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May 3[edit]

How to make my own do it yourelf leading indicator?[edit]

I have two different sets of time series data, A and B. I think that changes in B lead those in A, although there is not a perfect correlation and I do not know what the time lag is. What is the best way to use B to predict A? 89.242.97.56 (talk) 12:48, 3 May 2009 (UTC)[reply]

Your question sounds like a homework problem. We don't do homework problems here. Wikiant (talk) 13:16, 3 May 2009 (UTC)[reply]

It is not a homework question. I've just expressed it in the simplest clearest way I can. I'm not even a student of any kind - you do not get many students of my age. 89.242.97.56 (talk) 13:47, 3 May 2009 (UTC)[reply]

Calculate the correlation coefficient between A and B for each possible lag from 1 up to whatever you like, pick the highest (positive or negative) and calculate the linear regression equation. If this doesn't give anything useful, shove all tha data into a statistical analysis package.86.166.203.189 (talk) 15:47, 3 May 2009 (UTC)[reply]

Good answer, although using one data point in B may not make the best use of the information in other values of B. What statistical analysis package do you have in mind please - the more elaborate ones that could do this tend to require a lot of input and know-how from the user. 78.146.219.21 (talk) 09:40, 5 May 2009 (UTC)[reply]

Thanks. What about using multiple regression where the independant variables are B at different times? Would that be sensible? I'm thinking it may be invalidated by correlations between the different B values. But I remember being told that only matters if you are trying to build a model, that it does not matter for prediction. Is that correct? 78.146.31.176 (talk) 17:02, 3 May 2009 (UTC)[reply]

To predict you need to make a model using the best correlation that you can find. If you are using real world data (Stock prices? Audio recordings? Crime statistics? Horoscopes?) the existence of the correlation that you expect is still just a hypothesis. Cuddlyable3 (talk) 20:18, 3 May 2009 (UTC)[reply]

What I'm thinking of is using something like autoregression. But autoregression uses just past values of A, whereas I want to use past values of B. Although using past values of both A and B would not do any harm. I'm trying to find a method without having to read thick academic textbooks. Is there any software 'sausage-machine' that can do all this without much input from the user? 78.146.219.21 (talk) 09:36, 5 May 2009 (UTC)[reply]

Does anyone know how government leading indicators are discovered in practice? The British ones, which may also be called cyclical indicators were apparantly designed by someone called O'Dea or some similar name who wrote a book about them in 1975. 78.145.24.191 (talk) 11:01, 7 May 2009 (UTC)[reply]

What's going on here?[edit]

Using the standard thin lens formula, 1/i + 1/o = 1/f, i tried to figure out the dependence of i (image distance) on f (focal length) and o (object distance). In Mathematica, I plotted Plot3D[1/(1/f - 1/o), {o, 2, 100}, {f, 2, 10}, PlotRange -> {0, 100}] and Plot3D[1/(1/f - 1/o), {o, 2, 10}, {f, 2, 10}, PlotRange -> {0, 10000} ] and got some wacky results (apologies to those who don't have Mathematica. What's going on here with the sharp periodic peaks? I see some wacky non-monotonic things, and I'm not any more enlightened than I was just looking at the formula, from which you can tell i approaches f for sufficiently large o/f. Thanks. --VectorField (talk) 16:49, 3 May 2009 (UTC)[reply]

Well, I'm not sure that Mathematica is smart enough to handle the line of values at

f=o

(and the 'periodic' features are artifacts of this). If you plot it in 1D, a few cross sections will give you an idea of the behaviour. Martlet1215 (talk) 20:04, 3 May 2009 (UTC)[reply]

The lens formula rearranged to express i as a function of o and f is i = o f / (o - f)

Choose a value for f and plot the graph of i depending on o. When you see what that curve looks like on paper (a 2-D graph), do it again for a different value of f. By now you will have a good idea of the dependence of i on the two variables. Cuddlyable3 (talk) 20:35, 3 May 2009 (UTC)[reply]

Well, when I said "plot it in 1D" it was just my personal notation for 2D. Is that a credible excuse? Martlet1215 (talk) 21:53, 3 May 2009 (UTC) [reply]

Adjusting a statistical snapshot for duration[edit]

Not a homework question. Lets say that in a census of mothers in an imaginary country, 60% of them have one child, 30% have two children, and 10% have three children. I want to estimate the average number of children a mother will have in her lifetime. However, I cannot just take a weighted average of the census results because many of the mothers are young and have not yet had all their children yet. How should I adjust the census results to take this into account? Edit - replaced "parents" by "mothers" for simplicity. 78.146.31.176 (talk) 18:09, 3 May 2009 (UTC)[reply]

There's two principal problems I can see. You need a lot more data than that to get any meaning result. And secondly what do you mean exactly since the average number probably varies with time. You'd have to say if it is for people born at a particular time or died at a particular time or average across the various averages for the ones alive at this particular time - which means extrapolating behaviour into the future. Dmcq (talk) 18:26, 3 May 2009 (UTC)[reply]

A census of a whole country is quite a lot of data. If the census indicates ages of parents having various numbers of children, I suspect you'd want to take that into account. I'd have to think about it further to say much more than that. Michael Hardy (talk) 22:27, 3 May 2009 (UTC)[reply]

