极大似然估计例题——二项分布的极大似然估计

设总体X∼B(1,p)X \sim B(1, p)X∼B(1,p)X1,X2,⋯ ,XnX_1, X_2, \cdots, X_nX1​,X2​,⋯,Xn​是来自总体XXX的样本求ppp的最大似然估计量。求解总体XXX的分布律为P(Xx)px(1−p)1−x,x0,1. P(X x) p^x (1 - p)^{1 - x}, \quad x 0, 1.P(Xx)px(1−p)1−x,x0,1.设x1,x2,⋯ ,xnx_1, x_2, \cdots, x_nx1​,x2​,⋯,xn​是相应的样本值则似然函数为L(p)∏i1npxi(1−p)1−xip∑i1nxi(1−p)n−∑i1nxi. L(p) \prod_{i1}^{n} p^{x_i} (1 - p)^{1 - x_i} p^{\sum_{i1}^{n} x_i} (1 - p)^{n - \sum_{i1}^{n} x_i}.L(p)i1∏n​pxi​(1−p)1−xi​p∑i1n​xi​(1−p)n−∑i1n​xi​.对上式两边取对数得对数似然方程ln⁡L(p)ln⁡(∏i1npxi(1−p)1−xi)(∑i1nxi)ln⁡p(n−∑i1nxi)ln⁡(1−p). \ln L(p) \ln \left( \prod_{i1}^{n} p^{x_i} (1 - p)^{1 - x_i} \right) \left( \sum_{i1}^{n} x_i \right) \ln p \left( n - \sum_{i1}^{n} x_i \right) \ln (1 - p).lnL(p)ln(i1∏n​pxi​(1−p)1−xi​)(i1∑n​xi​)lnp(n−i1∑n​xi​)ln(1−p).令dln⁡L(p)dp∑i1nxip−n−∑i1nxi1−p0, \frac{{\rm d}\ln L(p)}{{\rm d}p} \frac{\sum_{i1}^{n} x_i}{p} - \frac{n - \sum_{i1}^{n} x_i}{1-p} 0,dpdlnL(p)​p∑i1n​xi​​−1−pn−∑i1n​xi​​0,可得参数ppp的最大似然估计值p^1n∑i1nxixˉ. \hat{p} \frac{1}{n} \sum_{i1}^{n} x_i \bar{x}.p^​n1​i1∑n​xi​xˉ.因此参数ppp的最大似然估计量为p^1n∑i1nXiX‾. \hat{p} \frac{1}{n} \sum_{i1}^{n} X_i \overline{X}.p^​n1​i1∑n​Xi​X.