College Math Teaching

November 1, 2016

A test for the independence of random variables

Filed under: algebra, probability, statistics — Tags: , — collegemathteaching @ 10:36 pm

We are using Mathematical Statistics with Applications (7’th Ed.) by Wackerly, Mendenhall and Scheaffer for our calculus based probability and statistics course.

They present the following Theorem (5.5 in this edition)

Let Y_1 and Y_2 have a joint density f(y_1, y_2) that is positive if and only if a \leq y_1 \leq b and c \leq y_2 \leq d for constants a, b, c, d and f(y_1, y_2)=0 otherwise. Then $Y_1, Y_2 $ are independent random variables if and only if f(y_1, y_2) = g(y_1)h(y_2) where g(y_1), h(y_2) are non-negative functions of y_1, y_2 alone (respectively).

Ok, that is fine as it goes, but then they apply the above theorem to the joint density function: f(y_1, y_2) = 2y_1 for (y_1,y_2) \in [0,1] \times [0,1] and 0 otherwise. Do you see the problem? Technically speaking, the theorem doesn’t apply as f(y_1, y_2) is NOT positive if and only if (y_1, y_2) is in some closed rectangle.

It isn’t that hard to fix, I don’t think.

Now there is the density function f(y_1, y_2) = y_1 + y_2 on [0,1] \times [0,1] and zero elsewhere. Here, Y_1, Y_2 are not independent.

But how does one KNOW that y_1 + y_2 \neq g(y_1)h(y_2) ?

I played around a bit and came up with the following:

Statement: \sum^{n}_{i=1} a_i(x_i)^{r_i} \neq f_1(x_1)f_2(x_2).....f_n(x_n) (note: assume r_i \in \{1,2,3,....\}, a_i \neq 0

Proof of the statement: substitute x_2 =x_3 = x_4....=x_n = 0 into both sides to obtain a_1 x_1^{r_1} = f_1(x_1)(f_2(0)f_3(0)...f_n(0)) Now none of the f_k(0) = 0 else function equality would be impossible. The same argument shows that a_2 x_2^{r_2} = f_2(x_2)f_1(0)f_3(0)f_4(0)...f_n(0) with none of the f_k(0) = 0.

Now substitute x_1=x_2 =x_3 = x_4....=x_n = 0 into both sides and get 0 = f_1(0)f_2(0)f_3(0)f_4(0)...f_n(0) but no factor on the right hand side can be zero.

This is hardly profound but I admit that I’ve been negligent in pointing this out to classes.

January 6, 2016

On all but a set of measure zero

Filed under: analysis, physics, popular mathematics, probability — Tags: — collegemathteaching @ 7:36 pm

This blog isn’t about cosmology or about arguments over religion. But it is unusual to hear “on all but a set of measure zero” in the middle of a pop-science talk: (2:40-2:50)

September 2, 2014

Using convolutions and Fourier Transforms to prove the Central Limit Theorem

Filed under: probability — Tags: , , — collegemathteaching @ 5:40 pm

I’ve used the presentation in the our Probability and Statistics text; it is appropriate given that many of our students haven’t seen the Fourier Transform. But this presentation is excellent.

Upshot: use the convolution to derive the density function for S_n = X_1 + X_2 + ....X_n (independent, identically distributed random variables of finite variance), assume mean is zero, variance is 1 and divide S_n by \sqrt{n} to obtain the variance of the sum to be 1. Then use the Fourier transform on the whole thing (the normalized version) to turn convolution into products, use the definition of Fourier transform and use the Taylor series for the e^{i 2 \pi x \frac{s}{\sqrt{n}}} terms, discard the high order terms, take the limit as n goes to infinity and obtain a Gaussian, which, of course, inverse Fourier transforms to another Gaussian.

May 22, 2013

In the news….and THINK before you reply to an article. :-)

Ok, a mathematician who is known to be brilliant self-publishes (on the internet) a dense, 512 page proof of a famous conjecture. So what happens?

The Internet exploded. Within days, even the mainstream media had picked up on the story. “World’s Most Complex Mathematical Theory Cracked,” announced the Telegraph. “Possible Breakthrough in ABC Conjecture,” reported the New York Times, more demurely.

