a,b

Uniform distribution
• Assume that X is random variable with uniform
distribution over interval (a,b)
• Calculate:
–
–
–
–
–
f(x),
F(x),
E[X],
E[X2],
V[X]
f(x)
x
a
b
Uniform distribution
f(x)
1
f ( x) 
ba
1
ba
a
b
x
F(x)
1
1
F ( x) 
( x  a)
ba
a
b
x
Uniform distribution
b
b
b
2
b
1
1
x
E[ X ]   x f ( x)   x

xdx
dx


ba a
2(b  a) a
ba
a
a
ab
b2  a2


2
2(b  a)
b
b
b
3 b
1 x
1
1
2
x dx 
E[ X ]   x f ( x)   x
dx 

ba 3
ba a
ba
a
a
2
2
2
b3  a 3
a 2  ab  b 2


3(b  a)
3
a
Uniform distribution
b
b
1
dx 
V [ X ]   ( x  E[ X ]) f ( x)dx   x  E[ X ]
ba
a
a
2
2
b
x
E[ X ]x
x  2 xE[ X ]  E[ X ]


dx 
3(b  a) a
ba
ba
a
b
2
2
3
2 b
2
b
E[ X ] x

ba a
a
b3  a 3 E[ X ](b 2  a 2 ) E[ X ]2 (b  a)



3(b  a)
ba
ba
b 2  ab  a 2 (a  b)(a  b) (a  b) 2



3
2
4
(a  b) 2 (a  b) 2 (a  b) 2 ab (a  b) 2 4ab (b  a) 2







3
2
4
3
12
12
12
Geometric distribution
• Number of failures before first
success
• Numer of trials to get one
success
• Calculate G(z)
• Calculate E[X]
• Prove the memoryless property
p(k )  p(1  p)
k
p(k )  p(1  p)
k 1
Geometric distribution
GX ( z )  ?


i 0
i 0
GX ( z )  E[ z X ]   p(1  p)i z i  p  ((1  p) z )i
p

1  (1  p) z

a0
q 

1 q
k  k0
k
Geometric distribution
1 p
p(1  p)


p

 
E[ X ]  G (1)  
2
p
 1  (1  p) z  1 (1  (1  p) z ) 1
'
'
• Quotient rule for derivatives
'
 f ( x) 
f ' ( x) g ( x)  f ( x) g ' ( x)

 
2
g
(
x
)
g
(
x
)


• Product rule for derivatives
 f ( x) g ( x)' 
f ' ( x) g ( x)  f ( x) g ' ( x)
Geometric distribution



i 0
i 0
i 1
E[ X ]   ip (1  p)i  p  i (1  p)i  p(1  p) i (1  p)i 1

d
d 
i
 p(1  p) (1  p)  p(1  p)  (1  p)i
dp i 1
i 1 dp
d 1 p
 p  (1  p) 1  p
 p(1  p)

 p(1  p)
2
p
p
dp p
Hint:

q
k  k0
k

ak 0
1 q
Geometric distribution
p(k )  p(1  p)
k 1
P{X  i  j | X  i}  P{X  j}
P{ X  i  j, X  i} P{ X  i  j}

P{X  i  j | X  i}
P{ X  i}
P{ X  i}


 p(1  p)
k i  j 1

k 1
k 1
p
(
1

p
)

k i 1
(1  p)
i j

(1  p)

 (1  p)
k 1

i
 (1  p)
k 1
 (1  p) j  P{X  j}
k 1
k 1
(1  p)i  j

i
(1  p)
Poisson distribution
• X is random variable with Poisson distribution
X~Poiss(a), calculate:
–
–
–
–
–
G(z)
E[X] from G(z)
E[X2] from G(z)
V[X] = E[X2] – E[X]2
E[X] from definition
a k a
p(k )  e
k!
Poisson distribution

k

a k a k
(
az
)
GX ( z )   e z  e a 
 e  a e az  e a ( z 1)
k!
k  0 k!
k 0
E[ X ]  G (1)  e
'

