As a slight modification of my previous problem:

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 $\theta$, the Cramer-Rao Lower Bound (CRLB) for unbiased estimators of $\theta$ and compute the variance of the UMVUE of $\theta$.

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

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

As for finding the UMVUE of $\theta$, since $\frac{1}{n}\sum \text{log}(1+X_i)$ is unbiased for $\frac{1}{\theta}$ then perhaps something similar to $\frac{n}{\sum\text{log}(1+X_i)}$ will be unbiased for $\theta$. After finding the expected value, I can hopefully make a slight adjustment to get an unbiased estimator. Let $T=\sum \text{log}(1+X_i)$

$$\mathsf E\left(\frac{n}{T}\right)=n\cdot\mathsf E\left(\frac{1}{T}\right)=n\int_0^{\infty}\frac{1}{t}f_T(t)dx$$

We must next find the distribution of $T$, but first let's find the distribution of $t=\text{log}(1+X)$. Let $Y=\text{log}(1+X)$. Then

$$\begin{align*} F_Y(y) &=\mathsf P(Y\leq y)\\\\ &=\mathsf P(\text{log}(1+X)\leq y)\\\\ &=\mathsf P(1+X\leq e^y)\\\\ &=\mathsf P(X\leq e^y -1)\\\\ &=F_X\left(-(1+e^y-1)^{-\theta}+1\right)\\\\ &=1-e^{-\theta y} \end{align*}$$

So $Y\sim \text{exp}\left(\frac{1}{\theta}\right)$ and hence $T\sim\text{Gamma}\left(\alpha=n,\beta=\frac{1}{\theta}\right)$


$$\begin{align*} \mathsf E\left(\frac{n}{T}\right) &=n\int_0^{\infty} \frac{1}{t} \frac{\theta^n}{\Gamma(n)}t^{n-1}e^{-\theta t}dt\\\\ &=\frac{n\theta}{n-1}\underbrace{\int_0^{\infty}\frac{\theta^{n-1}}{\Gamma(n-1)}t^{n-2}e^{-\theta t}dt}_{=1}\\\\ &=\frac{n}{n-1}\theta \end{align*}$$

It follows that $$\frac{n-1}{n}\cdot\frac{n}{\sum\text{log}(1+X_i)}=\boxed{\frac{n-1}{\sum\text{log}(1+X_i)}}$$ is an unbiased estimator of $\theta$ which is a function of the complete sufficient statistic $T$, and so by the Lehmann-Scheffe Theorem, it's the unique UMVUE of $\theta$.

As $\hat{\theta}\sim(n-1)\cdot\text{Inv-Gamma}(n,\theta)$


$$\mathsf{Var}\left(\frac{n-1}{T}\right)=(n-1)^2\cdot\mathsf{Var}\left(\frac{1}{T}\right)=(n-1)^2 \cdot \frac{\theta^2}{(n-1)^2\cdot(n-2)}=\boxed{\frac{\theta^2}{n-2}}$$

Are these valid solutions?

  • 2
    $\begingroup$ I think you're making a mistake with $E\left(\frac 1{X+Y}\right) \neq E(1/X) + E(1/Y)$ when you check your work $\endgroup$
    – jld
    Oct 25, 2018 at 19:07
  • 3
    $\begingroup$ With a change of variables $Y=\ln (1+X)$, I get an exponential density for $Y$ having mean $1/\theta$. This should help you to get a Gamma density for the complete sufficient statistic $T=\sum \ln(1+X_i)$, and hence an unbiased estimator of $\theta$ based on $T$. $\endgroup$ Oct 25, 2018 at 19:32
  • 2
    $\begingroup$ No. From the transformation formula, pdf of $Y=\ln(1+X)$ is $$f_Y(y)=f_X(e^y-1)\left|\frac{dx}{dy}\right|\mathbf1_{y>0}$$, where $f_X$ is the pdf of $X$. (Note that $x>0\implies y>0$) $\endgroup$ Oct 25, 2018 at 19:58
  • 1
    $\begingroup$ I used the CDF method, as I am more familiar with it, and came to the same conclusion. I will try to get get the unbiased estimator from here and update my answer. I appreciate your help. $\endgroup$
    – Remy
    Oct 25, 2018 at 20:09
  • 1
    $\begingroup$ The CRLB is correct. $\endgroup$ Oct 26, 2018 at 6:31

1 Answer 1


From your previous question, you already have the complete sufficient statistic:

$$T(\mathbf{X}) = \sum_{i=1}^n \ln(1+X_i).$$

The simplest way to find the UMVUE estimator for $\theta$ is to appeal to the Lehmann-Scheffé theorem, which says that any unbiased estimator of $\theta$ which is a function of $T$ is the unique UMVUE.

To find an estimator with these properties, let $T_i = \ln(1+X_i)$ and observe that $T_i \sim \text{Exp}(\theta)$ so that $T \sim \text{Gamma}(n,\theta)$. Hence, we can use the complete sufficient statistic to form the estimator:

$$\hat{\theta}(\mathbf{X}) = \frac{n-1}{T(\mathbf{X})} = \frac{n-1}{\sum_{i=1}^n \ln(1+X_i)} \sim (n-1) \cdot \text{Inv-Gamma}(n,\theta).$$

From the known moments of the inverse gamma distribution, we have $\mathbb{E}(\hat{\theta}) = \theta$ so we have found an unbiased estimator that is a function of the complete sufficient statistic. The Lehmann-Scheffé theorem ensures that our estimator is UMVUE for $\theta$. The variance of the UMVUE can easily be found by appeal to the moments of the inverse gamma distribution.

  • $\begingroup$ Thank you. I had not heard of the inverse gamma distribution until now. I have updated my answer attempting to solve for the variance of the UMVUE. Why is it not $\text{Inv-Gamma}(n,\frac{1}{\theta})$? $\endgroup$
    – Remy
    Oct 26, 2018 at 3:16
  • $\begingroup$ Never mind, my textbook has a different way of expressing the gamma distribution than wikipedia. My textbook has $f(x\mid\alpha,\beta)=\frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-\frac{x}{\beta}}$ $\endgroup$
    – Remy
    Oct 26, 2018 at 3:40
  • $\begingroup$ @Remy If your UMVUE is $\hat\theta=\frac{n-1}{T}$, then \begin{align} \operatorname{Var}(\hat\theta)&=(n-1)^2\operatorname{Var}\left(\frac{1}{T}\right) \\&=(n-1)^2\left[E\left(\frac{1}{T^2}\right)-\left(E\left(\frac{1}{T}\right)\right)^2\right] \end{align} , which you can calculate from the distribution of $T$. $\endgroup$ Oct 26, 2018 at 5:15
  • $\begingroup$ @StubbornAtom Is what I have after using the method suggested by Ben incorrect or are you just suggesting an alternative approach? $\endgroup$
    – Remy
    Oct 26, 2018 at 5:25
  • 1
    $\begingroup$ @Remy I was suggesting a direct approach (without using Inverse Gamma distribution) since you only need to calculate a couple of moments. $\endgroup$ Oct 26, 2018 at 5:34

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