3.泊松分布例3 设 X ～ P(a),且分布律为ak-2.P[X = k} :k = 0,1,2,., 2 > 0.二k!则有11k88Z2ZleE(X)=kek!(k -1)!k=lk=0= e-r .e~ = .沈阳师范大学ShenYangNoemal Univenit
3. 泊松分布 e , 0,1,2, , 0. ! { = } = = − k k P X k k 则有 = − = 0 e ! ( ) k k k E X k 1 1 e ( 1)! k k k − − = = − = e e − = . 例3 设 X P ~ ( , ) 且分布律为
EX? = E[X(X -1)+ X]= E[X(X -1)I+ EXk+80Zk(k-2P十k!k=02k~2+822Z+ = e-~e~ + =? +.e(k - 2)!k=2所以DX = EX?-(EX)2 = 2 +-? = .泊松分布的期望和方差都等于参数2沈阳师范大学ShentYangNoemal Unive
2 EX E X X X = − + [ ( 1) ] = − + E X X EX [ ( 1)] + = − = − + 0 e ! ( 1) k k k k k + = − − + − = 2 2 2 ( 2)! e k k k = + − e e 2 . 2 = + 所以 2 2 DX EX EX = − ( ) 2 2 = + − = . 泊松分布的期望和方差都等于参数.