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?

  • 1
    $\begingroup$ 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
    Commented Feb 5, 2021 at 14:24
  • 1
    $\begingroup$ Have a look at Theorem 3.3 of this: projecteuclid.org/euclid.bsmsp/1200500219 $\endgroup$
    – passerby51
    Commented Feb 7, 2021 at 15:53

1 Answer 1


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.

  • $\begingroup$ This partly repeats what I mentioned in my post and does not really address the problematic part. $\endgroup$ Commented 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
    Commented Feb 6, 2021 at 6:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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