expected_values_continuous.pdf

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Expected Value Variance

Expected Value and Variance forContinuous Random Variables

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Expected Value and Variance for Continuous Random Variables

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Introduction

1. The underlying ideas for expected value and variance arethe same as for discrete distributions.

2. The expected value gives us the expected long termaverage of measurements. (The Central Limit Theoremwill formally confirm this statement.)

3. The variance is a measure how spread out the distributionis.



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Introduction1. The underlying ideas for expected value and variance are

the same as for discrete distributions.

2. The expected value gives us the expected long termaverage of measurements. (The Central Limit Theoremwill formally confirm this statement.)




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the same as for discrete distributions.2. The expected value gives us the expected long term

average of measurements.

(The Central Limit Theoremwill formally confirm this statement.)




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the same as for discrete distributions.2. The expected value gives us the expected long term

average of measurements. (The Central Limit Theoremwill formally confirm this statement.)




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From Discrete to Continuous Probability

1. In the discrete expected value, the outcome x contributes asummand xP(X = x).

2. In the continuous setting, P(X = x) = 0, but

fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)

3. So an interval [x,x+dx] should contribute about xfX(x) dx.4. The summation becomes an integral.

Integrals are continuous sums.



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From Discrete to Continuous Probability1. In the discrete expected value, the outcome x contributes a

summand xP(X = x).

2. In the continuous setting, P(X = x) = 0, but

fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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summand xP(X = x).2. In the continuous setting, P(X = x) = 0

, but

fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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summand xP(X = x).2. In the continuous setting, P(X = x) = 0, but

fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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fX

-

x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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fX

-x

x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)

3. So an interval [x,x+dx] should contribute about xfX(x) dx.

4. The summation becomes an integral.Integrals are continuous sums.



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fX

-x x+dx

probability to

be in [x,x+dx] is

approximately

fX(x) dx (shaded)





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Definition.

The expected value or mean of a continuousrandom variable X with probability density function fX is

E(X) := X :=

xfX(x) dx.

This formula is exactly the same as the formula for the center ofmass of a linear mass density of total mass 1.

Cx =

x(x) dx.

Hence the analogy between probability and mass andprobability density and mass density persists.

As noted, the Central Limit Theorem will show that theexpected value gives the long term averages of sample values.



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Definition. The expected value or mean of a continuousrandom variable X with probability density function fX is

E(X) := X :=

xfX(x) dx.


Cx =

x(x) dx.





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E(X)

:= X :=

xfX(x) dx.


Cx =

x(x) dx.





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E(X) := X

:=

xfX(x) dx.


Cx =

x(x) dx.





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E(X) := X :=

xfX(x) dx.


Cx =

x(x) dx.





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The Expected Value Itself Need Not Be VeryLikely

qE(X)

So we should not necessarily expect measurements to givenumbers near the expected value. Instead, long term averageswill be near the expected value. (So maybe expected averagewould be more accurate, but expected value is customary.)



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q

E(X)




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qE(X)




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qE(X)

So we should not necessarily expect measurements to givenumbers near the expected value.

Instead, long term averageswill be near the expected value. (So maybe expected averagewould be more accurate, but expected value is customary.)



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qE(X)

So we should not necessarily expect measurements to givenumbers near the expected value. Instead, long term averageswill be near the expected value.

(So maybe expected averagewould be more accurate, but expected value is customary.)



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qE(X)




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Theorem.

The probability density function of a randomvariable UA,B that is uniformly distributed over the interval

[A,B] is f (x;A,B) ={ 1

BA ; for A x B,0; otherwise.

The expected

value is E(UA,B) =A+B

2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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Theorem. The probability density function of a randomvariable UA,B that is uniformly distributed over the interval

[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B)

=

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx

= B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

A

x1

BAdx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)

=B+A

2



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[A,B] is f (x;A,B) ={ 1


The expected


2.

Proof.

E(UA,B) =

xf (x;A,B) dx = B

Ax

1BA

dx

=1

2(BA)x2BA

=B2A2

2(BA)=

B+A2



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Visualization

A BuBA

2

AAA



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Visualization

A

BuBA

2

AAA



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Visualization

A B

uBA

2

AAA



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Visualization

A Bu

BA2

AAA



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Visualization

A BuBA

2

AAA



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Theorem.

The probability density function of an exponentiallydistributed random variable W is

f (x;) ={ 1

e x ; for x 0,

0; otherwise.

The expected value is

E(W) = .

Proof. Good exercise for integration by parts.

