Why Convolution Tables Help
When two independent discrete variables are added, a particular total may arise in several different ways. To find the probability of that total, each possible pair of component outcomes must be considered, its probability calculated, and the probabilities of all pairs producing the same total added. A convolution table organizes this work so that overlapping sums are not missed or counted twice.
This method is especially useful when the variables have small, finite sets of possible values but do not share the same support. It also makes gaps in the distribution visible: some totals between the smallest and largest possible sums may have probability zero.
The independence assumption matters: it gives \(P(X=x,Y=y)=p_X(x)p_Y(y)\). If \(X\) and \(Y\) are dependent, the probability of a pair generally cannot be obtained by multiplying the marginal probabilities. The same table structure can still be used, but each cell must contain the appropriate joint probability.
How to Build and Read the Table
Put the possible values of one variable down the rows and the values of the other across the columns. In each cell, record the sum of the row and column values and the probability of that pair. For independent variables, the cell probability is the product of the corresponding marginal probabilities. Finally, collect all cells with the same sum.
Write each variable’s possible values and probabilities. Use the stated supports, not an assumed common support.
For each pair \((x,y)\), record the sum \(x+y\) and, under independence, the probability \(p_X(x)p_Y(y)\).
Add the probabilities of every pair producing the same sum. Include zero-probability totals if the requested range includes them.
The resulting probabilities must be nonnegative and sum to one. This checks the arithmetic, though it does not by itself prove every cell was assigned correctly.
Worked Tables with Unequal Supports
Worked Example: Gaps in the Sum Distribution
Let \(X\) and \(Y\) be independent. Suppose \(X\) takes values 0 and 2 with probabilities \(1/3\) and \(2/3\), respectively, while \(Y\) takes values \(-1\), 1, and 4 with probabilities \(1/4\), \(1/2\), and \(1/4\), respectively. Find the pmf of \(Z=X+Y\), including any zero-probability totals from \(-1\) through 6.
Method. Use a pair table, multiplying the row and column probabilities. The supports are unequal, so retain all two \(X\)-values and all three \(Y\)-values.
| \(X\) | \(Y=-1\), probability \(1/4\) | \(Y=1\), probability \(1/2\) | \(Y=4\), probability \(1/4\) |
|---|---|---|---|
| \(X=0\), probability \(1/3\) | Sum \(-1\); probability \(1/12\) | Sum \(1\); probability \(1/6\) | Sum \(4\); probability \(1/12\) |
| \(X=2\), probability \(2/3\) | Sum \(1\); probability \(1/6\) | Sum \(3\); probability \(1/3\) | Sum \(6\); probability \(1/6\) |
For example, the cell for \((X,Y)=(2,-1)\) has sum \(2+(-1)=1\) and probability \((2/3)(1/4)=1/6\). The total 1 occurs in two cells, so \(P(Z=1)=1/6+1/6=1/3\). Grouping every cell by its sum gives:
| \(z\) | \(-1\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|
| \(P(Z=z)\) | \(1/12\) | 0 | \(1/3\) | 0 | \(1/3\) | \(1/12\) | 0 | \(1/6\) |
The unobserved totals 0, 2, and 5 have probability zero: no pair in the table produces them. As a check, \(1/12+1/3+1/3+1/12+1/6=1\). The table has six cells, and their probabilities also sum to one: \(1/12+1/6+1/12+1/6+1/3+1/6=1\).
Common mistake avoided: Listing the range of possible totals and assigning positive probability to every integer in that range. A total is possible only if at least one component-outcome pair produces it.
Why the Convolution Formula Holds
For any fixed value \(z\), the event \(\{X+Y=z\}\) consists of all the pair events \(\{X=x,Y=y\}\) satisfying \(x+y=z\). Distinct pairs are mutually exclusive: \(X\) and \(Y\) cannot simultaneously take two different specified values. Therefore, the probability of their union is the sum of their probabilities. Independence then factors each pair probability.
This argument explains both operations in the table: multiply within each pair because the variables are independent, then add across pairs because the pair events are disjoint. The finite supports ensure that only finitely many terms need to be considered.
Worked Example: Combining All Pairs for a Requested Total
Let \(C\) and \(D\) be independent. Suppose \(C\) takes values 1, 2, and 4 with probabilities \(1/4\), \(1/2\), and \(1/4\). Suppose \(D\) takes values \(-1\) and 1 with probabilities \(1/3\) and \(2/3\). Find the pmf of \(C+D\), and then find \(P(C+D=3)\).
Method. Multiply the probabilities for each of the six pairs, then combine cells with matching sums.
| \(C\) | \(D=-1\), probability \(1/3\) | \(D=1\), probability \(2/3\) |
|---|---|---|
| \(C=1\), probability \(1/4\) | Sum \(0\); probability \(1/12\) | Sum \(2\); probability \(1/6\) |
| \(C=2\), probability \(1/2\) | Sum \(1\); probability \(1/6\) | Sum \(3\); probability \(1/3\) |
| \(C=4\), probability \(1/4\) | Sum \(3\); probability \(1/12\) | Sum \(5\); probability \(1/6\) |
The distinct sum values are 0, 1, 2, 3, and 5. Only the two cells with sum 3 are combined:
The full pmf is \(P(C+D=0)=1/12\), \(P(C+D=1)=1/6\), \(P(C+D=2)=1/6\), \(P(C+D=3)=5/12\), and \(P(C+D=5)=1/6\). These add to \(1/12+1/6+1/6+5/12+1/6=12/12=1\). Thus the requested probability is \(5/12\).
