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Let $X_1,\ldots,X_n$ be i.i.d shifted exponential with pdf $f_{\theta}(x)=e^{-(x-\theta)}\mathbf1_{x> \theta}$, where $\theta\in \mathbb R$. I have to show that $X_{(1)}-\frac1n$ is the unique minimax estimator of $\theta$ under squared error loss.

I am basically trying to figure out the connections different estimators have with a minimax rule.

Since $\delta=X_{(1)}-\frac1n$ is unbiased for $\theta$, it cannot be a Bayes estimator under quadratic loss. So I cannot relate minimax with Bayes rule here. I think $\delta$ is also the Pitman estimator of $\theta$ as it is location invariant and UMVUE at the same time. The problem implies that $\delta$ is admissible but I don't know how to show this or if this is a property of Pitman estimators under quadratic loss. An admissible Pitman estimator is then minimax perhaps?

I also found a theorem stating that an admissible estimator with constant risk is minimax. And if the loss is strictly convex, the estimator is unique minimax. I could show that $\delta$ has constant risk but I am wondering how to argue the admissibility. I tried a proof by contradiction but could not proceed. Any suggestion is welcome.


Another theorem in Lehmann's Theory of Point Estimation roughly says that a Pitman estimator with finite variance in a one-parameter location family is indeed minimax under squared error loss. Using this result solves the problem then. There is also a sufficient condition for admissibility of a Pitman estimator in the book, but it is not easy to verify.

Is there a simpler way of solving this?

I considered estimators of the form $T=aX_{(1)}+b$ where $a,b$ are real constants free of $\theta$. Then a routine calculation shows that risk of $T$ is $$R(\theta,T)=\theta^2(a-1)^2+2\theta\left(\frac{a^2}{n}-\frac an+ab-b\right)+\frac{2a}n\left(b+\frac an\right)+b^2$$

This is constant only for $a=1$ and the constant risk is $b^2+\frac2n(b+\frac1n)$. Minimizing this further yields $b=-\frac1n$. I guess this makes $\delta$ admissible in the class of linear estimators based on $X_{(1)}$ with constant risk. Is this somehow enough to ensure admissibility of $\delta$ in general?

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    $\begingroup$ Before reading the entire question, I was going to suggest the constant risk argument. If $\delta$ is Pitman, it is Bayes against the Lebesque measure prior. Since this is a dimension one problem, it is admissible. (To prove admissibility, one could look at Stein's method: produce $\delta$ as the limit of a sequence of proper Bayes estimates, e.g. using uniform $(-n,n)$ priors.) $\endgroup$
    – Xi'an
    Feb 5, 2021 at 14:24
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    $\begingroup$ Have a look at Theorem 3.3 of this: projecteuclid.org/euclid.bsmsp/1200500219 $\endgroup$
    – passerby51
    Feb 7, 2021 at 15:53

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One of the results for minimax estimators is that if you can find an admissible estimator with a constant risk function, then that is the minimax estimator. Since $X_1,...,X_n \sim \theta + \text{IID Exp}(1)$ you should be able to show that:

$$X_{(n)} - \theta \sim \text{Exp}(\text{Rate} = n).$$

Thus, under squared error loss, this estimator gives the risk function:

$$\begin{align} R(\theta, \delta) &= \mathbb{E} \Big( (X_{(n)} - \tfrac{1}{n} - \theta)^2 \Big| \theta \Big) \\[12pt] &= \int \limits_{0}^{\infty} (x - \tfrac{1}{n})^2 \ \text{Exp}(x|n) \ dx \\[6pt] &= n \int \limits_{0}^{\infty} (x - \tfrac{1}{n})^2 \ \exp(-nx) \ dx \\[12pt] &= \Bigg[ - \frac{n^2 x^2 + 1}{n^2} \cdot \exp(-nx) \Bigg]_0^\infty \\[12pt] &= \Bigg[ 0 - - \frac{1}{n^2} \Bigg] \\[12pt] &= \frac{1}{n^2}, \\[6pt] \end{align}$$

which is constant. This gets you part-way to your result. If you can show that the estimator is admissible then you will estabish that this is a minimax estimator. Proving uniqueness is trickier, but I would suggest you try to show that every other admissible estimator has a risk function that goes above $1/n^2$ at some point.

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  • $\begingroup$ This partly repeats what I mentioned in my post and does not really address the problematic part. $\endgroup$ Feb 6, 2021 at 5:07
  • $\begingroup$ Okay, I'll have more of a think and come back and edit if I think of more. $\endgroup$
    – Ben
    Feb 6, 2021 at 6:02

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