On MathOverflow, an online math forum, mathematicians around the world began to debate and discuss Mochizuki’s claim. The question which quickly bubbled to the top of the forum, encouraged by the community’s “upvotes,” was simple: “Can someone briefly explain the philosophy behind his work and comment on why it might be expected to shed light on questions like the ABC conjecture?” asked Andy Putman, assistant professor at Rice University. Or, in plainer words: I don’t get it. Does anyone?

The problem, as many mathematicians were discovering when they flocked to Mochizuki’s website, was that the proof was impossible to read. The first paper, entitled “Inter-universal Teichmuller Theory I: Construction of Hodge Theaters,” starts out by stating that the goal is “to establish an arithmetic version of Teichmuller theory for number fields equipped with an elliptic curve…by applying the theory of semi-graphs of anabelioids, Frobenioids, the etale theta function, and log-shells.”

This is not just gibberish to the average layman. It was gibberish to the math community as well.

[…]

Here is the deal: reading a mid level mathematics research paper is hard work. Refereeing it is even harder work (really checking the proofs) and it is hard work that is not really going to result in anything positive for the person doing the work.

Of course, if you referee for a journal, you do your best because you want YOUR papers to get good refereeing. You want them fairly evaluated and if there is a mistake in your work, it is much better for the referee to catch it than to look like an idiot in front of your community.

But this work was not submitted to a journal. Interesting, no?

Of course, were I to do this, it would be ok to dismiss me as a crank since I haven’t given the mathematical community any reason to grant me the benefit of the doubt.

And speaking of idiots; I made a rather foolish remark in the comments section of this article by Edward Frenkel in Scientific American. The article itself is fine: it is about the Abel prize and the work by Pierre Deligne which won this prize. The work deals with what one might call the geometry of number theory. The idea: if one wants to look for solutions to an equation, say, x^2 + y^2 = 1 one gets different associated geometric objects which depend on “what kind of numbers” we allow for x, y . For example, if x, y are integers, we get a 4 point set. If x, y are real numbers, we get a circle in the plane. Then Frenkel remarked:

such as x2 + y2 = 1, we can look for its solutions in different domains: in the familiar numerical systems, such as real or complex numbers, or in less familiar ones, like natural numbers modulo N. For example, solutions of the above equation in real numbers form a circle, but solutions in complex numbers form a sphere.

The comment that I bolded didn’t make sense to me; I did a quick look up and reviewed that |z_1|^2 + |z_2|^2 = 1 actually forms a 3-sphere which lives in R^4 . Note: I added in the “absolute value” signs which were not there in the article.

This is easy to see: if z_1 = x_1 + y_1 i, z_2 = x_2 + y_2i then |z_1|^2 + |z_2|^2 = 1 implies that x_1^2 + y_1^2 + x_2^2 + y_2^2 = 1 . But that isn’t what was in the article.

Frenkel made a patient, kind response …and as soon as I read “equate real and imaginary parts” I winced with self-embarrassment.

Of course, he admits that the complex version of this equation really yields a PUNCTURED sphere; basically a copy of R^2 in R^4 .

Just for fun, let’s look at this beast.

Real part of the equation: x_1^2 + x_2^2 - (y_1^2 + y_2^2) = 1
Imaginary part: x_1y_1 + x_2y_2 = 0 (for you experts: this is a real algebraic variety in 4-space).

Now let’s look at the intersection of this surface in 4 space with some coordinate planes:
Clearly this surface misses the x_1=x_2 = 0 plane (look at the real part of the equation).
Intersection with the y_1 = y_2 = 0 plane yields x_1^2+ x_2^2 = 1 which is just the unit circle.
Intersection with the y_1 = x_2 = 0 plane yields the hyperbola x_1^2 - y_2^2 = 1
Intersection with the y_2 = x_1 = 0 plane yields the hyperbola x_2^2 - y_1^2 = 1
Intersection with the x_1 = y_1 = 0 plane yields two isolated points: x_2 = \pm 1
Intersection with the x_2 = y_2 = 0 plane yields two isolated points: x_1 = \pm 1
(so we know that this object is non-compact; this is one reason the “sphere” remark puzzled me)

