# Tutorial 6 by gegeshandong

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```									Tutorial 6

Mode and Median of Random
Variable, Gamma Distribution

1
Mode of random variable
 Discrete  R.V.:
If T = k is the mode, then,
P(T  k )  P(T  i) for i
 Continuous   R.V.:
If a is the mode, then
f(a)  f(b) for b
 Position   of the maximum in p.d.f / p.m.f
2
Median of random variable
 Discrete  R.V.:
If T = i is the median, then,
i                         i 1

 P (T  k )  1
k 1
2
and  P (T  k )  1
k 1
2
for i  1

P (T  i )  1                       for i  1
2
 Continuous   R.V.:
a           1
If a is the median, then F (a)   f ( x)dx 
                2
3
Example - Binomial R.V

X   is a random variable with parameter (n,p),
find the median and mode of X?
 Median: a minimum k such that
k
1
 P( X  r )  2
r 1
k
1
  nCr (1  p)   nr
p 
r

r 1                     2

4
Example - Binomial R.V

 Mode:   a maximum k such that
P{ X  k}
1
P{ X  k  1}
Ck (1  p ) n  k p k
n
n  k 1 k 1
1
nCk  1(1  p )         p
(n  k  1) p
1
k (1  p )
k  (n  1) p
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Example – Exponential R.V

X   is a random variable with p.d.f,
e  x if 0  x  
f ( x)  
0       otherwise
where is a positive constant.

 Find its mode and median.
 Find its generating function, and thereby its
skewness and kurtosis.
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Example – Exponential R.V

 Mode:
As f(x) is strictly decreasing function, its
maximum occurs at x=0.
 0 is the mode.
a           1
 Median:
0 f ( x)dx  2
a           1       ln 2
0 e dx  2  a  
 x

7
Example – Exponential R.V
   Moment Generating Function:
                       
M (t )  E[e ]   e e
tX             tx    x
dx 
0                        t
E[( X   ) 3 ]
   Skewness 
3
E[( X   ) 4 ]
Kurtosis 
4
   By finding E[X], E[X²], E[X³],…from M(t),
skewness and kurtosis can be found.
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Gamma Distribution
   A random variable is said to have a gamma
distribution with parameters (t, ), if its density
function is given by
 e  x (x)t 1
                  x0
f ( x)        (t )
0                 x0

Where (t ) is the gamma function,

define by  e  y y t 1dy
0                                     9
Gamma Distribution

Since (t )   y t 1e  y dy,
0

       
then (t )  e y    y   t 1
  e  y (t  1) y t  2 dy
0      0


 (t  1)  e  y y t 2 dy
0

 (t  1)(t  1)

(n)  (n  1)! for positive n
10
Gamma Distribution

 The   cumulative density function,
F ( x)  P{ X  x}
t 1
x    j
 1 e    x

j 0     j!
if x  0

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Mean of Gamma Distribution
 ConsiderX as a gamma random variable
with parameter t and .
1 
(t ) 0
E[ X ]           xe x (x) t 1 dx

1      

(t ) 0
            e x (x) t dx

1      

(t ) 0
            x x (x) t d (x)

12
Mean of Gamma Distribution
1     

(t ) 0
E[ X ]             e  y y t dy

(t  1)

(t )
t!

 (t  1)!
t


 Mean   of Gamma(t , ) = t/ 
13
Variance of Gamma Distribution
 First   calculate E[X2]:
1  2 x
(t ) 0
E[ X ] 
2
x e (x) t 1 dx

1       

 (t ) 0
 2           e x (x)t 1dx

1       

 (t ) 0
 2           e x (x) t 1d (x)

1       

 (t ) 0
 2           e  y t t 1d ( y )
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Variance of Gamma Distribution
 (t  2)
E[ X ]  2
2

 (t )
t (t  1)
      2

Var( X )  E[ X 2 ]  ( E[ X ])2
t (t  1)      t
                ( )2
2       
t

2
Variance       of Gamma(t , ) = t/ ²   15
Mode of Gamma Distribution
 Since
e  x (x) t 1
f ( x)                              if x  0
(t )
                                    x         t 1  x
f ' ( x)            [ (t  1)( x)   t 2
e            (x) e        ]
(t )

 x  0  )x ( ' f
)1  t(

 Mode    of Gamma(t, ) = (t-1)/
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HW3 Q10 Hints.
 Given   the mean and mode.
 i) First find the value of t and ’s.
 ii) Once you find the t, , you can plot the
income distributions.
P{ X  x}  1  G ( x;  , n)
n 1
(x) k
 1  e x            for 0  x  
k 0   k!
income distribution : Gamma(n, )
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Income Distribution
 x
e      ( x )   t 1
f ( x)                            if x  0
(t )

>> x=[0:60000];
>> t=4;
>> m=1/3000;
>> y=m*exp(-1*m*x).*(m*x).^(t-1)/gamma(t);
>> plot(y)
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Income Distribution

19
Gini Index

1    


G            G ( x)[1  G ( x)]dx
0

20

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