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设 $X_1, X_2, . . . , X_n$ 是具有pdf的iid随机变量

$f_{X} (x ∣ θ) = θ (1 + x)^{- (1 + θ)} I_{(0, \infty)} (x)$ $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$

其中 $\theta >0$ 。给出的UMVUE $\frac{1}{\theta}$ 并计算其方差

我了解了两种用于获得UMVUE的方法：

克莱默罗下界（CRLB）
莱曼-舍夫·特莱姆

我将尝试使用两者中的前者。我必须承认，我不完全了解这里发生的事情，而我的尝试解决方案是基于一个示例问题。我有一个 $f_X(x\mid\theta)$ 是一个完整的单参数指数族与

$h(x)=I_{(0,\infty)}$ ， $c(\theta)=\theta$ ， $w(\theta)=-(1+\theta)$ ， $t(x)=\text{log}(1+x)$

由于 $w'(\theta)=1$ 在 $\Theta$ 上不为零，因此适用CRLB结果。我们有

log f_{X} (x ∣ θ) = log (θ) - (1 + θ) \cdot log (1 + x)

$\text{log }f_X(x\mid\theta)=\text{log}(\theta)-(1+\theta)\cdot\text{log}(1+x)$

\frac{\partial}{\partial θ} log f_{X} (x ∣ θ) = \frac{1}{θ} - log (1 + x)

$\frac{\partial}{\partial \theta}\text{log }f_X(x\mid\theta)=\frac{1}{\theta}-\text{log}(1+x)$

\frac{\partial^{2}}{\partial θ^{2}} log f_{X} (x ∣ θ) = - \frac{1}{θ^{2}}

$\frac{\partial^2}{\partial \theta^2}\text{log }f_X(x\mid\theta)=-\frac{1}{\theta^2}$

所以

I_{1} (θ) = - E (- \frac{1}{θ^{2}}) = \frac{1}{θ^{2}}

$I_1(\theta)=-\mathsf E\left(-\frac{1}{\theta^2}\right)=\frac{1}{\theta^2}$

和CRLB为的无偏估计 $\tau(\theta)$ 是

\frac{[τ^{'} (θ)]^{2}}{n \cdot I_{1} (θ)} = \frac{θ^{2}}{n} [τ^{'} (θ)]^{2}

$\frac{[\tau'(\theta)]^2}{n\cdot I _1(\theta)} = \frac{\theta^2}{n}[\tau'(\theta)]^2$

由于

\sum_{i = 1}^{n} t (X_{i}) = \sum_{i = 1}^{n} log (1 + X_{i})

$\sum_{i=1}^n t(X_i)=\sum_{i=1}^n \text{log}(1+X_i)$

那么 $\sum_{i=1}^n \text{log}(1+X_i)$ 任何线性函数，或者等效地，任何线性函数 $\frac{1}{n}\sum_{i=1}^n \text{log}(1+X_i)$ ，将达到其期望的CRLB，因此将成为其期望的UMVUE。由于 $\mathsf E(\text{log}(1+X))=\frac{1}{\theta}$ 我们的UMVUE为 $\frac{1}{\theta}$ 是 $\frac{1}{n}\sum_{i=1}^n \text{log}(1+X_i)$

对于天然的参数，我们可以让 $\eta=-(1+\theta)\Rightarrow \theta=-(\eta+1)$

然后

V a r (log (1 + X)) = \frac{d}{d η} (- \frac{1}{η + 1}) = \frac{1}{(η + 1)^{2}} = \frac{1}{θ^{2}}

$\mathsf{Var}(\text{log}(1+X))=\frac{d}{d\eta}\left(-\frac{1}{\eta+1}\right)=\frac{1}{(\eta+1)^2}=\frac{1}{\theta^2}$

这是有效的解决方案吗？有没有更简单的方法？仅当 $\mathsf E(t(x))$ 等于您要估算的值时，此方法才有效吗？

— 雷米
source

在您表明pdf是单参数指数族的成员这一点上，很明显，该族的完整充分统计量为

因为正如您所说，

T (X_{1}, \dots, X_{n}) = \sum_{i = 1}^{n} \ln (1 + X_{i})

$T(X_1,\ldots,X_n)=\sum_{i=1}^n\ln(1+X_i)$

根据Lehmann-Scheffe定理，

，

是

的UMVUE。

E (T / n) = \frac{1}{θ}

$E(T/n)=\frac{1}{\theta}$

T / n

$T/n$

1 / θ

$1/\theta$

— StubbornAtom

那么，我有“自部分

是非零.....

w^{'} (θ) = 1

$w'(\theta)=1$

“是否无关紧要？

\frac{θ^{2}}{n} [τ^{'} (θ)]^{2}

$\frac{\theta^2}{n}[\tau'(\theta)]^2$

— Remy

并不是的; 使用CRLB更容易找到

的方差。因此，要一次解决两个问题，您的论点就足够了。

T

$T$

— StubbornAtom

要找到方差这样，我会采取

？因此，我以前做错了吗？

\frac{θ^{2}}{n} [τ^{'} (θ)]^{2} = \frac{θ^{2}}{n} {(- \frac{1}{θ^{2}})}^{2} = \frac{1}{n θ^{2}}

$\frac{\theta^2}{n}[\tau'(\theta)]^2=\frac{\theta^2}{n}\left(-\frac{1}{\theta^2}\right)^2=\frac{1}{n\theta^2}$

