I am not interested in football, neither as a player (a primary school trauma when I was the last being picked!) or as a fan, contrary to my dad (who was a football referee in his youth) and my kids, but Gareth Roberts (University of Warwick) and Jeff Rosenthal wrote a paper on football draws […]
a football post?! — Xi’an’s Og
Suppose an urn contains N coloured balls, out of which M are blue and the rest are red in color. Suppose we draw a sample of ‘n’ number of balls from the urn at random and without replacements. Then if X a random variable denoting the number of blue balls drawn, then the probability that there will be x number of blue balls drawn is given by:
p(x ; N,M) = Pr[X=x] \\ \hskip{1.8cm}= \frac{\binom M x \binom {N-M}{n-x}}{\binom N n} \\ \hskip{3cm}; x=0,1,\dots,MEXPECTATION
E(X)= \sum_{x=0}^{M} x p(x;n,N,M) = \sum_{x=1}^{M} x \frac{\binom M x \binom {N-M}{n-x}}{\binom N n} \\
= \sum_{x=1}^{M} x \frac{n!}{N!(N-n)!)} \frac{M!}{x!(M-x)!} \frac{(N-M)!}{(n-x)!(N-M-n+x)!} \\
= \sum_{x=1}^{M} \frac{n!}{N!(N-n)!)} \frac{M(M-1)!}{(x-1)!(M-x)!} \frac{(N-M)!}{(n-x)!(N-M-n+x)!} \\
= \frac{M}{\binom N n} \sum_{x=1}^{M} \binom {M-1}{x-1} \binom {(N-1)-(M-1)}{(n-1)-(x-1)} \\
= \frac{M}{\binom N n} \binom {N-1}{n-1} \sum_{x=1}^{M} p(x-1 ; n=n-1 ,N=N-1 , M =M-1) \\
= \frac{M}{\binom N n} \binom {N-1}{n-1} \sum_{x=0}^{M-1} p(x ; n=n-1 ,N=N-1 , M =M-1) \\
= \frac{M n! (N-n)! (N-1)!}{(n-1)!(N-n)!N!} = n \frac{M}{N}
FACTORIAL MOMENTS
For any integer, s=0,1,…,M, we have,
\mu_{[s]}=E(X(X-1)\dots(X-s+1)) \\= \sum_{x=0}^{M} \frac{x!}{(x-s)!}p(x; n,N,M) \\
= \sum_{x=s}^{M} \frac{x!}{(x-s)!} \frac{\binom M x \binom {N-M}{n-x}}{\binom N n} \\
= \sum_{x=s}^{M} \frac{x!}{(x-s)!} \frac{\binom M x \binom {N-M}{n-x}}{\binom N n} \\
= \sum_{x=s}^{M} \frac{x!}{(x-s)!} \frac{M!}{x!(M-x)!} \frac{ \binom {N-M}{n-x}}{\binom N n} \\
= \frac{M!}{(M-s)!} \sum_{x=s}^{M} \frac{(M-s)!}{(x-s)!(M-x)!} \frac{ \binom {N-M}{n-x}}{\binom N n} \\
= \frac{M!}{(M-s)!} \frac{1}{\binom N n} \sum_{x=s}^{M} \binom {M-s}{x-s} { \binom {N-M}{n-x}} \\
= \frac{M!}{(M-s)!} \frac{1}{\binom N n} \sum_{x=0}^{M-s} \binom {M-s}{x} { \binom {(N-s)-(M-s)}{n-x-s}} \\
= \frac{M!}{(M-s)!} \frac{\binom {N-s}{n-s}}{\binom N n} \sum_{x=0}^{M-s} p(x ; n'=n-s,N'=N-s,M'=M-s) \\
= \frac{M!}{(M-s)!} \frac{(N-s)! n! (N-n)!}{N! (n-s)! (N-n)!)} \\
= \frac{M!}{(M-s)!} \frac{(N-s)!}{N!} \frac{n!}{(n-s)!} \\ VARIANCE:
E(X(X-1)) = \frac{M!}{(M-2)!} \frac{(N-2)!}{N!} \frac{n!}{(n-2)!} = \frac{n(n-1)M(M-1)}{N(N-1)} \\
------------------------------ \\
E(X^2) = E(X(X-1)) +E(X) \\
= \frac{n(n-1)M(M-1)}{N(N-1)} + \frac{nM}{N} \\
= \frac{nM}{N(N-1)} \left[ (n-1)(M-1) + (N-1) \right] \\
------------------------------ \\
V(X) = \frac{nM}{N(N-1)} \left[ (n-1)(M-1) + (N-1) \right] - \frac{n^2M^2}{N^2} \\
= \frac{nM}{N^2 (N-1)} \left[ N(n-1)(M-1) + N(N-1) - nM(N-1)) \right] \\
= \frac{nM}{N^2 (N-1)} \left[ nNM-nN-NM+N + N^2-N - nMN+nM \right] \\
= \frac{nM}{N^2 (N-1)} \left[ -nN-NM + N^2 +nM \right] \\
= \frac{nM}{N^2 (N-1)} (N-n)(N-M) \\
= n \frac{M}{N} \frac{N-M}{N} \frac{N-n}{N-1}Provided that the specified moment exists, it is straightforward (though tedious) to show via the factorial moments that…
Univariate Discrete Distributions – Johnson, Kemp, Kotz
\mu_3= \frac{nM(N-M)(N-n)(N-2M)(N-2n)}{N^3(N-1)(N-2)} \\\mu_4 = \frac{nM(N-M)(N-2n)}{N^3(N-1)(N-2)(N-3)} \left[\frac{}{} N(N+1)-6n(N-n) \right. \\
+ \left. \frac{M(N-M)}{N^2} \left( N^2(n-2) -Nn^2 +6n(N-n) \right) \right]It’s really tedious!!!
