Show that $\frac{d}{d\theta}E_{\theta}W(\textbf{X})=\int_{x}\frac{\partial}{\partial \theta}[W(\textbf{X})f(\textbf{x}|\theta)]dx$

Question

Given that $f(x|\theta)=\theta x^{\theta-1},\quad 0<x<1, \quad \theta >0$ and $W(x)=-\sum_{i=1}^{n}\frac{\log x_{i}}{n}$ show that

$$\frac{d}{d\theta}E_{\theta}W(\textbf{X})=\int_{x}\frac{\partial}{\partial \theta}[W(\textbf{X})f(\textbf{x}|\theta)]dx$$ and $$Var_{\theta}W(\textbf{X})<\infty.$$ Then $$Var_{\theta}W(\textbf{X})\geq \frac{\left( \frac{d}{d\theta}E_{\theta}W(\textbf{X}) \right)^2}{E_{\theta}\left( \left( \frac{\partial}{\partial \theta}\log f(\textbf{X}|\theta) \right)^2 \right)}$$

I get that $\frac{d}{d\theta}E_{\theta}W(\textbf{X}) = -\frac{1}{\theta}$ but I am having a hard time showing that $\int_{x}\frac{\partial}{\partial \theta}[W(\textbf{X})f(\textbf{x}|\theta)]dx = -\frac{1}{\theta}$. Can someone help me?

Answer 1

Your $W(\boldsymbol X)=-\frac1n\sum\limits_{i=1}^n \ln X_i$ is unbiased for $\frac1\theta$ which can be verified directly or by observing that $-\theta \ln X_i$'s are distributed as $\text{Exp}(1)$. Hence, $$\frac{d}{d \theta}E_{\theta}W(\boldsymbol X) = -\frac1{\theta^2}$$

Density of $\boldsymbol X$ is $$f(\boldsymbol x\mid \theta)=\theta^n \left(\prod_{i=1}^n x_i\right)^{\theta-1}\mathbf1_{0<x_1,\ldots,x_n<1} \quad,\,\theta>0$$

Taking logarithm and differentiating both sides with respect to $\theta$, you have

$$\frac{\partial}{\partial\theta}\ln f(\boldsymbol x\mid \theta)=\frac1{f(\boldsymbol x\mid \theta)}\cdot\frac{\partial}{\partial\theta}f(\boldsymbol x\mid \theta)=\frac n{\theta}+\sum_{i=1}^n \ln x_i$$

That is, $$\frac{\partial}{\partial\theta}f(\boldsymbol x\mid \theta)=\left(\frac n{\theta}+\sum_{i=1}^n \ln x_i\right)f(\boldsymbol x\mid \theta)$$

So,

\begin{align} \int W(\boldsymbol x)\frac{\partial}{\partial\theta}f(\boldsymbol x\mid \theta)\,d\boldsymbol x&=\frac n{\theta}\int W(\boldsymbol x)f(\boldsymbol x\mid \theta)\,d\boldsymbol x -\frac1n \int \left(\sum_{i=1}^n \ln x_i\right)^2 f(\boldsymbol x\mid \theta)\,d\boldsymbol x \\&=\frac n{\theta}E_{\theta}[W(\boldsymbol X)]-\frac1n E_{\theta}\left[\sum_{i=1}^n \ln X_i\right]^2 \\&=\frac n{\theta}E_{\theta}[W(\boldsymbol X)]-n E_{\theta}\left[W(\boldsymbol X)\right]^2 \end{align}

The second expectation can be found by noting that $-2\theta \ln X_i$ are i.i.d $\chi^2_2$ variables, so that $$-2\theta\sum\limits_{i=1}^n \ln X_i=2n\theta W(\boldsymbol X)\sim \chi^2_{2n}$$