Wikipedia:Reference desk/Archives/Mathematics/2011 October 30

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October 30

Funding students without student loans

I have a proposal to fund students for their education needs throughout their youth. Instead of having students carrying crushing debts for the rest of their life, this system will allow the student to have zero debt regardless of their personal circumstances.

The system requires the student to obtain funding from the open market because the market knows best. At the age of 10, a child academic records is published for the whole world to see. Then an IPO (initial public offering) is made where 1000 shares for that child is sold in an open auction. A 5% commission is deducted from the sales, the rest of the money is held in an education fund which will fund the education of the child from the age of 10 to the age of 25.

In return for funding the education cost of the child, the owner of the shares will received 33.333% of the student's income (before tax) starting from when the student reach the age of 26 until the death of the student.

The question is this. How much is one share of the student worth. You may assumed there is zero inflation and the student earns an income of 30,000 dollars per year from the age of 26 until the age of 56 (thirty years of income).220.239.44.187 (talk) 08:09, 30 October 2011 (UTC)

I had thought of a similar system myself; but I think people will have a moral problem with issuing an IPO of a person.

If the ROI rate is r (by which I mean that a dollar today is worth r dollars in a year), then $30000 per year received starting 16 years from now and ending 46 years from now are worth ${\frac {30000(r^{30}-1)}{r^{46}\ln r}}$ today. Divide by 3000, and you get that each share is worth ${\frac {10(r^{30}-1)}{r^{46}\ln r}}$. For $r=1+\epsilon ,\ \epsilon \approx 0$ this is approximately $300(1-31\epsilon )$. If there is no return on investment ($r=1$) then this is simply $300. -- Meni Rosenfeld (talk) 11:53, 30 October 2011 (UTC)

It seems that your suggestion would be far more "crushing" than any current student loan system. There is a real public misconception about how these schemes work. The cost and the value of the loans are often very different. For example in the UK, students are allowed to make GB£ 15,000 ( ≈ US$ 24,000) per annum without repaying anything. Once they earn more than that, they pay 9% of the surplus. (Student Loans Company. "How and When you Repay"..) So if a graduate earned £ 40,000 ( ≈ $ 65,000) p/a then they would have to repay 9% of £ 25,000, i.e. £ 2,300 ( ≈ $ 3,600) p/a. Using the method above, a graduate would have to repay £ 13,300 ( ≈ $ 22,000) p/a. That's six times what a graduate would pay back under the current UK system, and God help the graduate that ends up running Microsoft and earns millions per year (and so has to pay back millions each year). Moreover, any student loan is wiped clear when a graduate reaches retirement age. The fact of the matter is that most graduates will never pay back all that they were given and so, in real terms, the value of any loan will far out-weight its cost. If a graduate does earn enough to pay back the full amount then their salary will be such that their contributions won't effect their standard of living (unlike a 33.333% until death contribution; which would be "crushing"!) — Fly by Night (talk) 17:18, 30 October 2011 (UTC)

Trisecting an angle

Our tutor devised a simple way of trisecting an angle using compasses and straightedge: 1.Take the compass and keeping its leg at the vertex O of the given angle, cut off an arc 2.Let the points of intersection be A and B. 3.As OA=OB; therefore ∆OAB is isosceles. 4.Line segment AB can be divided into 3 equal parts by drawing lines from A and B parallel to each other,cutting 3 equal arcs on each line and joining the lines. <ref>Reference, Math Open. "Dividing a line segment into N equal parts with compass and straightedge or ruler - Math Open Reference". Dividing an angle into n equal parts.</ref> 5.Let the two points which divide the triangle into 3 equal parts be P and Q. 6.Join O to P and Q. 7.Lines OP and OQ trisect the angle O.[It can be proved by congruence that in an isosceles triangle, the lines joining the points of trisection of the base trisect the vertex angle. Drop the altitude from the vertex angle.]--117.227.41.150 (talk) 12:57, 30 October 2011 (UTC)Baibhab Pattnaik

Everything is fine until step 7. These lines do not trisect the angle, angle AOP is unequal to angle POQ. -- Meni Rosenfeld (talk) 13:13, 30 October 2011 (UTC)

