Wikipedia:Reference desk/Archives/Mathematics/2012 March 12

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March 12

Quantifying heterogeneity

I'm looking for a formal mathematical/statistical measurement of heterogeneity within a set. Basically, what I'm asking is, how does one mathematically/statistically quantify heterogeneity within a set? Wikipedia's article on homogeneity and heterogeneity only talks about the concepts in general terms and from an applied science point of view, whereas I'm looking for an exact mathematical/statistical measurement.

To further explain, imagine a bucket consisting of 100 colored balls. In ascending order of heterogeneity:

• All 100 balls are blue. (Completely homogeneous; no heterogeneity.)
• 90 balls are blue, and the other 10 are 10 other different colors.
• 50 balls are blue, and the other 50 are 10 other different colors.
• 10 different colors are represented, each color by 10 balls.
• All 100 balls are different colors. (Completely heterogeneous; no homogeneity.)

While the ordering of the above is fairly intuitive, more complex situations lead to ambiguity. For instance:

• 50 balls are blue, and the other 50 are 10 other different colors.
• 5 different colors are represented, each color by 20 balls.

Which of these two is more "heterogeneous"? It is not clear: the former has greater color variation, but the latter has less concentration in one color. So I'm look for some sort of way to mathematically measure heterogeneity within a set, both of sets with a finite number of elements (like the above ball bucket examples) and of sets with non-countable constituent parts (for instance, a tank filled with various gases, each of which make up a certain percentage of the total mass or volume of the tank). I assume there must be such a measure in common usage, since this seems like a fairly basic kind of statistics problem?

—SeekingAnswers (reply) 00:55, 12 March 2012 (UTC)

I think the answer is that there are different ways to measure it, depending on what you're looking for. StuRat (talk) 04:02, 12 March 2012 (UTC)

So, could you give some of those ways? —SeekingAnswers (reply) 05:44, 12 March 2012 (UTC)

Perhaps total number of colors or average number of balls of each color. Percentage of balls in the biggest group would be another. StuRat (talk) 05:59, 12 March 2012 (UTC)

Borrowing from statistical mechanics and the concept of entropy, one measure of heterogeneity is the number of possible microstates a configuration allows. For example, 100 blue balls permits exactly 1 microstate (all balls are blue), but the 100 balls of different colors would allow you to have 100! ≈ 9×10¹⁵⁷ different ways to arrange the balls. Measuring the number of possible arrangements can be extended to all of your examples, though doing the math may be complicated in practice. Dragons flight (talk) 06:34, 12 March 2012 (UTC)

Very clever, but isn't the formula a simple one, ie. N!/(a!b!c!...) where N is the total number, and a, b, c,... are the numbers of each individual colour? N! is the total number of permutations (microstates), and then you divide by the number of microstates for each macrostate. The numbers might get horrendous, but I'm sure there are ways of estimating the total. IBE (talk) 07:36, 12 March 2012 (UTC)

The measure of heterogeneity that you are looking for is the Shannon formula for entropy (information theory):

$\sum _{i}-p_{i}\log _{2}(p_{i})$

bits, where $p_{i}$ is the probability that a random ball has color no i.

• All 100 balls are blue. $\sum _{i}-p_{i}\log _{2}(p_{i})=-1\times \log _{2}(1)=0$ bits.
• 90 balls are blue, and the other 10 are 10 other different colors. $-{\frac {90}{100}}\log _{2}\left({\frac {90}{100}}\right)-10\times {\frac {1}{100}}\log _{2}\left({\frac {1}{100}}\right)=0.801188$ bits.
• 10 different colors are represented, each color by 10 balls. $-10\times {\frac {10}{100}}\log _{2}\left({\frac {10}{100}}\right)=3.32193$ bits.
• All 100 balls are different colors. $-100\times {\frac {1}{100}}\log _{2}\left({\frac {1}{100}}\right)=6.64386$ bits.

Bo Jacoby (talk) 08:20, 12 March 2012 (UTC).

Information theory barnstar for hitting the nail right on the head :) Rschwieb (talk) 14:32, 12 March 2012 (UTC)

Need the answer of this question in algebra

how many integer solutions for y and x does the equation x^2 -y^2=91 have? I thought it was 4 but the answer key says its 16. — Preceding unsigned comment added by Thepurplelefant (talk • contribs) 03:10, 12 March 2012 (UTC)

Which 4 did you find ? StuRat (talk) 03:12, 12 March 2012 (UTC)

(10,3),(10,-3),(-10,3),(-10,-3) — Preceding unsigned comment added by Thepurplelefant (talk • contribs) 03:14, 12 March 2012 (UTC)

There's another set of 4 answers:

(-46,-45),(-46,45),(46,-45),(46,45)

