Hyperphronesis

Sunday, February 22, 2015

Valuation Theory

Valuation Systems

A valuation system (VS) is any system by which value is assigned to things. That is, the way in which terms like "better", and "worse", "good", and "bad", are given meaning or are understood. For example, in choosing a hinge for a door, one system of saying "hinge X is better than hinge Y" is to consider price (cheaper being better, for instance), or resistance to rust, or weight, or color, or size, etc. After all, there is no unqualified way to say "hinge A is better than hinge B", and any statement that does not explicitly state the way in which hinge A is deemed better than hinge B will have some implicit VS.

All VSs have a domain, which is the set of all things which can be valued by that VS. The VS used to compare hinges won't be able to compare the value of microprocessors, or political parties, or cake recipes. It is important to keep in mind the domain of a VS when discussing it. We will denote the domain of VS X as D_X.

Types of Valuation Systems

There are two general sorts of valuation systems:

Comparative Valuation Systems (CVSs): Determines only the ranking of value for the elements of a given, countable set. If X is a CVS and X values A above B, we will write that as \( (A>B)_X\), which we can read as "A is better than B, according to X". Note that CVSs don't have any notion of "good" or "bad", but only "better" and "worse", and possibly "best", if there is some element better than the rest.

A subset of CVSs are Bi-comparative VSs (bCVSs, or C₂VSs), which only rank sets with exactly two elements, either with one better and one worse, or with both equal. If the bCVS has the additional property of being transitive, then the system can be used to impose a partial ordering on the elements of its domain.

Evaluative Valuation Systems (EVSs): Determines the plain value of every element in its domain, like a function. Namely, we can symbolize "the value of A, according to EVS X" as \(V_X(A)\). Without loss of generality, we can take the values assigned to be real numbers. If only order is important, we can take the range to be the numbers in the interval \([-1,1]\). Note that EVSs can have a notion of "good" and "bad", in that we can define "A is bad, according to EVS X" as \(V_X(A)< c \), for some number c, which we can take to be 0. Similar statements can be similarly defined. To keep notation consistent, we will write \((A>B)_X\) iff \(V_X(A)>V_X(B)\), for some EVS X.

Indifferent Extensions

We can also define the indifferent extension of a valuation system X with domain D_X as the valuation system that is identical to X for any elements in D_X, and is indifferent to all other things. More exactly, we can define it for the cases of CVSs and EVSs as follows:

CVSs:
Let \(X\) be a CVS with domain \(D_X\). The CVS \(X'\) is the indifferent extension of \(X\), such that, for any \( a,b \notin D_X\) and \(c \in D_X\), \((a< c )_{X'} \), \((a=b)_{X'}\).

EVSs:
Let \(X\) be an EVS with domain \(D_X\). The EVS \(X'\) is the indifferent extension of \(X\), such that, for any \( a\notin D_X\), \(V_{X'}(a)=0\).

Optimal Elements

We can also give meaning to statements like "t is the best element in set S, according to X", in two senses. We can say that t is the optimal element of S according to VS X if, for every element s of S such that \(s \neq t\), then \( (t > s)_X \). We can say that t is an equi-optimal element of S according to VS X if, for every element s of S, \( (t \geq s)_X \). We can also say that "t is the best element in set S, according to set A", for some set A of VSs, if, for each VS X in A, s is the optimal element in X. We might also stipulate that for every VS in A there is an optimal element in S. Similarly for equi-optimal.

If we want to say something like "t is the best element in S" without qualifying it by a VS, it must be the case that all valuation systems agree (or perhaps there is some "best VS" which would deem s optimal, but we will get to that later). Namely, we say that s is the universo-optimal (UO) element of S if, for every VS X for which there is an optimal element in S, s is the optimal element of X. We also can say that s is a universo-equi-optimal(UEO) element of S if, for every VS X for which there is an equi-optimal element in S, s is an equi-optimal element of X. Note that for there to be a universo-optimal element, all relevant VSs must agree: if there is even one VS for which there is a different optimal element than another, then there is no universo-optimal element in S.

