Why a Sum Needs Its Own Distribution
Knowing the possible values and probabilities of two random variables does not immediately show how likely each value of their sum is. Several different pairs of outcomes can produce the same total. For example, when two fair dice are rolled, a total of 5 can come from \((1,4)\), \((2,3)\), \((3,2)\), or \((4,1)\). To find the probability of a total, we must account for every pair that produces it.
For independent variables, each pair's probability is the product of its two marginal probabilities. The convolution collects those products by total. A sum table makes the calculation visible: each row fixes a total, lists the pairs that produce it, and adds their probabilities.
Why the Convolution Formula Works
The formula follows from two facts: the different pairs that produce a fixed total cannot happen at the same time, and independence lets us multiply the probabilities of the two outcomes in each pair. More precisely, for a fixed \(s\), the event \(S=s\) consists of the possible events \((X=x,\ Y=s-x)\), one for each possible \(x\). These events are disjoint, so their probabilities add. Since \(X\) and \(Y\) are independent, each joint probability factors into a product.
This argument also explains the role of independence: the multiplication in the last line is justified by the model assumption that \(X\) and \(Y\) are independent. The marginal probability distributions alone do not generally determine the distribution of a sum when the variables are dependent.
If \(X\) and \(Y\) each have finitely many possible values, there are only finitely many nonzero terms. For discrete variables with countably many possible values, the sum can have infinitely many terms; the same formula applies, with the nonnegative probabilities summed over the possible values of \(X\).
A Reliable Sum-Table Method
Before calculating, identify the possible values and probabilities for each variable, and state that the variables are independent. Then, for each possible total \(s\), find every pair \((x,y)\) satisfying \(x+y=s\). Multiply \(P(X=x)\) by \(P(Y=y)\) for each pair, and add the products in that row. A pair must be included exactly once in the row for its total.
Record every possible value of \(X\) and \(Y\), with its probability. Check that each distribution's probabilities add to one.
Combine possible values from the two variables. A total is possible if at least one pair of outcomes produces it.
For each total \(s\), list all pairs \((x,y)\) satisfying \(x+y=s\).
For each pair, multiply its two probabilities, using independence. Add the products within each total's row to obtain \(P(S=s)\).
Worked Example: The Sum of Two Fair Dice
Suppose two fair six-sided dice are rolled independently. Let \(X\) and \(Y\) be their respective results, each with possible values 1 through 6 and probability \(1/6\) for each value. Find the distribution of \(S=X+Y\), including the probability of a total of 8.
The model conditions are satisfied: each die has a discrete distribution, and the rolls are stipulated to be independent. Thus every ordered pair \((x,y)\) has probability \((1/6)(1/6)=1/36\). The smallest total is \(1+1=2\), and the largest is \(6+6=12\).
| Total \(s\) | Contributing pairs \((x,y)\) | Number of pairs | \(P(S=s)\) |
|---|---|---|---|
| 2 | (1,1) | 1 | \(1/36\) |
| 3 | (1,2), (2,1) | 2 | \(2/36\) |
| 4 | (1,3), (2,2), (3,1) | 3 | \(3/36\) |
| 5 | (1,4), (2,3), (3,2), (4,1) | 4 | \(4/36\) |
| 6 | (1,5), (2,4), (3,3), (4,2), (5,1) | 5 | \(5/36\) |
| 7 | (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) | 6 | \(6/36\) |
| 8 | (2,6), (3,5), (4,4), (5,3), (6,2) | 5 | \(5/36\) |
| 9 | (3,6), (4,5), (5,4), (6,3) | 4 | \(4/36\) |
| 10 | (4,6), (5,5), (6,4) | 3 | \(3/36\) |
| 11 | (5,6), (6,5) | 2 | \(2/36\) |
| 12 | (6,6) | 1 | \(1/36\) |
For a total of 8, the five contributing pairs each have probability \(1/36\). Therefore,
The full distribution is a probability mass function: its numerators sum to \(1+2+3+4+5+6+5+4+3+2+1=36\), so its probabilities sum to \(36/36=1\).
Unequal Probabilities and Missing Values
When the variables have different probability distributions, counting pairs is not enough: pairs can have different probabilities. The convolution formula handles this by multiplying each pair's actual probabilities. It also handles values that are impossible for one variable: such pairs make no contribution.
Worked Example: A Three-Value Variable and a Two-Value Variable
Let \(X\) and \(Y\) be independent discrete random variables with the following distributions. Find the full distribution of \(S=X+Y\).
| Value of \(X\) | \(P(X=x)\) | Value of \(Y\) | \(P(Y=y)\) |
|---|---|---|---|
| 0 | 0.20 | 0 | 0.75 |
| 1 | 0.50 | 2 | 0.25 |
| 2 | 0.30 |
The probabilities for \(X\) add to \(0.20+0.50+0.30=1\), and those for \(Y\) add to \(0.75+0.25=1\). Independence is part of the stated model, so each pair probability is the product of its marginal probabilities. The possible totals are 0, 1, 2, 3, and 4.
| Total \(s\) | Contributing pair(s) \((x,y)\) | Probability product(s) | \(P(S=s)\) |
|---|---|---|---|
| 0 | (0,0) | \(0.20(0.75)=0.15\) | 0.15 |
| 1 | (1,0) | \(0.50(0.75)=0.375\) | 0.375 |
| 2 | (2,0), (0,2) | \(0.30(0.75)=0.225;\ 0.20(0.25)=0.05\) | 0.275 |
| 3 | (1,2) | \(0.50(0.25)=0.125\) | 0.125 |
| 4 | (2,2) | \(0.30(0.25)=0.075\) | 0.075 |
For example, a total of 2 can occur as \(X=2,Y=0\) or as \(X=0,Y=2\). The two pair events are disjoint, so
The distribution is complete because \(0.15+0.375+0.275+0.125+0.075=1\). The total of 2 has two contributors, while each other total has just one in this example. The table prevents either contributor from being missed.
A Short Check That Convolution Gives a Distribution
For finite supports, the probabilities produced by the formula must add to one. Each pair of possible values \((x,y)\) appears once in the row for its total \(s=x+y\). Summing the convolution across all totals therefore counts every pair probability once. Independence and the fact that each input distribution sums to one give the result:
In the second line, for any fixed \(x\), as \(s\) ranges over all totals, \(s-x\) ranges over all possible values of \(Y\), including values with probability zero. The inner sum is therefore the total probability for \(Y\). This verifies that the convolution accounts for all probability mass and assigns none twice.
Worked Example: The Sum of Two Independent Bernoulli Variables
Suppose two independent components each either function or fail during a test. Let \(X=1\) if the first functions and \(X=0\) otherwise, with \(P(X=1)=0.30\). Let \(Y=1\) if the second functions and \(Y=0\) otherwise, with \(P(Y=1)=0.60\). Find the distribution of the number \(S=X+Y\) of functioning components.
The variables are discrete and the test model specifies independence. Their remaining probabilities are \(P(X=0)=0.70\) and \(P(Y=0)=0.40\). The possible totals are 0, 1, and 2.
| Total \(s\) | Contributing pairs \((x,y)\) | Calculation | \(P(S=s)\) |
|---|---|---|---|
| 0 | (0,0) | \((0.70)(0.40)=0.28\) | 0.28 |
| 1 | (1,0), (0,1) | \((0.30)(0.40)+(0.70)(0.60)=0.12+0.42\) | 0.54 |
| 2 | (1,1) | \((0.30)(0.60)=0.18\) | 0.18 |
Thus, the probability that exactly one component functions is \(0.54\); it is the sum of the probabilities of the two distinct ways to obtain a total of 1. The distribution checks out because \(0.28+0.54+0.18=1\). The two ways to reach the same total need not have equal probabilities.
Common Mistakes and Why the Method Matters
A frequent error is to calculate \(P(X=x)P(Y=y)\) without checking independence. Multiplying marginal probabilities is justified only when the variables are independent. If a model specifies dependence, the pair probabilities must instead come from their joint distribution. For example, knowing only the separate chances that two components function is not enough to calculate the chance that both function if their outcomes are dependent.
Another error is to include only one pair for a total that has several contributors. The event \(S=s\) includes all possible pairs adding to \(s\); omitting one understates the probability. Conversely, listing a pair twice overstates it. A useful check is that each possible pair belongs to exactly one row, the row corresponding to its sum.
Convolution matters whenever a question concerns a total formed from independent discrete quantities. The inputs might represent separate measurements, counts, or outcomes. Their individual distributions describe each variable separately; convolution combines them into the distribution needed for a question about the total.
Check Your Understanding
Answer using the convolution formula and, where useful, a sum table.
- Two independent variables each take values 0 and 1 with probabilities \(P(0)=0.4\) and \(P(1)=0.6\). What pairs contribute to a total of 1?
- Suppose \(P(X=0)=0.3\), \(P(X=1)=0.7\), \(P(Y=0)=0.5\), and \(P(Y=2)=0.5\), with \(X\) and \(Y\) independent. Find \(P(X+Y=2)\).
- Why is it necessary to add the probabilities of different pairs that produce the same total?
- Two discrete variables have the marginal probabilities needed for a calculation, but they are dependent. What part of the convolution procedure cannot be assumed?
- What should the probabilities in a complete distribution for \(S=X+Y\) sum to, and why?