February 10

Hi there refdesk - I was wondering if I could get some advice on this problem, I fear it may be a bit above my current capabilities!

Let ${\displaystyle {\hat {\theta ))}$ be an unbiased estimator of ${\displaystyle \theta \in \Theta \subseteq \mathbb {R} }$ satisfying ${\displaystyle \mathbb {E} _{\theta }({\hat {\theta ))^{2})<\infty }$ for all ${\displaystyle \theta \in \Theta }$. We say ${\displaystyle {\hat {\theta ))}$ is a 'uniform minimum variance unbiased estimator' if ${\displaystyle {\text{Var))_{\theta }{\hat {\theta ))\leq {\text{Var))_{\theta }{\tilde {\theta ))}$ for all ${\displaystyle \theta \in \Theta }$ and any other unbiased estimator ${\displaystyle {\tilde {\theta ))}$. Prove a nec. and suff. condition for ${\displaystyle {\hat {\theta ))}$ to be a UMVU estimator is ${\displaystyle \mathbb {E} _{\theta }({\hat {\theta ))U)=0}$ for all ${\displaystyle \theta \in \Theta }$ and all estimators U with ${\displaystyle \mathbb {E} _{\theta }(U)=0}$ and ${\displaystyle \mathbb {E} _{\theta }(U^{2})<\infty }$ (i.e. '${\displaystyle {\hat {\theta ))}$ is uncorrelated with every unbiased estimator of 0').

May just be me but that seems exceedingly complicated, I don't think I can quite get my head around it I'm afraid! The more help/detail I can get, the better, so please do be generous :) Thankyou! Otherlobby17 (talk) 07:05, 10 February 2010 (UTC)[reply]

For ${\displaystyle \Leftarrow }$, use ${\displaystyle U={\hat {\theta ))-{\tilde {\theta ))}$ and the Cauchy–Schwarz inequality. For ${\displaystyle \Rightarrow }$, use ${\displaystyle {\tilde {\theta ))={\hat {\theta ))+\lambda U}$. -- Meni Rosenfeld (talk) 09:56, 10 February 2010 (UTC)[reply]

A div operator is mentioned at mediawiki. I expected it to be a modular division operator. However, ((#expr:3 div 2)) = 1.5. There is a ((div)) template that uses (a - (a mod n)) / n (humm). So now I'm curious. Does anyone know what the function of the div operator. Is it a simple division operator. I can't find any documentation on this. Thanks. –droll [chat] 08:41, 10 February 2010 (UTC)[reply]

I tried it for a few cases and it gave simple division every time, so it is highly unlikely that it is anything else. -- Meni Rosenfeld (talk) 09:25, 10 February 2010 (UTC)[reply]

Thanks. Things like this get to me. –droll [chat] 10:46, 10 February 2010 (UTC)[reply]

mw:Help:Extension:ParserFunctions##expr refers to meta:Help:Calculation for more details of the function of each operator. There it says about div: "Division, same as /, not integer division". PrimeHunter (talk) 00:38, 12 February 2010 (UTC)[reply]

I've been wondereing lately if there are distributive lattices of length n with more than 2ⁿ elements? If yes, are there distributive lattices of finite length with infinitely many elements?

In either case, proof or counterexample would be nice. An example for a distributive lattice with 2ⁿ elements is the Boolean lattice over some set with cardinality n. Yaan (talk) 13:17, 10 February 2010 (UTC)[reply]

The answer is no. First assume that the lattice is finite. If it has length 0, it clearly has size 1. If it has length n > 0, fix a chain a₀ < a₁ < ... < a_n. Since the chain cannot be expanded, 0 := a₀ must be the smallest element of the lattice, 1 := a_n the largest element, and a := a₁ an atom. Put ${\displaystyle b=\bigvee \{c\mid a\nleq c\}=\bigvee \{c\mid c\land a=0\))$. By distributivity, ${\displaystyle b\land a=0}$, thus b < 1. It follows that the lattice is a union of two sublattices [0,b] and [a,1], each of which has length at most n − 1. By the induction hypothesis, the sublattices have cardinality at most 2ⁿ⁻¹, thus the whole lattice has cardinality at most 2ⁿ. This completes the proof for finite distributive lattices. If the lattice were infinite, pick arbitrary 2ⁿ + 1 elements, and take the sublattice generated by these elements. This is a finite distributive lattice with length at most n, but more than 2ⁿ elements, contradicting the previous part of the proof. —Emil J. 13:59, 10 February 2010 (UTC)[reply]

Alternatively, any distributive lattice of length n can be embedded in a Boolean algebra of length n (and therefore of size 2ⁿ), but I'm not sure how to prove that in an elementary way. —Emil J. 14:10, 10 February 2010 (UTC)[reply]

It follows immediately from Birkhoff's representation theorem. If a distributive lattice is represented by a poset, the length is exactly the number of elements in the poset, and each lattice element is represented by a lower set in the poset, so the number of lattice elements is at most the number of subsets of the poset, which is 2ⁿ. —David Eppstein (talk) 17:17, 10 February 2010 (UTC)[reply]

Thanks a lot everybody. Yaan (talk) 21:47, 15 February 2010 (UTC)[reply]

Is there a distinguishing property of periodic functions limits? For example, a periodic function might have its limit undefined at ∞; but still has a defined boundary (i.e: [a,b]) at that limit.--Email4mobile (talk) 19:43, 10 February 2010 (UTC)[reply]

A periodic function has a limit at ∞ if and only if it is a constant function. A periodic function is bounded between a and b as x approaches ∞ if and only if it is bounded between a and b everywhere.

