定义
二项分布的期望和方差
期望
[EX = np
]
证明
设 (X sim B(n,p)), 求 (EX).
解
[EX = sum_{K = 0}^{n}x_kp_k = sum_{k = 0}^{n}kC_n^kp^k(1 - p)^{n - k} = sum_{k = 1}^{n}kdisplaystylefrac{n!}{k!(n - k)!}p^k(1 - p)^{n - k}
]
[= npsum_{k = 1}^{n}C_{n - 1}^{k - 1}p^{k - 1}(1 - p)^{n - k} = npsum_{k = 0}^{n - 1}C_{n - 1}^kp^k(1 - p)^{n - 1 - k}
]
[= np[p + (1 - p)]^{n - 1} = np, qquadqquadqquadqquadqquadqquadquad
]
即 (EX = np).
方差
[DX = np(1 - p)
]
证明
设 (X sim B(n, p)), 求 (DX).
解
[EX^2 = sum_{k = 0}^{n}x_k^2p_k = sum_{k = 0}^{n}k^2C_n^kp^k(1 - p)^{n - k} = sum_{k = 1}^{n}kdisplaystylefrac{n!}{(k - 1)!(n - k)!}p^k(1 -p)^{n - k}
]
[= sum_{k = 1}^{n}(k - 1)displaystylefrac{n!}{(k - 1)!(n - k)!}p^k(1 - p)^{n - k} + sum_{k = 1}^{n}displaystylefrac{n!}{(k - 1)!(n - k)!}p^k(1 - p)^{n - k}
]
[= np[(n - 1)p + 1], qquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadqquadquad
]
记 (1 - p = q), 由 (DX = EX^2 - (EX)^2) 可得
[DX = npq.
]