Problem: Find the UMVUE of $\frac1\theta$ for a random sample from the population distribution with density $$f(x;\theta)=\theta x^{\theta-1}$$ and show that its variance reaches the Cramér–Rao lower bound.

Progress: Knowing that the Cramér–Rao Lower Bound is $\frac{\gamma(\theta)^2}{nI(\theta)}$, where $\gamma(\theta)^2=\left(-\frac1{\theta^2}\right)^2=\frac1{\theta^4}$, I then compute the Fisher Information $I(\theta)$. This is where I get lost and become unsure about my work, mostly due to computation.

$$ \begin{align} I(\theta)&=\mathbb{E}\left(\frac{\partial\log{f(x;\theta)}}{\partial\theta}\right)^2\\ &=\int_0^1\left(\frac{\frac{\partial f(x;\theta)}{\partial \theta}}{f(x;\theta)}\right)^2f(x;\theta)\partial x \\ &=\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\right)^2}{f(x;\theta)}\partial x \\ &=\int_0^1\left(\frac{1}{\theta}+\log{x})(x^{\theta-1}+(\theta^2-\theta)x^{\theta-2}\right)\partial x \\ &=\left[\log{x}\frac{x^\theta}{\theta}-\frac{x^{\theta-1}}{\theta-1}+\theta(\log{x})x^{\theta-1}\right]\biggr\rvert_0^1\\ &=\frac{-1}{\theta-1} \end{align} $$

I computed this and consistently got a negative answer, which doesn't make sense as plugging that into the CRLB formula yields a negative variance. I'd really appreciate if someone could quickly help me out on this!

Edit: I found out my mistake! It's a rather stupid one at the fourth equal sign: $$=\int_0^1\left(\frac{\partial\log{f(x;\theta)}}{\partial\theta}\right)\left(\frac{\partial f(x;\theta)}{\partial\theta}\right)$$ where I partially integrate the second term with respect to $x$ rather to to $\theta$.

  • 1
    $\begingroup$ You don't need to find the Cramér-Rao bound because the score function is of the form $\frac{\partial}{\partial\theta}\ln f(x;\theta)=k(\theta)\left(T(x)-\frac1{\theta}\right)$. This is the equality condition of Cramér-Rao equality. If $T$ is unbiased for $1/\theta$ then it must be UMVUE with its variance attaining the C-R bound. $\endgroup$ Commented Apr 17, 2021 at 11:14
  • $\begingroup$ Hi @StubbornAtom, that's an interesting concept I think I either haven't learned or fully internalized yet. Are there any lecture notes or sources you can direct me to on this? $\endgroup$
    – mtcicero
    Commented Apr 17, 2021 at 20:01

1 Answer 1


I can't follow your working after $\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\right)^2}{f(x;\theta)}d x$, note that this expression is nonnegative as the function being integrated is nonnegative.

Let $u = 1 + \theta \log x, \frac{du}{dx}=\frac{\theta}x$,

\begin{align} &\int_0^1\frac{\left(\frac{\partial f(x;\theta)}{\partial \theta}\right)^2}{f(x;\theta)}d x \\ &=\int_0^1 \frac{\left( x^{\theta-1}+\theta x^{\theta -1} \log x\right)^2}{\theta x^{\theta -1}}\, dx \\ &=\frac1{\theta}\int_0^1 x^{\theta-1}(1+\theta \log x)^2 \, dx \\ &= \frac1{\theta} \int_{-\infty}^1 x^{\theta-1}u^2 \frac{x}{\theta}\, du \\ &= \frac1{\theta^2} \int_{-\infty}^1 x^\theta u^2 \, du\\ &= \frac1{\theta^2} \int_{-\infty}^1 e^{u-1}u^2 \, du\\ &= \frac1{\theta^2} \left[ e^{u-1}u^2|_{-\infty}^1 - 2\int_{-\infty}^1 ue^{u-1}\, du \right]\\ &=\frac1{\theta^2} \left[ 1-2\left[ e^{u-1}u|_{-\infty}^1-\int_{-\infty}^1 e^{u-1}\, du\right]\right]\\ &= \frac1{\theta^2} \ge 0 \end{align}


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