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