Does the weibull distribution has a sufficient statistic?

Question

When using the following definition of weibull:

$f(y) = \beta \alpha y^{\alpha - 1}e^{-\beta y ^ {\alpha}} $ , When $\beta>0 \alpha >0$.

I could only find (using the factorization theorem) the following two possible sufficient statistics:

$\sum ln(y_i) $ , $\sum (y_i^\alpha) $

The first one seems fine. However, the second one is not clear to me. From what I've learned, a statistic need to not be dependent on the parameters of the function. How then can the second statistic (depending on $\alpha$), be a statistic (let alone sufficient)? (I would guess that one can estimate $\alpha$ using the first statistic, and then use it for the second one for $\beta$, however - I did not see any nice theory explaining why such an approach would make sense).

Thanks!

Is this two parameter or 1 parameter, i.e. Are both of the parameters unknown? — Lost1, Jan 12 '14 at 20:33
Dear Lost1 - both parameters are unknown (for 1 parameter, the solution is simple) — Tal Galili, Jan 12 '14 at 22:05

score 2 · Accepted Answer · answered Jan 13 '14 at 15:16

Does the weibull distribution has a sufficient statistic?

Yes, trivially, the data itself is always sufficient. Notice order statistic are always sufficient also - i.e. ordering the $(x_1,x_2,...x_n)$ in ascending or descending order, this is always the case when the samples are iid, i.e. order does notmatter.

You probably want a something better than order statistics, minimal sufficient? The answer seems to be: there is none. The reason is you need

$\prod_{i=1}^n f_{\alpha,\beta}(x)/f_{\alpha,\beta}(y)$ is independent iff $S(x)=S(y)$. This needs to hold for ALL $\alpha>0$.

The only way I see that might happen is by permutation.

http://books.google.co.uk/books?id=6wdcTfiLNS4C&pg=PA313&lpg=PA313&dq=%27sufficient+statistics%27+weibull+distribution&source=bl&ots=7C6XVI1J-x&sig=-02V4thha7rYWIbMo3FluLdSocQ&hl=en&sa=X&ei=KgPUUv2UA4mp7Qb10oGYAQ&ved=0CEAQ6AEwAg#v=onepage&q=%27sufficient%20statistics%27%20weibull%20distribution&f=false seems to confirm my answer.

After talking about this with several colleges, I seem to get to the same conclusion. Thanks you Lost1 for looking into this. — Tal Galili, Jan 13 '14 at 15:47
In case anyone is looking for details, here's a proof sketch: Expand the given expression $\prod_{i=1}^n f_{\alpha,\beta}(x)/f_{\alpha,\beta}(y)$ and take its logarithm, then differentiate twice with respect to $\alpha$. This second derivative must be $0$ for all $\alpha > 0$, and you'll see that it follows that all moments of the two samples $x$ and $y$ must agree. Thus, $x = y$ up to permutation, so no dimension reduction is possible. — Yakov Shklarov, Mar 25 '19 at 04:24

score 2 · Answer 2 · answered Oct 27 '14 at 04:05

2

For a known $\alpha$ (and unknown $\beta$) it belongs to the exponential families and the sufficient (and complete) statistic can easily be derived.

answered Oct 27 '14 at 04:05

Daniel

2,630

Does the weibull distribution has a sufficient statistic?

2 Answers2