Let \(m_1(x)\)and \(m_2(x)\) represent the ditribution functions of \(X_1\) and \(X_2\), then the convolution of \(m_1(x)\) and \(m_2(x)\) is the distribution function \(m_3=m_1*m_2\) given by \(m_3(j)=\sum_k m_1(k)m_2(j-k)\), for j = 0, 1, 2, 3.
m0 <- 1/8
m1 <- 3/8
m2 <- 1/2\(m_3(0)=m_1(0)m_2(0)=(\frac{1}{8})(\frac{1}{8})=\frac{1}{64}\)
m0*m0## [1] 0.0156251/64## [1] 0.015625\(m_3(1)=m_1(0)m_2(1)+m_1(1)m_2(0)=(\frac{1}{8})(\frac{3}{8})+(\frac{3}{8})(\frac{1}{8})=\frac{6}{64}=\frac{3}{32}\)
m0*m1+m1*m0## [1] 0.093753/32## [1] 0.09375\(m_3(2)=m_1(0)m_2(2)+m_1(1)m_2(1)+m_1(2)m_2(0)=(\frac{1}{8})(\frac{1}{2})+(\frac{3}{8})(\frac{3}{8})+(\frac{1}{2})(\frac{1}{8})=\frac{17}{64}\)
m0*m2+m1*m1+m2*m0## [1] 0.26562517/64## [1] 0.265625\(m_3(3)=m_1(1)m_2(2)+m_1(2)m_2(1)=(\frac{3}{8})(\frac{1}{2})+(\frac{1}{2})(\frac{3}{8})=\frac{3}{8}\)
m1*m2+m2*m1## [1] 0.3753/8## [1] 0.375