Ch3-118 Poisson分布可加性的证明 X~P(1),Y~P(2),则 Z=X+Y的可能取值为0,1,2,, P(Z=k)=>P(X=i, Y=k-i), 大xe,2e2 ∑ o i!(k-i) 1-2k ∑ k! k k! i0i! (k-i) (41+2)e4b k=0.1.2 k!Ch3-118 X ~ P(1 ), Y ~ P(2 ), 则 Z = X + Y 的可能取值为 0,1,2, , ( ) ( , ), 0 = = = = = − k i P Z k P X i Y k i = − − − − = k i i k i k i e i e 0 1 2 ! ( )! 1 2 = − − − − = k i i k i i k i k k e 0 1 2 !( )! ! ! 1 2 ! ( ) 1 2 1 2 k e k − − + = k = 0,1,2, Poisson分布可加性的证明