Why Constraints Change the Sum
For a pair of integer-valued random variables, a joint probability calculation is a sum over lattice points: each possible pair \((i,j)\) contributes its joint mass. Constraints such as \(i+j=4\), \(i\leq j\), or \(i+j\leq 4\) determine which points are possible. A sum that includes points outside the support counts impossible outcomes; a sum that overlooks admissible points misses probability.
The practical task is therefore to translate the constraint into an exact set of integer indices before adding any probabilities. This is particularly important at boundaries: an inequality may include an endpoint, and an equality such as \(i+j=m\) usually leaves only a one-dimensional set of possible pairs.
For a fixed first coordinate \(i\), the admissible partners form the slice \(S_i=\{j\in\mathbb{Z}:(i,j)\in S\}\). The marginal mass at \(i\) is obtained by summing across that slice. More generally, the probability of a region is obtained by summing over the intersection of that region with the support.
These sums may be finite or countably infinite. For an infinite support, they are sums of nonnegative terms, so they are well-defined as limits of increasing finite partial sums. In practice, first writing down the admissible indices makes it easier to see whether the sum is finite, geometric, or another familiar series.
Reading the Integer Bounds
A constraint on the coordinates often becomes a simple bound once one coordinate is fixed. For example, suppose \(a\leq i\leq b\), \(c\leq j\leq d\), and \(r\leq i+j\leq s\), where all the bounds are integers. If \(i\) is fixed, the allowed values of \(j\) must satisfy both the coordinate bounds and the sum bounds. The slice is therefore given by the following range, provided its lower endpoint does not exceed its upper endpoint.
Indeed, \(j\geq c\) and \(i+j\geq r\) together require \(j\geq\max(c,r-i)\). Likewise, \(j\leq d\) and \(i+j\leq s\) require \(j\leq\min(d,s-i)\). If the lower bound exceeds the upper bound, there are no admissible points in that slice. For strict inequalities, remember that the coordinates are integers: for instance, \(j>r-i\) means \(j\geq r-i+1\).
Worked Example: A Uniform Triangular Support
Suppose \((X,Y)\) is uniform on the positive-integer pairs satisfying \(x+y\leq 4\). Find the marginal mass functions and calculate \(P(X\leq Y)\) and \(P(X+Y=3)\).
The constraints \(x\geq1\), \(y\geq1\), and \(x+y\leq4\) give the six points \((1,1),(1,2),(1,3),(2,1),(2,2),(3,1)\). Uniformity assigns mass \(1/6\) to each point.
For \(X=1\), there are three possible values of \(Y\); for \(X=2\), there are two; and for \(X=3\), there is one. Thus \(f_X(1)=3/6\), \(f_X(2)=2/6\), and \(f_X(3)=1/6\). The same counts apply to \(Y\), so \(f_Y(1)=3/6\), \(f_Y(2)=2/6\), and \(f_Y(3)=1/6\). Each marginal sums to \(1\).
The points with \(X\leq Y\) are \((1,1),(1,2),(1,3),(2,2)\). There are four, each with mass \(1/6\), so \(P(X\leq Y)=4/6=2/3\). The points with \(X+Y=3\) are \((1,2)\) and \((2,1)\), giving \(P(X+Y=3)=2/6=1/3\).
Summing Along Slices
The slice formula is an instance of countable additivity. It is useful to state the calculation explicitly because constraints often make one marginal slice shorter than another, or leave some slices empty.
Proof. The event \(\{(X,Y)\in A\}\) is the union of the events \(\{X=i,Y=j\}\) over \((i,j)\in A\cap S\). These events are pairwise disjoint: a single outcome \((X,Y)\) cannot equal two different pairs. Countable additivity therefore gives \(P((X,Y)\in A)=\sum_{(i,j)\in A\cap S}P(X=i,Y=j)\), which is the stated region formula. Taking \(A=\{(i,j):i=i_0\}\), for a fixed integer \(i_0\), gives \(f_X(i_0)=\sum_{j:(i_0,j)\in S}f(i_0,j)\). Taking \(A=\{(i,j):j=j_0\}\) gives the corresponding formula for \(f_Y(j_0)\). \(\square\)
For a support confined to a fixed-sum line, a slice contains at most one point. In that case the marginal calculation has a particularly direct form.
Proof. On the event of probability one where \(X+Y=m\), the value of \(Y\) must equal \(m-X\). For a fixed \(i\), the only point on the line \(x+y=m\) with first coordinate \(i\) is \((i,m-i)\). Summing the joint mass over that slice therefore gives \(f_X(i)=f(i,m-i)\). The argument for fixed \(j\) gives \(f_Y(j)=f(m-j,j)\). Finally, both expressions refer to the mass at the same point \((m-j,j)\), so \(f_Y(j)=f_X(m-j)\). \(\square\)
Worked Example: Marginals on a Fixed-Sum Line
Let \((X,Y)\) have support \(\{(i,4-i):i=0,1,2,3,4\}\), with \(f(i,4-i)=\binom{4}{i}/16\). Find both marginals and calculate \(P(X\leq1,\ Y\geq3)\).
