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：此问题显示为数字答案类型。
(一) 54
(乙) 55
(三) 56
(四) 57答案：(一)
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