Science and the media
This Guardian article points out that it is hard to do good science reporting that goes beyond information entertainment. Of course, one of the reasons is that many “groundbreaking” science findings turn out to be false, even if the scientists in question did their work carefully. If this sounds strange, consider the following “thought experiment”: suppose that there are, say, 1000 factors that one can study and only 1 of them is relevant to the issue at hand (say, one place on the genome might indicate a genuine risk factor for a given disease, and it makes sense to study 1000 different places). So you take one at random, run a statistical test at p = .05 and find statistical significance at p = .05 . So, if we get a “positive” result from an experiment, what is the chance that it is a true positive? (assume 95 percent accuracy)

So let P represent a positive outcome of a test, N a negative outcome, T means that this is a genuine factor, and F that it isn’t.
Note: P(T) = .001, P(F) = .999, P(P|T) = .95, P(N|T) = .05, P(P|F) = .05, P(N|F) = .95 . It follows P(P) = P(T)P(P \cap T)P(T) + P(F)P(P \cap F) = (.001)(.95) + (.999)(.05) = .0509

So we seek: the probability that a result is true given that a positive test occurred: we seek P(T|P) =\frac{P(P|T)P(T)}{P(P)} = \frac{(.95)(.001)}{.0509} = .018664. That is, given a test is 95 percent accurate, if one is testing for something very rare, there is only about a 2 percent chance that a positive test is from a true factor, even if the test is done correctly!

March 18, 2013

Odds and the transitive property

Filed under: media, movies, popular mathematics, probability — Tags: — collegemathteaching @ 9:51 pm

I got this from Mano Singham’s blog: he is a physics professor who mostly writes about social issues. But on occasion he writes about physics and mathematics, as he does here. In this post, he talks about the transitive property.

Most students are familiar with this property; roughly speaking it says that if one has a partially ordered set and a \le b and b \le c then a \le c . Those who have studied the real numbers might be tempted to greet this concept with a shrug. However in more complicated cases, the transitive property simply doesn’t hold, even when it makes sense to order things. Here is an example: consider the following sets of dice:

dice

What we have going here: Red beats green 4 out of 6 times. Green beats blue 4 out of 6 times. Blue beats red 4 out of 6 times. All the colored dice tie the “normal” die. Yet, the means of the numbers are all the same.

Note: that this can happen is probably not a surprise to sports fans; for example, in boxing: Ken Norton beat Muhammed Ali (the first time), George Foreman destroyed Ken Norton and, Ali beat Foreman in a classic. Of course things like this happen in sports like basketball but when team doesn’t always play its best or its worst.

But this dice example works so beautifully because this “impossibility of the dice obeying a transitive ordering relation is theoretically impossible, by design.

Movies
Since the wife has been gone on a trip, I’ve watched some old movies at night. One of them was the Cincinnati Kid, which features this classic scene:

Basically, the Kid has a full house, but ends up losing to a straight flush. Yes, the odds of the ten cards (in stud poker) ending up in “one hand a full house, the other a straight flush” are extremely remote. I haven’t done the calculations but this assertion seems plausible:

Holden states that the chances of both such hands appearing in one deal are “a laughable” 332,220,508,619 to 1 (more than 332 billion to 1 against) and goes on: “If these two played 50 hands of stud an hour, eight hours a day, five days a week, the situation would arise about once every 443 years.”

But there is one remark from this Wikipedia article that seems interesting:

The unlikely nature of the final hand is discussed by Anthony Holden in his book Big Deal: A Year as a Professional Poker Player, “the odds against any full house losing to any straight flush, in a two-handed game, are 45,102,781 to 1,”

I haven’t done the calculation but that seems plausible. But, here is the real point to the final scene: the Kid knows that he has a full house but The Man is showing 8, 9, 10, Q of diamonds. He knows that the only “down” card that can beat him is the J of diamonds but he knows that he has 3 10’s, 2 A’s. So there are, to his knowledge, 52 - 9 = 43 cards out, and only 1 that can beat him. So the Kid’s probability of winning is \frac{42}{43} which are pretty strong odds, but they are not of the “million to one” variety.

March 3, 2013

Mathematics, Statistics, Physics

Filed under: applications of calculus, media, news, physics, probability, science, statistics — collegemathteaching @ 11:00 pm

This is a fun little post about the interplay between physics, mathematics and statistics (Brownian Motion)

Here is a teaser video:

The article itself has a nice animation showing the effects of a Poisson process: one will get some statistical clumping in areas rather than uniform spreading.

