Questions: 设 x₁, x₂, ..., xₙ 是来自 Exp(λ) 的样本, 已知 x̄ 为 1 / λ 的无偏估计, 试说明 1 / x̄ 是否为 λ 的无偏估计。

Solution Steps

To determine if \( \frac{1}{\bar{x}} \) is an unbiased estimator for \( \lambda \), we need to calculate the expected value of \( \frac{1}{\bar{x}} \) and compare it to \( \lambda \). Given that \( x_i \) are i.i.d. from \( \operatorname{Exp}(\lambda) \), we know that \( \bar{x} \) is an unbiased estimator for \( \frac{1}{\lambda} \). We will use properties of the Gamma distribution to find \( E\left(\frac{1}{\bar{x}}\right) \).

We need to determine if \( \frac{1}{\bar{x}} \) is an unbiased estimator for \( \lambda \). Given that \( x_i \) are i.i.d. from \( \operatorname{Exp}(\lambda) \), we know that \( \bar{x} \) is an unbiased estimator for \( \frac{1}{\lambda} \).

We start by calculating the expected value of \( \frac{1}{\bar{x}} \). Since \( \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \) and \( \sum_{i=1}^n x_i \sim \operatorname{Gamma}(n, \lambda) \), we can use properties of the Gamma distribution.

For \( Y = \sum_{i=1}^n x_i \sim \operatorname{Gamma}(n, \lambda) \), the expected value of \( \frac{1}{Y} \) is given by: \[ E\left(\frac{1}{Y}\right) = \frac{\lambda}{n-1} \]

Since \( \bar{x} = \frac{Y}{n} \), we have: \[ \frac{1}{\bar{x}} = \frac{n}{Y} \] Thus, the expected value is: \[ E\left(\frac{1}{\bar{x}}\right) = E\left(\frac{n}{Y}\right) = n \cdot E\left(\frac{1}{Y}\right) = n \cdot \frac{\lambda}{n-1} = \frac{n \lambda}{n-1} \]

We compare \( E\left(\frac{1}{\bar{x}}\right) \) with \( \lambda \): \[ E\left(\frac{1}{\bar{x}}\right) = \frac{n \lambda}{n-1} \] Since \( \frac{n \lambda}{n-1} \neq \lambda \), \( \frac{1}{\bar{x}} \) is not an unbiased estimator for \( \lambda \).

Final Answer