These seem like very easy questions. Did you mean to ask something else, or were you just confused about the definition of a limit approaching ∞? --COVIZAPIBETEFOKY (talk) 21:26, 10 February 2010 (UTC)[reply]

• It might be very easy to you, but new to me. I mean by common sense I knew that but I didn't know there is a rule. Thanks a lot.--Email4mobile (talk) 21:48, 10 February 2010 (UTC)[reply]

I didn't quote any rule. It might be a well-known theorem with a name, for all I know, but I'd never heard that question before in my life. I just figured it out logically; they're really not hard questions to work out. A periodic function can't have a limit to ∞ unless it's constant, because if it isn't constant, then it has at least two distinct values all the way out to ∞ (since it's periodic, two distinct values at x and x' imply two distinct values infinitely far out). Similarly, if the function is bounded over its period, then it's bounded over the entire real numbers, since the function simply repeats that one portion (and unbounded over any portion implies unbounded infinitely far out). --COVIZAPIBETEFOKY (talk) 01:01, 11 February 2010 (UTC)[reply]

People don't usually bother giving names to results with one-line proofs, so I doubt it has one. --Tango (talk) 01:09, 11 February 2010 (UTC)[reply]

What about the Frattini argument? Of course, this is just an arbitrary example and I agree that in general, arguments which follow directly from definitions do not warrant a name. PST 04:25, 11 February 2010 (UTC)[reply]

I said "usually". --Tango (talk) 12:16, 11 February 2010 (UTC)[reply]

The question may not be that trivial. It should be possible and interesting to extend the definition of limits to the interval arithmetic framework, so the interval-limit of a periodic function at ∞ is the set of values it can take. Mathematica does this - Sin[∞] returns Interval[{-1, 1}]. -- Meni Rosenfeld (talk) 07:34, 11 February 2010 (UTC)[reply]

Isn't that basically limsup and liminf? --Tango (talk) 12:16, 11 February 2010 (UTC)[reply]

Depends. If the function is discontinuous then its "interval-limit" can be different from ${\displaystyle [liminf,limsup]}$ (in which case it is not an interval at all). But anyway I'm not implying there's any new information in such a definition, only that it is a nice way to look at it. -- Meni Rosenfeld (talk) 12:50, 11 February 2010 (UTC)[reply]

There are given 9999 sticks with lengths of 1, 2, ..., 9998, 9999.

The players A and B alternately remove one of the sticks, player A begins. The game ends when only three sticks remain.

Player A wins if it can be formed from these sticks a "non-degenerate" triangle, otherwise B. Who can force a win? —Preceding unsigned comment added by 85.178.25.121 (talk) 23:03, 10 February 2010 (UTC)[reply]

This is a problem of the current Bundeswettbewerb Mathematik, the German mathematical Olympiad, so please don't discuss it till 1. March. Sorry to all that answered this question. --Schnark (talk) 09:11, 11 February 2010 (UTC)[reply]

Oh! Thanks for telling us. Eric. 131.215.159.171 (talk) 12:01, 11 February 2010 (UTC)[reply]

I have read its defination as ,"If one arm of an angle is on the edge of a half plane,then though the end of this arm one and only one ray can be drawn to the half plan to construct an angle of certain measure. " I could not understand it,please explain it. Is this defination correct.if not tell me correct defination.-mks-True path finder (talk) 23:45, 10 February 2010 (UTC)[reply]

The main confusion is the use of "half-plane". What they are saying is that you have a plane. You draw a line on it. That line cuts the plane in half. Now, you have a half-plane and a line (along the edge of it). If I give you an angle from 0 degrees to 180 degrees and ask you to draw another line on that half place at that angle to the original line, there is only one line you can draw. The limit of "one line you can draw" is based on another fishy word they used - "end". That original line must have an end point. So, the line you draw has to go through that end point at a specific angle. The entire point of all of this is that two lines form a distinct angle. You can't have an angle be 45 degrees and 50 degrees at the same time. -- kainaw™ 23:50, 10 February 2010 (UTC)[reply]

Thank u very much. —Preceding unsigned comment added by True path finder (talk • contribs) 00:02, 11 February 2010 (UTC)[reply]