bayes

cancer testing scenario:

• • 1% of women have breast cancer

  • 80% of mammograms detect breast cancer when it is there

  • 9.6% of mammograms detect breast cancer when it’s not there

Denominator is Pr(X) - probability of positive test

And here’s the decoder key to read it:

• • Pr(A|X) = Chance of having cancer (A) given a positive test (X).

  • Pr(X|A) = Chance of a positive test (X) given that you had cancer (A). This is the chance of a true positive, 80% in our case.

  • Pr(A) = Chance of having cancer (1%).

  • Pr(not A) = Chance of not having cancer (99%).

  • Pr(X|not A) = Chance of a positive test (X) given that you didn’t have cancer (~A). This is a false positive, 9.6% in our case.

Statistics

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html

http://www.mathpages.com/home/iprobabi.htm

http://www.eknigu.com/lib/M_Mathematics/MV_Probability/MVspa_Stochastic%20processes/

http://teorver.mccme.ru/books/

http://en.wikipedia.org/wiki/Confidence_interval#Meaning_and_interpretation

http://www.mrc-bsu.cam.ac.uk/bugs/ - Bayesian inference Using Gibbs Sampling

Generating Data Analysis Programs from Statistical Models :

http://ase.arc.nasa.gov/people/fischer/papers/jfp-01.html

http://www.yudkowsky.net/bayes/bayes.html

http://www.cs.ualberta.ca/~greiner/BN-results.html

http://www.inference.phy.cam.ac.uk/mackay/itila/book.html

http://en.wikipedia.org/wiki/Probability_distribution

http://en.wikipedia.org/wiki/L%C3%A9vy_process Levy process

http://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/book.html

http://www.mathcs.carleton.edu/probweb/probweb.html

http://www.stat.berkeley.edu/~aldous/

http://it.stlawu.edu/~rlock/maa51/texts.html

http://en.wikipedia.org/wiki/Chernoff_bound

http://www-stat.wharton.upenn.edu/~steele/Rants/AdviceGS.html PhD Advice

http://www.bbn-school.org/us/math/ap_stats/applets/applets.html

http://arbuz.uz/t_paradox.html

http://ega-math.narod.ru/Books/Kac67.htm

http://ghmm.org General Hidden Markov Model library

http://www.math.tau.ac.il/~tsirel Boris Tsirelson

http://en.wikipedia.org/wiki/Convolution Convolution

http://en.wikipedia.org/wiki/List_of_convolutions_of_probability_distributions

-----------------------------------------------------

http://www.probability.net/

http://www.socr.ucla.edu/ Statistics Online Computational Resource

http://www.math.csusb.edu/faculty/stanton/calico/index.html

http://www.math.uah.edu/stat/index.xhtml Virtual Laboratories in Probability and Statistics

http://en.wikipedia.org/wiki/Comparison_of_statistical_packages

http://davidmlane.com/hyperstat/index.html

Dice http://en.wikipedia.org/wiki/Dice

A and B play a game. There is 1 die. A wins if A roll a 4. B wins if he rolls a 5. If A go first, what is the probability that A win? the game is over when A rolls a 4 or when B rolls a 5. Until that happens, the game is played.

http://stochastix.wordpress.com/2008/02/28/rolling-dice/#comments

First player A has a 1/6 chance of winning on the first toss. Otherwise (5/6 chance), we are playing the same game but now A is a player #2, so A have (1-p) chance of winning. Algebraically,

p = 1/6 + 5/6 (1-p)

11p/6 = 1

p = 6/11

This approach generalizes to the case of more than two outcomes (e.g. allowing ties). The proof for any given outcome A works like this:

Let p(A) be the probability that you eventually end up in outcome A.

Let p(A1) be the probability that you end up in outcome A on the first “iteration”, defined as the minimum number of rolls after which the game has either terminated or you are in an equivalent situation as you were at the beginning.

Let q be the probability that the game didn’t terminate on the first iteration. Then q = 1 - p(A1) - p(B1) - p(C1) - … for all the possible outcomes.

Following the above outline, you get an equation that looks like this:

p(A) = p(A1) + p(A)q

p(A) = p(A1) / (1-q)

p(A) = p(A1) / (p(A1) + p(B1) + p(C1) + …) or in other words, once you know the relative chance of a given outcome on the first iteration, you know the odds of that outcome overall.

