PROBABILITY AND STATISTICS
FOR ENGINEERING
Hossein Sameti
Department of Computer Engineering
Sharif University of Technology
 Let X represent a Binomial r.v ,Then from
P(" Occurrence s of A is between k1 and k2 " )
 n  k n k
 P( X k1  X k1 1    X k2 )   P( X k )     p q .
k  k1
k  k1  k 
k2
k2
 n  k n k
P k1  X  k2    Pn ( k )     p q .
k  k1
k  k1  k 
k2
k2
 for large n. In this context, two approximations are extremely useful.
The Normal Approximation (Demoivre-Laplace Theorem)
 If 𝑛𝑝𝑞 ≫ 1, then for k in the
npq neighborhood of np, we can approximate
 n  k n k

  p q
k 
1
2npq
e
( k  np ) 2 / 2 npq
.
(for example, when n   with p held fixed)
 And we have:
Pk1  X  k2   
k2
k1
1
2npq
 where
x1 
k1  np
npq
, x2 
k2  np
npq
.
e
dx  
( x np ) 2 / 2 npq
x2
1
x1
2
e  y / 2dy,
2
 n  k nk
P
(
k
)

 n
  k  p q
k  k1
k  k1  
k2
 As we know, Pk1  X  k 2  
k2


 If k1 and k2 are within np  npq , np  npq , with approximation:
Pk1  X  k2   
 where x1 
k1  np
npq
k2
k1
, x2 
1
2npq
k2  np
npq
e
dx  
( x np ) 2 / 2 npq
x2
1
x1
2
e  y / 2dy,
.
 We can express this formula in terms of the normalized integral
erf ( x ) 
1
2

x
0
 that has been tabulated extensively.
e
 y2 / 2
dy  erf (  x )
2
x
erf(x)
x
erf(x)
x
erf(x)
x
erf(x)
0.05
0.10
0.15
0.20
0.25
0.01994
0.03983
0.05962
0.07926
0.09871
0.80
0.85
0.90
0.95
1.00
0.28814
0.30234
0.31594
0.32894
0.34134
1.55
1.60
1.65
1.70
1.75
0.43943
0.44520
0.45053
0.45543
0.45994
2.30
2.35
2.40
2.45
2.50
0.48928
0.49061
0.49180
0.49286
0.49379
0.30
0.35
0.40
0.45
0.50
0.11791
0.13683
0.15542
0.17364
0.19146
1.05
1.10
1.15
1.20
1.25
0.35314
0.36433
0.37493
0.38493
0.39435
1.80
1.85
1.90
1.95
2.00
0.46407
0.46784
0.47128
0.47441
0.47726
2.55
2.60
2.65
2.70
2.75
0.49461
0.49534
0.49597
0.49653
0.49702
0.55
0.60
0.65
0.70
0.75
0.20884
0.22575
0.24215
0.25804
0.27337
1.30
1.35
1.40
1.45
1.50
0.40320
0.41149
0.41924
0.42647
0.43319
2.05
2.10
2.15
2.20
2.25
0.47982
0.48214
0.48422
0.48610
0.48778
2.80
2.85
2.90
2.95
3.00
0.49744
0.49781
0.49813
0.49841
0.49865
Example
 A fair coin is tossed 5,000 times.
 Find the probability that the number of heads is between 2,475 to 2,525.
Solution
 We need P(2,475  X  2,525).
 Since n is large we can use the normal approximation.
1
 p  , so that np  2,500 and npq  35.
2
 np  npq  2,465, and np  npq  2,535,
 So the approximation is valid for k1  2,475 and k 2  2,525.
Example - continued
P k1  X  k2   
x2
x1
 Here,
x1 
1  y2 / 2
e
dy.
2
k1  np
5
k  np 5
  , x2  2
 .
7
npq
npq 7
 Using the table, erf (0.7)  0.258 .
P2,475  X  2,525  erf ( x2 )  erf ( x1 )
 erf ( x2 )  erf (| x1 |)
5
 2erf    0.516
7
The Poisson Approximation
 For large n, the Gaussian approximation of a binomial r.v is valid only
if p is fixed, i.e., only if np  1 and npq  1.
 What if np is small, or if it does not increase with n?
- for example when p  0,
number.
n  , such that np   is a fixed
The Poisson Theorem
 If
p  0,
n  ,
np  
 Then
n!
Pn (k ) 
p k (1  p ) n  k
(n  k )!k!
k
𝑛→∞
k!
e 
The Poisson Approximation
 Consider random arrivals such as telephone calls over a line.
 n : total number of calls in the interval (0, T ).
 as T   we have n  
 Suppose n  T .
 Δ : a small interval of duration
0