Try extending the base of an equilateral triangle both ways the same amount so you get a line cut in three. Then the angles at the top of the equilateral triangle formed by the construction are 30, 60 and 30 degrees. Very much not trisected. The angles are 30 because the outer triangles are also isosceles and the two base angles equal the exterior angle at the vertex which is one of the 60 degree angles of the equilateral triangle. Dmcq (talk) 13:58, 30 October 2011 (UTC)

Càdlàg functions

Are there any càdlàg functions which aren't also Borel functions? --84.62.204.7 (talk) 18:03, 30 October 2011 (UTC)

Have you looked at our càdlàg and Borel functions articles? What are your thoughts? For example, what are the differences and the similarities between the definitions and the examples? — Fly by Night (talk) 20:52, 30 October 2011 (UTC)

Trisection

How can i trisect a line without measuring with a rule?--92.29.199.179 (talk) 19:47, 30 October 2011 (UTC)

Here is depicted one way you can do it. -- Meni Rosenfeld (talk) 20:07, 30 October 2011 (UTC)

Graph of sqrt(x)=sqrt(y)

Does the graph of ${\sqrt {x}}={\sqrt {y}}$ include only the first quadrant, or is it the full line? Wolfram Alpha/Mathematica shows the former, but doesn't that mean that Mathematica thinks $i\neq i$? --wj32 ^t/c 23:43, 30 October 2011 (UTC)

You need to be clear what exactly you mean by the square root. Typically the function $f(x)={\sqrt {x}}$ is only defined on the non-negative real numbers, and returns the principle square root (the positive one). If you want to define a square root function elsewhere you need a way to decide which of the two square roots you want to take. For example is f(-1) equal to i or -i? What about for other values? Rckrone (talk) 01:13, 31 October 2011 (UTC)

I'm taking the square root to mean what it conventionally means - the definition here - and I haven't suggested otherwise. Anyway, my question remains exactly the same regardless of whether you take the square root of -1 to be i or -i. --wj32 ^t/c 01:51, 31 October 2011 (UTC)

The solution set is going to be all points with x = y with x ranging over whatever set you choose to be the domain of your square root function. If you want the conventional definition absent of any context, it's not the one you linked to. Conventionally the function is only taken to have the non-negative reals as its domain, which is why Wolfram Alpha says what it says. You're certainly welcome to define it on a bigger set if you want to. I'm not going to tell you how to live your life. Rckrone (talk) 03:16, 31 October 2011 (UTC)

The article states "By convention, the principal square root of –1 is i, or more generally, if x is any positive number, then the principal square root of –x is ...". Am I misinterpreting the article? --wj32 ^t/c 07:55, 31 October 2011 (UTC)

Where the square root function is conventionally defined is only on the set of non-negative real numbers. Square_root#Properties explains this. If you decide to extend your square root function to all complex numbers then there is a convention for how the values are chosen which is what you described, but this is not the conventional square root function. It's a convention for how to extend it. If you don't specify the domain of your square root function and provide no context that might indicate what you mean, then people (or computer programs) will probably assume you mean the conventional square root function which is only defined on the non-negative reals. Rckrone (talk) 04:44, 1 November 2011 (UTC)

If you're drawing a graph of x,y then you're talking about reals, not complex numbers. if you want to plot complex numbers w and z against each other then you'd need a 4 dimensional plot - which we sort of do with colours in many of the articles here. Dmcq (talk) 06:51, 31 October 2011 (UTC)

Yes, that's right. I'm only interested in the graph where x and y are real numbers. But I don't see the relation to my question. Shouldn't (-1,-1) be a point on the graph since ${\sqrt {-1}}={\sqrt {-1}}$? Or maybe I'm being confused about the type of graph... I just can't see where exactly my error is. --wj32 ^t/c 08:15, 31 October 2011 (UTC)

