I am trying to work through the very short paper entitled "On Minimium Variance Estimators" by J. Kiefer (1952). In this work he gives two explicit examples on calculating the lower bound on the variance of the two random variables. In both cases, he calculates the bound two different ways, using equation (3) and (4) of his paper.
Eq. 4 of his paper is the Chapman-Robbins bound, whereas Eq. 3 is a more general Barankin type bound. I think I understand how he got his Chapman-Robbins result, but I am confused as to how he calculates the more general bound.
Ill now paraphrase his first example:
For n observations on a uniform distribution from $0$ to $\theta$, $(\Omega =\{\theta|\theta>0\})$, the random variable Y is the maximum of those observations with pdf $f(y;\theta)=ny^{n-1}/\theta^{n}$ for $0\leq y\leq \theta$ and is $0$ elsewhere. The Chapman Robbins bound on a pdf $f(y;\theta)$ is given by
\begin{align}
E_{\theta}(t-\theta)^2\geq[\inf_h\frac{1}{h^2}(\int_{\chi}\frac{(f(y;\theta+h))^2}{f(y;\theta)}\text{d}\mu-1)]^{-1}
\end{align},
for any $t(y)$ such that $E_{\theta}t=\theta$.
Kiefer finds that for $n=1$, $[(\int_{\chi}\frac{(f(y;\theta+h))^2}{f(y;\theta)}\text{d}\mu-1)]=-\frac{1}{h(\theta+h)}$, which in turn yields a bound of $\theta^2/4$. Lets start by deriving this result in more detail.
We have that $\frac{(f(y;\theta+h))^2}{f(y;\theta)}=\frac{\theta}{(\theta+h)^2}$, and $\chi$ has support from $0$ to $\theta+h$ (note that $-\theta<h<0$), so $\int_{\chi}\frac{(f(y;\theta+h))^2}{f(y;\theta)}\text{d}\mu=\int_{0}^{\theta+h}\frac{\theta}{(\theta+h)^2}\text{d}y=\frac{\theta}{(\theta+h)}$, and so we have that $[\inf_h\frac{1}{h^2}(\frac{\theta}{(\theta+h)}-1)]=\inf_h(-\frac{1}{h(\theta+h)})$ as desired. This term obtains its minimum at $h=-\theta/2$, and is $4/\theta^2$. The Chapman-Robbins bound is the inverse of this term, $\theta^2/4$. This coincides with what Kiefer found in his paper.
Now for the more general Barankin type bound, which Kiefer gives as \begin{align} E_{\theta}(t-\theta)^2\geq\sup_{\lambda_1}\left[(E_1(h))^2\left(\int_{\chi}\frac{[\int_{\Omega_{\theta}}f(y;\theta+h)\text{d}\lambda_{1}(h)]^{2}}{f(y;\theta)}\text{d}\mu-1\right)^{-1}\right] \end{align} where $E_{1}(h)=\int_{\Omega_{\theta}}h\text{d}\lambda_{1}(h)$, Kiefer claims the ideal measure $\text{d}\lambda_{1}(h)=\frac{n+1}{\theta}(h/\theta+1)^n\text{d}h$. Proceeding now for general $n$, we have that $E_{1}(h)=\int_{\Omega_{\theta}}h\frac{n+1}{\theta}(h/\theta+1)^n\text{d}h=\int_{\Omega_{\theta}}(h+\theta-\theta)\frac{n+1}{\theta^{n+1}}(h+\theta)^n\text{d}h=\int_{-\theta}^{0}\frac{n+1}{\theta^{n+1}}(h+\theta)^{n+1}\text{d}h-\int_{-\theta}^{0}\frac{n+1}{\theta^{n}}(h+\theta)^{n}\text{d}h=\frac{n+1}{n+2}\theta-\theta=-\frac{1}{n+2}\theta$. We now calculate the integral \begin{align} \int_{\Omega_{\theta}}f(y;\theta+h)\text{d}\lambda_{1}(h) = \int_{-\theta}^{0}\frac{n y^{n-1}}{(\theta+h)^n}\frac{n+1}{\theta^{n+1}}(\theta+h)^{n}\text{d}h = \frac{n(n+1)y^{n-1}}{\theta^{n}} \end{align} This means that \begin{align} \int_{\chi}\frac{[\int_{\Omega_{\theta}}f(y;\theta+h)\text{d}\lambda_{1}(h)]^{2}}{f(y;\theta)}\text{d}\mu = \int_{\chi}(n+1)^2 f(y;\theta) \text{d}\mu = (n+1)^2 \end{align} Since $(n+1)^2-1=n(n+2)$ we can combine our earlier results to find \begin{align} \left[(E_1(h))^2\left(\int_{\chi}\frac{[\int_{\Omega_{\theta}}f(y;\theta+h)\text{d}\lambda_{1}(h)]^{2}}{f(y;\theta)}\text{d}\mu-1\right)^{-1}\right] = \frac{1}{n(n+2)^3}\theta^{2}. \end{align} However, Kiefer finds the variance to be bounded by $\frac{1}{n(n+2)}\theta^{2}$, which equals $\frac{n+1}{n}Y$, so their is an unbiased estimator that reaches this bound (and clearly the answer I derived is wrong).
My questions are: where did I go wrong in my derivation? Also, is there an easy way to see how Kiefer knew to choose $\text{d}\lambda_{1}(h)$ the way that he did, other than the obvious answer that this gives the best bound?
Also, if anyone has general advice on calculation of tight bounds beyond Chapman-Robbins bound, I would be interested in hearing their strategies. Hopefully I have included enough notation to make the problem palatable, but if not I can copy more from the paper.
edit: fixed incorrect mathematical symbol.