Suppose that $X_1,...,X_n$ are iid with common pdf given by $$f(x;\theta)=2e^{2x}\theta^{-2}I( x<log(\theta)).$$ I am tasked with finding a complete-sufficient statistic for $\theta$, and I have already shown via Neyman-factorization that $X_{(n)}$ is sufficient.

To demonstrate completeness, let $h(x_{(n)})$ be a real-valued function of $x_{(n)}$ such that $E[h(x_{(n)})]=0$. We can now express this as the following integral: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})2n\theta^{-2n}e^{2nx_{(n)}}dx_{(n)},$$

where the expression being multiplied by $h$ is the pdf of the n-th order statistic. Now, this is where I am a little confused. First, note that because we are integrating w.r.t. the n-th order statistic, we can divide both sides by all expressions in the integral which do not contain the n-th order statistic. We now have the following: $$0=\int_{-\infty}^{log(\theta)}h(x_{(n)})e^{2nx_{(n)}}dx_{(n)}.$$

I am now inclined to use Leibnitz' rule, but in the last expression of Leibnitz' rule, where I take the partial w.r.t. $\theta$ of everything in the integral, it will be free of $h(x_{(n)})$ which is a problem, as I need to show that $h(x_{(n)})=0$.

My question is, should I not have eliminated the seemingly ancillary expressions inside the integral? Or am I making some other type of mistake?


The derivative of $$\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x$$ which is the composition of the two functions $$\theta \mapsto \log(\theta)\qquad\text{ and }\qquad \xi \mapsto \int_{-\infty}^{\xi}h(x)e^{2nx}\text{d}x$$ wrt $\theta$ is $$\frac{\text{d}}{\text{d}\theta}\int_{-\infty}^{\log(\theta)}h(x)e^{2nx}\text{d}x=\frac{1}{\theta}\,h(\log(\theta))e^{2n\log(\theta)}=\theta^{2n-1}h(\log(\theta))$$ which is uniformly null iff $h\equiv 0$. There is no reason for resorting to Leibniz's rule in that case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.