This is just a convention that Wolfram Alpha is using when it draws the graph. The process for drawing the graph must be something like: at various values of x and y, compute both sides of the equation and check if they are equal. Since we're graphing real values, the computation on both sides of the equation performs the real square root, which doesn't exist for negative x or y. Therefore those values are ignored, and so you don't get any line over there. Of course you are right that ${\sqrt {-1}}={\sqrt {-1}}$ when this is defined. But Wolfram is restricting itself to real numbers in the computation, so ${\sqrt {-1}}$ is not defined and so that equality isn't valid. Staecker (talk) 13:19, 31 October 2011 (UTC)

Coloured balls in a hat probability question

Hello all; I had this probability question come up in an interview, and though I apparently answered it sufficiently i wasn't satisfied with the solution, hoping I could get your thoughts:

The interviewer walked in with a hat, and pulled out 4 balls, without looking at which ones he pulled out. All 4 were yellow. He then said there was a 5th (final) ball in the hat, and asked me whether I would put on a 1:1 bet with him that the final ball was yellow. (He knew beforehand what colour(s?) all 5 balls were when putting them into the hat.)

So, my only useful information really was that he pulled out 4 balls -without looking-, that there were exactly 5 balls in the hat to start with, and all 4 pulled out were yellow, and also the assumption that he was a rational person and therefore didn't want to lose money. Now my eventual deduction was that since he didn't want to lose money, the game must have a negative expectation for me (positive for him), otherwise it would be skewed in my favour, and so I shouldn't play - he said this was correct. However, I got a bit confused about looking at the problem from a different perspective; in terms of probability, can we say how likely it is that the final ball is yellow given the previous 4 were (some sort of bayesian probability calculation?), without knowing anything such as how the balls were originally selected, etc?

I figured we probably can work out that probability (though didn't attempt to at the time), and that it was probably more likely the final ball was yellow given 4 yellow balls were pulled out without looking; and therefore I should assume that the 5th ball was yellow. However, while the interviewer doesn't control which balls he manages to pull out, he may assume that having pulled out 4 yellow balls I would deduce the last should be yellow, and therefore given that he knows what my deduction would be, and wants me to make the wrong choice, should that have any bearing on my final answer of whether or not to play? At the time I thought it did, but now I'm starting to think given that he had already pulled the balls out of the hat, he couldn't change the colour of the 5th ball even knowing what I would deduce after seeing 4 yellow balls, unless of course all 5 balls were yellow to begin with in which case he would know I would see 4 yellow balls and would also know what I would deduce from that and how I would act. Anyway, I have gotten myself into a state of thorough confusion, and would be very grateful for any thoughts you have on the problem! Many thanks, 86.26.13.2 (talk) 23:57, 30 October 2011 (UTC)

There's no way to calculate a probability like this without any information about how the colors of the balls were chosen. Getting at the psychology part of it, suppose there was a 5th ball and suppose it was not yellow. Then he only picks 4 yellow balls out 1/5 of the time. Is it likely that his setup for this interview question fails in 4/5 of his interviews and he just has to move on? Probably not. However it's possible that if he picks differently, then he asks a totally different question. You could even set up a problem like this where there are 5 objects each with a unique property different from the other 4, and after picking 4 objects you bet that the last object doesn't have that unique property (the bet depending on which object you didn't choose). On the other hand it could be that the last ball was yellow, or that there was no 5th ball and he just gauges your answer either way without going through with the bet. It's impossible to know. Rckrone (talk) 01:03, 31 October 2011 (UTC)

Yeah he's interested in how you answer, not in making money off you. The above is a very good answer. By the way overall I'd go with the bet but then again perhaps I'm a sucker for the bunko booth. Dmcq (talk) 06:42, 31 October 2011 (UTC)

See Bayesian_inference#Distribution of a parameter of the hypergeometric distribution. Bo Jacoby (talk) 07:41, 31 October 2011 (UTC).

