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September 23[edit]

Calculating risk for insurance[edit]

How do I work out whether a particular premium that an insurance company offers me is reasonable? e.g. I am keen to insure two new kittens against large vet fees. Can I reach a decision on how much I should pay for insurance based purely on mathematical probability? I guess there must be things that are hard to quantify like how much I love my cat or how likely I feel it is that things will go wrong, but is there a recognised process to quantify these things and analyse an insurance premium for how good it is for my particular circumstances? I have a degree in maths, so can cope with high level stuff, but I did no probability or stats beyond first year degree level. Darkhorse06 (talk) 09:27, 23 September 2008 (UTC)[reply]

I have no idea what a "good" premium would be for this type of insurance (which I've never heard of) but I can give you an idea of how the insurer might quantify the risk using probabilities. For life insurance, the method is covered in Actuarial present value. Basically, you need to quantify the amount you expect to claim each year (of the cat's life) and the probability that you will make that claim. If you have many many friends who have lots of cats then you can use their experience to quantify these. You'll also need to decide on an interest rate. If your premium is a lump sum then it would be

\sum _{k=1}^{\infty }v^{k}\,p_{k}\,X_{k}.

where p_k is the probability of a claim in year k and X_k is the amount you expect to claim in year k. If your premium is recurring then you need to equate the above value to

a_{x}*P

as defined in that article (where the probabilities for

a_{x}

are for your cats) and solve for P.

Note that

this is the cost to the insurer to cover your cats' risk only. You can't just add up the premium for each cat because their health is not independant (in fact, the above calc should be done for both cats together) so the premium for both will be more than the sum of the premiums for each.
the insurer will charge you for expenses (they employ actuaries so this can be a lot) and profits and for the opportunity cost of capital.
the insurer will charge you extra if it can't get a large enough portfolio of clients who want this policy (to diversify its risks).

Can't you just shop around and then take the cheapest one. Zain Ebrahim (talk) 10:27, 23 September 2008 (UTC)[reply]

The mean cost of the insurance is guaranteed to exceed the mean cost of the vet. So if you cannot afford to pay the vet, then you cannot afford to own the kittens in the first place. Bo Jacoby (talk) 11:16, 23 September 2008 (UTC).[reply]

What? You seem to be ignoring the whole point of insurance. You pay a bit more than the average cost (a sum you can afford) to avoid the risk of having to pay an unusually large cost (which you can't afford). Algebraist 12:46, 23 September 2008 (UTC)[reply]

Indeed. Insurance protects you against extreme events. Many of the risks that you can sensibly insure against have long-tailed distributions - there is a small but still significant chance of an event that could cost you perhaps 10 or 20 times the mean cost, or even more. Alternatively, you can think of insurance as a sort of enforced savings plan - it makes sure you put money aside in good times to cover the bad times.

Returning to the original question, there are many variables in pet insurance policies. There will be an upper limit on vet's fees that are covered - how high do you want to set this limit ? Do you want a policy that pays boarding fees as well as vet's fees ? Do you want a policy that covers other expenses that may be associated with a pet's illness e.g. holiday cancellation cover ? Do you want a policy that pays death benefits, and if so, how much ? Make sure you are comparing like with like when you look at the cost of policies. Gandalf61 (talk) 13:08, 23 September 2008 (UTC)[reply]

And also exclusions for specific illnesses or treatments. At least with dogs, these exclusions may be breed-specific, and naturally exclude problems that are common to those breeds. A trusted vet can help assess the exclusions. (Vets may be biased either way regarding the insurance; some may see it as a hassle, while others may appreciate that it can result in treatment instead of euthanasia.) -- Coneslayer (talk) 13:13, 23 September 2008 (UTC)[reply]

I have a degree in finance. I haven't read the other answer, but just skimmed. You are analyzing the wrong end of the equation. The mathematical odds and NPV and IRR and all that finance jargon is to be dealt with on their end... not yours! You buying insurance is a service. It is not a bet. If you think of it as a bet, it is always, in theory, a losing bet for your side of the gamble. If not, then you'll be going to jail as you engage in insurance fraud. The goods/services that you buy from an insurance company are:

piece of mind
avoidance of worry, fear
ability to pay a small predictable monthly cost, rather than an occasional, unpredictable large cost
ability to plan ahead 10-20 years with strong certainty

