Download Poisson Distribution

Handout Ch 5 Review
Bernoulli Distribution

A random variable X has a Bernoulli distribution if
Pr(X = 1) = p and Pr(X = 0) = 1– p = q

The p.m.f. of X can be written as
 p x q1 x for x  0, 1
f ( x | p)  
otherwise
 0
E ( X )  1 p  0  q  p

E ( X 2 )  12  p  0 2  q  p
Var( X )  E ( X 2 )  E ( X )  pq
2
 (t )  E (e tX )  pet  q for    t  
2
Jia-Ying Chen
Binomial Distribution



If the random variable X1, …, Xn form n Bernoulli trials with parameter
p, and if X  X 1    X n , then X has a binomial distribution.
The p.m.f. of X can be written as
 n  x n  x
for x  0, 1, 2,, n
  p q
f ( x | n, p)   x 

0
otherwise

n
E ( X )   E ( X i )  np
i 1
n
Var ( X )   Var ( X i )  npq
i 1
n
 (t )  E (e )   E (etXi )  ( pet  q) n
tX
i 1

3
If X1, …, Xk are independent random variables and if Xi has a binomial
distribution with parameters ni and p, then the sum X 1   X k has a
Jia-Ying Chen
binomial distribution with parameters n  n1    nk and p.
Example 1 (5.2.10)

4
The probability that each specific child in a given
family will inherit a certain disease is p. If it is known
that at least one child in a family of n children has
inherited the disease, what is the expected number of
children in the family who have inherited the disease?
Jia-Ying Chen
Solution
5
Jia-Ying Chen
Poisson Distribution

X has a Poisson distribution with mean l if the p.m.f.
of X has:
e l lx
for x  0,1, 2,

f ( x | l )   x!
 0
otherwise

lx
Recall that e  
x 0 x!
l

lx
Therefore,  f ( x | l )  e   e l e l  1
x 0
x 0 x!

6

 e  l l x 1
 e l l y
e l lx
E ( X )   xf ( x | l )   x
 l
l
l
x!
y!
x 0
x 1
x 1 ( x  1)!
y 0


l

Jia-Ying Chen
Poisson Distribution

 E X ( X  1)   x( x  1) f ( x | l )
x 0
 e l lx  2
 e l l y
e l lx
  x( x  1)
 l2 
 l2 
 l2
x!
x2
x  2 ( x  2)!
y 0 y!

E  X ( X  1)  E ( X 2 )  E ( X )  E ( X 2 )  l
E ( X 2 )  l2  l
Var ( X )  E ( X 2 )  [ E ( X )]2  l2  l  l2  l

The moment generating function
 (let ) x
t
t
etxe l lx
l
 (t )  E (e )  
e 
 e l ele  el ( e 1)
x!
x!
x 0
x 0
tX
7

Jia-Ying Chen
Poisson Distribution

If the random variables X1, …, Xk are independent and if Xi has a
Poisson distribution with mean li , then the sum X 1   X k has
a Poisson distribution with mean l1   lk .
Proof: Let  i (t ) denote the m.g.f. of Xi and  (t ) denote the m.g.f.
of the sum
X1    X k .
k
k
i 1
i 1
 (t )   i ( t )   e

8
li ( et 1)
e
( l1  lk )( et 1)
Example 5.4.1: The mean number of customers who visit the
store in one hour is 4.5. What is the probability that at least 12
customers will arrive in a two-hour period?
X = X1 + X2 has a Poisson distribution with mean 9.
So, Pr( X  12)  0.1970.
Jia-Ying Chen
Poisson Approximation to Binomial
Distribution


When the value of n is large and the value of p is close to 0,
the binomial distribution with parameters n and p can be
approximated by a Poisson distribution with mean np.
Proof: For a binomial distribution with l= np, we have
n(n  1)  (n  x  1) x
f ( x | n, p) 
p (1  p) n  x
x!
lx n n  1 n  x  1 l n
l

.
...
(1  ) (1  )  x
x! n n
n
n
n
n n 1 n  x 1
l x
.
...
(
1

) 1
As n  , then p  0 , then nlim
n
n
n
 n
l n l
(
1

) e
Also, nlim
n

e  l lx
 f ( x | n, p) 
(for every fixed positive integer x)
x!
9
l
For x  0, f ( x | n, p)  (1  p) n  (1  ) n .
n
Jia-Ying Chen
羅必達法則

