Download MGF and statistical distributions.

MOMENT GENERATING
FUNCTION AND
STATISTICAL
DISTRIBUTIONS
1
MOMENT GENERATING
FUNCTION
The m.g.f. of random variable X is defined as
 e tx f ( x )dx if X is cont .
 
all x
tX
M X ( t )  E (e )  
  e tx f ( x ) if X is discrete

all x
for t Є (-h,h) for some h>0.
2
Properties of m.g.f.
• M(0)=E[1]=1
• If a r.v. X has m.g.f. M(t), then Y=aX+b
has a m.g.f. e bt M(at )
E(X k )  M(k ) (0) where M(k ) is the k th derivative .
•
• M.g.f does not always exists (e.g. Cauchy
distribution)
3
Example
• Suppose that X has the following p.d.f.
f ( x )  xe  x for x  0
Find the m.g.f; expectation and variance.
4
CHARACTERISTIC FUNCTION
The c.h.f. of random variable X is defined as
 e itx f ( x)dx if X is cont.


all x
itX
 X (t )  E (e )  
itx
e
f ( x) if X is discrete


all x
for all real numbers t.
i 2  1, i   1
C.h.f. always exists.
5
Uniqueness
Theorem:
1. If two r.v.s have mg.f.s that exist and are
equal, then they have the same
distribution.
2. If two r,v,s have the same distribution,
then they have the same m.g.f. (if they
exist)
Similar statements are true for c.h.f.
6
STATISTICAL
DISTRIBUTIONS
7
SOME DISCRETE
PROBABILITY
DISTRIBUTIONS
Degenerate, Uniform, Bernoulli,
Binomial, Poisson, Negative
Binomial, Geometric,
Hypergeometric
8
DEGENERATE DISTRIBUTION
• An rv X is degenerate at point k if
1, X  k
P X  x  
0, o.w.
The cdf:
0, X  k
F  x  P X  x  
1, X  k
9
UNIFORM DISTRIBUTION
• A finite number of equally spaced values are
equally likely to be observed.
1
P(X  x )  ; x  1,2,..., N; N  1,2,...
N
• Example: throw a fair die.
P(X=1)=…=P(X=6)=1/6
N 1
E(X) 
;
2
( N  1)( N  1)
Var (X) 
12
10
BERNOULLI DISTRIBUTION
• A Bernoulli trial is an experiment with only
two outcomes. An r.v. X has Bernoulli(p)
distribution if
1 with probability p
X 
;0  p  1
0 with probability 1  p
P(X  x )  p x (1  p)1 x for x  0,1; and 0  p  1
11
BINOMIAL DISTRIBUTION
• Define an rv Y by
Y = total number of successes in n Bernoulli trials.
1. There are n trials (n is finite and fixed).
2. Each trial can result in a success or a failure.
3. The probability p of success is the same for all
the trials.
4. All the trials of the experiment are independent.
Y   X ~ Bin  n, p  where X ~ Ber  p  .
n
i 1
i
i
~ Bin  n , p . Then,
independent
Let X
i
i
 X ~ Bin  n  n 
k
i 1
i
1
2
 n , p .
k
12
BINOMIAL DISTRIBUTION
• Example:
• There are black and white balls in a box. Select
and record the color of the ball. Put it back and
re-pick (sampling with replacement).
• n: number of independent and identical trials
• p: probability of success (e.g. probability of
picking a black ball)
• X: number of successes in n trials
13
BINOMIAL THEOREM
• For any real numbers x and y and integer n>0
 n  i n i
( x  y )    x y
i 0  i 
n
n
14
BINOMIAL DISTRIBUTION
• If Y~Bin(n,p), then
E(Y)  np
Var(Y)  np(1 - p)
t
M Y ( t )  [pe  (1  p)]
n
15
POISSON DISTRIBUTION
• The number of occurrences in a given time
interval can be modeled by the Poisson
distribution.
• e.g. waiting for bus, waiting for customers to
arrive in a bank.
• Another application is in spatial distributions.
• e.g. modeling the distribution of bomb hits in an
area or the distribution of fish in a lake.
16
POISSON DISTRIBUTION
• If X~ Poi(λ), then E(X)= Var(X)=λ
M X (t )  exp{[exp( t ) 1]}
17
Relationship between Binomial and
Poisson
X ~ Bin  n, p  with mgf M
 t    pe  1  p 
t
X
n
Let =np.
lim M
n
 t   lim  pe  1  p 
t
X
n
n

