Discrete CAse

Z=X+Y

$$\{Z=z\}= \cup_{k=0}^\infty ((X=k),(Y=z-k))$$

Since the events are disjoint we just sum their probabilities:

$$P(Z=z)=\sum_{k=0}^\infty P(X=k) \dot P(Y=z-k)$$

Definition: For X and Y two random variables with mass distribution functions m₁ and m₂, the convolution of m₁ and m₂ is the distribution function m₃ denoted m₁*m₂ given by:

$$m_3(j)=\sum_k m_1(k) \dot m_2(j-k)$$

m₃ is the distribution function of the random variable X+Y.

This operation is commutative and associative. We use induction to find the distribution of the sum of n random variables. Suppose that S_k is the sum of k dice:

$$P(S_3=3)=P(S_2=2)P(X_3=1)=\frac{1}{36}\frac{1}{6}$$

See another example for the distribution of the sum of points in a bridge hand.