 Let 𝑃𝑛 𝑘 = 𝑃 𝑘 𝑖𝑛 ∆ be the probability of k calls in the interval Δ
 We have shown that
𝑛 𝑘 𝑛−𝑘
𝑃𝑛 𝑘 = 𝑃 𝑘 𝑖𝑛 ∆ =
𝑝 𝑞
𝑘
n
2
1
∆
𝑤ℎ𝑒𝑟𝑒 𝑝 =
𝑇
T
The Poisson Approximation

p .
T
 p: probability of a particular call occurring in Δ:
 p0
as T  .
np  T 

   
T
 Normal approximation is invalid here because 𝑛𝑝 ≫ 1 is not valid.
0
n
2
1


T
 : probability of obtaining k calls (in any order) in an interval of duration Δ ,
Pn (k ) 
n!
p k (1  p) n k
(n  k )! k!
The Poisson Approximation
n(n  1) (n  k  1) (np )k
n k
Pn (k ) 
(
1

np
/
n
)
nk
k!
1 
2 
k  1  k (1   / n )n

  1   1     1 
.

k
n 
n 
n  k! (1   / n )

 Thus,
lim
n  , p 0 , np  
Pn (k ) 
k
k!
e 
the Poisson p.m.f
Example: Winning a Lottery
 Suppose
- two million lottery tickets are issued
- with 100 winning tickets among them.
 a) If a person purchases 100 tickets, what is the probability of winning?
Solution
The probability of buying
a winning ticket
p
Total no. of winning tickets
100
5


5

10
.
6
Total no. of tickets
2 10
Winning a Lottery - continued
 X: number of winning tickets

n: number of purchased tickets ,
  np  100  5 105  0.005.
P( X  k )  e 
k
k!
,
 P: an approximate Poisson distribution with parameter 
 So, The Probability of winning is:
P( X  1)  1  P( X  0)  1  e   0.005.
Winning a Lottery - continued
 b) How many tickets should one buy to be 95% confident of having a winning
ticket?
Solution
 we need P( X  1)  0.95.
P ( X  1)  1  e    0.95 implies   ln 20  3.
 But   np  n  5  10 5  3 or n  60,000.
 Thus one needs to buy about 60,000 tickets to be 95% confident of having a
winning ticket!
Example: Danger in Space Mission
 A space craft has 100,000 components n  
 The probability of any one component being defective is 2  10 5 ( p  0).
 The mission will be in danger if five or more components become defective.
 Find the probability of such an event.
Solution
 n is large and p is small
 Poisson Approximation with parameter np    100,000  2 105  2
4
P( X  5)  1  P( X  4)  1   e
k 0

k
k!
 1 e
4 2

 1  e 2 1  2  2     0.052.
3 3

2
4
k
 k!
k 0
Conditional Probability Density Function
P( A  B )
P( A | B ) 
, P( B )  0.
P( B )
FX ( x)  P X ( )  x ,
P  X ( )  x   B 
FX ( x | B)  P X ( )  x | B 
.
P( B )
P  X ( )     B  P( B )
FX (  | B ) 

 1,
P( B )
P( B )
P  X ( )     B  P( )
FX (  | B ) 

 0.
P( B )
P( B )
Conditional Probability Density Function
 Further,
P( x1  X ( )  x2 | B ) 
P  x1  X ( )  x2   B 
P( B )
 FX ( x2 | B )  FX ( x1 | B ),
 Since for x2  x1 ,
 X ( )  x2    X ( )  x1   x1  X ( )  x2  .
f X ( x | B) 
dFX ( x | B )
,
dx
FX ( x | B)  
x

f X (u | B)du.
P  x1  X ( )  x2 | B  

x2
x1
f X ( x | B )dx.
Example
 Toss a coin and X(T)=0, X(H)=1.
 Suppose B  {H }.
 Determine FX ( x | B).
Solution

FX (x ) has the following form.
 We need FX ( x | B)
 For x  0,
for all x.
X ( )  x   ,
so that
  X ( )  x   B   ,
 and FX ( x | B)  0.
FX ( x | B)
FX (x)
1
1
q
1
(a)
x
1
(b)
x
Example - continued
 For 0  x  1, X ( )  x   T , so that
  X ( )  x   B  T   H  
 For x  1,
and FX ( x | B)  0.
X ( )  x  , and
  X ( )  x   B     B  {B} and
FX ( x | B ) 
FX ( x | B)
1
1
x
P( B )
1
P( B )
Example
 Given FX (x ), suppose B  X ( )  a. Find f X ( x | B).
Solution
 We will first determine FX ( x | B).
FX ( x | B ) 
 For x  a,
P  X  x    X  a  
.
P X  a 
X  x  X  a  X  x
FX ( x | B ) 
 For x  a,
so that
P X  x 
F ( x)
 X
.
P X  a 
FX ( a )
 X  x    X  a   ( X  a)
so that FX ( x | B)  1.
Example - continued
 Thus
 FX ( x )
, x  a,

FX ( x | B )   FX ( a )

1,
x  a,

 and hence
 f X ( x)
,
d

f X ( x | B) 
FX ( x | B )   FX ( a )
dx

 0,
FX ( x | B)
x  a,
otherwise.
f X ( x | B)
1
f X (x )
FX (x )
a
(a)
x
a
(b)
x
Example
 Let B represent the event  a  X ( )  b with b  a.
 For a given FX (x ), determine FX ( x | B) and f X ( x | B).
Solution
FX ( x | B )  P X ( )  x | B



P  X ( )  x   a  X ( )  b  
P a  X ( )  b 
P X ( )  x   a  X ( )  b 
FX (b)  FX ( a )
.
Example - continued
 For
x  a, we have
 X ( )  x  a  X ( )  b   ,
and hence FX ( x | B)  0.
 For a  x  b, we have  X ( )  x  a  X ( )  b  {a  X ( )  x}
and hence
FX ( x | B ) 
P a  X ( )  x  FX ( x )  FX ( a )

.
FX (b)  FX ( a )
FX (b)  FX ( a )
x  b, we have  X ( )  x  a  X ( )  b   a  X ( )  b
so that FX ( x | B)  1.
 For
 Thus,
f X ( x)

,

f X ( x | B )   FX (b)  FX ( a )

0,

f X ( x | B)
a  x  b,
f X ( x)
otherwise.
a
b
x
Conditional p.d.f & Bayes’ Theorem
 First, we extend the conditional probability results to random variables:
 We know that If U  [ A1 , A2 ,, An ] is a partition of S and B is an arbitrary event,
then:
P( B)  P( B | A1 ) P( A1 )    P( B | An ) P( An )
 By setting B  { X  x} we obtain:
P ( X  x)  P ( X  x | A1 ) P ( A1 )    P ( X  x | An ) P ( An )
Hence
F ( x)  F ( x | A1 ) P ( A1 )    F ( x | An ) P ( An )
f ( x)  f ( x | A1 ) P ( A1 )    f ( x | An ) P ( An )
A1 ,  An form a partition on S
Conditional p.d.f & Bayes’ Theorem
 Using:
 We obtain:
P( A | B ) 
P( B | A) P( A)
.
P( B )
PA | X  x ) 
 For B  x1  X ( )  x2  ,
PA | x1  X  x 2  

P  X  x | A
P X  x

P( A) 
P x1  X  x 2 | A
P  x1  X  x 2 
FX ( x 2 | A)  FX ( x1 | A)
FX ( x 2 )  FX ( x1 )
F x | A
F x

P( A)
P ( A)
P ( A) 

x2
f X ( x | A) dx
x1

x2
x1
P ( A).
f X ( x) dx
Conditional p.d.f & Bayes’ Theorem
 Let x1  x, x2  x   ,   0, so that in the limit as   0,
lim PA | ( x  X  x   )  P A | X  x  
 0
 or
f X | A ( x | A) 
f X ( x | A)
f X ( x)
P( A).
P( A | X  x ) f X ( x )
.
P( A)
 we also get

P( A)  f X ( x | A)dx 







P( A | X  x ) f X ( x )dx,
1
 or
P( A) 



P( A | X  x ) f X ( x )dx
(Total Probability Theorem)
Bayes’ Theorem (continuous version)
 using total probability theorem in
f X | A ( x | A) 
P( A | X  x) f X ( x)
,
P( A)
 We get the desired result
f X | A ( x | A) 
P( A | X  x ) f X ( x )



P( A | X  x ) f X ( x )dx
.
Example: Coin Tossing Problem Revisited

p  P(H ) probability of obtaining a head in a toss.
 For a given coin, a-priori p can possess any value in (0,1).
 f P ( p) : A uniform in the absence of any additional information
 After tossing the coin n times, k heads are observed.
 How can we update
f P ( p) this is new information?
Solution
 Let A= “k heads in n specific tosses”.
f P ( p)
 Since these tosses result in a specific sequence,
P( A | P  p )  p k q n  k ,
 and using Total Probability Theorem we get
1
1
0
0
P( A)   P( A | P  p) f P ( p)dp   p k (1  p)n k dp 
0
(n  k )! k!
.
(n  1)!
1
p
Example - continued
 The a-posteriori p.d.f f P| A ( p | A) represents the updated information given the
event A,
 Using
f X | A ( x | A) 
f P |A ( p | A ) 

f P| A ( p | A)
P( A | X  x) f X ( x)
,
P( A)
0
P ( A | P  p )f P ( p )
P (A )
(n  1)!
p k q n k , 0  p  1
(n  k )!k !
1
p
 ( n , k ).
 This is a beta distribution.
 We can use this a-posteriori p.d.f to make further predictions.
 For example, in the light of the above experiment, what can we say about the
probability of a head occurring in the next (n+1)th toss?
Example - continued
 Let B= “head occurring in the (n+1)th toss, given that k heads have occurred in n
previous tosses”.
 Clearly
P ( B | P  p )  p,
 From Total Probability Theorem,
 Using (1) in (2), we get:
1
P( B)   P( B | P  p) f P ( p | A)dp.
0
(n  1)!
k 1
k n k
P (B )   p .
p q dp 
0
(n  k )!k !
n 2
 Thus, if n =10, and k = 6, then
1
P( B ) 
 which is more realistic compared to p = 0.5.
7
 0.58,
12