a ( z 1) '
 ae a ( z 1)  a
E[ X ]  G (1)  G (1)  e
2
''
'
1
1

a ( z 1) ''
 a  a 2e a ( z 1)  a  a 2  a
1
1
V [ X ]  E[ X 2 ]  E[ X ]2  a
d  d 
E[ X ]   z 
dz  dz 
k
( k 1)
(k )
 d 
G( z )   z  G( z )
dz 

z 1
z 1
Poisson distribution



k
a
a a a
a
 e a 
E[ X ]   k e  e  k
k!
k!
k 1 ( k  1)!
k 0
k 0
k
k
k 1
k

a
a
 e a a
 e a a   e a aea  a
k 1 ( k  1)!
k  0 k!

Poisson distribution
• X and Y are independent random variables with
Poisson distributions X~Poiss(a), Y~Poiss(b), prove
that:
– Z=X+Y ~ Poiss(a+b)
– Hint: calculate G(z)
GZ ( z )  GX ( z)GY ( z)  e( z 1) a e( z 1)b  e( z 1)( a b)
Exponential distribution
• X is random variable with exponential distribution
X~Exp(), calculate:
–
–
–
–
–
–
–
F(x)= ∫f(t)dt
E[X] = ∫tλe-λtdt
E[X2] = ∫t2λe-λtdt
f*(s)= ∫e-st f(t)dt
E[X] = -f*’(0)
E[X2] = -f*”(0)
V[X] = 1/λ2
Exponential distribution
x
F ( x) 


x
x
f (t )dt   e dt    e dt  
 t
0
 t
0



0
1

x
e
 t
 e
0

d  e

 
d  
0
 1  e x
d
d  t
 t
e
dt
e dt  

d 0
d
0
E[ X ]   tf (t )dt   te t dt   
 t

 t x


1 1
d 1
  
    2 


d   
0
Exponential distribution




2
d
d
2
2
E[ X ]   t f (t )dt   t 2 e t dt   
te t dt    2 e t dt
d
d 

0
0
0

2
d
d
d  1 
2 2
1
 t
  2  e dt   2    
 2  4  2
d  
d 0
d   


2

f ( s)   e
*

 st


f (t )dt   e e dt    e
0
 st
 t
0
(   s ) t
dt 

s
Exponential distribution

1
E[ X ]   f (0)  

2
(  s ) 0 
'
  

E[ X ]  f (0)  
2 
 (  s ) 
2
'

''
V [ X ]  E[ X ]  E[ X ] 
2
2
0
2

2

 2(  s)
2

(  s) 4 0 2
1

2

1
2
Exponential distribution
• X1~Exp(), X2~Exp() independent, Z=min(X1,X2)
– Prove that Z~Exp( +)
– Hint – try to calculate P{Z> z}
– Interpretation – joint service intensity of two servers working in
parallel
Exponential distribution
P{Z  z}  1  P{Z  z}  1  P{min( X1 , X 2 )  z}
 1  P{X1  z  X 2  z} 1  P{X1  z}P{X 2  z} 1  ez e z
 1  e(   ) z
Exponential distribution
• X~Exp(), Y~Exp() independent, find P{X<Y}
• Use joint distribution: fX,Y = fX fY
• P{X<Y} = ∫ P{X<Y | X=t} fX(t)dt = ∫ P{Y>t} fX(t)dt =
= ∫ GY(t) fX(t)dt = /(+)
• interpretation: probability that next customer arrives before the
current customer service is finished
Exponential distribution
Y
X<Y
Y=X
y
f
XY
( x, y )dx
0
 y

dy
f XY ( x, y )dxdy
0 0
0
X
y
Exponential distribution
 y
P{ X  Y }   
0 0
 y x  y 
f XY ( x, y )dxdy     e e dx dy
00


y



 1

   e  y   e x dx dy    e  y  e x  dy

0
0
0
0


y




1
1


   e  y  e y  dy    e  y 1  e y dy    e  y  e (    ) y dy


0
0
0

  e

 y

dy    e
0
 1
(    ) y
0





dy  
1



e
 y
0
 1 (    ) y

e


0