answer: \(P(X=1)=0.60\), \(P(Y=1)=0.50\), the union probability is \(0.85\), and the variables are dependent.)
- A joint density is \(f_{X,Y}(x,y)=3x\) on \(0<y<x<1\), and zero elsewhere. Find \(f_X(x)\), \(f_Y(y)\), and \(P(Y<1/2)\). Are the variables independent? (Answer: \(f_X(x)=3x^2\) on \(0<x<1\), \(f_Y(y)=\frac32(1-y^2)\) on \(0<y<1\), \(P(Y<1/2)=\frac{11}{16}\), and the variables are dependent.)
- For \(X,Y\in\{0,1\}\), suppose \(F(0,0)=0.12\), \(F(1,0)=0.42\), \(F(0,1)=0.52\), and \(F(1,1)=1\). Find \(P(X=0,Y=1)\), \(P(X+Y\geq1)\), and decide independence. (Answer: \(P(X=0,Y=1)=0.40\), \(P(X+Y\geq1)=0.88\), and the variables are dependent.)
- A density is \(f_{X,Y}(x,y)=1\) on \(0<x<1,\ 0<y<1\), and zero elsewhere. Find both marginal densities and \(P(X\leq0.4,Y>0.7)\); test independence. (Answer: both marginal densities equal \(1\) on \((0,1)\), the probability is \(0.4(0.3)=0.12\), and the variables are independent.)
- A joint CDF gives \(F(2,3)=0.70\), \(F(1,3)=0.40\), \(F(2,2)=0.50\), and \(F(1,2)=0.30\). Find \(P(1<X\leq2,\ 2<Y\leq3)\). (Answer: \(0.70-0.40-0.50+0.30=0.10\).)