I have some more data about another imaginary country bordering the previous one. 50% of the children born between 2002 and 2004 were born to mothers who had no other children. Of this group of mothers, 15% of them now have another child in 2009. Is this any use in getting a rough estimate of the average number of children that mothers have over a lifetime? What about if I combined this data with the previous data? Very rough estimates are better than nothing. Reasonable guestimates can be made: for example that a mother could bear children over a thirty-year interval. 89.243.185.122 (talk) 10:42, 4 May 2009 (UTC)[reply]

You really do need to figure out what your question is first. In the real world birth rate can and has varied quite dramatically in countries with better medicine, more education or wealth, the sudden influx of poor people or wars. How is one to say what the average number of children over a lifetime is if the people haven't died and circumstances might change? The British government for instance keeps saying that people on pensions are getting better off with time, and they arer quite correct when you just consider the average at each time of the pensioners then alive. However each individual person gets considerably worse off with time. Figure that out and then see if you even want to choose whatever definition it is you want of averager number of children per woman in their lifetime. You need a purpose and meaning for a figure before you start calculating it, and figuring that out is nrmally much harder than just crunching th numbers at the end. Dmcq (talk) 09:39, 6 May 2009 (UTC)[reply]

I understand as mathematicians you find the idea of a very approximate answer based on several guestimatated assumptions to be difficult to accept, but thats what the situation pragmatically demands. The data above is all the data I have. If you could show me how to calculate the answers if you had the full information required, then I can make the guestimates myself, and play around with a range of values to see how that affects the results. We used to do this all the time in GCE "A" level Nuffield Physics. 78.145.24.191 (talk) 10:55, 7 May 2009 (UTC)[reply]

You have to come up with a model for the arrival time of children. Suppose, for example, we know that mothers choose in advance how many children to have and then have them as quickly as possible; then our sample is (nearly) representative of the "final" distribution, because almost none of our sample will be found during the (short) time between kids #1 and #N. If, on the other hand, it is traditional to have one child as soon as it's possible, and then to wait as long as possible before having any more, then a large adjustment must be made. A slightly more reasonable (but still very unrealistic) model would have each mother choose in advance the number of children (up to

N=3

), but then have them at times drawn independently and uniformly from an interval of b time after which they live another a time. The distribution of choices and the intervals are taken to be constant. Then let the proportion of mothers who will have n children be

x_{n}

(

x_{0}=0

); the expected proportion who will have n and currently have

i\geq 0

children is just

x_{ni}=x_{n}{\begin{cases}1-{\frac {nb}{(b+a)(n+1)}}&i=n\\{\frac {b}{(b+a)(n+1)}}&i<n\end{cases}}

. What we observe is

y_{i}={\frac {z_{i}}{Z}}

for

i>0

, where

z_{i}:=\sum _{n=i}^{N}x_{ni},\;Z:=\sum _{j=1}^{N}z_{j}=1-z_{0}=1-\sum _{n=0}^{N}x_{n0}=1-{\frac {b}{b+a}}\sum _{n=1}^{N}{\frac {x_{n}}{n+1}}

. We can then multiply through by Z and set up a set of simultaneous equations in the

x_{n}

. (The same general approach applies for other models, but the formula for

x_{ni}

(and thus

z_{i}

and Z) will differ.) Does that help? --Tardis (talk) 16:12, 7 May 2009 (UTC)[reply]

Or simply look up List of countries and territories by fertility rate. As far as I know the fertility calculated there assumes that the fertility is constant through time. This is the easiest thing to do and is the same as how I was saying the government says pensioners are getting better off with time whereas on average each particular one gets worse off. You'd really need some information about how long the women live and when they have children. If you have no extra information you might be able to match it to some country to get an estimate assuming that circumstances will produce a characteristic signature of probabilities. But generally this is the problem with people going and doing experiments or surveys without consulting a statistician first. They waste money and time. Dmcq (talk) 17:33, 7 May 2009 (UTC)[reply]

Applications of Conway's cosmological theorem[edit]

Does Conway's cosmological theorem (dealing with the look-and-say sequence) or his constant have any known or predicted practical applications? Neon Merlin 23:33, 3 May 2009 (UTC)[reply]

It's what's termed recreational maths and contributed to the sales of some books, and has probably made more money that way than most maths papers! When Euclid was asked what was the use of learning Geometry he had a servant give the man a coin so he could profit from it. Dmcq (talk) 07:56, 4 May 2009 (UTC)[reply]

"He" - would that be the man, the servant, or Euclid? 89.243.185.122 (talk) 09:03, 4 May 2009 (UTC)[reply]

In long: Euclid had a servant give the man a coin so the man could profit from Geometry. Dmcq (talk) 09:37, 4 May 2009 (UTC)[reply]

By the way I think the anecdote is reported by Proclus (Commentary on the first book of Euclid's "Elements of Geometry"). The man who asked Euclid what is the use of geometry was a student of him. Euclid's response, of course, intended to scorn that guy and show despise (he didn't even answer to him, but to the servant: "give this man a coin, since he must make a profit from what he learns"). --pma (talk) 23:17, 4 May 2009 (UTC)[reply]