Download Poisson distribution Poisson distribution

Random Points and Random
tools
Dr Zvi Lotker
Two that are almost the same
‫עודד‬
‫לוטקר‬
‫עודד‬
‫לוטקר‬
Using Poisson assumption



Assume we have an x~ Poisson[]
Let A[0,1]2 be an area.
Denote (A) to be the random variable
that counts the number of points in A.
P[   k ] 

e
 A
A
k
k!
The advantage of the Poisson assumption
is that the random variable are i.i.d if
(A), (B) if AB=.
Poisson distribution
Poisson distribution
P[   k ] 
e
 A
A
k
k!
Some fact on Poisson
Poisson property
Parameters
Support
Probability mass function (pmf)
Cumulative distribution function (cdf)
Mean
Median
Mode
Variance
Skewness
Excess kurtosis
Entropy
Moment-generating function (mgf)
Characteristic function
e  A A
P[   k ] 
k!
Poisson and the
binomial distribution.


The binomial dis is
Usually when p is fix


we use the normal dis
But when p~O(1/n)

we use Poisson
distribution
k
Tail of the Poisson[]

Let X be a Poisson dis with parameter 



If t> then P[Xt]≤e-(e)t/tt
Proof
For any a>0 and t>,




P[Xt]=P[Exp(aX)Exp(at)] ≤E[exp(aX)]/Exp(aX)
Now E[exp(aX)] is the moment generating
E[exp(aX)]=Exp((Exp(a)-1)
Choosing a=log(t/) give the
P[Xt] ≤e-t--tlog(t/) ≤e-(e)t/tt
Poisson and the
uniform distribution


Thm:
 If Yi~Pos[i] follow a Poisson distribution
with parameter i and Yi are independent,
then Y=Yi ~Pos[ i]
Thm:
 Let Yi~Pos[i], assume that Yi=k then
 (Y1,Y2,…,Yn)~ Uniform i.e.
k




n
k1 , k1 ,..., kn 

P[Y1  k1 ,, Yn  k n |  Yi  k] 
n
n
i 1
Two that are almost the same

Any event that takes place with
probability p in the Poisson case takes
place with probability at most pem in
the exact case.
Poisson and the
uniform distribution

Suppose that m balls are thrown into n bins i.u.



be the number of balls in the i bin
be the independent Poisson var with mean m/m
X im
Yi m
Theorem


Let
Let
Let f(xm1,xm2,…,xmn) be a non negative function then
E f X1m , X 2m ,..., X nm  e mE f Y1m , Y2m ,..., Ynm


,..., Y    E  f Y , Y ,..., Y  |  Y  k P[ Y  k ]


m
1
m
2
E f Y ,Y
m
n
k 1

n
m
1

m
2
m
n
n
m
i 1
i
m

i 1
i
n
 m m
 n m
m
m
 E  f Y1 , Y2 ,..., Yn |  Yi  m P[ Yi  m]
i 1

 i 1



 E f X , X ,..., X

m
1
m
2
 E f X , X ,..., X
m
1
m
2
m
n
P[ Y
m
n

n
i 1
i
m
 m]


m me m
1
 E f X 1m , X 2m ,..., X nm
m!
e m


Birthday Paradox

The birthday problem is: What is the
probability of two or more people out of
a group of m peoples do have the same
birthday?
Birthday Paradox

What is the prob that no two students have the
 365 
same Birthday is

m!


P
Note that if m>365 this is 0
m 
365m
We can also by considering one person at a time:

If we take k to be small we get that
We can use and get  1  j    e
m 1

 k
1    e
 n
k
 
n
j 1
m 1

n
 j
 
n
m 1
j

1




n
j 1 
j 1
 m 1  j 
 exp    
 j 1  n 
e
e
 m ( m 1) 


 2n 
 m2

 2n




Birthday Paradox

Hence the value for m at which
the prob that all people have
different birth is ½ is
approximately:
m2=ln
2
In case of n=365, m=22.5
 j
m 1
j  m 1  n 

1     e

n  j 1
j 1 
 m 1  j 
 exp    
 j 1  n 
e
e
 m ( m 1) 


 2n 
 m2

 2n




Balls into bins

Suppose we have n bins and we
randomly throw m balls into them

We consider the case n=m.
Balls into bins

Suppose we have n bins and we
randomly throw m balls into them



We consider the case n=m.
What is the maximum number of balls
in any bin.
Theorem: The prob that the max load
bin has more them 3 ln n/ln(ln n) is less
then 1/n for big n.
Balls into bins

Theorem: The prob that the max load bin has
more them 3 ln n/ln(ln n) is less then 1/n for
big n.
The prob that bin 1 have at list M ball is  Mn  1n 
 
Now we use the inequality
The second inequality follows from  Mn  1n   M1 !   Me 
M


M


k
i
k
k
   ek
k!
i i!

M
Balls into bins

Theorem: The prob that the max load bin has
more them 3 ln n/ln(ln n) is less then 1/n for
big n.

 n  1 
  
 M  n 
The prob that bin 1 have at list M ball is
 n  1 
1  e 
   
 
Now we use the inequality
M
n
M
!
M 
  
Applying union bound for M> 3 ln n/ln(ln n)
M


M
M
 e 
 e ln ln n 
n   n

M 
 3 ln n 
 ln ln n 
 n

 ln n 
1

n
3 ln n / ln ln n
3 ln n / ln ln n

 e ln n e ln ln ln n ln ln n

3 ln n / ln ln n
M
Coupon Collectors problem



For n types of coupons, at each trial a coupon
is picked at uniform random and independent
probability .
Find the earliest time at which all n objects
have been picked at least once.
We need nlog(n) time.
Coupon Collectors problem

Random coupon sequence a1, a2, …, an

Ei = time to visit i+1st new couponn.

a1 a2 … a3 … … ai … ai+1 … … an

Assume we have i coupons and we are missing, n-i.
(n-i)/n is probability the next node chosen is
unvisited
 n/(n-i) is the expected timed until an unvisted
node is chosen
Cover time is


E1  E2    En 1 
n
n
n
 1


    n
   1  n log n
n 1 n  2
1
 n 1

Coupon Collectors problem

Theorem: Let  be the random time we
have all the coupons. Than for all c>0,


P[ >nlog(n) + c n]< Exp[-c]
Proof:




Let Ai be the event that we did not have
the i coupon at time T=nlog(n) + c n
P[Ai]<(1-1/n)^T
Using union bound the theorem follows.
P[Ai]<nExp[-log [n]+c]
Conclusion

Birthday Paradox:
n
log( n)
log(log( n))

Balls into bins:

Coupon Collectors:
n log( n)