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

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

Hence

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

then

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

Are these valid solutions?

• 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 – jld Oct 25 '18 at 19:07
• 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 $T$. – StubbornAtom Oct 25 '18 at 19:32
• 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 $x>0\implies y>0$) – StubbornAtom Oct 25 '18 at 19:58
• 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. – Remy Oct 25 '18 at 20:09
• The CRLB is correct. – StubbornAtom Oct 26 '18 at 6:31

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

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.

• 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 $\text{Inv-Gamma}(n,\frac{1}{\theta})$? – Remy Oct 26 '18 at 3:16
• Never mind, my textbook has a different way of expressing the gamma distribution than wikipedia. My textbook has $f(x\mid\alpha,\beta)=\frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-\frac{x}{\beta}}$ – Remy Oct 26 '18 at 3:40
• @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 $T$. – StubbornAtom Oct 26 '18 at 5:15
• @StubbornAtom Is what I have after using the method suggested by Ben incorrect or are you just suggesting an alternative approach? – Remy Oct 26 '18 at 5:25
• @Remy I was suggesting a direct approach (without using Inverse Gamma distribution) since you only need to calculate a couple of moments. – StubbornAtom Oct 26 '18 at 5:34