   e  1 

 lim 1 
e
 M t 

n


t
n
n
et 1
Y
The mgf of Poisson()
The limiting distribution of Binomial rv is the
18
Poisson distribution.
NEGATIVE BINOMIAL DISTRIBUTION
(PASCAL OR WAITING TIME DISTRIBUTION)
• X: number of Bernoulli trials required to
get a fixed number of failures before the r
th success; or, alternatively,
• Y: number of Bernoulli trials required to
get a fixed number of successes, such as
r successes.
19
NEGATIVE BINOMIAL DISTRIBUTION
(PASCAL OR WAITING TIME DISTRIBUTION)
X~NB(r,p)
 r  x  1 r
p (1  p) x ;
P(X  x )  
x 

x  0,1,...;0  p  1
M X (t )  pr [1  (1  p)e t ]r
r (1  p)
E(X) 
p
Var (X) 
r (1  p)
p2
20
NEGATIVE BINOMIAL DISTRIBUTION
• An alternative form of the pdf:
 y  1 r
p (1  p) y  r ; y  r, r  1,...; 0  p  1
P(Y  y)  
 r 1 
Note: Y=X+r
r
r (1  p)
E ( Y)  E( X)  r 
Var (Y)  Var (X) 
p
p2
21
GEOMETRIC DISTRIBUTION
• Distribution of the number of Bernoulli trials
required to get the first success.
• It is the special case of the Negative Binomial
Distribution r=1.
X~Geometric(p)
P  X  x   p 1  p  , x  1,2,
x 1
1
(1  p)
E ( X) 
Var (X) 
p
p2
22
GEOMETRIC DISTRIBUTION
• Example: If probability is 0.001 that a light bulb
will fail on any given day, then what is the
probability that it will last at least 30 days?
• Solution:
P(X  30) 

x 1
30
0
.
001
(
1

0
.
001
)

(
0
.
999
)
 0.97

x  31
23
HYPERGEOMETRIC DISTRIBUTION
• A box contains N marbles. Of these, M are red.
Suppose that n marbles are drawn randomly
from the box without replacement. The
distribution of the number of red marbles, x is
 M  N  M  X~Hypergeometric(N,M,n)
 x  n  x 
 , x  0,1,..., n
P  X  x    
N
n
 
It is dealing with finite population.
24
SOME CONTINUOUS
PROBABILITY
DISTRIBUTIONS
Uniform, Normal, Exponential,
Gamma, Chi-Square, Beta
Distributions
25
Uniform Distribution
– A random variable X is said to be uniformly
distributed if its density function is
1
f ( x) 
a  x  b.
ba
– The expected value and the variance are
ab
(b  a) 2
E(X) 
V( X ) 
2
12
26
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,000 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(2500X3000) = (3000-2500)(1/3000) = .1667
1/3000
2000 2500 3000
5000
x
27
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X4000) = (5000-4000)(1/3000) = .333
1/3000
2000
4000
5000
x
28
Uniform Distribution
• Example 1
– The daily sale of gasoline is uniformly distributed
between 2,000 and 5,000 gallons. Find the
probability that sales are:
– Between 2,500 and 3,500 gallons
– More than 4,000 gallons
– Exactly 2,500 gallons
f(x) = 1/(5000-2000) = 1/3000 for x: [2000,5000]
P(X=2500) = (2500-2500)(1/3000) = 0
1/3000
2000 2500
5000
x
29
Normal Distribution
• This is the most popular continuous
distribution.
– Many distributions can be approximated by a
normal distribution.
– The normal distribution is the cornerstone
distribution of statistical inference.
30
Normal Distribution
• A random variable X with mean m and
variance s2 is normally distributed if its
probability density function is given by
 xm 
 (1 / 2)

