Finite-Support Convolution Tables

Why Use a Convolution Table?

When two independent random variables take only finitely many values, their sum is also finite-support. The calculation is conceptually simple: consider every possible pair of values, find its probability, and record the sum produced by that pair. The detail that most often requires care is that different pairs can produce the same sum. Their probabilities must be added.

A convolution table makes both parts visible. The row and column labels list the possible values of the two variables; each cell corresponds to one pair. Its probability is the product of the two marginal probabilities, by independence, and its sum is the value contributed by that pair. Grouping the cells with the same sum gives the distribution of the total.

Definition and Table Construction

Definition: Let \(X\) and \(Y\) be independent discrete random variables with finite supports \(A=\{x_1,\ldots,x_r\}\) and \(B=\{y_1,\ldots,y_s\}\). Write \(p_X(x)=P(X=x)\) and \(p_Y(y)=P(Y=y)\). For \(Z=X+Y\), the convolution table has one cell for each pair \((x,y)\in A\times B\). The cell records the sum \(x+y\) and the probability product \(p_X(x)p_Y(y)\). To obtain \(P(Z=z)\), add the products in all cells whose sums equal \(z\).

In formula form, the table calculation is

P(X+Y=z)=\sum_{\substack{x\in A,\ y\in B\\x+y=z}}p_X(x)p_Y(y).

If there are no pairs with sum \(z\), the sum on the right is empty and \(P(X+Y=z)=0\). It is usually helpful to write a cell as “sum; probability,” so the outcome and its weight are not confused. A table can have repeated sum labels in different cells: those cells remain distinct pairs until their probabilities are accumulated.

List the supports and probabilities.
Include each value with positive probability and check that each variable’s probabilities sum to one.

Fill every pair cell.
For the pair \((x,y)\), record the sum \(x+y\) and the product \(p_X(x)p_Y(y)\).

Group by sum.
Add the probabilities of all cells with a common sum; list each distinct attainable sum once.

Check the result.
Confirm that the support consists exactly of attainable sums and that the accumulated probabilities total one.

Worked Examples

Worked Example: Two Unequal Finite Supports

Suppose \(X\) takes values \(0,2,5\) with probabilities \(1/5,1/2,3/10\), respectively, and \(Y\) takes values \(-1,1\) with probabilities \(1/4,3/4\), respectively. Assume \(X\) and \(Y\) are independent. Find the distribution of \(Z=X+Y\).

The entries below show the sum in each cell, followed by the probability product in parentheses. For example, the upper-left cell corresponds to \(X=0,Y=-1\), so its sum is \(-1\) and its probability is \((1/5)(1/4)=1/20\).

\(X\) value	\(Y=-1\), probability \(1/4\)	\(Y=1\), probability \(3/4\)
\(X=0\), probability \(1/5\)	\(-1;\ 1/20\)	\(1;\ 3/20\)
\(X=2\), probability \(1/2\)	\(1;\ 1/8\)	\(3;\ 3/8\)
\(X=5\), probability \(3/10\)	\(4;\ 3/40\)	\(6;\ 9/40\)

The sum \(1\) occurs in two cells, so its probability is \(3/20+1/8=6/40+5/40=11/40\). The other sums each occur once. Thus the distribution is

\begin{array}{c|ccccc} z&-1&1&3&4&6\\ \hline P(Z=z)&1/20&11/40&3/8&3/40&9/40 \end{array}

The attainable sums are exactly \(\{-1,1,3,4,6\}\), as shown by the six table cells after repeated sums are listed only once. The probabilities sum to \[ \frac{2}{40}+\frac{11}{40}+\frac{15}{40}+\frac{3}{40}+\frac{9}{40} =\frac{40}{40}=1. \] This verifies that the table accounts for the full distribution.

Worked Example: Several Pairs Contribute to One Sum

Let \(X\) have values \(1,2,3\) with probabilities \(1/2,1/3,1/6\), and let \(Y\) have values \(0,1,2\) with probabilities \(1/4,1/2,1/4\). Suppose the variables are independent. Calculate the law of \(Z=X+Y\).

The nine cells are:

\(X\) value	\(Y=0\), probability \(1/4\)	\(Y=1\), probability \(1/2\)	\(Y=2\), probability \(1/4\)
\(X=1\), probability \(1/2\)	\(1;\ 1/8\)	\(2;\ 1/4\)	\(3;\ 1/8\)
\(X=2\), probability \(1/3\)	\(2;\ 1/12\)	\(3;\ 1/6\)	\(4;\ 1/12\)
\(X=3\), probability \(1/6\)	\(3;\ 1/24\)	\(4;\ 1/12\)	\(5;\ 1/24\)

The possible sums are all the integers from \(1\) through \(5\). Accumulating repeated sums gives

\begin{aligned} P(Z=1)&=\frac18,\\ P(Z=2)&=\frac14+\frac1{12}=\frac13,\\ P(Z=3)&=\frac18+\frac16+\frac1{24}=\frac13,\\ P(Z=4)&=\frac1{12}+\frac1{12}=\frac16,\\ P(Z=5)&=\frac1{24}. \end{aligned}

For the check, use denominator \(24\): \[ \frac18+\frac13+\frac13+\frac16+\frac1{24} =\frac3{24}+\frac8{24}+\frac8{24}+\frac4{24}+\frac1{24} =\frac{24}{24}=1. \] Notice that the largest individual cell probability is not automatically \(P(Z=3)\): three different pairs contribute to that value, and all three products are needed.

