Wikipedia:Reference desk/Archives/Mathematics/2015 September 3

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September 3

Linear Regression

Simple linear regression gives a good explanation of how to calculate the confidence intervals on a linear least squares fit.

I would like to be able to do the same thing for (i) the case where a point (x0,y0) is known a priori to sit on the line and (ii) the case where the errors are known to be drawn from distributions of differing width (themselves students-t distributions - but they can be treated in the large-n limit as normal distributions if it is simpler).

My problem is that the estimated gradient does not appear to be a simple t-statistic as in the simple case treated in the article.

Does anyone know where I can find a treatment of these problems that would allow me to calculate confidence intervals in these marginally more complex cases.

Thank you. — Preceding unsigned comment added by 128.40.61.82 (talk) 10:07, 3 September 2015 (UTC)

For (ii), I think bootstrapping the errors is your best bet if you truly want to use the family of t-distributions. If you can analytically specify the error structure (e.g. the variance increases proportionate to the magnitude of the observation, etc.), then you might be able to use some form of heteroskedastic regression, typically estimated with either Maximum_Likelihood_Estimation or Weighted Least Squares: here's a link to get you started: http://www.econ.uiuc.edu/~wsosa/econ471/GLSHeteroskedasticity.pdf. In either case, it's extremely likely that the R statistics software has packages to do the calculations.

For (i), I'm not sure what you mean: can you provide an example? You can constrain the line so that it passes through certain points; then, you are minimizing the sum of squared errors subject to this constraint. Be aware that the "nice" properties of linear regression won't necessarily hold: in particular, the estimate is not guaranteed to be unbiased, or to have minimum variance. If you are fitting a curve where the curve points are known exactly, then a cubic spline is a better method than regression. Good luck! OldTimeNESter (talk) 12:40, 4 September 2015 (UTC)

Domino cycles

The usual set of dominoes has 28 pieces, ranging from double blank to double six. There are many ways of arranging them in a circle so that adjacent ends have the same number of spots - I'm trying to determine how many, for a set of dominoes from double blank to double n. For example, for n=0 there is one trivial solution 00, for n=1 there is no solution, for n=2 there are just two, 00 01 11 12 22 20 or 00 02 22 21 11 10. I think that no solution exists for odd n, but can't think of a systematic way if n is an even integer greater than 2. I see that the problem can be represented by an (n+1)-vertex graph, with each connected to all others and to itself by a loop, but this doesn't help me to count the number of possible closed paths. No doubt there is a standard result, but I'd like to see a derivation. — Preceding unsigned comment added by 86.138.241.98 (talk) 19:51, 3 September 2015 (UTC)

Each digit must be used an even number of times as they abut in pairs. We can ignore doubles as they can be inserted at any convenient point later. Thus with a 6 piece set 01 02 03 12 13 23, we can start say with 01 12. But if we continue to 01 12 23 31 we are stuck - there is nothing to pair with the 1 from 31. So, as you are calling this set n=3, it has no solutions. The main thrust of your question requires further thought. -- SGBailey (talk) 22:30, 3 September 2015 (UTC)

Thanks - I'd already said that I saw no solution for n odd, which is obvious from the graph analogy as all vertices have an odd number of edges so no Eulerian cycle is possible. Likewise, all cases with n even do have such cycles.86.138.241.98 (talk) 16:30, 4 September 2015 (UTC)

With set n, n even, there are n+1 doubles. If you determine the number of loops without doubles (call it 'L'), then each double can be inserted in one of n/2 locations. There are n+1 doubles therefore the new number of loops is L * (n/2) ^ (n+1). With n=4 and dominoes 01 02 03 04 12 13 14 23 24 34; L = { 01 12 23 34 42 -- SGBailey (talk) 07:12, 4 September 2015 (UTC)

Take as an example n=4 { 01 02 03 04 12 13 14 23 24 34 } and assuming that we always start from the 01 piece with 0 on the left and 1 on the right, then one loop is [ 01 12 23 34 42 20 03 31 14 40 ]. We can obviously derive 6 loops by permuting 2 3 and 4 (234 243 324 342 423 432). Additionally the rightmost 4 pieces are a sub loop [ 03 31 14 40 ] and that loop can be reversed [ 04 41 13 30 ], so this produces 12 loops. Are there more? can this be generalized? -- SGBailey (talk) 07:22, 4 September 2015 (UTC)

Your two solutions for n=2 are essentially the same, one is just the other reversed. --CiaPan (talk) 08:04, 4 September 2015 (UTC)

Yes, I want to count clockwise and anticlockwise as distinct. I understand that there is no explicit formula for the number of cycles possible with an (n+1)-vertex complete graph (n even), so presumably that applies too to the extended case of each vertex being self-connected. My programming skills aren't up to generating all cycles for particular cases of n>2; is anyone here able to do this so that the first few numerical values can be found?86.138.241.98 (talk) 16:30, 4 September 2015 (UTC)

Here's a paper that might give some hints of where to look in the literature, or some keywords to search for in the OEIS: http://arxiv.org/abs/math/0701488 . --JBL (talk) 18:05, 4 September 2015 (UTC)

How can I extract value of s from Xi function?

what is derivation of riemann xi function with proof?

151.236.160.75 (talk) 17:34, 3 September 2015 (UTC)

I'm not sure what you mean by derivation and proof in this case. We have an article on Riemann_Xi_function. It is simply defined in terms of the Riemann zeta function. The article gives some properties of the Xi function. Are you looking for proofs of some of those properties and identities? SemanticMantis (talk) 17:37, 3 September 2015 (UTC)

How can I denote value of s by Xi function?

ξ(s)=(s/2)Γ(s/2)(s-1) pi^-(s/2)ζ(s)

Exmple

y=x^2+1

x=sqrt (y-1)

Which is meaning inverse function — Preceding unsigned comment added by 151.236.160.79 (talk) 06:16, 4 September 2015 (UTC)

See Inverse function. It's the function that reverses the original function, so if g(x) is the inverse of f(x), g(f(x)) = x. Rojomoke (talk) 12:23, 4 September 2015 (UTC)

I've brought this all back into a single section. Original intent was confusing due to the formatting of the question, but 151.236.160.79 is asking for the inverse of the Riemann Xi function. (Or is that Landau's lower-case xi function?) -- ToE 13:02, 4 September 2015 (UTC)