 s 
e
2
1
f (x) 
   x  ; s  0
s 2
where   3.14159... and e  2.71828...
31
The Shape of the Normal
Distribution
The normal distribution is bell shaped, and
symmetrical around m.
90
m
Why symmetrical? Let m = 100. Suppose x = 110.
f (110) 
1
s 2
 110 100 
(1/ 2)

s 

e
2

1
s 2
 10 
(1/ 2) 
s
e
110
Now suppose x = 90
2
f (90) 
1
s 2
 90 100 
 (1 / 2)

s


e
2

1
s 2
 10 
 (1 / 2)

s


e
2
The Effects of m and s
How does the standard deviation affect the shape of f(x)?
s= 2
s =3
s =4
How does the expected value affect the location of f(x)?
m = 10 m = 11 m = 12
33
Finding Normal Probabilities
• Two facts help calculate normal probabilities:
– The normal distribution is symmetrical.
– Any normal distribution can be transformed into
a specific normal distribution called…
“STANDARD NORMAL DISTRIBUTION”
Example
The amount of time it takes to assemble a computer
is normally distributed, with a mean of 50 minutes
and a standard deviation of 10 minutes. What is
the probability that a computer is assembled in a
time between 45 and 60 minutes?
34
STANDARD NORMAL
DISTRIBUTION
• NORMAL DISTRIBUTION WITH MEAN 0
AND VARIANCE 1.
• IF X~N(m , s2), THEN
Z
X m
s
~ N (0,1)
NOTE: Z IS KNOWN AS Z SCORES.
• “ ~ “ MEANS “DISTRIBUTED AS”
35
Finding Normal Probabilities
• Solution
– If X denotes the assembly time of a computer,
we seek the probability P(45<X<60).
– This probability can be calculated by creating a
new normal variable the standard normal
variable.
Every normal variable
with some m and s, can
be transformed into this Z.
X  mx
Z
sx
E(Z) = m = 0
Therefore, once probabilities for Z
are calculated, probabilities of any
normal variable can be found.
V(Z) = s2 = 1
36
Standard normal probabilities
Copied from Walck, C (2007) Handbook on Statistical Distributions for
experimentalists
37
Standard normal table 1
38
Standard normal table 2
39
Standard normal table 3
40
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-0.5 < Z < 1)
To complete the calculation we need to compute
the probability under the standard normal distribution
41
Using the Standard Normal Table
Standard normal probabilities have been
calculated and are provided in a table .
The tabulated probabilities correspond
to the area between Z=0 and some Z = z0 >0
z
0.0
0.1
.
.
1.0
.
.
1.2
.
.
0
0.0000
0.0398
.
.
0.3413
.
.
0.3849
.
.
0.01
0.0040
0.0438
.
.
0.3438
.
.
0.3869
.
.
…….
…….
.
.
0.05
0.0199
0.0596
.
.
0.3531
.
.
0.3944
.
.
P(0<Z<z0)
Z=0
0.06
0.0239
0.0636
.
.
0.3554
.
.
0.3962
.
.
Z = z0
42
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5 < Z < 1)
We need to find the shaded area
z0 = -.5
z0 = 1
43
Finding Normal Probabilities
• Example - continued
45 - 50
X m
60 - 50
P(45<X<60) = P(
<
<
)
s
10
10
= P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1
P(0<Z<1)
z
0.0
0.1
.
.
1.0
.
0
0.0000
0.0398
.
.
0.3413
.
0.1
0.0040
0.0438
.
.
0.3438
.
…….
0.05
0.0199
0.0596
.
.3413
.
0.3531
.
z0 =-.5z=0 z0 = 1
0.06
0.0239
0.636
.
.
0.3554
.
44
Finding Normal Probabilities
• The symmetry of the normal distribution
makes it possible to calculate
probabilities for negative values of Z
using the table as follows:
-z0
0
+z0
P(-z0<Z<0) = P(0<Z<z0)
45
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5
46
Finding Normal Probabilities
• Example - continued
0
0.0000
0.0398
.
.
0.1915
.
0.1
0.0040
0.0438
.
.
….
.
…….
.3413
.1915
.1915
.1915
z
0.0
0.1
.
.
0.5
.
-.5
0.05
0.0199
0.0596
.
.
….
.
0.06
0.0239
0.636
.
.
….
.
.5 1.0
P(-.5<Z<1) = P(-.5<Z<0)+ P(0<Z<1) = .1915 + .3413 = .5328
47
Finding Normal Probabilities
• Example - continued
0
0.5000
0.5398
.
.
0.6915
.
0.1
0.5040
0.5438
.
.
….
.
…….
.3413
z
0.0
0.1
.
.
0.5
.
0.05
0.5199
0.5596
.
.
….
.
0.06
0.5239
0.5636
.
.
….
.
P(Z<-0.5)=1-P(Z>-0.5)=1-0.6915=0.3085
By Symmetry
P(Z<0.5)
48
Finding Normal Probabilities
• Example
– The rate of return (X) on an investment is normally
distributed with a mean of 10% and standard
deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
0%
10%
0 - 10
(i) P(X< 0 ) = P(Z<
) = P(Z< - 2)
5
X
.4772
-2
=P(Z>2) = 0.5 - P(0<Z<2) = 0.5 - .4772 = .0228
0
2
Z
49
Finding Normal Probabilities
• Example
– The rate of return (X) on an investment is normally
distributed with mean of 10% and standard
deviation of (i) 5%, (ii) 10%.
– What is the probability of losing money?
X
0%
10%
0 - 10
(ii) P(X< 0 ) = P(Z<
)
10
.3413
-1
= P(Z< - 1) = P(Z>1) = 0.5 - P(0<Z<1) = 0.5 - .3413 = .1587
1
Z
50
AREAS UNDER THE STANDARD
NORMAL DENSITY
P(0<Z<1)=.3413
0
1
Z
51
AREAS UNDER THE STANDARD
NORMAL DENSITY
.3413
.4772
P(1<Z<2)=.4772-.3413=.1359
52
EXAMPLES
• P( Z < 0.94 ) = 0.5 + P( 0 < Z < 0.94 )
= 0.5 + 0.3264 = 0.8264
0.8264
0
0.94
53
EXAMPLES
• P( Z > 1.76 ) = 0.5 – P( 0 < Z < 1.76 )
= 0.5 – 0.4608 = 0.0392
0.0392
0
1.76
54
EXAMPLES
• P( -1.56 < Z < 2.13 ) =
= P( -1.56 < Z < 0 ) + P( 0 < Z < 2.13 )
Because of symmetry
P(0 < Z < 1.56)
= 0.4406 + 0.4834 = 0.9240
0.9240
-1.56
2.13
55
STANDARDIZATION
FORMULA
• If X~N(m , s2), then the standardized value Z
of any ‘X-score’ associated with calculating
probabilities for the X distribution is:
Z
X m
s
• The standardized value Z of any ‘X-score’
associated with calculating probabilities for
the X distribution is:
x  m  z.s (Converse Formula)
56
Finding Values of Z
• Sometimes we need to find the value of Z
for a given probability
• We use the notation zA to express a Z
value for which P(Z > zA) = A
A
zA
57
PERCENTILE
• The pth percentile of a set of measurements
is the value for which at most p% of the
measurements are less than that value.
• 80th percentile means P( Z < a ) = 0.80
• If Z ~ N(0,1) and A is any probability, then
P( Z > zA) = A
A
zA
58
Finding Values of Z
• Example
– Determine z exceeded by 5% of the population
– Determine z such that 5% of the population is below
• Solution
z.05 is defined as the z value for which the area on its
right under the standard normal curve is .05.
0.45
0.05
0.05
-Z0.05
0
Z0.05
1.645
59
EXAMPLES
• Let X be rate of return on a proposed
investment. Mean is 0.30 and standard
deviation is 0.1.
a) P(X>.55)=?
Standardization formula
b) P(X<.22)=?
c) P(.25<X<.35)=?
d) 80th Percentile of X is? Converse Formula
e) 30th Percentile of X is?
60
a)P(X  0.55)  P
b)P(X  0.22)  P


ANSWERS

X - 0.3
0.55 - 0.3
=Z>
= 2.5 = 0.5 - 0.4938 = 0.0062
0.1
0.1

X - 0.3
0.22 - 0.3
=Z
= 0.8 = 0.5 - 0.2881 = 0.2119
0.1
0.1
c) P(0.25  X  0.35)  P 0.25  0.3  0.5  X - 0.3 = Z  0.35 - 0.3 = 0.5

= 2.* (0.1915)  0.3830
0.1
0.1
0.1

d) 80th Percentile of X is x  m  s .z0.20  .3+(.85)*(.1)=.385
e) 30th Percentile of X is x  m  s .z0.70  .3+(-.53)*(.1)=.247
61
The Normal Approximation to the
Binomial Distribution
• The normal distribution provides a close
approximation to the Binomial distribution
when n (number of trials) is large and p
(success probability) is close to 0.5.
• The approximation is used only when
np  5 and
n(1-p)  5
62
The Normal Approximation to the
Binomial Distribution
• If the assumptions are satisfied, the Binomial random
variable X can be approximated by normal distribution
with mean m = np and s2 = np(1-p).
• In probability calculations, the continuity correction
improves the results. For example, if X is Binomial
random variable, then
P(X  a) ≈ P(X<a+0.5)
P(X  a) ≈ P(X>a-0.5)
63
EXAMPLE
• Let X ~ Binomial(25,0.6), and want to find P(X ≤ 13).
• Exact binomial calculation:
P(X  13) 
13  25 
x
25  x


(
0
.
6
)
(
0
.
4
)
 0.267
 x 
x  0 
• Normal approximation (w/o correction): Y~N(15,2.45²)
13  15
P(X  13)  P(Y  13)  P( Z 
)  P( Z  0.82)  0.206
2.45
Normal approximation is good, but not great!
64
EXAMPLE, cont.
Bars – Bin(25,0.6); line – N(15, 2.45²)
Copied from Casella & Berger (1990)
65
EXAMPLE, cont.
• Normal approximation (w correction):
Y~N(15,2.45²)
P(X  13)  P(X  13.5)  P(Y  13.5)  P( Z  0.61)  0.271
Much better approximation to the exact value:
0.267
66
Exponential Distribution
• The exponential distribution can be used
to model
– the length of time between telephone calls
– the length of time between arrivals at a
service station
– the lifetime of electronic components.
• When the number of occurrences of an
event follows the Poisson distribution, the
time between occurrences follows the
exponential distribution.
67
Exponential Distribution
A random variable is exponentially
distributed if its probability density function
is given by
f
1
X
 x  e
 x/
, x  0,   0
 = e-x,
f(x)
x>=0.
 is a
distribution
parameter.
is the
distribution parameter
>0).
E(X) = 
V(X) = 2
The cumulative distribution function is
F(x) =1e-x/, x0
68
Exponential distribution for 1 = .5, 1, 2
2.5
f(x) = 2e-2x
2
f(x) = 1e-1x
1.5
f(x) = .5e-.5x
1
0.5
0
0
1
2
3
4
5
2.5
2
1.5
1
P(a<X<b) = e-a/ e-b/
0.5
0
a
b
69
Exponential Distribution
• Finding exponential probabilities is
relatively easy:
– P(X < a) = P(X ≤ a)=F(a)=1 – e –a/ 
– P(X > a) = e–a/ 
– P(a< X < b) = e – a/ – e – b/ 
70
Exponential Distribution
• Example
The service rate at a supermarket checkout
is 6 customers per hour.
– If the service time is exponential, find the
following probabilities:
• A service is completed in 5 minutes,
• A customer leaves the counter more than 10
minutes after arriving
• A service is completed between 5 and 8 minutes.
71
Exponential Distribution
• Solution
– A service rate of 6 per hour =
A service rate of .1 per minute (1 = .1/minute).
– P(X < 5) = 1-e-.lx = 1 – e-.1(5) = .3935
– P(X >10) = e-.lx = e-.1(10) = .3679
– P(5 < X < 8) = e-.1(5) – e-.1(8) = .1572
72
Exponential Distribution
• If pdf of lifetime of fluorescent lamp is
exponential with mean 0.10, find the life
for 95% reliability?
The reliability function = R(t) = 1F(t) = e-t/
R(t) = 0.95  t = ?
What is the mean time to failure?
73
GAMMA DISTRIBUTION
• X~ Gamma(,)
1
 1  x / 
f  x 
x
e , x  0,   0,   0

   
E  X    and Var  X   
M  t   1   t  , t 

2
1

74
GAMMA DISTRIBUTION
• Gamma Function:

     x e dx
 1
x
0
where  is a positive integer.
Properties:
   1     ,  0
  n    n  1! for any integer n  1
1

   
 2
75
GAMMA DISTRIBUTION
• Let X1,X2,…,Xn be independent rvs with
Xi~Gamma(i, ). Then,
 X ~ Gamma   , 

n
i 1
i
n
i 1
i

•Let X be an rv with X~Gamma(, ). Then,
cX ~ Gamma  , c  where c is positive constant.
• Let X1,X2,…,Xn be a random sample with
Xi~Gamma(, ). Then,
n
X


X
~ Gamma  n , 
n
n

i 1
i
76
GAMMA DISTRIBUTION
• Special cases: Suppose X~Gamma(α,β)
– If α=1, then X~ Exponential(β)
– If α=p/2, β=2, then X~ 2 (p) (will come back in a min.)
– If Y=1/X, then Y ~ inverted gamma.
77
Integral tricks
• Recall: h(t) is an odd function if h(-t)=-h(t);
It is an even function if h(-t)=h(t).
• Ex: h(x)=x exp{-x²/2} odd;
h(x)= exp{-x²/2-x} neither odd nor even

•
 odd function =0


•

 even function = 2* 0 even function

78
CHI-SQUARE DISTRIBUTION
Chi-square with  degrees of freedom
• X~ 2()= Gamma(/2,2)
1
f ( x)   / 2
x / 21e  x / 2 , x  0,   1,2,...
2 ( / 2)
E  X    and Var  X   2
M(t )  (1  2t )p / 2 t  1 / 2
79
DEGREES OF FREEDOM
• In statistics, the phrase degrees of freedom is
used to describe the number of values in the
final calculation of a statistic that are free to vary.
• The number of independent pieces of
information that go into the estimate of a
parameter is called the degrees of freedom (df) .
• How many components need to be known
before the vector is fully determined?
80
CHI-SQUARE DISTRIBUTION
• If rv X has Gamma(,) distribution, then
Y=2X/ has Gamma(,2) distribution. If 2
2
is positive integer, then Y has  2
distribution.
•Let X be an rv with X~N(0, 1). Then,
2
2
X ~ 1
•Let X1,X2,…,Xn be a r.s. with Xi~N(0,1). Then,
X ~ 
n
i 1
2
2
i
n
81
BETA DISTRIBUTION
• The Beta family of distributions is a
continuous family on (0,1) and often used
to model proportions.
1
 1
 1
f  x 
x 1  x  , 0  x  1,   0,   0.
B  ,  
where
      