However, that only gives 8 answers. I suspect that they doubled that since there are 2 numbers in each answer, but they really should be more clear than that. StuRat (talk) 03:31, 12 March 2012 (UTC)

You can use x^2 - y^2 = (x+y)×(x-y) to show that these are the only 8 answers. PrimeHunter (talk) 03:33, 12 March 2012 (UTC)

To clarify what StuRat said, the answer key is probably counting (3,10) and (10,3) as distinct answers, which would double the count. Dragons flight (talk) 04:24, 12 March 2012 (UTC) Okay, I'm dumb... Dragons flight (talk) 05:18, 12 March 2012 (UTC)

No. 3² - 10² = -91, so that's not a correct solution. They seem to be counting the (10,3) as two answers, being 10 and 3, which is a weird way to put it. StuRat (talk) 04:56, 12 March 2012 (UTC)

$x,y\in \mathbb {Z} \implies (x+y),(x-y)\in \mathbb {Z}$ so $(x+y)$ and $(x-y)$ must be factors of 91. That gives you eight systems of linear equations:

$x+y=1$

$x-y=91$

$x+y=7$

$x-y=13$

$x+y=13$

$x-y=7$

$x+y=91$

$x-y=1$

$x+y=-1$

$x-y=-91$

$x+y=-7$

$x-y=-13$

$x+y=-13$

$x-y=-7$

$x+y=-91$

$x-y=-1$

Widener (talk) 05:29, 12 March 2012 (UTC)

Extending theorem to uncountable sets

One can show easily that bounded countable subsets of $\mathbb {R}$ have a supremum given that all bounded nondecreasing sequences have a limit. I'm trying to think of a way to extend this to uncountable subsets. One way I've thought of it is to show that you can take any bounded countable subset of $S\subset \mathbb {R}$ and add an uncountable number of reals to that set, and as long as each element added is less than the supremum of $S$, the supremum of $S$ will also be the supremum of this new set. It is obviously important to show that all bounded subsets of $\mathbb {R}$ can be generated in this way. I'm sure they can, but I'm not sure how to prove it. Given a bounded subset $U\subset \mathbb {R}$, one can find a countable subset $C\subset U$ which has a supremum greater than or equal to every element in $U$, but why? Widener (talk) 06:00, 12 March 2012 (UTC)

The short answer is that $\mathbb {R}$ is a hereditarily separable space. If you want to prove it directly, you can exploit the density of the rationals. Let $\langle q_{i}\rangle$ be a listing of the rationals. Fix some $u_{0}\in U$. Define $c_{i}$ to be an element of $U$ bigger than $q_{i}$, if there is such an element, and $c_{i}=u_{0}$ otherwise. Let $C$ be the collection of $c_{i}$. Now verify that $C$ is as desired.--121.74.121.82 (talk) 06:38, 12 March 2012 (UTC)

Tensors

$T_{ij}$ is a Cartesian tensor with respect to A and also a Cartesian tensor with respect to B. Show that it is a Cartesian tensor with respect to BA. (A and B are orthogonal but that might not be relevant).
This is what I did:
$B_{ik}B_{jl}A_{km}A_{ln}T_{mn}$
$=B_{ik}B_{jl}{\bar {T_{kl}}}$
$={\bar {\bar {T_{ij}}}}$
But now that I think about it, this depends on ${\bar {T_{ij}}}$ being a Cartesian tensor with respect to B, not $T_{ij}$ Widener (talk) 10:46, 12 March 2012 (UTC)

"depends on ${\bar {T_{ij}}}$ being a Cartesian tensor with respect to B, not $T_{ij}$": Well, $T_{ij}$ and ${\overline {T}}_{ij}$ are not different tensors, they are different components for the same tensor. What you have above looks fine, as long as you can see that the B_ij are components for B with respect to the image of your original basis in A. Specifically if you start in basis $x^{i}$, go to the basis ${\overline {x}}^{i}=Ax^{i}$ and then ${\overline {\overline {x}}}^{i}=B{\overline {x}}^{i}$. Orthogonality will ensure that the image of a basis is a basis. Rschwieb (talk) 11:53, 12 March 2012 (UTC)

who sets racetrack odds and how?

who sets racetrack odds and how? How do bookies have a good enough knowledge of who is likely to win to set favorite/longshot odds to begin with, and how can this knowledge be SO good that a fluke doesn't leave them broke? If I were a bookie, I would have to have an 10x broker's fee on any bets placed (so that only if a 10:1 underdog wins, I lose) before I could offer any odds I could live with.... 84.1.193.119 (talk) 13:38, 12 March 2012 (UTC)

They don't try and forecast the race or bet against you, they try to set their odds as new bets come in so whoever wins they always make a profit. They do of course do some forecasting to set initial odds and can lose because they can't balance the books if everyone bets on the favourite and it wins for instance - but in those case people don't make much anyway and the betting shop can reject the bet or offer lower odds for very large bids in advance of the day. There's also systems where people just get a cut of the total bet and the winnings from them tend to be pretty similar to the fixed price odds. There the betting shop takes no risk at all. Dmcq (talk) 14:03, 12 March 2012 (UTC)

are you telling me that for the most part it's like a market, and odds are really just 'prices' that float with some transaction cost? (the fees you mention) 84.1.193.119 (talk) 14:06, 12 March 2012 (UTC)

Also, why do you say "in those case people don't make much anyway" -- if a 100:1 underdog wins, wouldn't the people win the MOST in those cases? So what if, in setting those initial odds, too many people bet on the underdog and it wins? 84.1.193.119 (talk) 14:07, 12 March 2012 (UTC)

Yeas a comparison to a market is good. I see we have an article on all this Mathematics of bookmaking. It is of course possible for someone to come in and win big on a 100:1 bet, but the bookmaker will change the odds pretty quick if everyone started betting on a particular underdog that way. Dmcq (talk) 14:11, 12 March 2012 (UTC)

your last sentence is particularly interesting. I wonder if a crowd of people could come in as a flash mob and start placing (cheap) bets on a particular, longshot horse - making it, at spot prices, not in the forecast sense, the "favorite" - and pushing the current (forecast meaning) favorite out of that position. Then the person who orchestrated all this could simple bet on the (forecast-meaning) favorite at favorable prices. How do bookies protect themselves from this, if they don't forecast? And if, as you say, they will "change the odds pretty quick if everyone started betting on a particular underdog that way". --84.1.193.119 (talk) 14:56, 12 March 2012 (UTC)

Presumably, the amount that you'd win by getting slightly better odds for betting on the favorite would be offset by the cost of all those long-shot bets. For an example of a distillation of this system, see Intrade, which structurally operates like a bookie, but merely facilitates bets between users (they don't stand to gain or lose money on bets, so they charge a monthly fee). Paul (Stansifer) 15:42, 12 March 2012 (UTC)

wow, private betting, all this sounds terribly smoke-filled! I'm just a lowly mathematician interested in economics, but I feel like I'm smoking a cigar just reading this! --84.1.193.119 (talk) 16:20, 12 March 2012 (UTC)

There are other bets, where they do try predict the chance of an event happening. There's the 20 million-to-1 if Elvis crashed a UFO on top of the Loch Ness Monster. Bets on where skylab would crash, first man on the moon, and other bets that for example William Hill has accepted (the chance of the olympic flame blowing out..) 84.197.178.75 (talk) 17:01, 12 March 2012 (UTC)

Any ONE of these things are far less likely than one in twenty million: Elvis is alive (he couldn't be seen by anybody; EVER); the Loch Ness Monster is real (the lake's been plumbed); and a UFO crashes. That these things happen should independently at once has way less than 8000 million million million in one odds, and should pay much better. You are getting ripped off! --84.1.193.119 (talk) 12:05, 13 March 2012 (UTC)

applications of linear functions

What are some direct applications of knowing how to solve linear functions--148.241.97.247 (talk) 14:36, 12 March 2012 (UTC)

depends what you mean, solving a linear equation in one variable is quite trivial, that's for example calculating how many apples you can buy at 20 cent a piece if you got 200 cent, or how fast you ran when you did 25 km in three hours. If you mean a real linear function, one for which f(x)+f(y)=f(x+y) and f(ax) = a*f(x), then I'm not sure what's there to solve. Solving a System of linear equations is used a lot in science, economics etc.. a direct application , hmm.. solving riddles? Like if you count 50 heads and 120 legs, how many people and how many dogs are there ... 84.197.178.75 (talk) 17:57, 12 March 2012 (UTC)

Here's a more complex example than your usual one to really get your teeth into. Having hated school work and never doing any after leaving you get a job cleaning offices, you join another person who has been cleaning the offices in five hours and together you do the job in three. However they want a night off to escape and get a life - how long will it take you to clean the offices on your own? ;-) Dmcq (talk) 18:10, 12 March 2012 (UTC)

40,10 & 7.5, WTP? :)Naraht (talk) 21:50, 12 March 2012 (UTC)

I think one of the commonest uses outside of scientific disciplines is in linear programming. There is an example in that article of an application by a farmer in planting grain. They also use it o decide what feedstock to buy for their animals. Dmcq (talk) 13:47, 13 March 2012 (UTC)

in 7 hours and a half?--148.241.97.247 (talk) 20:33, 13 March 2012 (UTC)

That's right, very good. You can see why people don't like that happening. Having a group of cleaners go round few offices means they can cope with holidays and illness better. Dmcq (talk) 23:49, 13 March 2012 (UTC)