Given a set of N random numbers between 0 and 1, determine

the expected value of the maximum of the N number. Solution

Two Dices http://mathforum.org/library/drmath/sets/select/dm_dice.html

You NEED to roll at least one 6 using 2 dices. A six appearing on any one (or more) of the 2 dice will win the game for you! What are your chances?

if the chances of rolling a 6 with two dice is 2/6 and the chances of rolling a 6 with three dice is 3/6, then the chances of rolling a 6 with six dice would be 6/6 !! 100%?? Of course, this is obviously incorrect.

When two dice are rolled, there are now 36 different and unique ways the dice can come up.

The chances of a six NOT appearing on the first die= 5/6 multiply this by the chances of a six NOT appearing on the second die: 5/6 * 5/6 =25/36

бросают два кубика. на одном из них выпала четвёрка. какая вероятность того, что на другом кубике тройка? Answer: 2/11

Two ordinary dice are rolled. If it is known that one shows a 5, what is the probability that they total 8?

http://mathcentral.uregina.ca/QQ/database/QQ.09.01/owen1.html

I have two different but (seemingly) correct solutions.

The first and most logical one is that if there is 5 showing, then there's only one possible 'show' to make 8 - and that's 3. There are six sides to a dice, and therefore your answer would be one in six.

But consider all the possible combinations that involve one die showing 5.

(1,5);(2,5);(3,5);(4,5);(5,1);(5,2);(5,3);(5,4);(5,5);(5,6);(6,5)for a total of 11 combos. Now there are only two of those that add to 8, being (3,5) and (5,3). Therefore the probability being 2/11.

The second solution is correct.

The flaw in the first solution is that in it we do not distinguish the dice. If you assume that one die is red and the other is green then in the first solution we do not take in account which die shows 5 - red or green.

среднее расстояние между любыми двумя точками на отрезке [0, 1]

интеграл от |x-y| по единичному квадрату:

обозначим через Z растояние между двумя точками отрезок длины a

Х -- координата первой, У - координата второй.

P(Z=z)dz=P(|X-Y|=z)dz=[P(X-Y=z)+P(Y-X=z)]dz=2[P(X-Y=z)]dz

P(X-Y=z)dz=

∫( P(Y=t-z|X=t)P(X=t)dt (from: 0 to: a-z)=

P(Y=t-z|X=t)∫(P(X=t)dt (from: 0 to: a-z)=

P(Y=t-z|X=t)P(X>a-z)=

=P(Y=t-z|X=t)[(a-z)/a]=(1/a)*[(a-z)/a]

P(Z=z)dz=2(1/a)*[(a-z)/a]dz

отсюда несложно и мат ожидание посчитать

Можно без интегралов:

Пусть d это среднее расстояние. Если мы вместо отрезка [0 1] будем рассматривать отрезок [0, 0.5], то в этом случае среднее расстояние будет d/2

С вероятностью 1/2 обе точки отрезка попадают в одну и ту же его половину. В этом случае среднее расстояние между ними d/2

С вероятностью 1/2 точки попадают в разные половины отрезка. Тогда среднее расстояние между ними это среднее расстояние от точки х до 0.5 + среднее расстояние от 0.5 до у. То есть 1/4+1/4=1/2

Итого: d=(d/2)*1/2+ 1/2*1/2

d=1/3

в кубической коробке со стороной 1 м летают два шмеля. каково среднее расстояние между ними?

http://mathworld.wolfram.com/HypercubeLinePicking.html

На отрезке [0,1] случайным образом размещено N точек

Найти вероятность того, что все точки находятся на расстоянии не больше чем L (L<1) друг от друга.

a) равномерная вероятность на отрезке [0,1]

b) общий случай: известна функция плотности распределения f(x) на отрезке [0,1]

"joint distribution of order statistics"

http://en.wikipedia.org/wiki/Order_statistic#Joint_distributions

http://mcs.une.edu.au/~stat354/notes/node42.html

для n=2 ответ: 1-(1-L)^2

Statistics of extreme

the sum of a large number of independent identically distributed (i.i.d) random variables tends to a normal distribution

the maximum of a large number of i.i.d. random variables tends to an extreme value distribution: http://en.wikipedia.org/wiki/Extreme_value_distribution

Statistics of sum of iid

http://en.wikipedia.org/wiki/List_of_convolutions_of_probability_distributions

Suppose I have n samples from a normal(μ, σ²) distribution, say n = 16, and σ is unknown. What is the distribution of the average of the samples?

if σ is known, the sample mean has a normal distribution,

otherwise it has a t distribution.

