Download Notes 20 - Wharton Statistics

Statistics 510: Notes 20
Reading: Sections 7.1-7.3
In Chapter 7, we study more properties of expected values.
I. Expectations of Sums of Random Variables (Section
7.2)
Recall Proposition 4.1 of Chapter 4:
E[ g ( X )] 
 g ( x) P( X  x)
possible values
of x
A two-dimensional analog of this proposition is Proposition
2.1 of Chapter 7: If X and Y have a joint probability mass
function p( x, y ) , then
E[ g ( X , Y )]   g ( x, y ) p( x, y )
y
x
Similarly, for continuous random variables,
E[ g ( X , Y )]  



 
g ( x, y) f ( x, y)dxdy
An important application of Proposition 2.1 is to sums of
random variables. Suppose E ( X ), E (Y ) are both finite and
let g ( X , Y )  X  Y . Then, in the continuous case,
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E[ X  Y ]  


 ( x  y) f ( x, y)dxdy
   xf ( x, y )dxdy    yf ( x, y )dxdy
   xf ( x, y )dydx    yf ( x, y )dxdy
  xf ( x) dx   yf ( y )dy
 



 

 


 




 

X

Y
 E[ X ]  E[Y ]
The same result holds for discrete random variables that
E[ X  Y ]  E[ X ]  E[Y ] .
By induction, if E[ X i ] is finite for all i  1, , n, then
E[ X1   X n ]  E[ X1 ]   E[ X n ]
(1.1)
Example 1: Let X be a binomial random variable with
parameters n and p , i.e., X is the number of successes in
n independent trials when each trial has probability p of
being a success. Find E ( X ) .
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Example 2: The Sample Mean. Let X 1 , , X n be
independent and identically distributed random variables
having cdf F and expected value  . Such a sequence of
random variables is said to constitute a random sample
from F. The quantity X , defined by
n
X
X  i ,
i 1 n
is called the sample mean. Compute E ( X ) .
Example 3: A group of N people throw their jackets into
the center of a room. The jackets are mixed up and each
person randomly selects one. Find the expected number of
people that select their own jacket.
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Example 4: Shuffle an ordinary deck of 52 playing cards.
Then turn up cards from the top until the first ace appears.
What is the expected value of the number of cards required
to produce the first ace?
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II. Moments of the Number of Events That Occur (Chapter
7.3)
Example 1 continued: Consider n independent trials, with
each trial being a success with probability p. Let X be the
number of successes in the n trials; X is a binomial
( n, p ) random variable.
In Section 7.2, we used the following strategy to find
E ( X ) . We noted that X is the number of the events
A1 , , An that occur where Ai is the event that the ith trial is
a success. We then defined indicator variables
1, if Ai occurs
Ii  
0, otherwise
We then computed E ( X ) via the formula
E ( X )  E ( I1   I n )  E ( I1 )   E ( I n ) .
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The use of indicator random variables for events to
compute the expected value can be extended to compute
higher order moments (e.g., variances) in the following
way.
Consider the number of pairs of events A1 , , An that occur.
Because I i I j will equal 1 if both Ai and A j occur, and will
equal 0 otherwise, it follows that the number of pairs is
equal to  I i I j . But as X is the number of events that
i j
occur, it also follows that the number of pairs of events that
X
occur is  2  . Consequently,
 
X
    Ii I j
 2  i j
Taking expectations yields,
 X 
E     E[ I i I j ]   P( Ai  Aj )
i j
 2  i j
(1.2)
or
 X ( X  1) 
E
  P( Ai  Aj )

2

 i j
giving that
E  X 2   E[ X ]  2 P( Ai  Aj )
i j
(1.3)
2
2
2
which yields E[ X ] and thus Var ( X )  E ( X )  ( E[ X ]) .
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Moreover, by considering the number of distinct subsets of
k events that all occur, we see that
X
    I i1 I i 2 I ik .
 k  i1 i2  ik
Taking expectations gives the identity
 X  
E      E[ I i1I i 2 I ik ]   P( Ai1  Ai 2  Aik )
i1 i2  ik
 k   i1 i2  ik
Example 1 continued: Ai is the event that the ith trial is a
2
success. When i  j, P( Ai  Aj )  p . Consequently, by
(1.2)
 X  
 n 2
2
E      p    p
2
 2   i  j
or equivalently by (1.4)
E[ X 2 ]  E[ X ]  n(n  1) p 2 .
n
Using that E[ X ]   P( Ai )  np , this yields that
i 1
Var[ X ]  E[ X 2 ]  ( E[ X ])2  n(n  1) p 2  np  (np) 2  np(1  p)
Example 3 continued: The Matching Problem. A group of
N people throw their jackets into the center of a room. The
jackets are mixed up and each person randomly selects one.
We showed that the expected number of people that select
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their own jacket is 1. Find the variance of the number of
people that select their own jacket.
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