-------------------------------\\
\\
\mu_{[3]}=E(X(X-1)(X-2)) = \frac{n(n-1)(n-2) M(M-1)(M-2)}{N(N-1)(N-2)} \\
------------------------------ \\
\\
\mu'_3 = E(X^3) = \mu_{[3]} + 3 \mu'_2 - 2 \mu'_1 \\
= \frac{n(n-1)(n-2) M(M-1)(M-2)}{N(N-1)(N-2)} + 3 \frac{n(n-1)M(M-1)}{N(N-1)} -2 \frac{nM}{N} \\
------------------------------
\mu_3 = \mu'_3 -3 \mu'_2 \mu_1 +2 (\mu'_1)^3 \\
= \mu_{[3]}+3(\mu_2 + \mu^2) -2\mu-3 (\mu_2 + \mu^2) \mu +2 (\mu'_1)^3 \\
= ( \mu_{[3]} +3 \mu^2 -2\mu + \mu^3) +3\mu_2 -3\mu_2 \mu \\ \mu_{[3]} +3 \mu^2 -2\mu + \mu^3 = \frac{n(n-1)(n-2)M(M-1)(M-2)}{N(N-1)(N-2)} \\ \hskip{3cm}+ \frac{3n^2M^2-2nMN^2-n^3M^3}{N^3} \\
=\frac{nM}{N^3(N-1)(N-2)} \left( n^2M^2N^2-3nM^2N^2+2M^2N^2 -3n^2MN^2 + 9nMN^2 \right. \\ \left.-6MN^2 +2n^2N^2 -6nN^2 +4N^2 + 3nMN^3-9nMN^2 +6nMN -2N^4 \right. \\ \left. +6N^3 -4N^2 -n^2M^2N^2 +3n^2M^2N -2n^2M^2 \right) \\
=\frac{nM}{N^3(N-1)(N-2)} \left[ -3nM^2N(N-n)+3nMN^2(N-n) \right. \\ \left. -6MN(N-n) +6N^2(N-n) \right. \\
\left. +2M^2(N^2-n^2)-2N^2(N^2-n^2) \right)\\
=\frac{nM}{N^3(N-1)(N-2)} \left( (N-n) \left[ 3nMN(N-M) +6N(N-M) \right. \right. \\ \left. \left.-2(N+n)(N^2-M^2) \right] \right) \\
=\frac{nM(N-n)(N-M)}{N^3(N-1)(N-2)} \left( 3nMN +6N - 2(N+n)(N+M) \right)\mu_3 = \frac{nM(N-n)(N-M)}{N^3(N-1)(N-2)} \left( 3nMN +6N - 2(N+n)(N+M) \right) \\
+ 3n \frac{M}{N} \frac{N-M}{N} \frac{N-n}{N-1} -3 n^2 \frac{M^2}{N^2} \frac{N-M}{N} \frac{N-n}{N-1} \\
= \frac{nM(N-n)(N-M)}{N^3(N-1)(N-2)} \left[ 3nMN +6N - 2N^2 -2nN -2NM -2nM \right. \\ \left. +3N^2 -6N -3nMN +6nM \right] \\
= \frac{nM(N-M)(N-n)(N-2M)(N-2n)}{N^3(N-1)(N-2)} \\
--------------------------------\mu_4 = \frac{nM(N-M)(N-2n)}{N^3(N-1)(N-2)(N-3)} \left[\frac{}{} N(N+1)-6n(N-n) \right. \\
+ \left. \frac{M(N-M)}{N^2} \left( N^2(n-2) -Nn^2 +6n(N-n) \right) \right]Definition 1 (Vector Space). A vector space over a field F is a quadruple (V, +, ., F )) satisfying the following axioms for all α, β ∈ F and x, y, z ∈ V :
1. (V, +) is a commutative group, that is,
(a) ’+’ is map from V x V to V. [Closure]
(b) (x + y) + z = x + (y + z). [Associative]
(c) there exists an element 0 of V such that x + 0 = 0 + x = x. [Existence of 0]
(d) for each x in V there exists an element −x in V such that x + (−x) = (−x) + x = 0.
[Existence of Negative]
(e) x + y = y + x. [Commutative]
2. ’.’ is a map from F xV to V. [Closure wrt . ]
3. α.(β.x) = (α.β).x.
4. 1.x=x
5. (α + β).x = (α.x) + (β.x). [Distributivity]
6. α.(x + y) = (α.x) + (α.y). [Distributivity]
Remark 1. The elements of V are called vectors and the elements of F are called scalars. F is called the base field or ground field of the vector space. ’0’ in axiom 1(c) is called the null vector or the zero vector. It is to noted that, the bold faced Roman letters are treated as vectors whereas the lower case Greek letters are treated as scalars throughout the course.
Suppose a normal QQ-plot is drawn using a reasonably large sample x1,x2,…xn from an unknown probability distribution. For which of the following distribution would you expect the QQ-plot to be convex (J-shaped)?
1. Beta(5,1)
2. Exponential(1)
3. Uniform(0,1)
4. Geometric(1/2)
The following facts would help to guess the answer: