Starting from this section, we will investigate the distribution of the leading statistic of the bounded deviated permutation.
Assume the permutation Sn+1`;r are uniformly distributed. We de…ne the random variable Xn on the set of all (`; r)-bounded deviated permutations Sn+1`;r by Xn = k if
1 = k + 1 for = 1 2 n2 Sn+1`;r : We set
n;k = n
2 Sn+1`;r : 1 = k + 1o Enumeration on Sn+1`;r yields the following counting formula:
n;k = n likewise for the sequence (bn) :
Now we de…ne a bivariate generating function
A(z; u) =
and we will use (2) to compute the expected value and the variance for the random variable Xn:
The expected value and variance of Xn can be computed as
n= [zn]@u@ A(z; u)ju=1
4 Convergence to a Normal Distribution
Due to the number n
2 Sn+1`;r : 1 = k + 1o
is a natural combinatoric object, we guess the the random variable Xn = kif 1 = k + 1for = 1 2 n2 Sn+1`;r converges to the Gaussian distribution. And we need the following theorem:
Theorem 6 (Generalized quasi-powers, [3], p.690)
Assume that, for u in a …xed neighbourhood of 1, the generating function pn(u) of a non-negative discrete random variable (supported by Z 0) Xn admits a representation of the form
pn(u) = exp (hn(u)) (1 + o (1)) ;
uniformly with respect to u; where each hn(u) is analytic in : Assume also the conditions, h0n(1) + h00n(1) ! 1 and h000n (u)
(h0n(1) + h00n(1))32 ! 0;
uniformly for u 2 : Then, the random variable Xn= Xn h0n(1)
(h0n(1) + h00n(1))12
converges in distribution to a Gaussian with mean 0 and variance 1.
In generally, considering the exact form pn(u) = exp (hn(u)) ; we have p0n(u) = h0n(u) exp (hn(u)) ;
p00n(u) = h00n(u) exp (hn(u)) + (h0n(u))2exp (hn(u)) ;
p000n(u) = h000n (u) exp (hn(u)) + 3h0n(u) h00n(u) exp (hn(u)) + (h0n(u))3exp (hn(u)) ; hence
2
n = h0n(1) + h00n(1) = p0n(1)
pn(1) +p00n(1) pn(1)
p0n(1) pn(1)
2
and
h000n (u) = p000n(u)
pn(u) 3 p0n(u)p00n(u)
(pn(u))2 + 2 p0n(u) pn(u)
3
We will show that n;k = nk akbn k satisfy the desire form of the theorem on the next section.
5 Main Results
In this section, we will show three cases:(`; r) = (1; 2); (1; 3) ; and (2; 2) .
As a motivation, let us look at the case (`; r) = (1; 1) …rst. Given a permutation
= 1 2 n+12 Sn+1;if we assign the value k + 1 to the position 1; then we can pick k positions in the remainnig to put the numbers from 1 to k: The way is nk and each arrangement is unique. So we have n;k = nk . When n tends to in…nity, the random variable Xn= k really converges to Gaussian distribution.
In our next three cases, the combinatoric number n;k = nk akbn k has no closed form, so we need to calculate its asymptotic. And the generating functions are all admissible in the sense of Hayman’s theorem, because they are of the form f (z) = eF (z);where F (z) is a polynomial whose coe¢ cient are all positive integers. The Hayman’s theorem is stated as following:
Theorem 7 (Hayman formula, [4], [5], p.183) Let f (z) =P
anzn be an admissible function. Let rn be the positive real root of the equation a (rn) = n, for each n = 1; 2; ; where a (rn) is given by a (r) = rff (r)0(r). Then
an f (rn) rnnp
2 b (rn) as n ! 1;
where b (rn) is given by b (r) = ra0(r) :
5.1 Case 1 (`; r) = (1; 2):
We set
A(z; u) = exp((1 + u)z + z2 2);
the expected value n can be computed as
n = [zn]@u@ A(z; u)ju=1
[zn]A(z; 1) = [zn]z exp(2z + z22)
[zn] exp(2z +z22) = [zn 1] exp(2z + z22) [zn] exp(2z +z22) :
Now we use the Hayman formula to compute the asymptotic formula for the coe¢ cients of the formula exp(2z +z22) :
let
f (z) = exp 2z + z2 2 ; then
f0(z) = (2 + z) exp 2z +z2 2 ; then
a (r) = rf0(r)
f (r) = 2r + r2; 6
thus
a0(r) = 2 + 2r;
we …rst solve the equation
2rn+ rn2 = n;
The asymptotic formula for the coe¢ cients of the formula exp(2z+z22)can be computed
where C is the counterclockwise oriented contour within center 0 and radius R.
Note that
Since
h000n (u) = p000n(u)
pn(u) 3 p0n(u)p00n(u)
(pn(u))2 + 2 p0n(u) pn(u)
3
= G (u; R) ; h000n (u)
(h0n(1) + h00n(1))32
= O n45 ! 0 as n ! 1:
Hence by generalized quasi-powers theorem, we have Theorem 8 On Sn+11;2 ; the leading statistics
Xn= 1 1 has the mean n p
n 1 and the variance 2n p
n 32; and it admits a limit Gaussian law.
We verify above consequences by computer plotting:
The red points are P (Xn) in the cases (`; r) = (1; 2) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.
