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MATHEMATICAL EXPECTATION

Module by: Ewa Paszek. E-mail the author

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Summary: This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by Ewa Paszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

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MATHEMATICAL EXPECTIATION

Definition 1: MATHEMATICAL EXPECTIATION
If f( x ) f( x ) is the p.d.f. of the random variable X of the discrete type with space R and if the summation
∑ R u( x )f( x )= ∑ x∈R u( x )f( x ) . ∑ R u( x )f( x )= ∑ x∈R u( x )f( x ) . (1)

exists, then the sum is called the mathematical expectation or the expected value of the function u( X ) u( X ) , and it is denoted by E[ u( x ) ] E[ u( x ) ] . That is,

E[ u( X ) ]= ∑ R u( x )f( x ) . E[ u( X ) ]= ∑ R u( x )f( x ) . (2)

We can think of the expected value E[ u( x ) ] E[ u( x ) ] as a weighted mean of u(x) u(x) , x∈R x∈R , where the weights are the probabilities f( x )=P( X=x ) f( x )=P( X=x ) .

REMARK:

The usual definition of the mathematical expectation of u( X ) u( X ) requires that the sum converges absolutely; that is, ∑ x∈R | u( x ) |f( x ) ∑ x∈R | u( x ) |f( x ) exists.

There is another important observation that must be made about consistency of this definition. Certainly, this function u( X ) u( X ) of the random variable X is itself a random variable, say Y. Suppose that we find the p.d.f. of Y to be g( y ) g( y ) on the support R 1 R 1 . Then, E( Y ) E( Y ) is given by the summation ∑ y∈ R 1 yg( y ) ∑ y∈ R 1 yg( y ) .

In general it is true that ∑ R u( x )f( x ) = ∑ y∈ R 1 yg( y ) ∑ R u( x )f( x ) = ∑ y∈ R 1 yg( y ) .

This is, the same expectation is obtained by either method.

Example 1

Let X be the random variable defined by the outcome of the cast of the die. Thus the p.d.f. of X is

f( x )= 1 6 f( x )= 1 6 , x=1,2,3,4,5,6. x=1,2,3,4,5,6.

In terms of the observed value x, the function is as follows

u( x )={ 1,x=1,2,3, 5,x=4,5, 35,x=6. u( x )={ 1,x=1,2,3, 5,x=4,5, 35,x=6.

The mathematical expectation is equal to

∑ x=1 6 u( x )f( x ) =1( 1 6 )+1( 1 6 )+1( 1 6 )+5( 1 6 )+5( 1 6 )+35( 1 6 )=1( 3 6 )+5( 2 6 )+35( 1 6 )=8. ∑ x=1 6 u( x )f( x ) =1( 1 6 )+1( 1 6 )+1( 1 6 )+5( 1 6 )+5( 1 6 )+35( 1 6 )=1( 3 6 )+5( 2 6 )+35( 1 6 )=8.

Example 2

Let the random variable X have the p.d.f.

f( x )= 1 3 , f( x )= 1 3 , x∈R x∈R ,

where, R=( −1,0,1 ) R=( −1,0,1 ) . Let u( X )= X 2 u( X )= X 2 . Then

∑ x∈R x 2 f( x )= ( −1 ) 2 ( 1 3 )+ ( 0 ) 2 ( 1 3 )+ ( 1 ) 2 ( 1 3 )= 2 3 . ∑ x∈R x 2 f( x )= ( −1 ) 2 ( 1 3 )+ ( 0 ) 2 ( 1 3 )+ ( 1 ) 2 ( 1 3 )= 2 3 .

However, the support of random variable Y= X 2 Y= X 2 is R 1 =( 0,1 ) R 1 =( 0,1 ) and

P( Y=0 )=P( X=0 )= 1 3 P( Y=1 )=P( X=−1 )+P( X=1 )= 1 3 + 1 3 = 2 3 . P( Y=0 )=P( X=0 )= 1 3 P( Y=1 )=P( X=−1 )+P( X=1 )= 1 3 + 1 3 = 2 3 .

