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November 26[edit]

Lognormal vs Pareto[edit]

When a compound Poisson process is used to model an insurance portfolio, it is typical to use lognormally distributed random variables or Pareto-ly distributed random variables to represent the probability distribution of the amount that the insurance company has to pay when an accident occurs (if you look at the article compound Poisson process, the insurance payouts are represented by the variables $\{\,D_{i}:i\geq 1\,\}$ ). What is the logic behind the use of the lognormal distribution or the Pareto distribution, and are there circumstances wherein it would be preferable to use a particular one of these distributions rather than the other? 220.253.217.130 (talk) 09:44, 26 November 2010 (UTC)[reply]

To say that D has a Pareto distribution is to say that

\Pr(D\geq x)=\left({\frac {x_{0}}{x}}\right)^{\alpha }{\text{ for }}x\geq x_{0}

for some value of α > 0 and x₀ > 0. This entails that

\Pr(\log _{e}D\geq w)=x_{0}^{\alpha }e^{-\alpha w}{\text{ for }}w\geq \log _{e}x_{0}.

That is the same as saying log_eD has an exponential distribution supported on the interval [log_ex₀, ∞). If the only constraint on x₀ is being positive, then log_ex₀ could be any real number. Exponential distributions are the only ones that are memoryless. Thus D has a sort of "multiplicative memorylessness": for positive numbers a, b,

\Pr(D\geq ab|D\geq a)=\Pr(D\geq b).\,

I think something like that would have to be the rationale. Michael Hardy (talk) 16:08, 26 November 2010 (UTC)[reply]

One reason for applying the lognormal distribution to financial markets is that a small change in stock prices (or orange juice prices or whatever....) mutliplies the value of your holding by something. E.g. if I own a share of stock and you own 100 shares and the price doubles from $1 per share to $2 per share, then then the value of my holdings and yours both get multiplied by 2. So the effect of a large number of small changes is the effect of multiplying by a large number of numbers. This has the effect of adding their logarithms, and it is to addition that the central limit theorem applies, so we get normal distributions of the logarithms of the prices.

(But normal distributions don't account for booms and crashes—their tails are not fat enough.) Michael Hardy (talk) 16:14, 26 November 2010 (UTC)[reply]

is there a way to calculate right numbers using completed sequences?[edit]

Is there a formula using completed sequences that yields (only) right numbers? 92.230.69.163 (talk) 12:26, 26 November 2010 (UTC)[reply]

Maybe I am missing something obvious, but I think we will need some clarification of the terms here. By "completed sequence" do you mean complete sequence ? And what is a "right number" ? Gandalf61 (talk) 13:03, 26 November 2010 (UTC)[reply]

My guess is that the question should be read "Is there a reliable way to (uniquely) answer 'complete the sequence' problems?". The answer to that is of course no: such problems almost never provide enough information to identify a unique solution, and one must then resort to guessing what sort of sequences the questioner had in mind. --Tardis (talk) 19:16, 26 November 2010 (UTC)[reply]

See Wikipedia:Reference_desk/Archives/Mathematics/2010_October_30#Algorithm_for_next_term_in_series. Bo Jacoby (talk) 22:07, 26 November 2010 (UTC).[reply]

fill the missing .[edit]

question is ,"fill the missing number.
  47(76)323
  857(133)591
  642(? )274

Answer on a book is 184 .But I do not know how we can get this number . plz help me —Preceding unsigned comment added by 119.154.134.230 (talk) 17:10, 26 November 2010 (UTC)[reply]

If row 1 said 475(76)323 then there would be a good linear system. Are you sure it says 47? PrimeHunter (talk) 17:23, 26 November 2010 (UTC)[reply]

I agree with PrimeHunter. Consider the second row: 857-591 is 266. Half 266 is 133. Dolphin (t) 02:08, 1 December 2010 (UTC)[reply]

Incenter of triangle[edit]

Hello all, I have been given equations of three sides of a triangle, L1=0, L2=0 and L3=0. I have to find the coordinates of Incenter without finding the coordinates of vertices. So, I start by finding the equation of angle bisector of two lines, let's say L1 and L2. But there would be two such angle bisectors for any pair of lines. How would I chose the right one which goes through the triangle? The other one will obviously not pass through the triangle. Thanks. - DSachan (talk) 20:07, 26 November 2010 (UTC)[reply]

This might not be a helpful reply, and doesn't answer your question, but a couple of alternative methods are suggested by the equations in our article Distance from a point to a line. Dbfirs 21:08, 26 November 2010 (UTC)[reply]

See Incircle and excircles of a triangle. The four centres of the three excircles and the incircle cannot be distinguished by algebraic means alone. If the signed angle ∠ADB + ∠BDC + ∠CDA = 0 then point D is outside the triangle ABC. (The angles are transcendental functions of the point coordinates). Bo Jacoby (talk) 22:48, 26 November 2010 (UTC).[reply]