Boy or Girl

You meet a friend on the street, and she happens to mention that she has given birth to two children. You ask: "Is at least one of your children a boy?" The friend says, "Yes, he is."

What is the probability that she has two boys? If you assume that the prior probability of a child being a boy is 1/2, then the probability that she has two boys, on the information given, is 1/3. The prior probabilities were: 1/4 two boys, 1/2 one boy one girl, 1/4 two girls. The mathematician's "Yes" response has probability ~1 in the first two cases, and probability ~0 in the third. Renormalizing leaves us with a 1/3 probability of two boys, and a 2/3 probability of one boy one girl.

But suppose that instead you had asked, "Is your eldest child a boy?" and the mathematician had answered "Yes." Then the probability of the mathematician having two boys would be 1/2. Since the eldest child is a boy, and the younger child can be anything it pleases.

Likewise if you'd asked "Is your youngest child a boy?" The probability of their being both boys would, again, be 1/2.

Now, if at least one child is a boy, it must be either the oldest child who is a boy, or the youngest child who is a boy. So how can the answer in the first case be different from the answer in the latter two

http://blogs.wsj.com/numbersguy/probability-quiz-results-and-winners-321/?mod=WSJBlog

1. Suppose 1,000 athletes are tested for drugs. One in 10 have used the drugs, and the test has a 1% false-positive rate (and the false-negative rate is negligible). If an athlete from this group tests positive, what is the probability that she has used the drugs, to the nearest percentage point?

Answer: One hundred athletes — one in 10 out of the 1,000 athletes — have used the drugs. So 900 haven’t. Because the false-negative rate is negligible, we can assume that all 100 of those athletes who have used the drugs test positive. Of those who haven’t used drugs, nine of them — 1% of 900 — will also test positive. In all, then, 109 will test positive. If you choose any one of them at random you have a 91.7% chance — 100 in 109 — of choosing a drug user. Rounded to the nearest percentage point, that’s 92%.

2. You know that a certain family has two children, and that at least one is a girl. But you can’t recall whether both are girls. What is the probability that the family has two girls — to the nearest percentage point?

Answer: Assume that there are an equal number of boys and girls, and that the gender of each child in the family is independent of the other’s (more on that below). Then there are four, equally likely possibilities for a two-child family’s history of procreation: Either a girl was born first, and then a boy; or a girl was born first, and then another girl; or boy then boy; or boy then girl. But you know that there is at least one girl in the family, so you can eliminate the (boy, boy) possibility. That leaves three scenarios, and only in one are there two girls. So the probability of two girls is one in three — 33%, to the nearest percentage point.

Comments: There’s an assumption I should have made explicit for this problem and the next problem: that there are an equal number of boys and girls (although in 2005, there were nearly 5% more boys than girls 14 and under, according to the Census Bureau). Kudos to Messrs. Newcombe and Plourde, and to Glenn Tippy and Gregg Skinner, for noting the importance of gender ratio at birth.

3. You know that a certain family has two children, and you remember that at least one is a girl with a very unusual name (that, say, one in a million females share), but you can’t recall whether both children are girls. What is the probability that the family has two girls — to the nearest percentage point?

Answer: Use the same logic as above, only this time there are three possibilities for each child: Boy, girl with the specific unusual name (let’s for the sake of argument make it Florida, the one used in Mr. Mlodinow’s book), and girl with a different name. Then there are five possibilities for a family with two children, one of them named Florida:

(boy, girl-F)

(girl-NF, girl-F)

(girl-F, boy)

(girl-F, girl-NF)

(girl-F, girl-F)

Unlike in question No. 2, these are not all equally likely. The last scenario is particularly unlikely, assuming the two children’s names are independent, because Florida is such an unusual name. So for the sake of this calculation, we can ignore it. The other four scenarios are, approximately, equally likely, because we’ve assumed that there are the same number of boys as girls, and nearly all girls have names other than Florida. In two of those four scenarios, the family has two girls. So the probability of two girls is about two in four — 50%, to the nearest percentage point. (There’s ample discussion of this question, and a more-detailed explanation from me, in the comments.)

Comments: It seems paradoxical that the girl’s name would make a difference, and in fact 75% of readers thought the answers to No. 2 and No. 3 were the same, including 68% of those who got No. 3 right. Mr. Mlodinow suggested I reward points for No. 3 only to those who also correctly answered No. 2. I disagreed, pointing out that his book is, after all, about the role of randomness in our lives. The final chapter makes a convincing case that much of what society defines as success is due to luck.

