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Discrete CAse

Z=X+Y

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

Since the events are disjoint we just sum their probabilities:

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

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:

\begin{displaymath}m_3(j)=\sum_k m_1(k) \dot m_2(j-k)
\end{displaymath}

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 Sk is the sum of k dice:

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

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



 

Susan Holmes
1998-12-07