# Find UMVUE of $\frac{1}{\theta}$ where $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$

Let $$X_1, X_2, . . . , X_n$$ be iid random variables having pdf

$$f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$$

where $$\theta >0$$. Give the UMVUE of $$\frac{1}{\theta}$$ and compute its variance

I have learned about two such methods for obtained UMVUE's:

• Cramer-Rao Lower Bound (CRLB)
• Lehmann-Scheffe Thereom

I am going to attempt this using the former of the two. I must admit that I do not completely understand what is going on here, and I'm basing my attempted solution off of an example problem. I have that $$f_X(x\mid\theta)$$ is a full one-parameter exponential family with

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

Since $$w'(\theta)=1$$ is nonzero on $$\Theta$$, the CRLB result applies. We have

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

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

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

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

and the CRLB for unbiased estimators of $$\tau(\theta)$$ is

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

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

then any linear function of $$\sum_{i=1}^n \text{log}(1+X_i)$$, or equivalently, any linear function of $$\frac{1}{n}\sum_{i=1}^n \text{log}(1+X_i)$$, will attain the CRLB of its expectation, and thus be a UMVUE of its expectation. Since $$\mathsf E(\text{log}(1+X))=\frac{1}{\theta}$$ we have that the UMVUE of $$\frac{1}{\theta}$$ is $$\frac{1}{n}\sum_{i=1}^n \text{log}(1+X_i)$$

For a natural parameterization we can let $$\eta=-(1+\theta)\Rightarrow \theta=-(\eta+1)$$

Then

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

Is this a valid solution? Is there a more simple approach? Does this method only work when the $$\mathsf E(t(x))$$ equals what you're trying to estimate?

• At the point where you showed that the pdf is a member of the one-parameter exponential family, it is immediately clear that a complete sufficient statistic for the family is $$T(X_1,\ldots,X_n)=\sum_{i=1}^n\ln(1+X_i)$$ Since, as you say, $E(T/n)=\frac{1}{\theta}$, $T/n$ is the UMVUE of $1/\theta$ by the Lehmann-Scheffe theorem. Oct 24 '18 at 18:41
• So the part where I have "Since $w'(\theta)=1$ is nonzero.....$\frac{\theta^2}{n}[\tau'(\theta)]^2$" is irrelevant?
– Remy
Oct 24 '18 at 18:46
• Not really; the variance of $T$ is easier to find using the CRLB. So to solve both questions at once, your argument is sufficient. Oct 24 '18 at 18:48
• To find the variance that way, would I take $\frac{\theta^2}{n}[\tau'(\theta)]^2=\frac{\theta^2}{n}\left(-\frac{1}{\theta^2}\right)^2=\frac{1}{n\theta^2}$? Hence, I did it incorrectly previously?
– Remy
Oct 24 '18 at 18:56
• Yes, that is the variance of $T$. Precisely. Oct 24 '18 at 18:58

The joint density of the sample $$(X_1,X_2,\ldots,X_n)$$ is

\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

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

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

It is not difficult to verify that $$E(T)=\frac{1}{n}\sum_{i=1}^n\underbrace{E(\ln(1+X_i))}_{=1/\theta}=\frac{1}{\theta}\tag{2}$$

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

This implies that the information function for the whole sample is $$I(\theta)=-nE_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]=\frac{n}{\theta^2}$$

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

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

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 $$\frac{\partial}{\partial\theta}\ln f_{\theta}(x)=k(\theta)\left(T(x)-g(\theta)\right)$$, then $$T$$ must be the UMVUE of $$g(\theta)$$. 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.

• One could also use the distribution of $T$ to find its mean,variance. Apr 2 '20 at 7:23