4. In baseball, suppose the American League champion is better than the National League champion, such that it has a 55% probability of winning each game against the NL champ. Then the NL champ nonetheless will win a best-of-seven-games series four in 10 times. What is the smallest odd number, X, for which a World Series between these two league champs that is best-of-X will ensure that there’s a 95% probability of a just result — the superior AL champ winning?

Answer: The probability of the AL team winning a certain number of games can be represented by a graph of the binomial distribution, explained here. Calculating the probability that the AL team wins at least n games in a series of 2n-1 games is equivalent to graphing the probability of winning a certain number of games, and then adding the area under the part of the graph equivalent to the AL team winning at least n games. (This assumes the series continues even after one team clinches, which makes the calculation easier and doesn’t change the result.)

In Excel or another spreadsheet program, a single formula handles this. In Excel, that’s =BINOMDIST([number of games needed to win minus one], [number of games in the series],0.45,1). This calculates the probability that the NL team wins no more than n-1 games, which is necessary for the AL team to win the series. Applying this formula to every possible number of games, from seven and up, you eventually find that the AL has a 94.9% chance of winning a best-of-267 game series and a slightly greater than 95% chance of winning a best-of-269 game series. So the right answer is 269 games.

Comments: A best-of-269 game series is 107 games longer than the regular season. And yet that’s how many games it would take to feel 95% confident that the better team won, if the better team has a 55% probability of winning each game. Lower that to 53%, and it would take 751 games to properly sort them out — a doubleheader each day of the year wouldn’t suffice. Probability really varies each game, depending on home field and pitching matchups, but this approximation does illustrate how likely it is for the better team to lose. “This is why the baseball postseason is so ridiculous,” reader Joe Lurito commented. “It’s basically a crapshoot at seven games.”

5. Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind each of the other two, a cow. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, to reveal a cow. He then says to you, “Do you want to change your choice to door No. 2?” Is it to your advantage to switch your choice, assuming you prefer cars to cows?

Answer: Yes. When you initially chose door No. 1, you had a one in three chance of picking the car. The other scenario, in which the car is behind one of the other two doors and so only one of those two doors hides a cow, has a probability of two-thirds. Let’s focus on that scenario, and assume — since we don’t have any information to the contrary — that the host always opens a door that reveals a cow. In that case, the door he didn’t open and that you didn’t choose always contains the car. So if you switch, there’s a two-thirds chance that you’ll win the car. If you don’t, you’re stuck with a one-third chance.

Comments: If you’d like to write in to complain about this answer, you might want to first read Marilyn vos Savant’s record of her angry mail when she printed the answer to this problem in Parade Magazine in 1990. Unexpectedly, during this quiz the New York Times ran a column on this very problem, which several readers cited in their answers. Other readers pointed out that a variation of this problem appears in the poker film “21.”

Several readers pointed out that the answer depends on the host’s motives. For instance, it’s possible he only reveals a door when you’ve made the right choice, to try to lead you astray. I gave credit to readers who said the answer depends and explained why, but not to those who gave a blanket no because they made certain assumptions. (I also gave credit to several readers who correctly said you should switch, but incorrectly said the probability the car was behind the door you’d switch to was 50%, rather than 67%.) Kudos to Mr. Birnbaum — as well as Anthony Cox, Robert Simmons and Gavin Moore — for pointing out the issues concerning the host’s motives.

6. Suppose that 60% of eligible voters in Scranton, Pa., prefer Hillary Clinton to Barack Obama. If you randomly select five of these prospective voters and they tell you their true preferences, what is the probability that exactly three of these five support Clinton?

Answer: There are 10 ways to scatter three Clinton voters among the five, when chosen one by one:

CCCOO

CCOCO

CCOOC

COCCO

COCOC

COOCC

OCCCO

OCCOC

OCOCC

OOCCC

Each of these scenarios has a probability of 0.6 × 0.6 × 0.6 × 0.4 × 0.4 = 3.456%

Multiply that by 10 and you get 34.56% (34.6% and 35% were also accepted).

Comments: In an effort to avoid repetitive wording, I made this question unclear, though it didn’t seem to affect the answers. I meant to ask for the probability that three of these five support Sen. Clinton over Sen. Obama; Scranton voters may also support John McCain or another candidate, but for the purposes of this question such preferences were ignored. Kudos to Frank Benson and Ms. Murphy for noting this ambiguity. Messrs. Grossinger and Plourde, as well as David Gomel, noted another complicating factor: This question assumes there are a lot of eligible voters in Scranton. For instance, if there are exactly five (because there are “lots of people in prison in Scranton,” Mr. Plourde hypothesizes), then the probability is 100%.