Meta-Valuation Systems, Optimal Valuation Systems, and Recommendation

We can also have VSs whose domain includes some subset of the set of all VSs. We can call these meta-valuation systems (MVS). We can also define the set of totally meta-VSs (TMVS), which is the set of all VSs whose domain includes the set of all VSs.
Now, if there is to be some VS that can be called "the best VS", it must be the case that it is UO (or at least UEO) in the set of all VSs. Thus we define:
a VS X is the objectively best VS iff, for ever VS Y in the set TMVSs for which there is an optimal element, X is the optimal element of Y in the set of all VSs.
However, it seems not hard to very strongly suggest if not prove that there is no such VS, for all it takes are two TMVSs with optimal elements that disagree as to this optimal element, and this seems very easy to construct. Thus there simply is no such objectively best VS. We can call this the Universo-Optimality Absence Theorem.

Also, we can say that VS A recommends VS B if \((B>A)_A\). We denote this by \(A \rightarrow B\). Clearly A must be a MVS, as it includes the VS B in its domain. The relevance is that, if we hold to VS A, and A recommends B, then we should discard A and take up B instead. We may have some issues if A recommends multiple VSs, but then the solution would then be to follow the recommendation that is outranks the rest. For example, if \(A \rightarrow B\) and \(A \rightarrow C\), and \((B>C)_A\), then we should choose B, rather than C. However, we will say that a VS A is a consistent recommender if it is the case that if \(A \rightarrow B\), and \(A \rightarrow C\), and \((B>C)_A\), then \(C \rightarrow B\), and it is not the case that \(B \rightarrow C\).

Antagonist Valuation Systems and the Universo-Optimality Absence Theorem

Take any VS \(X\). We define the antagonist valuation system to \(X\) (denoted \(X^A\)) as follows: If \(X\) is a CVS, and \( (P>Q)_X\), then \( (P<Q)_{X^A}\). Similarly, if \(X\) is an EVS, then \(V_X(P)=-V_{X^A}(P)\). It is clear that is any VS is specifiable, its antagonist will likewise be specifiable merely by reversing all the valuations. It also clearly follows that a VS and its antagonist can never recommend the same thing, as a VS and its antagonist never agree (except in the case of indifference). Thus, for a given valuation, however many VS can be found that agree with that valuation, precisely the same number of antagonists can be found or formed which disagree with the valuation. It follows that there cannot possibly be any universo-optimal valuation system, and thus the theorem is proven.

Some Implications for Morality

Morality is always associated with valuation. Specifically, every moral system corresponds to a valuation system, which we may call a moral valuation system. The domain of a moral valuation system would be decisions made in response to a scenario (the scenario may be implicit, but it is always there. Murder, for instance, is incoherent unless there is someone to murder and some way to murder them). For the same scenario, one decision may be better than another, or one may be good while another bad. A decision (or subset of decisions) may be called "obligatory" if that decision is the only one that is good while the complement is bad. A decision may be supererogatory if the decision is better than other good (or not-bad) decisions. Similarly, a moral systems may recommend another moral system if the first system deemd the second better than the first. However, by the universo-optimality absence theorem, there cannot be any one moral system universally judged better than all others. As no valuation can be made apart from some (implicit) valuation system, and there is no intrinsically or universally preferred valuation system, there cannot be such a thing as "objective moral value". Even a deity could not have access to something like that. Nothing, even a deity, could not be called "good" except in reference to a VS which must of necessity be, at base, arbitrary (why pick out that VS instead of its antagonist? Or that VS rather than any other VS? The answer cannot possibly be that the VS is "the best" as that has no meaning apart from some VS).

Thus, there may be three general projects for moral philosophy:

1) Descriptive ethics: what moral valuation system do people use? Can we form a description that fully/maximally captures the way the person/population actually values things?

2) Educing and following recommendations: given a certain VS, what VS does it recommend? Can we iterate this until we have a self-recommending VS?

3) Axiomatics/foundations: what minimal set of axioms sufficiently characterize a VS? How can we simplify a VS so that it is maximally described with the smallest set of principles? Can this be algorithmized in some way? What information is morally relevant/irrelevant to a given VS?

Tuesday, July 1, 2014

Probability Problems

There are many interesting problems that can be studied with probability theory. Here I will discuss a few of my favorites.

