4.2 Variances of random variables. A useful further characteristic to consider is the degree of...
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4.2 Variances of random variables
A useful further characteristic to consider is the de
gree of dispersion in the distribution, i.e. the spread
of the possible values of X.
Example :There are two batch of bulbs, the average lifetime is E(X)=1000 hours.
O x
O x
1000
1000
Definition 4.2-P76
Let X be a r.v. and . The variance of X, denoted by D(X) or Var(X), is definedby
( )E X
2( ) ( ) [( ) ]D X Var X E X
1) If X is discrete with pmf pk, then
2
1
( ) ( )k kk
D X x p
2) If X is continuous with pdf f(x), then
2( ) ( ) ( )D X x f x dx
(1)we often use variance to consider the degree of d
ispersion in the distribution of r.v. X. If the value of
D(X) is large, it means the degree of dispersion of X
is large.
(2)D(X) ≥0.
(3) Standard deviation标准差 :
Notes
( ) ( ) ( )X Var X D x
Example 4.6-P76
.)]([)()( 22 XEXEXD
Proof })]({[)( 2XEXEXD
})]([)(2{ 22 XEXXEXE 22 )]([)()(2)( XEXEXEXE
22 )]([)( XEXE
Theorem 4.2
).()( 22 XEXE
Example 4.7,4.8-P78
Example Suppose the pmf of X is
P(X=k) = p(1- p)k-1, k=1, 2, …,
where 0<p<1, Find Var(X).
Solution Let q=1-p， then
1
1)(k
kkpqXE
1
)'(k
kqp
1
)'(k
kqp )'1
(q
qp
,1
p
1
122 )(k
kpqkXE
1
1
1
1)1(k
k
k
k kpqpqkk
)()(1
XEqpqk
k
pq
qpq
1)
1(
,2
2p
p
. 12
)]([][)(222
22
p
q
pp
pXEXEXVar
Find Var(X).
Example Suppose the pdf of X is
].1 ,0[ ,0
],1 ,0[ ,2)(
x
xxxf
.18
1
3
2
2
1
22
)()(
)]([)()(
2
21
0
1
0
2
22
22
xdxxxdxx
dxxxfdxxfx
XEXEXVarSolution
Theorem 4.3-P79
Y=g(X)
2( ) [ ( )] {[ ( ) ( ( ))] }D Y D g X E g X E g X
1) If X is discrete with pmf pk, then
2) If X is continuous with pdf f(x), then
2( )
1
( ) [ ( ) ]k g X kk
D Y g x p
2( )( ) [ ( ) ] ( )g XD Y g x f x dx
Example 4.9,4.10-P79
Proof 22 )]([)()( CECECD
Properties -P80
(1) If C is a constant, then .0)( CD
22 CC .0
(2) Suppose X is a r.v., C is a constant, then
).()( 2 XDCCXD
Proof )(CXD
})]({[ 22 XEXEC
).(2 XDC
})]({[ 2CXECXE
).()()( YDXDYXD
(3) Suppose X and Y are independent, D(X), D(Y) exist, then
Proof
})](){[()( 2YXEYXEYXD 2)]}([)]({[ YEYXEXE
)]}()][({[2
)]([)]([ 22
YEYXEXE
YEYEXEXE
).()( YDXD
Generally, suppose X1,X2,…,Xn mutually Independent, then
).()()()( 2121 nn XDXDXDXXXD
(4) ( ) 0 ( ) 1, ( ).D X P X C where C E X
1. 0-1 distribution
If X ~ B(1, p) ， then D(X) = p(1-p) ；
2. Binomial distribution
If X ~ B(n, p) ， then D(X) = n p(1-p)=npq ； 3. Poisson distribution
If X ~ P(λ) ， then D(X) = λ ；
,!
)1(
!)1(
)]1([
])1([)(
0
22
0
2
ek
kk
ek
kk
XXE
XXXEXE
k
k
k
k
∵E(X) =λ， then
since
So 2 2( )E X
. )]([)()( 22 XEXEXVar
4. Uniform distribution
,3
)(
)()(
22
2
22
baba
dxab
x
dxxfxXE
b
a
Suppose X U(∼ a, b) ， then
Since E(X)=(a+b)/2 ， so2( )
( ) .12
b aD X
5. Exponential distribution –P81
).0( 0 ,0
,0 ,)(
x
xexf
x
2 2
220
2 22
( ) ( )
2 ,
1 ( )
1 ( ) ( ) [ ( )] .
x
E X x f x dx
x e dx
E X
Var X E X E X
So
，
6. Normal distribution
.
2
1/)(
2
1)(
)]([)(
2
222
2
)(2
2
2
2
2
dtetuxt
dxeux
XEXEXVar
t
x
Suppose X N(∼ , 2)， then
Homework:
P89: 3,8,10