7. In a large set of financial data — for instance, the revenue of Fortune 500 companies in each of the last 50 years — what proportion of the financial figures should you expect to begin with the digit 1, to the nearest 5 percentage points?

Answer: Such a distribution is — or should be — governed by Benford’s Law, which holds that lower digits predominate in certain kinds of data. In particular, numbers starting with 1 should represent about 30.1% of all such data.

Comments: Benford’s Law is explained here. Its applicability to financial figures of companies is an interesting area. Mark Nigrini, a Benford’s Law scholar, has applied the principle to financial numbers, and suggests it may be used to detect fraudulent accounting data. Most of our winners assumed Benford’s Law would apply. Mr. Plourde took it a step further, analyzing the data itself. He found that the correct answer was 28%.

That 50% of marriages end in divorce is not the same as telling a newly wed couple that they have a 50% chance of divorcing (because some people can divorce several times).

The real difference in practice between Bayesian and frequentist approaches is the Bayesian use of priors. But if you repeat the same experiment many times, the sensitivity to your prior goes away. That's actually the nice thing about Bayesian probability theory, it has a subjective element, but as you acquire more data, it becomes more and more objective (in the sense that the subjective element becomes less and less important).

В большом множестве вероятность наличия признака А у одного из элементов равна 30%. Сколько элементов надо выбрать из множества, чтобы вероятность что среди этих элементов будет хотя бы один с признаком А была не менее 95%?

(1-0.3)^n=1-0,95

n=8,4

9 штук!

Bayes’ Rule http://yudkowsky.net/bayes/bayes.html

Let A1,A2, . . . , An be disjoint events that form a partition of the sample space,

and assume that P(Ai) > 0, Then, for any event B such that P(B) > 0, we have

P(Ai) * P(B|Ai)

P(Ai|B) = ------------------

P(B)

Here P(B) = P(A1)*P(B |A1) + · · · + P(An)*P(B |An)

Bayes’ rule is often used for inference. There are a number of “causes” that may result in a certain “effect.” We observe the effect, and we wish to infer the cause. The events A1, . . . , An are associated with the causes and the event B

represents the effect. The probability P(B |Ai) that the effect will be observed

when the cause Ai is present amounts to a probabilistic model of the cause-effect

relation. Given that the effect B has been observed, we wish to evaluate the probability P(Ai |B) that the cause Ai is present.

Confidence

After a particular drug is on the market, it will cause a particularly serious adverse effect to happen to one of every 3,000 patients in an epidemiological, i. e., post-hoc analysis. In retrospect, how many patients must be tested in the randomized, double-blinded, placebo-controlled, clinical trial to achieve 95% confidence that the side effect will show up?

Given a sample of data, a confidence interval is a way of representing the resulting uncertainty about the value of a parameter (such as the number of adverse events that would be expected if a given number of patients take a drug). But the question posed does not concern a sample. It does, however, refer to “95% confidence”, which is a bit misleading.

If the probability of an adverse event is 1 in 3000, then the probability of no adverse event is 2999 in 3000. The probability that, say 8985 patients have no adverse events is (2999/3000)^8985, or about 5%. In other words, you need at least 8985 people to take the drug to have 95% probability of seeing at least 1 adverse event.

But this is a randomized placebo-controlled clinical trial. Assuming that equal numbers of patients are randomized to receive the drug and the placebo, the trial would need to enroll 17970 patients (that is, twice 8985).

Time series analysis

http://chaos.ssu.runnet.ru/kafedra/edu_work/textbook/khovanovs-01/pos.html

www.gistatgroup.com/gus

"Predictability of Complex Dynamical Systems", J.B.Kadtke, Yu.A.Kravtsov, Eds., Springer, 1997.

Общая характеристика методов восстановления дана также в представительном обзоре Павлова и Янсон, посвященном задачам восстановления динамики из электрокардиограмм [18], в нашем обзоре [92] и в недавно вышедших книгах:

Анищенко B.C., Астахов В.В., Вадивасова Т.Е., Нелинейная динамика хаотических и стохастических систем.,

Anishchenko V.S., Astakhov V.V., Neiman A.B., Vadi-vasova Т.Е., L. Schimansky-Geier, Nonlinear Dynamics of Chaotic and Stochastic Systems. Tutorial and Modern Development. Springer, Berlin, Heidelberg, 2002.