— 雷米

是的，这就是

的方差。精确地

T

$T$

— StubbornAtom

您的推理基本上是正确的。

样品 $(X_1,X_2,\ldots,X_n)$ 的联合密度为

\begin{aligned} f_{θ} (x_{1}, x_{2}, \dots, x_{n}) & = \frac{θ^{n}}{{(\prod_{i = 1}^{n} (1 + x_{i}))}^{1 + θ}} 1_{x_{1}, x_{2}, \dots, x_{n} > 0}, θ > 0 \\ ⟹ \ln f_{θ} (x_{1}, x_{2}, \dots, x_{n}) & = n \ln (θ) - (1 + θ) \sum_{i = 1}^{n} \ln (1 + x_{i}) + \ln (1_{min_{1 \leq i \leq n} x_{i} > 0}) \\ ⟹ \frac{\partial}{\partial θ} \ln f_{θ} (x_{1}, x_{2}, \dots, x_{n}) & = \frac{n}{θ} - \sum_{i = 1}^{n} \ln (1 + x_{i}) \\ = - n (\frac{\sum_{i = 1}^{n} \ln (1 + x_{i})}{n} - \frac{1}{θ}) \end{aligned}

$\begin{align} f_{\theta}(x_1,x_2,\ldots,x_n)&=\frac{\theta^n}{\left(\prod_{i=1}^n (1+x_i)\right)^{1+\theta}}\mathbf1_{x_1,x_2,\ldots,x_n>0}\qquad,\,\theta>0 \\\\\implies \ln f_{\theta}(x_1,x_2,\ldots,x_n)&=n\ln(\theta)-(1+\theta)\sum_{i=1}^n\ln(1+x_i)+\ln(\mathbf1_{\min_{1\le i\le n} x_i>0}) \\\\\implies\frac{\partial}{\partial \theta}\ln f_{\theta}(x_1,x_2,\ldots,x_n)&=\frac{n}{\theta}-\sum_{i=1}^n\ln(1+x_i) \\\\&=-n\left(\frac{\sum_{i=1}^n\ln(1+x_i)}{n}-\frac{1}{\theta}\right) \end{align}$

Thus we have expressed the score function in the form

\begin{matrix} (1) & \frac{\partial}{\partial θ} \ln f_{θ} (x_{1}, x_{2}, \dots, x_{n}) = k (θ) (T (x_{1}, x_{2}, \dots, x_{n}) - \frac{1}{θ}) \end{matrix}

$\frac{\partial}{\partial \theta}\ln f_{\theta}(x_1,x_2,\ldots,x_n)=k(\theta)\left(T(x_1,x_2,\ldots,x_n)-\frac{1}{\theta}\right)\tag{1}$

, which is the equality condition in the Cramér-Rao inequality.

It is not difficult to verify that

\begin{matrix} (2) & E (T) = \frac{1}{n} \sum_{i = 1}^{n} \underset{= 1 / θ}{\underset{⏟}{E (\ln (1 + X_{i}))}} = \frac{1}{θ} \end{matrix}

$E(T)=\frac{1}{n}\sum_{i=1}^n\underbrace{E(\ln(1+X_i))}_{=1/\theta}=\frac{1}{\theta}\tag{2}$

From $(1)$ and $(2)$ we can conclude that

The statistic $T(X_1,X_2,\ldots,X_n)$ is an unbiased estimator of $1/\theta$ .
$T$ satisfies the equality condition of the Cramér-Rao inequality.

These two facts together imply that $T$ is the UMVUE of $1/\theta$ .

The second bullet actually tells us that variance of $T$ attains the Cramér-Rao lower bound for $1/\theta$ .

Indeed, as you have shown,

E_{θ} [\frac{\partial^{2}}{\partial θ^{2}} \ln f_{θ} (X_{1})] = - \frac{1}{θ^{2}}

$E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]=-\frac{1}{\theta^2}$

This implies that the information function for the whole sample is

I (θ) = - n E_{θ} [\frac{\partial^{2}}{\partial θ^{2}} \ln f_{θ} (X_{1})] = \frac{n}{θ^{2}}

$I(\theta)=-nE_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]=\frac{n}{\theta^2}$

So the Cramér-Rao lower bound for $1/\theta$ and hence the variance of the UMVUE is

Var (T) = \frac{{[\frac{d}{d θ} (\frac{1}{θ})]}^{2}}{I (θ)} = \frac{1}{n θ^{2}}

$\operatorname{Var}(T)=\frac{\left[\frac{d}{d\theta}\left(\frac{1}{\theta}\right)\right]^2}{I(\theta)}=\frac{1}{n\theta^2}$

Here we have exploited a corollary of the Cramér-Rao inequality, which says that for a family of distributions $f$ parametrised by $\theta$ (assuming regularity conditions of CR inequality to hold), if a statistic $T$ is unbiased for $g(\theta)$ for some function $g$ and if it satisfies the condition of equality in the CR inequality, namely

\frac{\partial}{\partial θ} \ln f_{θ} (x) = k (θ) (T (x) - g (θ))

$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=k(\theta)\left(T(x)-g(\theta)\right)$ , then

T

$T$ must be the UMVUE of

g (θ)

$g(\theta)$ . So this argument does not work in every problem.

Alternatively, using the Lehmann-Scheffe theorem you could say that $T=\frac{1}{n}\sum_{i=1}^{n} \ln(1+X_i)$ is the UMVUE of $1/\theta$ as it is unbiased for $1/\theta$ and is a complete sufficient statistic for the family of distributions. That $T$ is compete sufficient is clear from the structure of the joint density of the sample in terms of a one-parameter exponential family. But variance of $T$ might be a little tricky to find directly.

— StubbornAtom
source

One could also use the distribution of

T

$T$ to find its mean,variance.

— StubbornAtom