如果随机变量X具有均值为5的泊松分布，则表达式E [(X + 2) 2 ]等于_____。
注意：此问题显示为“数值答案类型”。
(A) 54
(B) 55
(C) 56
(D) 57答案： (A)
解释：
Using Linearity of Expectation, we can write,
E[(X+2)2] = E[X2] + E[4X] + E[4]
In Poisson distribution, mean and variance are same.
Mean is given as 5. So variance should also be 5.
Also, variance = E[X2] – (E[X])2
5 = E[X2] – 25
E[X2] = 30
Thus E[(X+2)2] = 30 + 4*5 + 4 = 54. 这个问题的测验