Mathematical Expectation:

Mean (or) expectation is a significant number representing the behavior of a random

variable.

Mathematical expectation of ‘X’ is denoted by E(X)


i) E( X )   xf ( x ) dx , [for a single dimensional continuous random variable]


ii) E ( X )  
x
xp ( x ) , [for a single dimensional discrete random variable]
2
iii) E ( X )   x 2 p ( x ) , [for a single dimensional discrete random variable]
x

 
iv) E( X )    xf ( x, y ) dxdy [ for a two dimensional continuous random variable]
 
 
v) E( X 2 )  
2
 x f ( x, y ) dxdy [ for a two dimensional continuous random variable]
 
 
vi) E ( XY )    xyf ( x, y ) dxdy , [ for a two dimensional continuous random variable]
 
vii) Var ( X )  E ( X 2 )  [ E ( X )] 2

Note:
If X and Y are independent random variables then E ( XY )  E ( X ).E (Y )

Two R.V’s X and Y have joint pdf

 xy
 ,0  x  4,1  y  5
f ( x, y )   96
 0 , elsewhere
 Find (i) E(X) (ii) E(Y) (iii) E(XY) (iv) E(2X + 3Y)
(v) Var(X)
(vi) Cov(X,Y).

Solution:
 
i) E( X )    xf ( x, y ) dxdy
 
54
 xy 
=   x dxdy
1 0  96 
8
=
3
 
ii) E (Y )    yf ( x, y ) dxdy
 
54
 xy 
=   y dxdy
1 0  96 
31
=
9
 
iii) E ( XY )    xyf ( x, y ) dxdy
 
54  xy 
=   xy dxdy
10  96 
248
=
27
8 31 47
iv) E[2 X  3Y ]  2 E ( X )  3E (Y ) = 2. + 3. =
3 9 3
v) We know that, Var ( X )  E ( X 2 )  [ E ( X )]2

x
2 2
Now, E ( X )  f ( x)dx =8


 Var ( X )  E ( X 2 )  [ E ( X )]2
8 2
= 8 -  
8
=
3 9
vi) Cov( X , Y )  E ( XY )  E ( X ).E (Y )
248  8  31 
= -   
27  3  9 
=0

1. If the joint pdf of (XY) is given by f ( x, y )  24 y (1  x), 0  y  x  1 ,

then find E(XY).

Solution:
  
We know that, E ( XY )    xyf ( x, y ) dxdy
 
11
=   xyf ( x, y ) dxdy {since x varies from y to
0y
1,
y varies from 0 to 1}

11
2
= 24   xy (1  x) dxdy
0y
1 1 y2 y 3 
= 24  y  
2
 dy
0 6 2 3 
1
 y3 y5 y6 
= 24    
 18 10 18 
0
4
=
15

2. If X and Y is a two dimensional R.V uniformly distributed over the

4x
triangular region R bounded by y  0, x  3 and y  . Find f (x),
3
f ( y ), E ( X ), Var(X), E (Y ),  XY .

Solution:
Given X and Y are uniformly distributed .
Therefore, f ( x, y )  k (a cons tan t )
We know that,   f ( x, y )dxdy  1
4 3
kdxdy  1
That is, 0 3y
4
4
 k  [ x ]33 y dy  1
0
4
4 3y 
 k  3  dy  1
0 4 
1
 6k  1  k 
6
3 3 1
f ( y )   f ( x, y ) dx dx 1
3y = 3y 6 = (4  y ),0  y  4
8
4 4
4x
3 1 2
f ( x)   dy = 9 x,0  x  3
0 6
  32
E( X )   xf ( x ) dx =  x 2 dx = 2
 09
 4y 4
E (Y )   yf ( y ) dy =  ( 4  y ) dy 
 08 3
 9
E ( X 2 )   x 2 f ( x )dx 
 2
 8
E (Y 2 )   y 2 f ( y ) dy 
 3
1
Var ( X )  E ( X 2 )  [ E ( X )]2 =
2
8
Var (Y )  E (Y 2 )  [ E (Y )] 2 =
9
1 4 3
E ( XY )    xydxdy
6 0 3y =3
4
E ( XY )  E ( X ) E (Y ) 1
Now,  XY  =
 X . Y 2

Moments are a set of statistical parameters to measure a distribution. Four

moments are commonly used:

 1st, Mean: the average

 2d, Variance:
o Standard deviation is the square root of the variance: an indication
of how closely the values are spread about the mean. A small
standard deviation means the values are all similar. If the
distribution is normal, 63% of the values will be within 1 standard
deviation.
 3d, Skewness: measure the asymmetry of a distribution about its peak; it
is a number that describes the shape of the distribution.
o It is often approximated by Skew = (Mean - Median) / (Std dev).
o If skewness is positive, the mean is bigger than the median and
the distribution has a large tail of high values.
o If skewness is negative, the mean is smaller than the median and
the distribution has a large tail of small values.
 4th: Kurtosis: measures the peakedness or flatness of a distribution.
o Positive kurtosis indicates a thin pointed distribution.
o Negative kurtosis indicates a broad flat distribution.
o