Worked Example: A Sum Distribution with Gaps

Let \(X\) take values \(-2,1,4\) with probabilities \(1/2,1/3,1/6\), and let \(Y\) take values \(0,3\) with probabilities \(2/3,1/3\). Assume independence. Find the support and probabilities of \(Z=X+Y\).

\(X\) value	\(Y=0\), probability \(2/3\)	\(Y=3\), probability \(1/3\)
\(X=-2\), probability \(1/2\)	\(-2;\ 1/3\)	\(1;\ 1/6\)
\(X=1\), probability \(1/3\)	\(1;\ 2/9\)	\(4;\ 1/9\)
\(X=4\), probability \(1/6\)	\(4;\ 1/9\)	\(7;\ 1/18\)

The distinct sums in the cells are \(-2,1,4,7\); values such as \(0,2,\) and \(3\) cannot occur. Combining the two cells with sum \(1\), and the two cells with sum \(4\), gives

\begin{aligned} P(Z=-2)&=\frac13,\\ P(Z=1)&=\frac16+\frac29=\frac7{18},\\ P(Z=4)&=\frac19+\frac19=\frac29,\\ P(Z=7)&=\frac1{18}. \end{aligned}

Their total is \(6/18+7/18+4/18+1/18=18/18=1\). This example shows why the support should be found from the actual pair sums, not guessed by filling every value between the minimum and maximum.

Why the Table Formula Is Correct

Theorem: If \(X\) and \(Y\) are independent discrete random variables with finite supports \(A\) and \(B\), then for every real number \(z\), \[ P(X+Y=z)=\sum_{\substack{x\in A,\ y\in B\\x+y=z}}p_X(x)p_Y(y). \] Moreover, the support of \(X+Y\) is exactly \(\{x+y:x\in A,\ y\in B\}\).

Proof. The event \(\{X+Y=z\}\) is the union, over pairs \((x,y)\in A\times B\) satisfying \(x+y=z\), of the events \(\{X=x,Y=y\}\). These events are disjoint: two different value pairs cannot both be the realized pair \((X,Y)\). Additivity of probability over this finite disjoint union gives

P(X+Y=z)=\sum_{\substack{x\in A,\ y\in B\\x+y=z}}P(X=x,Y=y).

Independence gives \(P(X=x,Y=y)=P(X=x)P(Y=y)=p_X(x)p_Y(y)\) for every pair, proving the formula. By definition of support, each \(x\in A\) and \(y\in B\) has positive probability. If \(z=x+y\) for such a pair, then the corresponding product is positive, so \(P(X+Y=z)>0\). Conversely, if \(P(X+Y=z)>0\), the formula has a positive finite sum and therefore at least one contributing pair, which satisfies \(z=x+y\). This proves the support statement. \(\square\)

The same argument also explains the total-probability check. Every cell corresponds to exactly one possible pair \((x,y)\), and the probabilities of all pair events sum to one:

\sum_{x\in A}\sum_{y\in B}p_X(x)p_Y(y) = \left(\sum_{x\in A}p_X(x)\right) \left(\sum_{y\in B}p_Y(y)\right) =1\cdot1=1.

Grouping these cell probabilities according to their sum does not change their total. Thus a correctly constructed and fully accumulated convolution table must yield probabilities totaling one. If it does not, a cell may have been omitted, a product may be incorrect, or a repeated sum may have been handled improperly.

Common Mistakes and Practical Checks

A frequent error is to report the probability of one pair as the probability of its sum. For example, in the first example, the pair \((0,1)\) has probability \(3/20\), but the sum \(1\) can also arise from \((2,-1)\). Therefore \(P(X+Y=1)=3/20+1/8\), not \(3/20\).

Another error is to list a continuous-looking interval between the smallest and largest attainable sums as the support. Finite supports need not fill such an interval. The support is obtained by adding every value in one support to every value in the other and retaining the distinct results. The table makes missing values apparent.

Check independence. The cell probability is the product \(p_X(x)p_Y(y)\) only under independence. Without independence, the joint probabilities \(P(X=x,Y=y)\) are needed instead.
Keep pair cells distinct at first. Even if two cells have the same sum, they represent different events and their probabilities are added only when forming the sum distribution.
Check both support and total mass. A total of one is a useful arithmetic check, but it does not by itself prove that each sum has been assigned correctly.

Key takeaway: In a finite-support convolution table, multiply probabilities within each value pair, then add products across all pairs producing the same sum. The attainable pair sums give the exact support, and the accumulated probabilities must total one.

Check Your Understanding

Try these questions by sketching a table and grouping its cells by sum.

If \(X\) has support \(\{0,3\}\) and \(Y\) has support \(\{1,2\}\), what is the support of \(X+Y\)? Which sum can arise from more than one pair?
Why does independence allow a cell for \((x,y)\) to be assigned probability \(p_X(x)p_Y(y)\)?
In a table, three cells have sum \(z\) and probabilities \(1/10,1/5,\) and \(1/4\). What is \(P(X+Y=z)\)?
Can the support of a sum omit a value between its minimum and maximum? Explain using the pair-sum rule.
What does a total probability of \(0.9\) indicate you should check in your convolution table?