The masses along the line, in order of \(i=0,1,2,3,4\), are \(1/16,4/16,6/16,4/16,1/16\). They sum to \((1+4+6+4+1)/16=16/16=1\). Since each \(X=i\) corresponds to exactly one supported point, \(f_X(i)=\binom{4}{i}/16\) for \(i=0,\ldots,4\), and is zero otherwise.
The relation \(Y=4-X\) reverses the order of these masses. Thus \(f_Y(j)=\binom{4}{4-j}/16=\binom{4}{j}/16\) for \(j=0,\ldots,4\), using \(\binom{4}{4-j}=\binom{4}{j}\).
On the support, \(Y\geq3\) means \(4-X\geq3\), or \(X\leq1\). The event therefore consists of \((0,4)\) and \((1,3)\), whose masses are \(1/16\) and \(4/16\). Hence \(P(X\leq1,\ Y\geq3)=1/16+4/16=5/16\).
An Infinite Ordered Support
Constraints need not produce a finite list of points. The same method applies to an infinite support, provided the joint masses sum to one. A geometric form makes it possible to carry out every sum explicitly.
Worked Example: A Geometric Mass on an Ordered Lattice
For nonnegative integers \(i,j\) with \(i\leq j\), define \(f(i,j)=\frac{3}{8}2^{-(i+j)}\), and set \(f(i,j)=0\) elsewhere. Verify the total mass, find both marginal mass functions, and calculate \(P(X=Y)\).
To sum the joint masses, first fix \(i\). The admissible values of \(j\) are \(i,i+1,\ldots\). Using the geometric-series identity \(\sum_{k=0}^{\infty}r^k=1/(1-r)\) for \(0\leq r<1\), \[ \sum_{i=0}^{\infty}\sum_{j=i}^{\infty}\frac{3}{8}2^{-(i+j)} =\frac{3}{8}\sum_{i=0}^{\infty}2^{-i}\frac{2^{-i}}{1-1/2} =\frac{3}{4}\sum_{i=0}^{\infty}4^{-i} =\frac{3}{4}\cdot\frac{1}{1-1/4}=1. \] Thus the displayed masses define a probability distribution.
For fixed \(i\), summing over \(j\geq i\) gives \[ f_X(i)=\frac{3}{8}2^{-i}\sum_{j=i}^{\infty}2^{-j} =\frac{3}{8}2^{-i}\frac{2^{-i}}{1-1/2} =\frac{3}{4}4^{-i},\qquad i=0,1,2,\ldots. \] For fixed \(j\), the constraint allows \(i=0,\ldots,j\), so \[ f_Y(j)=\frac{3}{8}2^{-j}\sum_{i=0}^{j}2^{-i} =\frac{3}{8}2^{-j}\frac{1-2^{-(j+1)}}{1-1/2} =\frac{3}{4}2^{-j}\bigl(1-2^{-(j+1)}\bigr),\qquad j=0,1,2,\ldots. \] Both marginals are zero at integers outside the stated ranges.
The event \(X=Y\) consists of the diagonal points \((i,i)\). Therefore \[ P(X=Y)=\sum_{i=0}^{\infty}\frac{3}{8}2^{-2i} =\frac{3}{8}\sum_{i=0}^{\infty}4^{-i} =\frac{3}{8}\cdot\frac{4}{3}=\frac{1}{2}. \] The remaining probability \(1/2\) lies on the strictly ordered points \(X<Y\).
Common Mistakes and a Reliable Method
A frequent error is to replace a lattice sum by an area calculation. For a discrete pair, the probability of a region is the sum of the masses at its supported integer points; the continuous area of the region does not enter. Another common error is to sum over all integer pairs satisfying the event while forgetting that some of those pairs are outside the support.
Boundary conditions matter as well. The constraints \(i+j\leq4\) and \(i+j<4\) differ at the lattice points with \(i+j=4\). Likewise, when converting a strict inequality into an integer range, the endpoint must be adjusted by one. Always check which points are included before summing.
List its constraints using integer coordinates, including nonnegativity or any lower and upper bounds.
For a marginal, fix one coordinate. For an event, combine the event’s inequalities with the support constraints.
Use inclusive or strict integer bounds carefully, and omit empty slices.
For an infinite range, identify a convergent series or use increasing finite partial sums.
Marginal masses should sum to one, and an event probability must lie between zero and one.
The central discipline is to keep the support visible throughout the calculation. A fixed-sum constraint reduces a slice to one point, an ordering constraint creates finite or infinite ranges, and a triangular constraint changes the available range from slice to slice. Once those indices are correct, the probability calculation is a direct sum of the relevant joint masses.
Check Your Understanding
For each question, identify the admissible lattice points before calculating.
- If \((X,Y)\) is uniform on the nonnegative integer pairs with \(x+y\leq2\), how many support points are there, and what is \(P(X=Y)\)?
- If \(P(X+Y=6)=1\), express \(f_Y(2)\) in terms of the joint mass function and in terms of \(f_X\).
- For support \(0\leq i\leq3,\ 1\leq j\leq4,\ i+j\leq4\), what values of \(j\) are admissible when \(i=2\)?
- On the support \(0\leq i\leq j\), describe the points in the event \(X<Y\) and explain how their probability can be found from the joint mass function.
- Why is the area of a triangular region not, by itself, the probability for an integer-valued pair supported on that region?