Figure 5.1 (`; r) = (1; 2) ; n = 10000
5.2 Case 2 (`; r) = (1; 3): We set
A(z; u) = exp((1 + u)z +z2 2 +z3
3);
the expected value n can be computed as
n= [zn]@u@ A(z; u)ju=1
[zn]A(z; 1) = [zn]z exp(2z + z22 + z33)
[zn] exp(2z + z22 +z33) = [zn 1] exp(2z + z22 +z33) [zn] exp(2z +z22 +z33) :
Now we use the Hayman formula to compute the asymptotic formula for the coe¢ cients of the formula exp(2z +z22 + z33) :
let
f (z) = exp 2z + z2 2 + z3
3 ; then
f0(z) = 2 + z + z2 exp 2z + z2 2 +z3
3 ; then
a (r) = rf0(r)
f (r) = 2r + r2 + r3; thus
a0(r) = 2 + 2r + 3r2: We …rst solve the equation
2rn+ rn2 + rn3 = n:
The Hayman formula is useful, but we face a problem at …rst: solve the equation a (rn) = n: It always in the form C1z1+ C2z2+ + Ckzk = n; but we have no formula to solve it. So we transform the equation to the speci…c form u = t (u), by proper substitution, then by the Lagrange inversion formula :
Theorem 9 (Lagrange Inversion Formula, [1], p.117) Suppose F (z) = zG (F (z)) ; G (0) 6= 0: Then
[zn] F (z) = 1
n zn 1 G (z)n (n 1) : Now we can get the asymptotic we want:
Let
u = (rn) 1; then
2u 1+ u 2+ u 3 = n;
implies
2u2+ u + 1 = u3n:
10
Thus Now we applied the Lagrange inversion formula
[tn] u = 1
Thus
where C is the counterclockwise oriented contour within center 0 and radius R.
Note that
bounded with respect to n and R, so does p0n(u); p00n(u);and p000n(u):
Since
h000n (u) = p000n(u)
pn(u) 3 p0n(u)p00n(u)
(pn(u))2 + 2 p0n(u) pn(u)
3
= G (u; R) ; h000n (u)
(h0n(1) + h00n(1))32
= O n23 ! 0 as n ! 1:
Hence by generalized quasi-powers theorem, we have Theorem 10 On Sn+11;3 ;the leading statistics
Xn= 1 1 has the mean n p3
n 13 and the variance 2n p3
n 13;and it admits a limit Gaussian law.
We verify above consequences by computer plotting:
The red points are P (Xn) in the cases (`; r) = (1; 3) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.
Figure 5.2 (`; r) = (1; 3) ; n = 10000
5.3 Case 3 (`; r) = (2; 2) :
We set
A(z; u) = exp (1 + u) z + 1 + u2 z2 2 ; the expected value n can be computed as
n = [zn]@u@A(z; u)ju=1 [zn]A(z; 1)
= [zn] (z + z2) exp(2z + z2) [zn] exp(2z + z2) :
Here we use the Heyman formula again to get the asymptotic formula for the coe¢ cients of the formula exp(2z + z2) :
We solve the equation at …rst,
2rn+ 2r2n= n;
Hence
The asymptotic formula for the coe¢ cients of the formula exp(2z + z2) can also be
Thus
where C is the counterclockwise oriented contour within center 0 and radius R.
Note that
bounded with respect to n and R, so does p0n(u); p00n(u);and p000n(u):
Since
h000n (u) = p000n(u)
pn(u) 3 p0n(u)p00n(u)
(pn(u))2 + 2 p0n(u) pn(u)
3
= G (u; R) ; h000n (u)
(h0n(1) + h00n(1))32
= O n21 ! 0 as n ! 1:
Hence by generalized quasi-powers theorem, we have Theorem 11 On Sn+12;2 ;the leading statistics
Xn= 1 1
has the mean n n2 and the variance 2n 12n p42n; and it admits a limit Gaussian law.
We verify above consequences by computer plotting.
The red points are P (Xn) in the cases (`; r) = (2; 2) ; and the blue curve is Gaussian distribution with n = 10000, centered at its peak.
Figure 5.3 (`; r) = (2; 2) ; n = 10000
6 Conclusion and Further Discussion
In this thesis, we have already use di¤erent ways to prove that the leading statistic of three special cases Sn+11;2 ; Sn+11;3 ; and Sn+12;2 , converge to normal distributions. For general cases, such like Sn+12;3 , we believe that they also converge to normal distributions in the same way. We will keep verifying these cases. In the future, a good direction to similar question is to prove other statistics, such like the second statistic, the third statistic, are all converge to normal distributions, or other types of distributions.
References
[1] M. Aigner, A Course in Enumeration, Springer-Verlag Berlin Heidelberg 2007
[2] Sen-Peng Eu, Yen-Chi R. Lin, Yuan-Hsun Lo, Bounded deviated permutations, preprint
[3] Philippe Flajolet, Robert Sedgewick, Analytic Combinatorics, Cambridge University Press 2009
[4] Hayman, Walter, A generalisation of Stirling’s formula, Journal für die reine und angewandte Mathematik 196 (1956), 67-95
[5] Herbert S. Wilf, Generatingfunctionology, 2nd edition, 1994
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