Treat yourself to the whole article; it is entertaining.

January 17, 2013

Enigma Machines: some of the elementary math

Note: this type of cipher is really an element of the group S_{26} , the symmetric group on 26 letters. Never allowing a letter to go to itself reduced the possibilites to products of cycles that covered all of the letters.

Math and Probability in Pinker’s book: The Better Angels of our Nature

Filed under: elementary mathematics, media, news, probability, statistics — Tags: , — collegemathteaching @ 1:01 am

I am reading The Better Angels of our Nature by Steven Pinker. Right now I am a little over 200 pages into this 700 page book; it is very interesting. The idea: Pinker is arguing that humans, over time, are becoming less violent. One interesting fact: right now, a random human is less likely to die violently than ever before. Yes, the last century saw astonishing genocides and two world wars. But: when one takes into account how many people there are in the world (2.5 billion in 1950, 6 billion right now) World War II, as horrific as it was, only ranks 9’th on the list of deaths due to deliberate human acts (genocides, wars, etc.) in terms of “percentage of the existing population killed in the event”. (here is Matthew White’s site)

But I have a ways to go in the book…but it is one I am eager to keep reading.

The purpose of this post is to talk about a bit of probability theory that occurs in the early part of the book. I’ll introduce it this way:

Suppose I select a 28 day period. On each day, say starting with Monday of the first week, I roll a fair die one time. I note when a “1” is rolled. Suppose my first “1” occurs Wednesday of the first week. Then answer this: “what is the most likely day that I obtain my NEXT “1”, or all days equally likely?”

Yes, it is true that on any given day, the probability of rolling a “1” is 1/6. But remember my question: “what day is most likely for the NEXT one?” If you have had some probability, the distribution you want to use is the geometric distribution, starting on Thursday of the next week.

So you can see, the mostly likely day for the next “1” is Thursday! Well, why not, say, Friday? Well, if Friday is the next 1, then this means that you got “any number but 1” on Thursday followed by a “1” on Friday, and the probability of that is \frac{5}{6} \frac{1}{6} = \frac{5}{36} . The probability of the next one being Saturday is \frac{25}{196} and so on.

The point: if one is studying the distribution of events that have a Poisson distribution (probability p ) on a given time period, the overall distribution of such events is likely to show up “clumped” rather than evenly spaced. For an example of this happening in sports, check this out.

Anyway, Pinker applies this principle to the outbreak of wars, mass killings and the like.

August 27, 2012

Why most “positive (preliminary) results” in medical research are wrong…

Filed under: editorial, pedagogy, probability, research, statistics — collegemathteaching @ 12:53 am

Suppose there is a search for a cure (or relief from) a certain disease.  Most of the time, cures are difficult (second law of thermodynamics at work here).  So, the ratio of “stuff that works” to “stuff that doesn’t work” is pretty small.  For our case, say it is 1 to 1000.

Now when a proposed “remedy” is tested in a clinical trial, there is always a possibility for two types of error: type I which is the “false positive” (e. g., the remedy appears to work beyond placebo but really doesn’t) and “false negative” (we miss a valid remedy).

Because there is so much variation in humans, setting the threshold for accepting the remedy too low means we’ll never get cures.  Hence a standard threshold is .05, or “the chance that this is a false positive is 5 percent”.

So, suppose 1001 different remedies are tried and it turns out that only 1 of them is a real remedy (and we’ll assume that we don’t suffer a type II error).  Well, we will have 1000 remedies that are not actually real remedies, but about 5 percent, or about 50 will show up as “positive” (e. g. brings relief beyond placebo).  Let’s just say that there are 49 “false positives”.

Now saying “we tried X and it didn’t work” isn’t really exciting news for anyone other than the people searching for the remedy.  So these results receive little publicity.  But “positive” results ARE considered newsworthy.  Hence the public sees 50 results being announced: 49 of these are false positive and 1 is true.   So the public sees 50 “this remedy works! (we think; we still need replication)” announcements, and often the medial leaves off the “still needs replication” part..at least out of the headline.

And….of the 50 announcements …..only ONE (or 2 percent) pans out.

The vast majority of results you see announced are…wrong. 🙂

Now, I just made up these numbers for the sake of argument; but this shows how this works, even when the scientists are completely honest and competent.

May 14, 2012

Probability in the Novel: The Universal Baseball Association, Inc. J. Henry Waugh, Prop. by Robert Coover

Filed under: books, editorial, elementary mathematics, pedagogy, popular mathematics, probability, statistics — collegemathteaching @ 2:31 am

The Robert Coover novel The Universal Baseball Association, Inc. J. Henry Waugh, Prop. is about the life of a low-level late-middle aged accountant who has devised a dice based baseball game that has taken over his life; the books main character has a baseball league which has played several seasons, has retired (and deceased!) veterans, a commissioner, records, etc. I talked a bit more about the book here. Of interest to mathematics teachers is the probability theory associated with the game that the Henry Waugh character devised. The games themselves are dictated by the the result of the throws of three dice. From pages 19 and 20 of the novel:

When he’d finally decided to settle on his baseball game, Henry had spent the better part of two months just working on the problem of odds and equilibrium points in an effort to approximate that complexity. Two dice had not done it. He’d tried three, each a different color, and the 216 different combinations had provided the complexity all right, but he’d nearly gone blind trying to sort the three colors on each throw. Finally, he compromised, keeping the three dice, but all white reducing the number of combinations to 56, though of course the odds were still based on 216.

The book goes on to say that the rarer throws (say, triples of one numbers) triggered a referral to a different chart and a repeat of the same triple (in this case, triple 1’s or triple 6’s (occurs about 3 times every 2 seasons) refers him to the chart of extraordinary occurrences which includes things like fights, injuries, and the like.

Note that the game was very complex; stars had a higher probability of success built into the game.

So, what about the probabilities; what can we infer?

First of all, the author got the number of combinations correct; the number of outcomes of the roll of three dice of different colors is indeed 6^3 = 216 . What about the number of outcomes of the three dice of the same color? There are three possibilities:

1. three of the same number: 6
2. two of the same number: 6*5 = 30 (6 numbers, each with 5 different possibilities for the remaining number)
3. all a different number: this might be the trickiest to see. Once one chooses the first number, there are 5 choices for the second number and 4 for the third. Hence there are 20 different possibilities. Or put a different way, since each choice has to be different: this is {{6}\choose{3}} = \frac{6!}{3! 3!} = \frac{120}{6} = 20

However, as the author points out (indirectly), each outcome in the three white dice set-up is NOT equally likely!
We can break down the potential outcomes into equal probability classes though:
1. Probability of a given triple (say, 1-1-1): \frac{1}{216} , with the probability of a given throw being a triple of any sort being \frac{1}{36} .
2. Probability of a given double (say, 1-1-2) is \frac{{{3}\choose{2}}}{216} = \frac{3}{216} = \frac{1}{72} So the probability of getting a given pair of numbers (with the third being any number other than the “doubled” number) would be \frac{5}{72} hence the probability of getting an arbitrary pair would be \frac{30}{72} = \frac{5}{12} .
3. Probability of getting a given trio of distinct numbers: there are three “colors” the first number could go, and two for the second number, hence the probability is: \frac{3*2}{216} = \frac{1}{36} . So there are {{{6}\choose{3}}} = 20 different ways that this can happen so the probability of obtaining all different numbers is \frac{20}{36} = \frac{5}{9} .

We can check: the probability of 3 of the same number plus getting two of the same number plus getting all distinct numbers is \frac{1}{36} + \frac{5}{12} + \frac{5}{9} = \frac{1 + 15 + 20}{36} = 1 .

Now, what can we infer about the number of throws in a season from the “three times every two seasons” statement about triple 1’s or triple 6’s?
If we use the expected value concept and figure that double triple 1’s has a probability of \frac{1}{216^2} = \frac{1}{46656} and getting either triple 1’s or triple 6’s would be \frac{1}{23328} and using E = np , we obtain \frac{n}{23328} = 3 which implies that n = 69984 throws per two seasons, or 34992 throws per season. There were 8 teams in the league and each played 84 games which means 336 games in a season. This means about 104 throws of the dice per game, or about 11.6 throws per inning or 5.8 throws per half of an inning; perhaps that is about 1 per batter.

Evidently, Robert Coover did his homework prior to writing this novel!

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