Your math degree only obfuscates your ability to solve this problem. Its like a cell phone warranty. It is a "losing bet" per se, but the benefits are that if you drop your phone in a pool, you can still have an enjoyable evening, you have no worry or regret. Even a $10/month plan on a $150 phone may be worth it. In theory, if every risk in your life was insurable, then you could buy insurance on everything and live a risk free, worry free life. If you're old enough, think of the few things in your life you wish that were unfortunate, and very improbable but still happened and caused you duress. If you put a price on the protection from these types of duress, then it doesn't really matter about the odds, and the npv, and the irr, does it? That's for the business-side of the contract, not for the consumer side. Think of insurance as a way to reduce utility volatility in your life. You never have bad days but you never have great days either. Every day is worry-free, and simplified, allowing you to focus on the stuff in your life which isn't outside of your control (like if someone runs a redlight and totals your car and puts you out of a job for 6 months).

If you really want to know the odds stuff, then I'd recommend reading on the differences in accounting rules between insurance companies and normal companies. You can pretty much understand the logic to everything, if you understand the lingo that insurance companies use--such as hazard, risk, EV, npv, irr, expected payout per policy, etc... Its probably the most reputable field in business, since it creates a service, instead of leeching off of existing markets by taking a 1% cut off customers. If you chose to look at it in terms of an investment then its a losing argument, since every dollar you send off, is guaranteed negative expected value (unless one is a scammer) which increases everyone's rates and increases the overhead costs of these companies. Its a service, not an investment! The math equations are not relevant to your utility derived by purchasing their "risk spreading service"Sentriclecub (talk) 07:12, 24 September 2008 (UTC)[reply]

Rows and columns[edit]

I have a grid of numbered boxes, of any given number of rows and columns:

00, 01, 02, 03, 04, 05,
06, 07, 08, 09, 10, 11,
12, 13, 14, 15, 16, 17,
...

I know the number of columns and the number of rows.

Given a box number, say, 13, how can I determine the number of the column and row is it on?

For rows, ROUNDDOWN((box_number/num_of_columns),0)+1 returns the right number (and the +1 is not really needed if you don't mind the first row being 0 instead of 1) but I'm stuck on a similar formula for columns.

195.195.236.129 (talk) 10:30, 23 September 2008 (UTC)[reply]

The column number is the remainder when (box_number+1) is divided by num_of_columns (taking remainders in the set {1, 2, ... , num_of_columns}). Algebraist 10:35, 23 September 2008 (UTC)[reply]

In other words, you need a modulo function. If this is Excel, you have MOD, so the column number is MOD(box_number,num_of_columns) (plus one or not). -- Jao (talk) 10:36, 23 September 2008 (UTC)[reply]

Thanks guys, that must have seemed so trivial to you! The Excel MOD function is exactly what I needed. 195.195.236.129 (talk) 10:51, 23 September 2008 (UTC)[reply]

(edit conflict). Well, you don't really need the modulo function. You calculated the row number (0,1,2) by dividing the cell number (00,01,02,03,04,05,06,07,08,09,10,11,12,13,14,15,16,17), by the number of columns (6), and round down to a whole number: (0,0,0,0,0,1,1,1,1,1,2,2,2,2,2). Then again multiply this row number by the number of columns to get (0,0,0,0,0,6,6,6,6,6,12,12,12,12,12), the cell number of the first cell in the row. Subtract it from the cell number to get the column number (0,1,2,3,4,5,0,1,2,3,4,5,0,1,2,3,4,5). Bo Jacoby (talk) 11:08, 23 September 2008 (UTC).[reply]

Of course, this is simply a homemade modulo function. Maelin (Talk | Contribs) 14:17, 23 September 2008 (UTC)[reply]

Yes, but it solves the problem with existing tools. Bo Jacoby (talk) 13:28, 24 September 2008 (UTC).[reply]

Seeking lightweight modal logic book[edit]

A few years ago I read a rather thin, paperback introduction to modal logic, which I liked a lot. I want to read it again, but I can't remember the title or the authors. Here's what I do remember:

Published recently—after 1988, say
"Modal Logic" is probably in the title
Informal tone and introductory level
Probably less than 300 pages
Exposition made heavy use of the tableau method
It is not any of:
- Modal Logic by Chagrov and Zakharyaschev
- Modal Logic by Blackburn, Rijke, Venema
- A New Introduction to Modal Logic by Hughes and Cresswell
- Modal Logic by J. Jay Zeman
- First Steps in Modal Logic by Sally Popkorn

Thanks for any help you can offer. -- Dominus (talk) 16:44, 23 September 2008 (UTC)[reply]

inverse function[edit]

Can somebody please show how the inverse of f(x) = x^2 + 2x, x > -1 is sqrt(x+1) - 1, x > -1.