That is, g( y )={ 1 3 ,y=0, 2 3 ,y=1; g( y )={ 1 3 ,y=0, 2 3 ,y=1; and R 1 =( 0,1 ) R 1 =( 0,1 ) . Hence

∑ y∈ R 1 yg( y )=0( 1 3 )+1( 2 3 ) = 2 3 , ∑ y∈ R 1 yg( y )=0( 1 3 )+1( 2 3 ) = 2 3 ,

which illustrates the preceding observation.

Theorem 1:

When it exists, mathematical expectation E satisfies the following properties:

  1. If c is a constant, E( c )=c, E( c )=c,
  2. If c is a constant and u is a function, E[ cu( X ) ]=cE[ u( X ) ], E[ cu( X ) ]=cE[ u( X ) ],
  3. If c 1 c 1 and c 2 c 2 are constants and u 1 u 1 and u 2 u 2 are functions, then E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= c 1 E[ u 1 ( X ) ]+ c 2 E[ u 2 ( X ) ]. E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= c 1 E[ u 1 ( X ) ]+ c 2 E[ u 2 ( X ) ].
Proof

First, we have for the proof of (1) that E( c )= ∑ R cf( x )=c ∑ R f( x ) =c, E( c )= ∑ R cf( x )=c ∑ R f( x ) =c, because ∑ R f( x )=1. ∑ R f( x )=1.

Proof

Next, to prove (2), we see that E[ cu( X ) ]= ∑ R cu( x )f( x )=c ∑ R u( x )f( x )=cE[ u( X ) ] . E[ cu( X ) ]= ∑ R cu( x )f( x )=c ∑ R u( x )f( x )=cE[ u( X ) ] .

Proof

Finally, the proof of (3) is given by E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= ∑ R [ c 1 u 1 ( x )+ c 2 u 2 ( x ) ] f( x )= ∑ R c 1 u 1 ( x )f( x )+ ∑ R c 2 u 2 ( x )f( x ) . E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= ∑ R [ c 1 u 1 ( x )+ c 2 u 2 ( x ) ] f( x )= ∑ R c 1 u 1 ( x )f( x )+ ∑ R c 2 u 2 ( x )f( x ) .

By applying (2), we obtain E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= c 1 E[ u 1 ( x ) ]+ c 2 E[ u 2 ( x ) ]. E[ c 1 u 1 ( X )+ c 2 u 2 ( X ) ]= c 1 E[ u 1 ( x ) ]+ c 2 E[ u 2 ( x ) ].

Property (3) can be extended to more than two terms by mathematical induction; that is, we have (3') E[ ∑ i=1 k c i u i ( X ) ]= ∑ i=1 k c i E[ u i ( X ) ] . E[ ∑ i=1 k c i u i ( X ) ]= ∑ i=1 k c i E[ u i ( X ) ] .

Because of property (3’), mathematical expectation E is called a linear or distributive operator.

Example 3

Let X have the p.d.f. f( x )= x 10 ,x=1,2,3,4 f( x )= x 10 ,x=1,2,3,4 , then

E( X )= ∑ x=1 4 x( x 10 )=1 ( 1 10 )+2( 2 10 )+3( 3 10 )+4( 4 10 )=3, E( X )= ∑ x=1 4 x( x 10 )=1 ( 1 10 )+2( 2 10 )+3( 3 10 )+4( 4 10 )=3,

E( X 2 )= ∑ x=1 4 x 2 ( x 10 )= 1 2 ( 1 10 )+ 2 2 ( 2 10 )+ 3 2 ( 3 10 )+ 4 2 ( 4 10 )=10, E( X 2 )= ∑ x=1 4 x 2 ( x 10 )= 1 2 ( 1 10 )+ 2 2 ( 2 10 )+ 3 2 ( 3 10 )+ 4 2 ( 4 10 )=10,

and E[ X( 5−X ) ]=5E( X )−E( X 2 )=( 5 )( 3 )−10=5. E[ X( 5−X ) ]=5E( X )−E( X 2 )=( 5 )( 3 )−10=5.

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