题目
3.设随机变量X的概率密度为-|||-f(x)= ) (theta +1)(x)^theta ,0lt xlt 1 0 .-|||-其中, theta gt -1 是未知参数,X1,X2,···Xn是来自总体X的样本,求θ的矩估计和极大似然估计.

题目解答
答案

解析
步骤 1:求解总体X的期望
为了求解θ的矩估计,我们首先需要求出总体X的期望E(X)。根据概率密度函数f(x),我们可以写出期望的积分表达式:
\[ E(X) = \int_{0}^{1} x \cdot f(x) dx = \int_{0}^{1} x \cdot (\theta + 1) x^{\theta} dx = (\theta + 1) \int_{0}^{1} x^{\theta + 1} dx \]
\[ = (\theta + 1) \left[ \frac{x^{\theta + 2}}{\theta + 2} \right]_{0}^{1} = \frac{\theta + 1}{\theta + 2} \]
步骤 2:求解θ的矩估计
根据矩估计的定义,我们用样本均值$\bar{X}$来估计总体期望E(X),即$\bar{X} = \frac{\theta + 1}{\theta + 2}$。解这个方程,得到θ的矩估计:
\[ \bar{X} = \frac{\theta + 1}{\theta + 2} \Rightarrow \theta = \frac{2\bar{X} - 1}{1 - \bar{X}} \]
步骤 3:求解θ的极大似然估计
为了求解θ的极大似然估计,我们首先写出似然函数L(θ):
\[ L(\theta) = \prod_{i=1}^{n} f(x_i) = \prod_{i=1}^{n} (\theta + 1) x_i^{\theta} = (\theta + 1)^n \prod_{i=1}^{n} x_i^{\theta} \]
取对数似然函数:
\[ \ln L(\theta) = n \ln (\theta + 1) + \theta \sum_{i=1}^{n} \ln x_i \]
对θ求导并令导数等于0,得到极大似然估计:
\[ \frac{d}{d\theta} \ln L(\theta) = \frac{n}{\theta + 1} + \sum_{i=1}^{n} \ln x_i = 0 \Rightarrow \theta = -1 - \frac{n}{\sum_{i=1}^{n} \ln x_i} \]
为了求解θ的矩估计,我们首先需要求出总体X的期望E(X)。根据概率密度函数f(x),我们可以写出期望的积分表达式:
\[ E(X) = \int_{0}^{1} x \cdot f(x) dx = \int_{0}^{1} x \cdot (\theta + 1) x^{\theta} dx = (\theta + 1) \int_{0}^{1} x^{\theta + 1} dx \]
\[ = (\theta + 1) \left[ \frac{x^{\theta + 2}}{\theta + 2} \right]_{0}^{1} = \frac{\theta + 1}{\theta + 2} \]
步骤 2:求解θ的矩估计
根据矩估计的定义,我们用样本均值$\bar{X}$来估计总体期望E(X),即$\bar{X} = \frac{\theta + 1}{\theta + 2}$。解这个方程,得到θ的矩估计:
\[ \bar{X} = \frac{\theta + 1}{\theta + 2} \Rightarrow \theta = \frac{2\bar{X} - 1}{1 - \bar{X}} \]
步骤 3:求解θ的极大似然估计
为了求解θ的极大似然估计,我们首先写出似然函数L(θ):
\[ L(\theta) = \prod_{i=1}^{n} f(x_i) = \prod_{i=1}^{n} (\theta + 1) x_i^{\theta} = (\theta + 1)^n \prod_{i=1}^{n} x_i^{\theta} \]
取对数似然函数:
\[ \ln L(\theta) = n \ln (\theta + 1) + \theta \sum_{i=1}^{n} \ln x_i \]
对θ求导并令导数等于0,得到极大似然估计:
\[ \frac{d}{d\theta} \ln L(\theta) = \frac{n}{\theta + 1} + \sum_{i=1}^{n} \ln x_i = 0 \Rightarrow \theta = -1 - \frac{n}{\sum_{i=1}^{n} \ln x_i} \]