Just show the steps that takes you there. Thank you. —Preceding unsigned comment added by 85.225.94.182 (talk) 17:21, 23 September 2008 (UTC)[reply]

As it says at the top of the page, do your own homework. We won't do it for you, but we might be able to help if there's any specific problems you're having. You'll want the quadratic formula. Algebraist 17:26, 23 September 2008 (UTC)[reply]

I'll give you a hint - try rewriting it like this: Set f(x)=y, and then write x^2+2x-y=0. That should look a little more familiar and hopefully you'll know what to do with it to get it in form x=something. --Tango (talk) 17:30, 23 September 2008 (UTC)[reply]

Tango: Thanks a lot for the hint, but even with that I can't solve it for the life of me.

Algebraist: I'm actually not doing homework (I'm just trying to refresh my high school math), but your point is both fair and well taken. So I'll rephrase my question: Does anybody know a good website where people like me could get help with problems like this?

Thanks again. —Preceding unsigned comment added by 85.225.94.182 (talk) 18:46, 23 September 2008 (UTC)[reply]

If you take Tango's equation and apply completing the square, you should get somewhere ...81.154.106.183 (talk) 19:54, 23 September 2008 (UTC)[reply]

Well look at that, how trivial it was. Man I wish I hadn't gone to law school; maybe then I'd have actually learned something since high school. Anyways, thanks a lot guys. —Preceding unsigned comment added by 85.225.94.182 (talk) 20:37, 23 September 2008 (UTC)[reply]

Logarithms[edit]

(This is just random curiosity, not a homework question.) Ok, I can't really figure out where to go with this equation: $2^{x}=11.50x$
This is what I have so far:
${\begin{aligned}2^{x}&=11.50x\\\log _{2}(2^{x})&=\log _{2}(11.50x)\\x&=\log _{2}(11.50x)\\\end{aligned}}$

How now can I get the x out of the logarithm to solve the problem? TIA, Ζρς ι'β' ^¡hábleme! 21:49, 23 September 2008 (UTC)[reply]

{\begin{aligned}e^{xln2}&=11.50x\\1&=11.50xe^{-xln2}\\-{\frac {ln2}{11.50}}&=-x(ln2)e^{-xln2}\\W(-{\frac {ln2}{11.50}})&=-xln2\\-{\frac {W(-{\frac {ln2}{11.50}})}{ln2}}&=x\end{aligned}}

where

W

is the Lambert W function. But that's not likely to be considered a simplification, so barring that, the short answer is "you can't". -- Jao (talk) 22:03, 23 September 2008 (UTC)[reply]

Well, my teacher said something about changing the bases, but I was wondering if someone could explain that better or give a better way to do it. Ζρς ι'β' ^¡hábleme! 22:12, 23 September 2008 (UTC)[reply]

No, you can't solve it simply by logarithms. The solution with the Lambert W function is the only algebraic one, and

x

cannot be expressed in terms of elementary functions (

W

is usually not considered one). But of course, it can still be solved numerically to arbitrary exactness. -- Jao (talk) 22:19, 23 September 2008 (UTC)[reply]

You should be able to solve it numerically rather than algebraically. Maybe by Newton's method or maybe by finding a function with attractive fixed points where the solutions are. Just drawing a graph or two shows you that there are two solutions. Michael Hardy (talk) 01:43, 24 September 2008 (UTC)[reply]

Indeed, the two real solutions are given by the 0'th and −1'th branches of the Lambert W function. There are infinitely many complex solutions, given by the remaining branches. Fredrik Johansson 08:33, 24 September 2008 (UTC)[reply]

Use the J programming language. Type _1{>1{p.(2&^-11.5&*)t. i.5 and receive the answer 0.0927292. It is the smallest, _1{>, root, 1{p., of the function (2&^-11.5&*) 's Taylor expansion, t., to degree 4, i.5 . Check that the function value (2&^-11.5&*)0.0927292 is small: _2.03071e_7, meaning that 2^0.0927292−11.5× 0.0927292 ≈ −2.03071×10⁻⁷ ≈ 0. My advice is that you spend time studying the J programming language rather than the Lambert W function. Bo Jacoby (talk) 14:15, 24 September 2008 (UTC).[reply]

Limit Question[edit]

$\lim _{x\rightarrow u^{-}}\left(\lim _{u\rightarrow 0^{+}}{\frac {1}{x}}\right)$