Common mistake avoided: Counting only \((2,1)\) as a way to obtain 3 gives \(1/3\), but \((4,-1)\) also gives 3 and contributes \(1/12\). Both disjoint possibilities must be included.
From the Sum Pmf to an Event Probability
Once the sum pmf is assembled, an event involving the sum can be answered by adding the probabilities of all included totals. For an inequality such as \(X+Y\leq c\), the event includes every pair whose total is at most \(c\), including pairs that produce the same total. It is often clearest to group by total first and then add the relevant pmf entries.
Worked Example: A Cumulative Probability with Unequal Supports
Let \(K\) and \(L\) be independent. Suppose \(K\) takes values 0 and 2 with probabilities \(1/3\) and \(2/3\), while \(L\) takes values 1 and 3 with probabilities \(1/4\) and \(3/4\). Find \(P(K+L\leq 3)\).
Method. Compute all four pair probabilities, group pairs with the same sum, and add the pmf entries for totals no greater than 3.
| \(K\) | \(L=1\), probability \(1/4\) | \(L=3\), probability \(3/4\) |
|---|---|---|
| \(K=0\), probability \(1/3\) | Sum \(1\); probability \(1/12\) | Sum \(3\); probability \(1/4\) |
| \(K=2\), probability \(2/3\) | Sum \(3\); probability \(1/6\) | Sum \(5\); probability \(1/2\) |
There are two pairs with sum 3. Their combined probability is \(1/4+1/6=3/12+2/12=5/12\). The sum pmf is therefore \(P(K+L=1)=1/12\), \(P(K+L=3)=5/12\), and \(P(K+L=5)=1/2\). Hence:
The probabilities check: \(1/12+5/12+1/2=1\). A direct pair check gives the same result: the qualifying pairs are \((0,1)\), \((0,3)\), and \((2,1)\), with probabilities \(1/12\), \(1/4\), and \(1/6\); their sum is \(1/2\).
Common mistake avoided: Including only \((0,1)\) and obtaining \(1/12\), or counting \((0,3)\) but overlooking \((2,1)\), fails to include every pair meeting the inequality. The boundary value 3 is included because the condition is \(\leq 3\).
Exam Checks and Common Traps
A convolution table is a finite enumeration, so accuracy depends on keeping the pair outcomes distinct until their probabilities have been calculated. In particular, do not combine matching sums before computing the probability of each contributing pair. Multiplying the probability of one pair by the number of pairs is valid only when those pair probabilities are equal.
- Forgetting independence: The product \(p_X(x)p_Y(y)\) is justified only when \(X\) and \(Y\) are independent. Otherwise, use the joint probabilities.
- Dropping a second route to a total: Scan every cell for each requested sum. Different pairs can produce the same total.
- Assuming every intermediate total is possible: A total with no matching pair has probability zero, even if it lies between the minimum and maximum sums.
- Misreading an inequality: For \(X+Y\leq c\), include the boundary total \(c\); for \(X+Y<c\), do not include it.
- Skipping the total-probability check: If the output probabilities do not sum to one, revisit the cell products and the grouping of equal sums.
This technique is aligned with the SOA Exam P (Probability) syllabus. In a timed problem, a compact table is usually safer than trying to reason directly from a single target event, especially when several pairs yield the same sum or the supports have gaps.
Check Your Understanding
For each problem, first list the pairs that produce the requested total or event, then add their probabilities.
- Independent \(A\) takes values 0 and 3 with probabilities \(1/2\) each. Independent \(B\) takes values \(-1\) and 2 with probabilities \(1/4\) and \(3/4\). Find \(P(A+B=2)\). (Answer: \(1/2\))
- Independent \(U\) takes values 1 and 2 with probabilities \(2/5\) and \(3/5\). Independent \(V\) takes values 0 and 4 with probabilities \(3/4\) and \(1/4\). Find \(P(U+V\leq 3)\). (Answer: \(3/4\))
- Independent \(X\) takes values 0 and 2 with probabilities \(1/4\) and \(3/4\). Independent \(Y\) takes values \(-2\) and 1 with probabilities \(2/3\) and \(1/3\). Find \(P(X+Y\geq 1)\). (Answer: \(1/3\))
- Independent \(M\) takes values \(-1\) and 2 with probabilities \(1/4\) and \(3/4\). Independent \(N\) takes values 0, 1, and 3 with probabilities \(1/2\), \(1/4\), and \(1/4\). Find \(P(M+N>2)\). (Answer: \(3/8\))
- Independent \(R\) takes values \(-2\) and 1 with probabilities \(1/3\) and \(2/3\). Independent \(S\) takes values 0 and 2 with probabilities \(3/4\) and \(1/4\). Find the pmf of \(R+S\), including any zero-probability integer totals from \(-2\) through 3. (Answer: \(P(R+S=-2)=1/4\), \(P(R+S=-1)=0\), \(P(R+S=0)=1/12\), \(P(R+S=1)=1/2\), \(P(R+S=2)=0\), \(P(R+S=3)=1/6\))