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January 25

Integrable sets of functions

I will define a set of functions to be complete if any mathematical expression using functions from the set has an antiderivative that also be expressed using only functions in the set. Given a set of functions, is it always possible to extend that set (defining new functions where necessary) so that the set is complete? 149.169.122.82 (talk) 00:03, 25 January 2011 (UTC)[reply]

No, because there are functions that have no antiderivative at all. If one of these are in your given set, then you're out of luck.

In order to get out of that problem, you need to specify explicitly that you're only considering certain "sufficiently nice" functions in the first place. The question then reduces to whether the set of all "sufficiently nice" functions is (a) closed under antidiffentiation, and also (b) closed under formation of "mathematical expression", which you also need to define explicitly.

One way to reach a "yes" answer would be to define a "sufficiently nice" function as a continuous real-valued function on an open subset of the real line, and "mathematical expression" as including constants and the four basic arithmetic operations.

On the other hand, if you include complex functions, you will need a very restricted notion of "mathematical expression"; otherwise you'll immediately end up defining 1/z on C\{0}, which has no antiderivative. –Henning Makholm (talk) 01:19, 25 January 2011 (UTC)[reply]

Basic probability question

Resolved

I have long had this niggling doubt and would appreeciate it if someone could explain it to me. Suppose the probability of recovery of a patient (in some fixed time t) is p and there are N number of patients. Why do mathematical models assume that after time t, pN people would have recovered? What is the justification behind this? Thanks-Shahab (talk) 04:48, 25 January 2011 (UTC)[reply]

The definition of probability. Saying that something has probability t is the same as saying that if an experiment is repeated N times then it will expect that something about tN times. Taemyr (talk) 05:53, 25 January 2011 (UTC)[reply]

Well, that's the frequentist interpretation of probability, anyway. A more rigorous (though more technical) explanation comes in the form of the law of large numbers: in a large number of trials, the average of the results should be close to the expected value (and there is a mathematically precise meaning of the word "close"). If we assign a value of 1 when a patient recovers and 0 otherwise, then the average of the results will be the fraction of the patients who recover; this should be close to the expected value, which (for this 0-1 valuation of the results) is equal to the probability that one patient recovers. —Bkell (talk) 06:02, 25 January 2011 (UTC)[reply]

Yes, and still ... what the law of large numbers gives us is a high probability that the total comes out close to Np. However, if our object was to understand "probability" primitively, it is not evident that we've gotten anywhere. In order to understand what "high probability of being close to Np" means we'd have to regress to another application of the law of large numbers, and then another one and another one. Or, put another way, we can learn something from the law of large numbers once we have already embraced a frequentist interpretation, but by that time we already have the desired result without it. I think the law of large numbers is more a demonstration that frequency probability is internally consistent (which is a good and worthy endeavor in itself), than a rigorous underpinning of it. –Henning Makholm (talk) 13:14, 25 January 2011 (UTC)[reply]

It's also useful to consider the binomial distribution. If there are N patients, each has a probability of p to recover, and they are all independent, then the number of recovering patients will be distributed binomially with N and p. This has an expectation of Np. The variance will be proportionally small when N is large, thus the value will not be too far from the expectation.

As an aside, if the patients are not independent then the distribution will not be binomial, but it will still have expectation Np. -- Meni Rosenfeld (talk) 09:07, 25 January 2011 (UTC)[reply]

Oh, that's no problem; they can wash and dress themselves, and most also do their own shopping and cooking. –Henning Makholm (talk) 17:05, 25 January 2011 (UTC)[reply]

You do not need the law of large numbers, nor do you need the binomial distribution. The matter can be explained in a more elementary and more general way.

Consider a population consisting of N patients, K of which recover, and take a sample consisting of n patients, k of which recover. The integers N,n,K,k satisfy 0≤k≤n≤N, and k≤K≤N.

Fixing N and K, the 2^N samples are classified according to the values of n and k. The number of samples in the class is $\scriptstyle {\binom {K}{k}}{\binom {N-K}{n-k}}$ .

Fixing N, n, and K, the mean value of k is $\scriptstyle {\frac {\sum _{k}k{\binom {K}{k}}{\binom {N-K}{n-k}}}{\sum _{k}{\binom {K}{k}}{\binom {N-K}{n-k}}}}={\frac {Kn}{N}}=np$ where $\scriptstyle p={\frac {K}{N}}$ is the probability of recovery. (See Hypergeometric distribution). This means that the mean value of $\scriptstyle {\frac {k}{n}}$ is $\scriptstyle {\frac {K}{N}}$ .

Fixing N, n, and k, the mean value of K is $\scriptstyle {\frac {\sum _{K}K{\binom {K}{k}}{\binom {N-K}{n-k}}}{\sum _{K}{\binom {K}{k}}{\binom {N-K}{n-k}}}}={\frac {(k+1)(N+2)}{(n+2)}}-1$ . This means that the mean value of $\scriptstyle {\frac {K+1}{N+2}}$ is $\scriptstyle {\frac {k+1}{n+2}}$ . Bo Jacoby (talk) 18:08, 25 January 2011 (UTC).[reply]

Thanks all.-Shahab (talk) 08:19, 29 January 2011 (UTC)[reply]

Random function on the sphere

In what sense is it meaningful to talk about a random function on a sphere? (Continuous, say, or in some other convenient class of functions. Functions that differ by a rotation should naturally be "equally likely".) Blanche, Blanche DuBois (talk) 23:16, 25 January 2011 (UTC)[reply]

It has about the same problems as talking about a "random real number" or "random function on the unit interval". One can certainly find probability distributions that are invariant under rotations (spherical harmonics would be my first idea for constructing one in practice), but that property is far from enough to characterize one distribution uniquely. –Henning Makholm (talk) 00:19, 26 January 2011 (UTC)[reply]

To clarify: random function on the unit interval can be characterized by the Wiener measure. Is there an analogous such "natural" definition of a random function on the sphere? Blanche, Blanche DuBois (talk) 00:45, 26 January 2011 (UTC)[reply]

It's possible to have a Gaussian process defined on a sphere, although you have to be careful to use positive definite covariance function. HTH, Robinh (talk) 08:27, 26 January 2011 (UTC)[reply]

Unfortunately, I don't think this applies to what I am curious about. I'd like to understand what a "random function" from the sphere to the real line should mean. (I know this isn't a very precise question to start with...) ;-) Blanche, Blanche DuBois (talk) 12:48, 26 January 2011 (UTC)[reply]

Ok, so a different route to the same question. Getting back to Henning's suggestion to use spherical harmonics, there are probabilistic constraints on the Fourier coefficients of a 1D Brownian motion. I think that one must have something like

c_{n}\sim 1/n

as $n\to \infty$ . If I had to guess wildly, I would say that the Fourier series of the Wiener process (say on $[0,\pi /2]$ ) is

\sum {\frac {X_{n}}{n}}\sin(n\theta )

where $X_{n}$ are iid Gaussian with mean zero. If this is true, then it certainly suggests how to approach the problem on the sphere. Of course, I haven't the foggiest idea how to prove it, so: (1) Is it true? (2) Is it some standard fact that I can find in a book somewhere? Blanche, Blanche DuBois (talk) 12:48, 26 January 2011 (UTC)[reply]

Our Brownian bridge article quotes that very Fourier series (modulo some differences in scaling and normalization). A slightly different expression for the actual Wiener process is given in the Karhunen–Loève theorem article. –Henning Makholm (talk) 14:14, 26 January 2011 (UTC)[reply]

Cool! That is most helpful. Merci beaucoup, Blanche, Blanche DuBois (talk) 12:50, 27 January 2011 (UTC)[reply]