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For sequences $a,b$ indexed by the integers, discrete convolution adds products whose indices sum to the output index:For finite sequences, treat every unspecified entry as zero; the sum then has finitely many nonzero terms. For infinite sequences, a convergence hypothesis is needed, for example absolute summability of both sequences.

Take $a=(1,2,1)$ at indices $0,1,2$ and $b=(1,1)$ at indices $0,1$. Their convolution is $(1,3,3,1)$ at indices $0,1,2,3$:Equivalently, $c_n=a_n+a_{n-1}$: add an unshifted and a one-step shifted copy of $a$. This explains the output by superposition, with every index included.

The same products appear in polynomial multiplication and generating functions:A term $a_m{z}^m$ times $b_k{z}^k$ contributes to coefficient $m+k$, precisely the index condition in discrete convolution. In probability, $a$ counts the outcomes of two independent fair coin tosses by their number of heads, while $b$ counts the outcomes of one more toss. Dividing the output by eight gives the probability mass function $(1,3,3,1)/8$ of the binomial distribution with $n=3$ and $p=1/2$. No new arithmetic is needed to pass from coefficient multiplication to counting independent sums.

The full finite convolution is different from circular convolution, where indices wrap modulo a chosen length. Wrapping this four-term output to length three adds its last entry to its first and gives $(2,3,3)$. When using a discrete Fourier transform to compute the full convolution, zero padding to length at least the sum of the input lengths minus one prevents this wraparound. ^[ref]