To me this is more a problem in game theory than in probability, though the usual assumption of game theory, that both players have access to the same information, does not hold here. The assumption that person offering the bet is acting rationally also seems questionable. That person, Player A, packed the hat and so knows the probability of a ball being yellow. From what you (Player B) know, that probability is at least 4/5. From A's perspective, the probability that the last ball is yellow does not change when the first four balls are picked, since only B knows those outcomes, so basically A is offering a even money bet when the chances of winning are at most 1 in 5 and I don't see how that can be rational. The conclusion I'm drawing is that if A is acting rationally then there is a rule in this game that A must offer a bet. The whole question of whether are are unstated assumptions or rules can make any kind of analysis meaningless, so at this point I'd say there isn't enough information to salve the problem rationally.--RDBury (talk) 13:22, 31 October 2011 (UTC)

The likelihood ratio of the hypothesis "last ball is yellow" vs "last ball is not yellow" is 5. Likelihood ratios (as opposed to probability) are often a useful way of thinking about such situations. HTH, Robinh (talk) 19:17, 31 October 2011 (UTC)

Likelihood ratio is a disambiguation page that list several (competing?) theories (Bayesian vs frequentist). I believe that you are referring to the likelihood-ratio test and its related likelihood function. -- 110.49.227.102 (talk) 13:09, 1 November 2011 (UTC)

I think you're misinterpreting "without looking at which ones he pulled". I take it to mean the interviewer was pulling them at random, but of course he saw what they were after they'd been pulled. So A does know the outcomes.--121.74.125.249 (talk) 20:51, 31 October 2011 (UTC)

Good point though I think the wording is ambiguous. This is exactly the kind of vagueness that needs to be resolved before a meaningful answer can be determined.--RDBury (talk) 01:15, 1 November 2011 (UTC)

Was this for maths related course or a general interview or what? For a maths course I would say a question like this is probably answered best without considering that there is an intelligent man there, i.e. with a mathematical and very easily fooled answer. Being able mathematically would normally be more important than being socially aware. Dmcq (talk) 21:50, 31 October 2011 (UTC)

Was the 5th ball half the size, or fuzzy vs not, or firmer vs softer, or warmer vs cooler; your explanation of his explanation of the criteria doesn't assure you that we was offering you a fair bet.  Just a thought. — Who R you? Talk 08:00, 1 November 2011 (UTC)

The purpose of this problem is to confuse the student. If you toss a coin four times and get heads all the times, then the probability of getting heads the fifth time is 0.50, because the probability of heads or tails does not depend on the history, and so the one to one bet that the fifth toss is heads, is fair. But if you take out four yellow balls off a hat, then the probability that the fifth ball is yellow too is 0.83, because if there were five yellow balls in the hat then the observed result could be obtained in five ways, while if only four balls in the hat were yellow then the observed result could be obtained in only one way. So a five to one bet that the fifth ball is yellow, is fair. Bo Jacoby (talk) 12:40, 1 November 2011 (UTC).

No. If you were presented with two hats, one with five yellow balls and the other with four yellow balls & one non-yellow ball (and you don't know which hat is which), and then chose one hat at random and drew four balls out of it, all yellow, then yes, the probability of that being the all yellow hat is 5/6 (as pointed out above in Robinh's mention of the likelihood ratio). But if instead you were given a hat which was loaded with five balls, each chosen at the time of loading with a 0.5 probability of being yellow (say there are 5 yellow/non-yellow pairs of balls, and one ball is chosen at random from each pair to be placed in the hat), then after drawing four yellow balls out the the hat the probability of the final ball being yellow is 0.5. This is because that even thought the likelihood of the four yellow ball draw happening with the 5 yellow ball hat is five times that of it happening with a 4 yellow / 1 non-yellow ball hat, the likelihood of being presented, before the start of the draw, with the 4 yellow / 1 non-yellow ball hat was five times that of being presented with a 5 yellow ball hat (5/32 vs. 1/32). Not knowing the likelihood of the hat having 4 or 5 yellow balls before the start of the draw is not the same thing as the likelihoods being equal, is it? -- 110.49.227.102 (talk) 14:16, 1 November 2011 (UTC)

Yes, not knowing the likelihood of the hat having 4 or 5 yellow balls before the start of the draw is the same thing as the likelihoods being equal. The likelihoods not being equal means that you know something about the matter, which you don't. Bo Jacoby (talk) 14:57, 1 November 2011 (UTC).

