Timeline for UMVUE as an optimization problem
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2022 at 9:36 | vote | accept | Ilia | ||
| Jun 19, 2022 at 2:24 | comment | added | Iosif Pinelis | @Ilia : No, calculus of variations doesn’t work here, because, as I said, what you want to minimize here (that is, the risk function) is in general an infinite-dimensional vector. In general, even a two-dimensional set (say, a unit disk) does not have a minimum (with respect to the partial coordinate-wise ordering). In a certain class of lucky settings, though, we do have a uniformly MVUE -- but the methods to get it are not at all variational. | |
| Jun 17, 2022 at 23:14 | comment | added | Ilia | I see, well, I can’t figure out how to modify the computation to minimize the risk for all values of 𝜃 at once. So calculus of variations just doesn’t work here? | |
| Jun 17, 2022 at 22:22 | comment | added | Iosif Pinelis | Previous comment continued: Since the risk function (as a function of $\theta$) is in general an inifinite-dimensional vector, such a "minimization" is not always possible. When it is possible, you get a UMVUE. Of course, your pointwise application of Lagrange multipliers is wrong: again, you want to minimize the risk for all values of $\theta$ at once, not just for one value of $\theta$. | |
| Jun 17, 2022 at 22:21 | comment | added | Iosif Pinelis | @Ilia : I did understand what you were doing. You minimize $L$ for each given value of $\theta$. So, your minimizer depends on $\theta$. Therefore, you minimizer is not an estimator. When you restrict the minimization to estimators, you want to "minimize" the entire risk function, which is a function of the estimator and of the parameter $\theta$; you should want to minimize the risk function with respect to the (unbiased) estimator. | |
| Jun 17, 2022 at 22:14 | comment | added | Ilia | And do you agree with the computation? I don’t understand how to interpret it. | |
| Jun 17, 2022 at 21:57 | comment | added | Ilia | I realised it is totally unclear what I want to say. Sorry for wasting your time. I completely rewrote the question and added a computation that confused me. Hopefully it is at least clear now what I was trying to do. | |
| Jun 17, 2022 at 20:32 | comment | added | Iosif Pinelis | Previous comment continued: The function $T$ must be the same for all values of $\theta$. E.g., $\bar X:=(X_1+\dots+X_n)/n$ is an estimator, but $\bar X-\theta$ is not an estimator, since it depends on $\theta$. This is explained in any decent textbook on mathematical statistics. | |
| Jun 17, 2022 at 20:32 | comment | added | Iosif Pinelis | @Ilia : What you call "a constant estimator" is apparently what is defined by the formula $\hat\theta(X_1,\dots,X_n):=\th$. As I said, this is not an estimator at all, -- because an estimator of $\th$ cannot depend on $\th$; it may depend only on the sample $X_1,\dots,X_n$. Mathematically, given a random sample $(X_1,\dots,X_n)$, where each $X_i$ is a real-valued random variable, a (real-valued) estimator is a random variable of the form $T(X_1,\dots,X_n)$, where $T\colon\mathbb R^n\to\mathbb R$ is a Borel-measurable function. | |
| Jun 17, 2022 at 20:20 | comment | added | Ilia | Mathematically theta is a parameter. But here an estimator cannot depend on theta. To put it differently, a constant estimator is ‘not allowed’. But how could we express it more formally? For example, it seems like Lagrange multipliers method does’t work here (leads to a constant estimator). | |
| Jun 17, 2022 at 20:14 | comment | added | Iosif Pinelis | @Ilia : So, what is the problem? That the correct solution "looks a bit different from what [you are] used to"? | |
| Jun 17, 2022 at 20:09 | comment | added | Ilia | Right, given this fact my question sound a bit stupid. But I don’t care about the bound itself. Let’s say we just consider the problem to find UMVUE. It is basically the optimisation problem I am taking about. The only problem is that mathematically it looks a bit different from what I am used to. In particular, a constant estimator looks like a mathematically correct solution, and that is absurd. | |
| Jun 17, 2022 at 19:56 | comment | added | Iosif Pinelis | @Ilia : The problem is that, even when a uniformly minimum variance unbiased estimator exists, the Cramér--Rao lower bound is usually not attained at such an estimator. This fact is found in any more or less decent textbook on mathematical statistics. Given this fact, I am wondering what your hope for "a meaningful opt. problem" is based on. | |
| Jun 17, 2022 at 19:37 | comment | added | Ilia | 1. I see, thanks. 2. I understand but I think it should be possible to formulate a meaningful opt. problem even if there is no (in general) solution to it. I don’t know to forbid a constant estimator. | |
| Jun 17, 2022 at 19:26 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |