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Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-1
1036: Probability & Statistics
1036: Probability & 1036: Probability & StatisticsStatistics
Lecture 4 Lecture 4 –– Mathematical Mathematical ExpectationExpectation
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-2
Mean of a Random Variable• Let X be a random variable with probability
distribution f (x). The mean or expected value of Xis
• Example: – If 2 coins are tossed 16 times. The outcomes are 0 head:
4 times; 1 head: 7 times; 2 heads: 5 times. The average number of heads per toss?
( ) ∑==x
xxfXE )(µ
( ) ∫∞
∞−
== dxxxfXE )(µ
if X is discrete, and
if X is continuous
06.1165)2(
167)1(
164)0(
16)5)(2()7)(1()4)(0(
=⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
++
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-3
Example 4.3• Let X be the random variables that denotes the life in
hours of a certain electronic device. The probability density function is
Find the expected life of this type of device.Solution
⎪⎩
⎪⎨⎧ >=
eleswhere,0
100,000,20)( 3 x
xxf
( ) 20020000
1003 === ∫
∞
dxx
xXEµ
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-4
Expectation of g (x)• Let X be a random variable with probability distribution f (x).
The mean or expected value of the random variable g (x) is
∫
∑∞
∞−
==
==
dxxfxgXgE
xfxgXgE
Xg
Xg
)()()]([
)()()]([
)(
)(
µ
µ If X is discrete
If X is continuous
Ex: X is a RV with pdf:elsewhere
2x1-
,0
,3)(
2<<
⎪⎩
⎪⎨⎧
=x
xf
The mean of g(X)=4X+3???
[ ] ( ) 83
34)()(342
1
2=∫ +=∫=+
−
∞
∞−dxxxdxxfxgXE
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-5
Expectation of g (X,Y)• Let X and Y be two random variables with joint probability
distribution f (x, y). The mean or expected value of the random variable g (X,Y) is
∫ ∫
∑∑∞
∞−
∞
∞−
===
==
dxdyyxfyxgYXgE
yxfyxgYXgE
YXg
x yYXg
),(),()],([
),(),()],([
),(
),(
µ
µ If X,Y are discrete
If X,Y are continuous
Ex: Find E(Y/X) for the density function( )
elsewhere10,20
,0
,431
),(2
<<<<
⎪⎩
⎪⎨⎧ +
=yxyx
yxf
( )85
431 21
0
2
0
=+
⋅=⎟⎠⎞
⎜⎝⎛ ∫ ∫ dxdyyx
xy
XYESol:
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-6
Example • Let X and Y be random variables with joint density
function
elsewhere.1y1,0x0
,0
,4),(
<<<<
⎩⎨⎧
=xy
yxf
Find the expected value of 22 YXZ +=
( ) ∫ ∫ ∫∫∞
∞−
∞
∞−
=+==1
0
1
0
22 4),( xydxdyyxdxdyyxzfZE
Solution:
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-7
Remark
( )⎪⎩
⎪⎨
⎧
∫ ∫=∫
∑ ∑=∑∑=∑
= ∞
∞−
∞
∞−
∞
∞−dxxxgdxdyyxxf
xxgyxfxyxxfXE
x xyxy
)(),(
)(),(),( discrete
continuous
( )⎪⎩
⎪⎨
⎧
∫ ∫=∫
∑ ∑=∑∑=∑
= ∞
∞−
∞
∞−
∞
∞−dyyyhdxdyyxxf
yyhyxfyyxyfYE
x yxyy
)(),(
)(),(),(
E(X) & E(Y) calculated by joint pdf or marginal pdf
discrete
continuous
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-8
Why Variance• To measure the variability !!• The mean of a random variable X (in statistics) describes where
the probability distribution is centered.
• The mean does not give adequate description of the shape of the distribution.
x x
2 2
f(x) f(x)
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-9
Variance and Standard Deviation• Let X be a random variables with probability distribution
f (x) and mean µ. The variance of X is
• The positive square root of the variance, σ, is called the standard deviation of X
• The quantity x−µ is called the deviation of an observationfrom its mean.
( )[ ] ( )∑ −=−=x
xfxXE )(222 µµσ
( )[ ] ( )∫∞
∞−
−=−= dxxfxXE )(222 µµσ
If X is discrete
If X is continuous
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-10
Example 4.8• Let X, a random variable, represent the number of
automobiles that are used for official business purpose on any given weekday. We have the following distributions:
x
f(x)
1 2 3
0.3 0.4 0.3
for company B
x
f(x)
1 2 3
0.2 0.1 0.3
0
0.1
4
0.1
for company A
µ=E(X)=(1)(0.3)+(2)(0.4)+(3)(0.3)=2.0Company A
( )[ ] ( ) 6.0)(23
1
222 =−=−= ∑=
xfxXEx
µσ
Company B: 2)( =XE ( )[ ] 6.122 =−= µσ XE
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-11
Variance[ ] 222 µσ −= XE
22
22
22
2222
)( 2)(
)()(2)(
)()2()()(
µ
µµµ
µµ
µµµσ
−=
+⋅−=
+−=
+−=−=
∑∑∑
∑∑
XEXE
xfxxfxfx
xfxxxfx
xxx
xx
Proof:This often simplifies the calculation
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-12
Example 4.10• X is a random variable with the pdf
( )elsewhere.
2x1
,0,12
)(<<
⎩⎨⎧ −
=x
xf
Mean & variance?
( )3512)()(
2
1
=−=== ∫∫∞
∞−
dxxxdxxxfXEµ
617)()( 22 == ∫
∞
∞−
dxxfxXE
[ ] 18/1222 =−= µσ XE
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-13
Remarks• Let X be a random variable with probability distribution f(x).
The variance of the random variable g(X) is
• Since g(X) is itself a random variable with mean µg(X)
( )[ ] [ ] )()()( 2)(
2)(
2)( xfxgXgE
xXgXgXg ∑ −=−= µµσ
( )[ ] [ ]∫∞
∞−
−=−= dxxfxgXgE XgXgXg )()()( 2)(
2)(
2)( µµσ
If X is discrete
If X is continuous
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-14
Example 4.11• Calculate the variance of g(X)=2X+3, where X is a
random variable with probability distribution:x
f(x)0
1/4
11/8
21/2
31/8
( ) 6)(32)32(3
0)( =+=+= ∑
=xXg xfxXEµ
( )[ ] ( ) 4)(632)(3
0
22)(
2)( =−+=−= ∑
=xXgXg xfxXgE µσ
Solution
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-15
Covariance• Let X and Y be random variables with joint probability
distribution f (x,y). The covariance of X and Y is
• When X and Y are statistically independently, σXY = 0. – The inverse is not generally true
• The sign of the covariance indicates whether the relationship between two dependent random variables is positive or negative.
( )( )[ ] ( )( )∑∑ −−=−−=x y
YXYXXY yxfyxYXE ),(µµµµσ
( )( )[ ] ( )( )∫ ∫∞
∞−
∞
∞−
−−=−−= dxdyyxfyxXXE YXYXXY ),(µµµµσ
If discrete
If continuous
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-16
CovarianceYXXY XYE µµσ −= )(
YX
YXYXYX
x y x yYXY
x y x yX
x yYXYX
x yYXXY
YXEYXE
yxfyxxf
yxyfyxxyf
yxfxyxy
yxfyx
µµµµµµµµ
µµµ
µ
µµµµ
µµσ
−=+−−=
+−
−=
+−−=
−−=
∑∑ ∑∑
∑∑ ∑∑
∑∑
∑∑
),( ),(
),(),(
),(),(
),()(
),())((Proof:
This often simplifies the calculation
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-17
Example 4.14• The fraction X of male runners and the fraction Y of female
runners who compete in marathon races is described by the joint density function
elsewhere010
,0
,8),(
xy,xxyyxf
≤≤≤≤
⎩⎨⎧
=
∫∞
∞−
== dxxxgXEX )()(µ elsewhere10
0
4),()(
3 ≤≤
⎩⎨⎧
== ∫∞
∞−
xxdyyxfxg
∫∞
∞−
== dyyyhYEY )()(µ ( )elsewhere
10
014
),()(2 ≤≤
⎩⎨⎧ −
== ∫∞
∞−
yyydxyxfyh
( ) 255/4=−= YXXY XYE µµσ
Find the covariance of X and Y
Sol.
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-18
Correlation Coefficient• The covariance does not indicate anything regarding the
strength of the relationship, since the value σXY is not scale free, which depends on the units measured for both X and Y.
• Let X and Y be random variables with covariance σXY and standard deviations σX and σY, respectively. The correlation coefficient X and Y is
YX
XYXY σσ
σρ =
11 ≤≤− XYρ
0=XYρ
X & Y linear dependence, i.e.Y=a+bX
X & Y independent
1±=XYρ
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-19
Linear Combinations of RVs.• If a and b are constant, then E(aX±b)=aE(X)±bProof.
• If b is constant, then E(b)=b• If a is constant, then E(aX)=aE(X)
• The expected value of the sum or difference of two or more functions of a random variable X is the sum or difference of the expected values of the functions.
[ ] [ ] [ ])()()()( XhEXgEXhXgE ±=±
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-20
Linear Functions of RVs• The expected value of the sum or difference of two or more
functions of random variable X and Y is the sum or difference of the expected values of the functions.
[ ] [ ] [ ]),(),(),(),( YXhEYXgEYXhYXgE ±=±
[ ] [ ]
[ ] [ ]),(),(
),(),(),(),(
),(),(),(),(),(
YXhEYXgE
dxdyyxfyxhdxdyyxfyxg
dxdyyxfyxhyxgYXhYXgE
±=
±=
±=±
∫ ∫∫ ∫
∫ ∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
∞
∞−
∞
∞−
)]([)]([)]()([ YhEXgEYhXgE ±=±•
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-21
Theorem 4.8• Let X and Y be two independent random variables. Then
)()()( YEXEXYE =
)()(
)()(),()(
YEXE
dxdyygxxyhdxdyyxxyfXYE
=
== ∫ ∫∫ ∫∞
∞−
∞
∞−
∞
∞−
∞
∞−
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-22
Linear Combinations of RVs.• If a and b are constant, then Proof.
• If b is constant, then • If a is constant, then
• If X and Y are random variables with joint probability distribution f (x,y), then
XYYXbYaX abba σσσσ 222222 ±+=±
222XbaX a σσ =+
22XbX σσ =+
222XaX a σσ =
XYYX
YXYX
YX
YXbYaX
abba
YXabEba
YbXaE
babYaXE
σσσ
µµσσ
µµ
µµσ
2
)])([(2
})]()({[
})](){[(
2222
2222
2
22
±+=
−−±+=
−±−=
±−±=±Proof
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-23
Remarks• If X and Y are independent, then
• If X and Y are independent, then
• If X1, X2, …, Xn are independent
22222YXbYaX ba σσσ +=±
22222
221
2212211 nnn XnXXxaxaxa aaa σσσσ +++=+++ LL
0=XY
σ
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-24
Variance ?• If a random variable has a small variance, we would expect
most of the values to be grouped around the mean.• A large value of σ indicates a greater variability and,
therefore, we would expect the spread distribution.• Since the total area under a probability distribution curve is
1, the area between any two numbers is then the probability of the random variable assuming a value between these numbers.
• Chebyshev (1821--1894) discovered that the fraction of the area between any two values symmetric about the mean is related to the standard deviation.
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-2522
22222
1)(1)()(
)()( isThat
kkxP
kdxxfdxxf
dxxfkdxxfk
k
k
k
k
≤≥−⇔≤+⇒
+≥
∫∫
∫∫∞
+
−
∞−
∞
+
−
∞−
σµ
σσσ
σµ
σµ
σµ
σµ
Chebyshev’s Theorem• The probability that any random variable X will assume a
value within k standard deviations of the mean is at least 1−1/k2. That is
Proof
( ) 2
11k
kXP −≥<− σµ
)(y probabilit heconsider tfirst We σµ kXP ≥−
∫∫∫∞
+
−
∞−
∞
∞−
−+−≥−=σµ
σµ
µµµσk
k
dxxfxdxxfxdxxfx )()()()()()( 2222
∫∫∞
+
−
∞−
+≥σµ
σµ
σσk
k
dxxfkdxxfk )()( 2222
Prob. & Stat. Lecture04 - mathematical expectation ([email protected])
4-26
Chebyshev’s Theorem• ¾ or more of the observations of any distribution lie in the
interval µ±2σ (1-1/22 = ¾ ), lower bound only (weak result)• The use of Chebyshev’s theorem is relegated to situations
where the form of the distribution is unknown.
Example 4.22 X a RV, with mean of 8 and variance of σ2=9,
( ) [ ] 2411)3)(4(8)3)(4(8204 . −≥+<<−=<<− XPXPa
µ σk µ σk
( ) ( )( ) ( )
4/1)211(1
)3)(2(8)3)(2(816861 68168 .
2 =−−≤
+<<−−=<−<−−=
<−−=≥−
XPXPXPXPb