當x→a時，函數f(x)及g(x)都趨於零；
在點a的附近鄰域內，f’(x)及g’(x)都存在，且g’(x) ≠0

lim

x a
f ' ( x)
存在(或為無窮大)，
g ' ( x)
 則 lim
x  a

10
f ( x)
f '( x)
 lim
g ( x) x a g '( x)
各種形式：0/0，∞/∞，0× ∞， ∞- ∞，00， ∞0，1∞
Jia-Ying Chen
Example 2 (5.4.8)

11
Suppose that X1 and X2 are independent random
variables and that Xi has a Poisson distribution with
mean li (i=1,2). For each fixed value of k (k=1,2,…),
determine the conditional distribution of X1 given that
X1+X2=k
Jia-Ying Chen
Solution
12
Jia-Ying Chen
Example 3 (5.4.14)

13
An airline sells 200 tickets for a certain flight on an
airplane that has only 198 seats because, on the
average, 1 percent of purchasers of airline tickets do
not appear for the departure of their flight. Determine
the probability that everyone who appears for the
departure of this flight will have a seat
Jia-Ying Chen
Solution
14
Jia-Ying Chen
Geometric Distribution

Suppose that the probability of a success is p, and the
probability of a failure is q=1 – p. Then these experiments
form an infinite sequence of Bernoulli trials with
parameter p.

Let X = number of failures to first success.
f ( x | p ) = pqx
for x = 0, 1, 2, …

Let Y = number of trials to first success.
f ( y | p ) = pqy–1
q
1
E( X ) 
E (Y ) 
p
p
q
q
Var( X )  2
Var(Y )  2
p
p

15
Jia-Ying Chen
The m.g.f of Geometric Distribution

If X1 has a geometric distribution with parameter p,
then the m.g.f.

 1 (t )  E (e )  p  (qet ) x
tX1
x 0

It is known that
 1 (t ) 

p
1  qet
E ( X 1 )   1(0) 
 x 
x 0
1
1
for t  log (1/q)
q
p
Var ( X 1 )   1(0)  [ 1(0)]2 
16
(for 0    1)
q
p2
Jia-Ying Chen
Example 4
假設某工廠產出不良品的機率為0.1，請問
(1)在發現第一個不良品前有10個良品產出的機
率為何？
(2)直到產出10 個或10個以上產品才發現第一個
不良品之機率
為何？
17
Jia-Ying Chen
Solution
令X 表示發現第一個不良品時檢查出良品的產品數，則
(1)發現第一個不良品前有10個良品產出的機率為
P( X  10)  0.1 (0.9)10  0.03487
(2)直到產出10 個或10個以上產品才發現第一個不良品之
機率， P( X  9)  0.1 (0.9)9  0.1 (0.9)10  0.1 (0.9)11 
 0.1 (0.9)9 (1  0.9  0.92 
1
9
 0.1 (0.9) 
1  0.9
 (0.9)9  0.3874
Or
18
)
P( X  9)  P(前 9 個產品均為良品)
 (0.9)9  0.3874
Jia-Ying Chen
Normal Distribution


19
There are three reasons why normal distribution is important
 Mathematical properties of the normal distribution have simple forms
 Many random variables often have distributions that are approximately
normal
 Central limit theorem tells that many sample functions have
distributions which are approximately normal
2


1
1
x




2
exp  
 
The p.d.f. of a normal distribution f ( x |  , ) 
1/ 2
2

(2 ) 

 

Jia-Ying Chen
極座標



直角座標與極座標的轉換
∫ ∫dxdy=rdrdθ
X=r*cosθ，y=r*sinθ
(x,y)

20
Ex:
x2+y2≦1
r
Jia-Ying Chen
The m.g.f. of Normal Distribution


 (t )  E (e )  
tX

1
( x   )2 
exp tx 
dx
2 
2 
2 

( x  u)2
1 2 2 [ x  (    2t )]2
since tx 
t  t 
2
2
2
2 2
 [ x  (    2t )]2 
1 22  1
so  (t )  exp( t   t ) 
exp 
 dx
2
2
2 
2


1
 exp(t   2t 2 )
2
E (X )   (0)  
Var ( X )   (0)  [ (0)]2   2
21
Jia-Ying Chen
Properties of Normal Distribution
22

If the random variables X1, …, Xk are independent and if Xi has a
normal distribution with mean mi and variance si2, then the sum
X1+ . . .+ Xk has a normal distribution with mean m1 + . . .+ mk and
variance s12 + . . .+ sk2.
k
k
1 2 2
 k   k 2  2 

Proof:  (t )   i (t )   exp  i t   i t   exp    i t     i t 
2
i 1
i 1


 i 1   i 1  

The variable a1x1 + . . .+ akxk+ b has a normal distribution with mean
a1m1 + . . .+ akmk + b and variance a12s12 + . . .+ ak2sk2

Suppose that X1, …, Xn form a random sample from a normal
distribution with mean m and variance s2 , and let X n denote the
sample mean. Then X n has a normal distribution with mean m and
s2/n.
Jia-Ying Chen
Example 5 (5.6.11)

Suppose that a random sample of size n is to be taken from
a normal distribution with mean μ and standard deviation 2.
Determine the smallest value of n such that
Pr( X n    0.1)  0.9
23
Jia-Ying Chen
Solution
24
Jia-Ying Chen
Example 6

Suppose that the joint p.d.f. of two random variables
X and Y is as follows. Show that these two random
variable X and Y are independent.
1 (1/ 2)( x2  y2 )
f ( x, y) 
e
for -   x  
2
and - <y<
25
Jia-Ying Chen
Solution

26
Recall that suppose X and Y are random variables that have a
continuous joint p.d.f. Then X and Y will be independent if
and only if, for    x   and    y  , f ( x, y )  g1 ( x) g 2 ( y )
1 (1/ 2)( x2  y2 )
1  12 x2
1  12 y2
f ( x, y) 
e
(
e )(
e
)  g1 ( x) g 2 ( y)
2
2
2

And

Therefore, X and Y are independent
Jia-Ying Chen
Exponential Distribution

A gamma distribution with parameters a = 1 and b is an exponential
distribution. f  x |    e   x (for x  0)

A random variable X has an exponential distribution with parameters b has:
1
1

E  X   , Var X   2 ,  t  
(for t   )

 t


Pr  X  t   t e   x dx  e   t

27

Pr  X  t  h  e   t  h 
Pr  X  t  h | X  t  
   t  e   h  Pr  X  h 
Pr  X  t 
e
Memoryless property of exponential distribution
Jia-Ying Chen
Life Test

Suppose X1, …, Xn denote the lifetime of bulb i and form a random
sample from an exponential distribution with parameter β. Then the
distribution of Y1=min{X1, …, Xn} will be an exponential distribution
with parameter n β.
Proof: Pr Y1  t   Pr  X 1  t , ..., X n  t 
 Pr  X 1  t     Pr  X n  t   e   t    e   t  e  n t

Determine the interval of time Y2 between the failure of the first bulb and
the failure of a second bulb.



28
Y2 will be equal to the smallest of (n-1) i.i.d. r.v., so Y2 has an
exponential distribution with parameter (n-1) β.
Y3 will have an exponential distribution with parameter (n-2) β.
The final bulb has an exponential distribution with parameter β.
Jia-Ying Chen
Physical Meaning of Exponential
Distribution
29

Following the physical meaning of gamma distribution, an
exponential distribution is the time required to have for the 1st
event to occur, i.e.,
Pr  X  t   e   t where β is rate of event.

In a Poisson process, both the waiting time until an event occurs
and the period of time between any two successive events will
have exponential distributions.

In a Poisson process, the waiting time until the nth occurrence
with rate b has a gamma distribution with parameters n and b.
Jia-Ying Chen
Example 7 (5.9.10)

30
Suppose that an electronic system contains n
similar components that function independently
of each other and that are connected in series so
that the system fails as soon as one of the
components fails. Suppose also that the length of
life of each component, measured in hours, has an
exponential distribution with mean μ. Determine
the mean and the variance of the length of time
until the system fails.
Jia-Ying Chen
Solution
31
Jia-Ying Chen
Example 8 (5.9.11)

32
Suppose that n items are being tested simultaneously,
the items are independent, and the length of life of
each item has an exponential distribution with
parameter β. Determine the expected length of time
until three items have failed. Hint: The required
value is E(Y1+Y2+Y3)
Jia-Ying Chen
Solution
33
Jia-Ying Chen