Really? Wow, that blows me away! Initial reading suggests that you are saying that the principle of indifference should always be applied to non-informative priors. That seems rather strange to me, but I've still got a lot of reading to do. Is this a common source of objection to Bayesian inference? (not the time to ask) -- 110.49.227.102 (talk) 16:26, 1 November 2011 (UTC)

I think that's what he's saying, yes. And I think he's wrong. The principle of indifference as stated in the article clearly doesn't apply here as the alternatives are distinguishable by more than just their names. -- Meni Rosenfeld (talk) 19:27, 1 November 2011 (UTC)

As this is an interview situation, I imagine the real but hidden question is "How many ways can you think of in which I could reliably and repeatably draw four yellow balls from a hat and then produce a fifth non-yellow ball from the same hat ?". I can think of four ways of doing this trick:

1. There are only four balls in the hat to start with; the fifth non-yellow ball appears to be come from the hat, but is actually producd by sleight of hand.
2. The interviewer has some way of distinguishing the non-yellow ball from the yellow balls by touch.
3. The interviewer has some way of altering the colour of a ball while it is out of sight in the hat e.g. with removable coloured shells.
4. There are two separate compartments in the hat, one containing yellow balls and the other containing the non-yellow ball.

I am sure there are other possible methods. If the interviewer lets you examine the balls but not the hat after the trick is done, the method could be 4; if he lets you examine the hat but not the balls the method could be 2 or 3; if he lets you examine both balls and hat then the method could be 1. Anyway, the whole thing isn't a probability question at all ; it is an exercise in lateral thinking. Gandalf61 (talk) 15:45, 1 November 2011 (UTC)

I'm with you on that; and I'd have made the bet (and paid the buck) just to figure out after the fact what the game was.   And of course I'd only pay the dollar if a) I could check the balls, b) I could inspect the hat, and c) I lost; kind of like poker, you pay to see the hand, but you get to see the hand once you pay. — Who R you? Talk 16:11, 1 November 2011 (UTC)

Meni - The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n. Clearly the condition that the n possibilities are indistinguishable except for their names is sufficient, but not necessary. In the case of five colored balls in the hat the 6 possibilities, 0 1 2 3 4 5 yellow balls, are equally likely unless or until further knowledge is provided. Bo Jacoby (talk) 07:31, 2 November 2011 (UTC).

Clearly these 6 possibilities are not a priori equally likely. When nothing else is known, a universal prior should be used (though that's harder in practice than in theory). -- Meni Rosenfeld (talk) 09:03, 2 November 2011 (UTC)

Why are the 6 possibilities not a priori equally likely? What then is the correct prior distribution? To me it is quite clear that when 0 balls are picked the likelihood distibution is 1 1 1 1 1 1. When 1 yellow ball is picked it is 0 1 2 3 4 5. When 2 yellow balls are picked it is 0 0 1 3 6 10. When 3 yellow balls are picked it is 0 0 0 1 4 10. When 4 yellow balls are picked it is 0 0 0 1 5, and when 5 yellow balls are picked it is 0 0 0 0 1. Bo Jacoby (talk) 09:42, 2 November 2011 (UTC).

It is difficult in practice to determine the "correct" prior, so uniform may be a suitable approximation here. But in a more general setting you need to consider the complexity of different alternatives. For example, let's say you are given a sequence from $\{H,T\}^{50}$ and know nothing else about it. Would you say the probably to get HHTTHHTTHHHTTTHTHTHTTTHHHHTHTTHTHHTHHTHTTHHTTHTHHH is equal to the probability to get HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH? Of course not, the latter can be obtained if whoever gave you the sequence decided to just write down all H (or the natural process generating the string always results in H), while the former can only be obtained by a complex process that matches it exactly, or a random process that happened to generate it with probability $2^{-50}$. -- Meni Rosenfeld (talk) 10:38, 2 November 2011 (UTC)

If a uniform prior is a suitable approximation here, then let us stick to it and leave the more general settings to more general problems. If it is difficult in practice to determine the "correct" prior, then forget it - practical questions need practical answers. The 2⁵⁰ different strings of 50 H's and T's are a priori equally likely - otherwise we do know something about it, and you assumed knowing nothing about it. (But it is difficult to provide nice examples of which we know nothing). Bo Jacoby (talk) 11:36, 2 November 2011 (UTC).