Maxima in Data

Suppose we have a sequence of values of random variables \(\left(X_{k} \right )_{k=1}^{n}\)that are independent and identically distributed with density function \(f_{X}(x)\). We wish to find the expected number of local maxima in the data, that is, values of \(X_{k}\) such that \(X_{k-1} < X_{k} > X_{k+1}\). Let \(y=X_k\). Then, the probability \(p\) that a certain value y is a maximum is that of having \(X_{k-1} < y > X_{k+1}\). As all the \(X\)s are independent and identically distributed, we can calculate this probability by finding \[p= \int_{-\infty}^{\infty}f_{X}(x)F_{X}^2(x)dx\] Where \(F_{X}(x)=\int_{-\infty}^{x}f_{X}(y)dy\) is the cumulative distribution function of \(f_{X}\). The form above is like \(\sum_y P(X_k=y)P(X_{k-1} < y)P(X_{k+1} < y)\), except we use the continuous formulation. However, clearly \(F_{X}'(x)=f_X(x)\), and so the formula becomes \[p= \int_{-\infty}^{\infty}F_{X}'(x)F_{X}^2(x)dx\] But, by elementary calculus, we can change this to \[p= \int_{0}^{1}F^2 dF=\frac{1}{3}\] (the change in bounds arises from the fact that \(F_{X}(-\infty)=0\) and \(F_{X}(+\infty)=1\)). Thus, we expect one-third of the values in the sequence to be local maxima. Likewise, we expect one-third to be local minima, and the remaining third to be neither.

By the same method, we can look for other patterns. For instance, the fraction of data points that are higher than all four of their closest neighbors is \(\frac {1}{5}\). The fraction of data points that are bigger than their closest neighbors and smaller than their next-closest neighbors is \(\frac {1}{30}\). In fact, all the calculations can be made by evaluating integrals of the form \( \int_{0}^{1} x^m (1-x)^n dx\). We can also use results like this to test for non-independent or non-identically distributed data. It may even be possible to use it in fraud or bias detection. Based on next-to-nothing-at-all, I would expect human generated data to fail some of these tests.

We can also find that the distribution of the number of maxima in \(n\) data points, regardless of the probability distribution \(f_{X}\), is approximately normally distributed with mean \(\frac{n}{3}\) and variance \(\frac{2 n}{45}\). Thus, if we found fewer than 2960 or more than 3040 maxima in a list of 9000 data points, we could be 95% confident that the list was not of independent and identically distributed values. We can also run the same test for minima, but for non-maxima-non-minima, the variance is instead \(\frac{2 n}{15}\).
The values for the variances were found empirically. I don't really know how one would go about finding them analytically.

We can also find the distribution of the values of the maxima, which is easily found to be \[g(x)=3 f_{X}(x)F_{X}^2\] Other distributions are similarly found.

Joint Lives

Suppose we stat with \(N\) couples (\(2N\) people), and at a later time, \(M\) of the original \(2N\) people remain. We want to find the expected number of intact couples remaining. Let \(C(M)\) and \(W(M)\) be the expected remaining number of couples and widows respectively when M total people are left. We then note that, as any remaining person is equally likely to be eliminated next, we have: \[ C(M-1)=C(M)-2 \frac{C(M)}{M} \\ W(M-1)=W(M)-\frac{W(M)}{M}+2 \frac{C(M)}{M}\] We can solve this recurrence relation, subject to the constraints \(W(M)+2 C(M)=M\) and \(C(2N)=N, W(2N)=0\), and find that \[ C(M)=\frac{M(M-1)}{2(2N-1)} \\ W(M)=\frac{M(2N-M)}{2N-1} \] If we express M as a fraction of the total starting population: \(M=2xN\), and express \(C\) and \(W\) as fractions of the total population, we find, for \(N\) big: \[ C(x)=x^2 \\ W(x)=x(1-x) \] Also, for the general case of starting out with \(kN\) \(k\)-tuples, the expected number of intact \(k\)-tuples when \(M\) individuals remain is given by: \[K(M)=N \frac{\binom{M}{k}}{\binom{kN}{k}}\] For the case of triples, we have the number of triples, doubles and singles when M individuals remain is given by: \[K_3 (M)= \frac{M(M-1)(M-2)}{3(3N-1)(3N-2)} \\ K_2 (M)= \frac{M(M-1)(3N-M)}{(3N-1)(3N-2)} \\ K_1 (M)= \frac{M(3N-M)(3N-M-1)}{(3N-1)(3N-2)} \] Generally, with the same sense as discussed above, the fraction of the population in a m-tuple, beginning with only k-tuples, when fraction \(x\) of the population remains, is given by: \[K_m (x)=\binom{k-1}{m-1} x^m (1-x)^{k-m}\] In fact, the general form for the expected number can be given as \[ K_m (M)= N \frac{\binom{M}{m}\binom{kN-M}{k-m}}{\binom{kN}{k}} \]

