Download 2 Introduction to Probability Stirling numbers of the second kind

Introduction to Probability
MSIS 575
Class notes : September 25, 2000
Scribe : Daniel Josiah-Akintonde
Stirling numbers of the second kind
Stirling numbers of the second kind are defined as the number .Of ways of partitioning a
set of n elements into exactly r
None empty subsets, denoted by the following symbol :
n 
 
r 
Example 1: If we start with the set {a,b,c,d} and we wish to partition this into two
subsets we have
a | bcd , b | acd , c | abd , d | abc , ab | cd , ac | bd , ad | bc
Therefore,
4 
 = 7
2 
Note : These numbers are also analogous to the binomial coefficients.
Result 1: Recurrence relations for Stirling numbers :
n
n 
n 
   0 ,    1,   =
0
1 
2 
 n   n  n 

 =   ,   = 1
n 1  2  n
n 1
2
 1,
Result 2: Decomposition rule for Stirling numbers :
n 
n  1 n  1
  =r 
 + 

r 
 r  r  1
Definition : Onto functions and Stirling numbers :
Let D be a set (actually, domain of some set) and R the range of D under some
function f. The function f is said
to be an “onto” function if every element in R is an image of some element in
D.
f:DR
Let |D| = n and |R| = m. Number of onto functions is :
n
  m!
m 
Result 3: Number of ways of distributing n labeled (distinct) balls into m unlabeled
boxes with no box left empty is :
n
 
m 
Result 4: Number of ways of distributing n labeled (distinct) balls into m labeled boxes
with no box left empty is :
n
  m!
m 
Conditional Probability
Consider two events A and B within the universal set U :
Definition:
The probability of event A occurring , given that the event B has already
occurred is defined as follows:
P(A | B) =
P( A  B)
| A B |
=
P( B)
|B|
Note:
Intuitively, given that the event B has occurred, then, the universe is reduced
to set B.
Definition: Events A and B are said to be independent if
P( A | B)  P( A)
thus :
P(A  B) = P(A)P(B)
Note:
Independent events are not necessarily disjoint events.
Lemma 1:
P( A1  A2 | B) =
Lemma 2:
P ( A1 | B ) + P ( A2 | B ) - P( A1  A2 | B)
P(A  B) = P(A|B)P(B)
Repeated application of lemma 2 gives the following result :
P( A1  A2  A3  A4 .... An) =
P( A1 |
A  A  A .... A ) . P( A | A  A .... A )
2
3
4
n
2
3
4
n
…..
Lemma 3: If , A and B are sets, and set A is partitioned as follows:
A
A  A .........  A
1
2
n
, and
AA
i
j

for all
i j
Then,
P( A | B)  P( A1 | B)  P( A2 | B)  .......  P( An | B)
Example 1: A bag contains m balls of which r balls are red, and the remaining m-r
balls are blue. Two balls are drawn at
Random without replacement. What is the probability that both balls are
red?
Solution :
Apply lemma 2 stated above, thus :
Prob (first ball is red) =
r
m
Prob (second ball is red) =
r 1
m 1
Prob (both balls are red) =
r r 1
m m 1
Example 2: A bag contains n balls of which r balls are red. A selection of k balls is
made at random without replacement. What
is the probability that exactly p of the balls are red?
Solution :
The total number of ways of choosing k balls out of n balls is (the sample
space) given by :
n
 
k 
Number of ways of choosing a red ball is :
r
 
 p
Number of ways of NOT choosing a red ball is :
nr


k  p
The required probability is given by :
 r  n  r 
 

p  k  p 

Prob (n,r,k,p) =
=
n
 
k 
 k  n  k 
 

 p  n  p 
n
 
r
Note: The formula above is the so-called Hypergeometric Distribution.
Example 3 : Consider example 3 again, but now with replacement. Choose a ball, note
its color, and return ball to bag.
Solution :
Number of ways of choosing a single ball (sample space) is:
n
k
A favorable outcome is a sequence of R’s and B’s with exactly p R’s
(where R = red ball, and B = blue ball). The
probability Prob(favor) of a favorable outcome is given by:
r
Prob ( favor) =  
 n
 
But the total number of favorable outcomes is :
P
 nr 


 n 
K P
k
 
 p
Thus, the required probability is :
Total Prob (favor) =
 =
By letting :
k  r
   
 p  n
 
P
 nr 


 n 
K P
r
n
We have :
Total Prob (favor) =
K P
k P
  (1 )
 B ( p, k ,  )
 p
Note : The above is the so-called Binomial Distribution.
Random Variables
A random variable X on a sample space S is a function from S into the set R of real
numbers such that the pre-image of any interval of R is an event in S.
Definition : Let X be a finite random variable on a sample space S, that is X assigns
only a finite number of values to S. Let f be a
function which assigns probabilities to the points in the image of S under
X i.e.
f ( xk )  Pr ob( X  xk )  Pr ob({s  S : X (s)  xk})
This function f is called the probability distribution function of the random
variable X; and it satisfies the following two
conditions :
(1)
f ( xk )  0 ,
and
n
(2)
 f (x )  1
k 1
k
Example 1: Let S = {a , b , c , d , e} be a universal set (sample space) and Z = {1,2,3}
with X(a) = X(b) = 1, X(c) = X(e) = 2,
and X(d) = 3.
Let f : R  [0, 1]. f(k) = Prob(X=k) and
[note: f(1) + f(2) + f(3) = 1]
f(1) = 0.4, f(2) = 0.4, f(3) = 0.2
Example 2: Bernoulli random variable : Tossing a single coin. Define f such that:
f(H) = 1 and f(T) = 0. f(1) = p , f(0) = q = 1-p
Example 3: Binomial random variable : Tossing a coin n times . The sample space has
a total number of distinct points equal to :
n
2
Let X be a random variable giving number of heads (H’s). Then the
probability distribution function f on X is given by :
f(k) = Prob[X = k]
f(k) = B(k; n, p) = Prob[X = k]
2
n
B(k ; n, p)   
 p
K
p q
n K
k = 0, 1,