Convolving finite distributions

What you will learn

Derive an unequal-weight finite sum law by convolution, audit it with tails and means, and construct a marginal-preserving joint-table change that invalidates the product formula.

Prerequisites: Transforming distributions and losing information · Testing event independence · Repeated bets and dependence

Two marginal laws need a joint mechanism

Assume X is a signed first-stage adjustment with values −1, 1 and 2 and probabilities 1/4, 2/4 and 1/4. Assume Y is a second-stage adjustment taking 0 or 2 with probabilities 1/4 and 3/4. These finite laws are stipulated, exact model inputs. They are not estimated from measurements and represent numerical scores rather than real money. Define S=X+Y for the same two-stage case. We initially assume X and Y are independent. They need not have identical distributions, and negative values are allowed. Nothing here assumes a normal approximation or an infinite sequence of trials.

The separate laws are called marginals. A joint law gives the probability of every pair (x,y), including which first-stage values can occur with which second-stage values. Under the independence assumption, multiply the corresponding marginal probabilities to fill each joint cell. With a common denominator of sixteen, the row for X=−1 has weights 1 and 3; the row for X=1 has weights 2 and 6; the row for X=2 has weights 1 and 3. The six cells are not equally likely. Their weights sum to sixteen, row totals are 4,8,4 and column totals are 4,12. Those checks verify compatibility with both stated marginals.

Collect pairs along a constant-sum diagonal

For a target total s, every contributing pair must satisfy y=s−x. Different pairs are disjoint events, so add their joint masses. The general rule is P(S=s)=Σx P(X=x,Y=s−x). Independence permits replacing each joint term by pX(x)pY(s−x); this specialized sum is discrete convolution. Values outside a variable’s support have probability zero. Consequently a negative input or a gap in the support causes no problem: enumerate the supported pairs instead of guessing a range from zero.

Our six pairs produce totals −1,1,1,3,2,4 in row order. The pair (−1,2) and the pair (1,0) collide at total 1, giving weight 3+2=5. The complete sorted law is P(S=−1)=1/16, P(S=1)=5/16, P(S=2)=1/16, P(S=3)=6/16 and P(S=4)=3/16. There is no mass at zero. Listing all integers between the minimum and maximum as equally likely would invent an outcome and discard the joint weights. Equal sum labels do not identify equal numbers of cases, and equal numbers of cells do not imply equal probability.

Audit the law with a tail and an independent mean

The normalization check gives (1+5+1+6+3)/16=1. For the inclusive tail S≥3, collect totals 3 and 4, so the probability is (6+3)/16=9/16. A strict S>3 tail would contain only total 4 and have probability 3/16. This distinction is especially consequential in finite models because boundary points can carry substantial mass. In the activity you must place every joint cell, including the two cells at total 1; a plausible tail probability cannot excuse a missing or misrouted cell.

The mean from the resulting law is [−1×1+1×5+2×1+3×6+4×3]/16=36/16=2.25. Check it by a different calculation: E[X]=(−1+2+2)/4=3/4 and E[Y]=(0+6)/4=3/2, hence E[S]=3/4+3/2=9/4. Linearity of expectation holds even without independence. Agreement of these two means is a useful arithmetic audit, but it cannot certify that the complete sum distribution or independence assumption is right. Many different joint tables can preserve the same marginals and mean.

Move probability without changing either marginal

Now deliberately alter the joint mechanism. Add d units of sixteenth-mass to cells (−1,0) and (1,2), and subtract d from (−1,2) and (1,0). Leave the X=2 row untouched. Each affected row gains and loses the same mass. Each column also gains and loses the same mass. Thus the new table has exactly the original marginals, regardless of d, as long as every cell remains nonnegative. The four affected weights become 1+d, 3−d, 2−d and 6+d. They permit −1≤d≤2; our activity explores the integer settings −1,0,1,2.

At d=2 the rows are [3,1], [0,8] and [1,3]. A zero joint cell is allowed even though its two marginal probabilities are positive: that pair is impossible under this dependent mechanism. The sum law now has weights 3,1,1,8,3 at the same ordered totals −1,1,2,3,4. Its inclusive upper tail is 11/16, while its mean remains [−3+1+2+24+12]/16=36/16. Convolution of the unchanged marginals would still return the original law and the incorrect tail 9/16. Preserving row sums, column sums and the mean is therefore insufficient to justify multiplying marginal probabilities.

Separate a computation from an assumption certificate

In this family, independence holds exactly at d=0. For example, independence requires the (−1,0) cell to equal (1/4)(1/4)=1/16. At any nonzero d that cell differs, furnishing a direct witness against independence. At d=0 every cell equals its marginal product, which verifies the full finite condition. Checking just one cell can refute independence, but a single matching cell generally cannot prove it. Here the entire displayed table can be compared with all six products.

Use the workshop in two stages of reasoning. First choose a coupling, route every joint cell into its exact sum bucket and inspect the accumulated masses. Then enter the inclusive upper-tail probability and mean, and classify whether the product formula is justified for that coupling. Change d and compare what moves and what stays fixed. The controls model a family of hypothetical joint mechanisms, not evidence that a real data pipeline has that dependence. If only the two marginals were supplied without a joint mechanism or independence assumption, the full law of the total would remain undetermined. Report that missing information instead of silently choosing the product table.

Hands-on: Route joint cells and move mass between couplings

Build an unequal-weight sum law, vary a marginal-preserving rectangle transfer, audit an inclusive tail and mean, and test independence.

Common mistakes

• Adding marginal PMFs at matching labels instead of multiplying joint contributions and grouping by sum.
• Assigning equal weights to six pairs even though the independent cell weights differ.
• Dropping negative supports or inventing mass at zero because it lies between the extremes.
• Treating an unchanged mean as proof of independence or of an unchanged tail.
• Allowing d outside [−1,2], which creates a negative probability.

Keep these ideas

• Sum joint masses for all pairs with x+y=s.
• Multiply marginals only with a justified independence assumption.
• Check support, total mass, a boundary-sensitive tail and a separately calculated mean.
• A balanced rectangle move preserves both marginals while changing the sum law.

Further reading

• Data 140: Exact Distribution of a Sum, section 14.1 ↗

  Read the opening general joint-mass formula and its independent discrete convolution specialization. Original signed supports, rectangle transfers and diagnostic exercises; no source exercises or video copied or reviewed. Checked 2026-10-24.