B  ,     x 1  x  dx 
    
1
1
0


EX  
and Var  X  
 
        1 82
2
CAUCHY DISTRIBUTION
• It is a symmetric and bell-shaped
distribution on (,) with pdf
f (x) 
1
s 1 (
1
x 
s
)2
,s  0
Since E X  , the mean does not exist.
• The mgf does not exist.
•  measures the center of the distribution
and it is the median.
• If X and Y have N(0,1) distribution, then Z=X/Y
has a Cauchy distribution with =0 and σ=1.
83
LOG-NORMAL DISTRIBUTION
• An rv X is said to have the lognormal
distribution, with parameters µ and s2, if
Y=ln(X) has the N(µ, s2) distribution.
•The lognormal distribution is used to model
continuous random quantities when the
distribution is believed to be skewed, such as
certain income and lifetime variables.
84
STUDENT’S T DISTRIBUTION
• This distribution will arise in the study of
population mean when the underlying
distribution is normal.
• Let Z be a standard normal rv and let U be
a chi-square distributed rv independent of
Z, with  degrees of freedom. Then,
Z
X
~ t
U /
When n, XN(0,1).
85
F DISTRIBUTION
• Let U and V be independent rvs with chisquare distributions with 1 and 2
degrees of freedom. Then,
U /
X
~ F 
V /
1
1, 2
2
86
MULTIVARIATE
DISTRIBUTIONS
87
EXTENDED HYPERGEOMETRIC
DISTRIBUTION
• Suppose that a collection consists of a finite
number of items, N and that there are k+1 different
types; M1 of type 1, M2 of type 2, and so on. Select
n items at random without replacement, and let Xi
be the number of items of type i that are selected.
The vector X=(X1, X2,…,Xk) has an extended
hypergeometric distribution and the joint pdf is
 M 1  M 2   M k  M k 1 




...
x
x
x
x
f  x1 , x2 ,..., xk    1  2   k  k 1  ,  xi  {0,1,..., M i }
N
 
n
k
k
i 1
i 1
where M k 1  N   M i and x k 1  n   xi .
88
MULTINOMIAL DISTRIBUTION
• Let E1,E2,...,Ek,Ek+1 be k+1 mutually exclusive
and exhaustive events which can occur on any
trial of an experiment with P(Ei)=pi,i=1,2,…,k+1.
On n independent trials of the experiment, let Xi
be the number of occurrences of the event Ei.
Then, the vector X=(X1, X2,…,Xk) has a multinomial
distribution with joint pdf
n!
f  x1 , x2 ,..., xk  
p1x1 p2x2 ... pkxk 11 ,  xi  {0,1,..., n}
x1! x2 !...xk 1!
k
k
i 1
i 1
where x k 1  n   xi and p k 1  1   pi .
89
BIVARIATE NORMAL
DISTRIBUTION
• A pair of continuous rvs X and Y is said to
have a bivariate normal distribution if it has
a joint pdf of the form

1
1

f x, y  
exp 
2
 2 1  
2s1s 2 1   2


2 
2

y

m
y

m







y
y 
 x  m x   2   x  m x 


 








 s1 
s1  s 2   s 2   



   x  ,  y  , s1  0, s 2  0, 1    1.
90
If
BIVARIATE NORMAL
DISTRIBUTION
2
2
 X ,Y  ~ BVN m x ,m y ,s 1 ,s 2 ,   , then
2
2
X ~ N m x ,s 1  and Y ~ N m y ,s 2 
and  is the correlation coefficient btw X and Y.
1. Conditional on X=x,





s2
Y x ~ N m y   (x  mX ), s 22 1   2
s1

2. Conditional on Y=y,

s1
X y ~ N m x  
( y  mY ), s12 1   2
s2







91