$\lim _{u\rightarrow 0^{+}}\left(\lim _{x\rightarrow u^{-}}{\frac {1}{x}}\right)$

I've stumped myself thinking of these--I can't think begin to think of how to approach them. I think they are both positive, just because the negative becomes a neighborhood to a neighborhood, and would become much smaller. —Preceding unsigned comment added by Dooglius (talk • contribs) 22:51, 23 September 2008 (UTC)[reply]

The second is positive infinity. In the first, u appears as a free variable, and the value of the limit depends on the value of u. Algebraist 22:54, 23 September 2008 (UTC)[reply]

In the second, the inner limit evaluates to 1/u, which makes the outer limit evaluate to positive infinity. The first cannot be evaluated because the inner limit treats u as a free variable, but the outer one treats it as some fixed constant. Confusing Manifestation(Say hi!) 23:02, 23 September 2008 (UTC)[reply]

...and the first one is nonsense, since it depends on the existence of some quantity called u independent of x, and upon which the answer depends, but then in the inside limit, u is treated as a bound variable! So the question doesn't make sense. Michael Hardy (talk) 01:47, 24 September 2008 (UTC)[reply]

It's not necessarily nonsense to have a variable both bound and free in the same formula (at least the way I was taught logic). It is very very very bad practice though. Algebraist 13:07, 24 September 2008 (UTC)[reply]

Indeed, the formula makes perfect sense. The inner limit evaluates to 1/x (it's a limit of a constant function), and thus the outer limit evaluates to 1/u if u is nonzero, and to

-\infty

if u = 0. — Emil J. 13:58, 24 September 2008 (UTC)[reply]

Using only the numbers 1,2,4,8 and do any operation (+*- () ) to them except for powers to make 27 ? YOU MAY ONLY USE EACH NUMBER ONCE![edit]

. —Preceding unsigned comment added by 81.159.216.26 (talk) 23:04, 23 September 2008 (UTC)[reply]

On this reference desk we won't do your homework or solve logic puzzles for you, but we will give you hints. Like this: Do the numbers have to be in order? Or this: what do you get if you write two numbers together without any operation between them? Confusing Manifestation(Say hi!) 23:12, 23 September 2008 (UTC)[reply]

Using only your IP address we know more about you you than you would like us to know. Be honest. -23:33, 23 September 2008 (UTC)

What are you talking about? --Tango (talk) 23:54, 23 September 2008 (UTC)[reply]

Trick question, 27 is the least positive integer that can't be formed by a combination of the operations +, -, * on integers 1, 2, 4, and 8. Dragons flight (talk) 00:00, 24 September 2008 (UTC)[reply]

Like I said, you can do it if you don't restrict yourself to n # n # n # n where n is one of the numbers and # is an operator. In fact, you only need one operator to do it. And I, too, am confused by hydnjo's comment. Confusing Manifestation(Say hi!) 04:56, 24 September 2008 (UTC)[reply]

4! + (8/2) - 1 = 27.^[1] Though this obviously depends on whether doing a factorial counts as a mathematical operation, which is a bit unclear. --MZMcBride (talk) 05:37, 24 September 2008 (UTC)[reply]

2 * (8 + 4) + 1 = 27 (in base 9). This type of puzzle is always so ambiguous. I suppose it might motivate a bright pupil to think about the meaning and usage of mathematical operations, difference between "number" and "digit", how to set up an exhaustive search etc. But an average pupil will be left with the impression that mathematics is a set of arbitrary and confusing rules. Gandalf61 (talk) 10:00, 24 September 2008 (UTC)[reply]

What is tenty eight minus one? Oops I did it again! 122.107.176.54 (talk) 11:43, 25 September 2008 (UTC)[reply]

Come on folks - this is an easy one. 48-21 = 27 :) Grutness...wha? 13:24, 25 September 2008 (UTC)[reply]
Wouldn't the question need to specify "digits" (or something) instead of "numbers" for that to be the answer? Zain Ebrahim (talk) 13:27, 25 September 2008 (UTC)[reply]
Possibly - hence the smiley. But I'm pretty sure that this solution is what CM was driving at with his comments above: " what do you get if you write two numbers together..." and "...you only need one operator to do it." Grutness...wha? 13:35, 25 September 2008 (UTC)[reply]

Yep, that's the one I found fairly quickly. I'm fairly sure it's what the question was asking for, but the only way to be sure would be to ask the OP's teacher. Confusing Manifestation(Say hi!) 23:24, 25 September 2008 (UTC)[reply]