Download Continuous distributions

Exponential Distribution.


= mean interval between consequent events
= rate = mean number of counts in the unit interval > 0
X = distance between events >0
X ~ e( )
f ( x)  e
 x
F ( x)  1  e
f(x)
E( X )  1/ 
 x
F(x)
V ( X )  1 / 2
Memoryless property
P( x  t1  t2 | x  t1 )  P( x  t2 )
Exponential and Poisson relationship
Unit Matching between x and
IME 312
!
  1/ 
Exponential Dist.
  1/ 
X ~ Expo(  )
f ( x)  e
Y ~ Pois ( )
 x
F ( x)  1  e
E( X ) 
 x
e 
f ( y) 
y!

1

V (X ) 
IME 312
Poisson Dist.
1
2
E (Y )  V (Y )  
y
Relation between
Exponential distribution ↔ Poisson distribution
Xi : Continuous random variable, time between arrivals,
has Exponential distribution with mean = 1/4
X 
X1=1/4
X2=1/2
0
X3=1/4
X4=1/8 X5=1/8
X6=1/2
1:00
Y1=3
X7=1/4
X8=1/4
i 1
12
X9=1/8 X10=1/8
2:00
Y2=4
Yi : Discrete random variable, number of arrivals per unit
of time, has Poisson distribution with mean = 4. (rate=4)
Y ~ Poisson (4)
IME 301 and 312

12
Xi

X11=3/8
1
4
X12=1/8
3:00
Y3=5
Y 
Y1  Y2  Y3
4
3
Continuous Uniform Distribution
a xb
x ~ u ( a, b)
1
f ( x) 
ba
0 xa

x a
F ( x)  
a xb
b  a
1 bx

ab
E( X ) 
2
IME 312
(b  a) 2
V (X ) 
12
f(x)
a
b
F(x)
a
b
Gamma Distribution
K = shape parameter >0
 = scale parameter >0
f ( x) 
 x
k

 x
e
( k )
(k )   x
0
k 1
k 1  x
e dx
For Gamma Function, you can use:
and if K is integer (k’) then:
IME 312
E( X )  k / 
Var( X )  k / 2
(1)  1
(k ' )  (k '1)!
Application of Gamma Distribution
K = shape parameter >0 = number of Yi added
= scale parameter >0 = rate

if X ~ Expo (  )
and Y = X1 + X2 + ……… + Xk
then Y ~ Gamma (k,
)

i.e.: Y is the time taken for K events to occur and X is the
time between two consecutive events to occur
IME 312
Relation between
Exponential distribution ↔ Gamma distribution
Xi : Continuous random variable, time between arrivals,
has Exponential distribution with mean = 1/4
X 
X1=1/4
0
X2=1/2
X3=1/4
X4=1/8 X5=1/8
X6=1/2
X7=1/4
1:00
X8=1/4

12
i 1
Xi
12
X9=1/8 X10=1/8
X11=3/8
2:00
Y1=1
Y2=7/8
Y3=1/2
Y4=3/4
Yi : Continuous random variable, time taken for 3 customers to arrive,
has Gamma distribution with shape parameter k = 3 and scale=4
IME 312

1
4
X12=1/8
3:00
Weibull Distribution
a = shape parameter >0
 = scale parameter >0
a 1  ( x ) a
f ( x)  a x e
a
f ( x)  0
F ( x)  1  e
IME 312
 ( x ) a
for
x0
for
x0
Normal Distribution
  x  
x ~ N ( , )
1
f ( x) 
e
 2
( x )2
z
~ N (0,1)

F ( x)   ( z )
IME 312
 0
    
V (X )  
E( X )  
(x  )
2 2
2
Standard Normal
Use the table in the Appendix
Normal Approximation to the Binomial
Use Normal for Binominal if n is large
X~Binomial (n, p)
  np

2
 np (1  p )
where : X   X  or  0.5 ~ Norm
X   np
Z
, where, Z ~ Norm(0,1)
np(1  p)
Refer to page 262
IME 312
Central Limit Theorem
x1 , x2 , x3 ,....., xn : random sample from a
population with
Then
Z
X
X 

n
as
and 
: sample mean
has standard normal distribution N(0, 1)
n   commonly n  30
IME 312

2
What does Central Limit Theorem mean?
Consider any distribution (uniform, exponential,
normal, or …). Assume that the distribution has a
mean of
and a standard deviation of  .
Pick up a sample of size “n” from this distribution.
Assume the values of variables are: x1 , x2 , x3 ,....., xn
Calculate the mean of this sample
. Repeat this
X
process and find many sample means. Then our
sample means will have a normal distribution
with a mean of
and a standard deviation of 
n
.


IME 312
u , v = degrees of freedom

= probability
Distribution
Definition
v
Chi-Square
X  Z
i 1
t dist.
F dist.
Z
T

 u2
F

2
v
Where: Z ~ N (0,1)
IME 312
2
i
2
v
u
v
Notation
 ,  ,v
2
v
2
t v , t ,v
v
Fu ,v , F ,u ,v