Talk:Stein's unbiased risk estimate

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Improvements?[edit]

I think this article has the following problems:

  • Something needs to be said about what is meant by "deterministic estimation" etc.;
  • It looks as if some of the results are only approximate (eg. unbiasedness) unless only linear models are condsidered;
  • A dimensional analysis of the equation for SURE makes it look wrong;
  • It is not said what norm is being used.
Melcombe (talk) 09:33, 12 June 2008 (UTC)[reply]

Incorrect Expression[edit]

Melcombe's comment on the dimensional analysis of the expression prompted me to take a closer look, and it turns out that the expression as it stands is incorrect. The expression as written is

.

The correct expression is

,

which when reduces to

Both sources cited give only the version, so even though it's less general I will opt to keep that for now and hunt for a source that provides the more general version. (I would like to also include the derivation, which is simple and instructive.)

Mygskr (talk) 02:25, 20 July 2012 (UTC)[reply]

OK, I found a reference for the generalized case with variance and have updated the article accordingly. Mygskr (talk) 16:56, 19 September 2012 (UTC)[reply]

Unbiasedness Claim[edit]

I can see that some defects have been corrected in the last 5 years. I am not an expert at all on this estimator, but I am inclined to accept that Melcombe's second bullet point is surely correct and the article still requires further correction on that account. Surely the "formal statement" cannot really be that there is perfect unbiasedness, under the constraint merely of weak differentiability on function . The real property must be some kind of approximation to unbiasedness whose proof would be based on a Taylor expansion of function . So my guess is that must be an affine function in order for the claimed unbiasedness to become an exact property. But is there also perhaps some correct result in the literature about asymptotic unbiasedness as increases, under some weaker (than affine) constraints on ? — Preceding unsigned comment added by 138.40.68.40 (talk) 16:20, 12 September 2013 (UTC)[reply]

It sounds incredible, but it is true. There is nothing asymptotic about this result, and the only condition required is the weak differentiability of , which allows for one application of integration by parts. There is no Taylor series involved.
For a proof, see Proposition 1 in this note, which states that the expectation of SURE is exactly equal to the MSE. I've been putting it off, but maybe within the next week or so I'll wikify the proof and include it in the article. Mygskr (talk) 16:53, 12 September 2013 (UTC)[reply]