Random Finite Discrete Distribution

Suppose we have a discrete random variable that can take on the values \(1,2,3,...,n\) with probabilities \(p_1,p_2,p_3,...p_n\) respectively, subject to the constraint \(\sum_{k=1}^n p_k=1\). Let \(p\) be an arbitrary value among the \(p\)s. We will take any combination of values for the \(p\)s as equally likely. By looking at the cases of n=2 and n=3, we find that the probability density function of \(p\) is given by \[ f_P(p)=(n-1)(1-p)^{n-2} \] And the cumulative distribution function is given by \[ F_P(p)=1-(1-p)^{n-1} \] The average value of \(p\) is then \[\int_{0}^{1} p(n-1)(1-p)^{n-2}dp=\frac{1}{n}\] And the variance is \[\int_{0}^{1} p^2 (n-1)(1-p)^{n-2}dp-\frac{1}{n^2}=\frac{n-1}{n^2 (n+1)}\] We thus find that the chance that \(p\) is above the average value is \[P\left ( p > \frac{1}{n} \right )=\left ( 1-\frac{1}{n} \right )^{n-1}\] In the limit as n becomes large, this value tends to \(\frac{1}{e}\).
A confidence interval containing fraction x of the total probability, for large n, is given by: \[ \frac{1}{n} \ln \left(\frac{e}{e x +1-x} \right) \leq p \leq \frac{1}{n} \ln \left(\frac{e}{1-x} \right) \] For instance, a \(50 \%\) confidence interval is given by \(\frac{1}{n}\ln \left(\frac{2 e}{1+e}\right) \leq p \leq \frac{1}{n}\ln(2 e)\).

We can also extend this to continuous distributions with finite support if we only consider the net probability of landing in equally-sized bins. While the calculation may break down if the number of possible values is actually infinite, it can be used to get some information about distributions with an arbitrarily large number of possible values.

Maximum of Exponential Random Variables

Suppose we have \(N\) independent and identically distributed exponential random variables \(X_1,X_2,...X_N\) with means \(\mu\). That is, \(f_X (x_k)=\frac{1}{\mu} e^{-\frac{x}{\mu}}\) when \(x \geq 0\) and zero otherwise. Let us interpret the random values as lifetimes for \(N\) units. The exponential distribution has the interesting property of memorylessness, which means that \(P(x > a+b| x > b)=P(x > a)\). We can show this by using the definition: \[ P(x>a+b|x>b)=\frac{P(x>a+b \cap x>b)}{P(x>b)}=\frac{P(x>a+b)}{P(x>b)} \\ P(x>a+b|x>b)=\frac{\int_{a+b}^{\infty}e^{-\frac{x}{\mu}}dx}{\int_{b}^{\infty}e^{-\frac{x}{\mu}}dx}=\frac{e^{-\frac{a+b}{\mu}}}{e^{-\frac{b}{\mu}}}=e^{-\frac{a}{\mu}}=P(x>a) \] In other words, given that a unit lasted \(b\) minutes, the chance that it will last another \(a\) minutes is the same as that it would last \(a\) minutes. We now calculate the probability distribution of the minimum of the \(N\) random variables. The probability that the minimum of \(X_1,X_2,...X_N\) is no less than \(x\) is the same as the probability that \(X_1 \geq x \cap X_2 \geq x \cap...X_N \geq x \). As all the \(X\)s are independent, this can be simplified to a product, and as all the \(X\)s are identically-distributed, we can simplify this further: \[ P(\min(X_1,X_2,...)\geq x)=\left ( P(X_1 \geq x) \right )^N= \left ( \int_{x}^{\infty}\frac{1}{\mu} e^{-\frac{x}{\mu}}\right )^N \\ P(\min(X_1,X_2,...)\geq x)= e^{-\frac{xN}{\mu}} \\ P(\min(X_1,X_2,...)\leq x)= 1-e^{-\frac{xN}{\mu}} \\ f_{\min(X)}(x)=\frac{N}{\mu}e^{-\frac{xN}{\mu}} \] Thus, the average of the minimum of \(X_1,X_2,...X_N\) is \(\frac{\mu}{N}\). We now combine these two facts, the mean minimum vale and the memorylessness. We start with all units operational, and we have to wait an average of \(\frac{\mu}{N}\) until the first one fails. However, given that the first one fails, the expected additional wait time until the next one fails is just \(\frac{\mu}{N-1}\), that is, the expected minimum of \(N-1\) units. Thus, the expected time that the \(m\)th unit fails is given by \[\mu\sum_{k=0}^{m-1}\frac{1}{N-k}\] Thus, the expected maximum time, when the \(N\)th unit fails is \[\mu\sum_{k=1}^{N}\frac{1}{k}\]

More generally, we can look at the distributions of the kth order statistic of \(X_1,X_2,...X_N\). The kth order statistic, denoted \(X_{(k)}\), is defined as the kth smallest value, so that \(X_{(1)}\) is the smallest (minimum) value, and \(X_{(N)}\) is the largest (maximum) value. The pdf is easily found to be: \[ f_{X_{(k)}}(x)=k {N \choose k} F_X^{k-1}(x)\left[1-F_X(x)\right]^{N-k}f_X(x) \] Where \(F_X(x)\) is the cdf of X, and \(f_X(x)\) is the pdf of X. So, in this case, \[ f_{X_{(k)}}(x)=\frac{k}{\mu} {N \choose k} e^{-(N-k+1)x/\mu}\left[1-e^{-x/\mu}\right]^{k-1} \] Thus, the moment generating function is given by: \[ g(t)=\frac{k}{\mu} {N \choose k} \int _0 ^\infty e^{-(N-k+1-\mu t)x/\mu}\left[1-e^{-x/\mu}\right]^{k-1} dx \] By a simple transformation, we find that: \[ g(t)=k {N \choose k} \int _0 ^1 u^{N-k-\mu t}(1-u)^{k-1} du \] This puts the integral in a well-known form, which has the value \[ g(t)=\frac{N!}{\Gamma(N+1-\mu t)}\frac{\Gamma(N-k+1-\mu t)}{(N-k)!} \] By a simple calculation, the cumulants are then given by the surprisingly simple form: \[ \kappa_n=\mu^n(n-1)!\sum_{j=N-k+1}^{N} \frac{1}{j^n} \] Several interesting results follow from this:

For the Nth order statistic (the maximum), we already know that the mean value goes as \(\sum_{j=1}^{N} \frac{1}{j}\). But now we see that the other cumulants go as \((n-1)!\sum_{j=1}^{N} \frac{1}{j^n}\). Thus, the variance converges, in the limit, to \(\mu^2 \frac{\pi^2}{6}\). The skewness converges, in the limit, to \(\frac{12 \sqrt{6}}{\pi^3}\zeta(3)\), and the excess kurtosis converges to \(\frac{12}{5}\). In fact, if we shift to take into account the unbounded mean, the distribution of the maximum converges to a Gumbel distribution. This is a special case of a fascinating result known as the extreme value theorem.
For any given, fixed, finite \(k\geq 0\), \(X_{(N-k)}\) converges, as N goes to infinity, to a non-degenerate distribution with finite, positive variance, if we shift it to account for the unbounded mean.
For k of the form \(k=\alpha N\) (or the nearest integer thereto), for some fixed alpha between 0 and 1, for \(\alpha \neq 1\), in the limit at N goes to infinity, the distribution of \(X_{(\alpha N)}\) become degenerate distributions with all the probability density located at \(\mu\ln\left(\frac{1}{1-\alpha}\right)\). These are, of course, the locations of the \(100\alpha \%\) quantiles, and so \(X_{(\alpha N)}\) is a consistent estimator for the \(100\alpha \%\) quantile.

As a more general result, let us find the cdf of \(X_{(\alpha N)}\) for an arbitrarily distributed X, in the limit as N goes to infinity. The cdf of \(X_{(\alpha N)}\) is given by: \[ F_{X_{(\alpha N)}}(y)=\alpha N {N \choose \alpha N} \int _{-\infty} ^{y} F_X^{\alpha N-1}(x)\left[1-F_X(x)\right]^{N-\alpha N}f_X(x) dx \] As \(f_X(x)=\frac{d}{dx}F_X(x)\), we then have, by a simple substitution: \[ F_{X_{(\alpha N)}}(y)=\alpha N {N \choose \alpha N} \int _{0} ^{F_X(y)} u^{\alpha N-1}\left[1-u\right]^{N-\alpha N} du \] This is the cdf of a Beta distributed random variable, with mean \(\mu=\frac{\alpha N}{N+1}\) and variance \(\sigma^2=\frac{\alpha N (N-\alpha N+1)}{(N+1)^2(N+2)}\). Thus, as N goes to infinity, this will converge in distribution to a degenerate distribution with all the density at \(y=F_X^{-1}(\alpha)\), that is, at the \(100\alpha \%\) quantile of the distribution.

Choosing a Secretary

Suppose we need to hire a secretary. We have \(N\) applicants arrive and we interview them sequentially: once we interview and dismiss an applicant, we cannot hire her. The applicants all have differing skill levels, and we want to pick as qualified an applicant as we can. We want to find the optimal strategy for choosing whom to hire. We easily see that the optimal strategy is something like the following. We consider and reject the first \(K\) applicants. We then choose the first applicant who is better than all the preceding ones. Thus, our problem reduces to finding the optimal value for \(K\). We will do so in a way that maximizes the probability that the most qualified secretary is selected. We thus have the probability: \[ P(\mathrm{best\, is\, chosen})=\sum_{n=1}^{N}P(\mathrm{n^{th}\, is\, chosen} \cap \mathrm{n^{th}\, is\, best}) \\ P(\mathrm{best\, is\, chosen})=\sum_{n=1}^{N}P(\mathrm{n^{th}\, is\, chosen}| \mathrm{n^{th}\, is\, best})P(\mathrm{n^{th}\, is\, best}) \] We then note that each applicant in line is the best applicant with equal probability. That is, \(P(\mathrm{n^{th}\, is\, best})=\frac{1}{N}\). Also, we can find the conditional probabilities. If \(M \leq K\), then \(P(\mathrm{M^{th}\, is\, chosen}| \mathrm{M^{th}\, is\, best})=0\). If the \((K+1)\)th applicant is best, she will certainly be chosen, that is \(P(\mathrm{(K+1)^{th}\, is\, chosen}| \mathrm{(K+1)^{th}\, is\, best})=1\). Also, we find that \(P(\mathrm{(K+m)^{th}\, is\, chosen}| \mathrm{(K+m)^{th}\, is\, best})=\frac{K}{K+m}\), as that is the chance that the second-best applicant among the first \(K+m\) applicants is in the first \(K\) applicants. We thus have \[ P(\mathrm{best\, is\, chosen})=\frac{K}{N}\sum_{n=K+1}^{N}\frac{1}{n} \] Let us assume we are dealing with a relatively large number of applicants. In that case, we can approximate \(\sum_{n=A+1}^{B}\frac{1}{n} \approx \ln \left(\frac{B}{A} \right )\). Thus \[ P(\mathrm{best\, is\, chosen})=\frac{K}{N}\ln \left(\frac{N}{K}\right )=-\frac{K}{N}\ln \left(\frac{K}{N}\right ) \] If we then let \(x=\frac{K}{N}\), we just need to maximize \(-x\ln(x)\), which happens at \(x=e^{-1}\). From this, we find that \(P(\mathrm{best\, is\, chosen})=e^{-1}\). Thus, the best strategy is to interview and reject the first \(36.8 \%\) of the applicants, and then choose the next applicant who is better than all the preceding ones. This will get us the best applicant with a probability of \(36.8 \%\).

A related problem involves finding a strategy that minimizes the expected rank of the selected candidate (the best candidate has rank 1, the second best rank 2, etc.). Chow, Moriguti, Robbins and Samuels have found that the optimal strategy involves the following (in the limit of large \(N\)): skip the first \(c_0 N\) applicants, then, for all applicants before the number \(c_1 N\), we stop looking if the applicant is the best so far. If we have not yet selected an applicant, we choose the best or second best so far before the number \(c_2 N\). If we have not yet selected an applicant, we choose the best or second best or third best so far before the number \(c_3 N\). And so on. By choosing the \(c_n\) optimally, we can get an expected rank of \(3.8695\). This is quite surprising: we can expect an applicant in the top 4, even among billions of applicants!
The optimal values for the \(c_n\) are \[ c_0=0.2584... \\ c_1=0.4476... \\ c_2=0.5639... \] The general formula for \(c_n\) is \[ c_n=\prod_{k=n+2}^{\infty}\left ( \frac{k-1}{k+1} \right )^{1/k}=\frac{1}{3.86951924...}\prod_{k=2}^{n+1}\left ( \frac{k+1}{k-1} \right )^{1/k} \]