Warning. Exponential distributions are also often given using

the parameter =1

.



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Theorem. The probability density function of an exponentiallydistributed random variable W is

f (x;) ={ 1

e x ; for x 0,

0; otherwise.


E(W) = .



the parameter =1

.



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f (x;) ={ 1

e x ; for x 0,

0; otherwise.


E(W) = .

Proof.

Good exercise for integration by parts.


the parameter =1

.



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f (x;) ={ 1

e x ; for x 0,

0; otherwise.


E(W) = .



the parameter =1

.



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f (x;) ={ 1

e x ; for x 0,

0; otherwise.


E(W) = .


Warning.

Exponential distributions are also often given using

the parameter =1

.



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f (x;) ={ 1

e x ; for x 0,

0; otherwise.


E(W) = .



the parameter =1

.



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Visualization

tLLL



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Visualization

t

LLL



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Visualization

tLLL



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Theorem.

The probability density function of a normallydistributed random variable N, with parameters and is

f (x; ,) =1

2e

(x)2

22 .

The expected value is E(N, ) = .

Proof. Substitution z :=x

leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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Theorem. The probability density function of a normallydistributed random variable N, with parameters and is

f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .


Proof.

Substitution z :=x

leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads todzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, )

=

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx

=

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz

= 0+ = .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+

= .



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f (x; ,) =1

2e

(x)2

22 .



leads to

dzdx

=1

or dx = dz.

E(N, ) =

x1

2e

(x)2

22 dx =

x1

2e

( x )2

2 dx

=

(z + )1

2e

z22 dz

=

z12

ez22 dz+

12

ez22 dz = 0+ = .



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Visualization

sBBB



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Visualization

s

BBB



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Visualization

sBBB



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Theorem.

If X is a continuous random variable with densityfunction fX and g() is a function, then

E(g(X)

)=

g(x)fX(x) dx.

Theorem. If X is a continuous random variable, g() and h()are functions, and a,b,c are numbers, then the expected valueof ag(X)+bh(X)+ c is

E(ag(X)+bh(X)+ c

)= aE

(g(X)

)+bE

(h(X)

)+ c



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Theorem. If X is a continuous random variable with densityfunction fX and g() is a function, then

E(g(X)

)=

g(x)fX(x) dx.


E(ag(X)+bh(X)+ c

)= aE

(g(X)

)+bE

(h(X)

)+ c



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E(g(X)

)=

g(x)fX(x) dx.

Theorem.

If X is a continuous random variable, g() and h()are functions, and a,b,c are numbers, then the expected valueof ag(X)+bh(X)+ c is

E(ag(X)+bh(X)+ c

)= aE

(g(X)

)+bE

(h(X)

)+ c



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E(g(X)

)=

g(x)fX(x) dx.


E(ag(X)+bh(X)+ c

)= aE

(g(X)

)+bE

(h(X)

)+ c



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The Spread of Measurements

handmeasurement

-

average

+ ++ + ++ ++ ++

electronicmeasurement

-

average

+ +++ + +++ ++



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The Spread of Measurementshandmeasurement

-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+

++ + ++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ +

+ + ++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++

+ ++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ +

++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ + +

+ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++

++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++ +

+ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++ ++

++


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++ ++ +

+


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+

+++ + +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +

++ + +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ ++

+ + +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++

+ +++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ +

+++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +

++ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + ++

+ ++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +++

++



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +++ +

+



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-

average

+ ++ + ++ ++ ++


-

average

+ +++ + +++ ++



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Definition.

The variance of a continuous random variable Xwith probability density function fX is

V(X) := E((

XE(X))2) =

(xE(X)

)2fX(x) dx.The standard deviation of X is X :=

V(X).

Theorem. V(X) = E(X2)(E(X)

)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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Definition. The variance of a continuous random variable Xwith probability density function fX is

V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X)

:= E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2)

=

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)

)2fX(x) dx.

The standard deviation of X is X :=

V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).

Theorem.

V(X) = E(X2)(E(X)

)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.

V(X) = E((

XE(X))2) = E(X22XE(X)+ (E(X))2)

= E(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X)

= E((

XE(X))2) = E(X22XE(X)+ (E(X))2)

= E(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2)

= E(

X22XE(X)+(E(X)

)2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)

= E(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2

= E(X2)(E(X)

)2.



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V(X) := E((

XE(X))2) =

(xE(X)


V(X).


)2.

Proof.V(X) = E

((XE(X)

)2) = E(X22XE(X)+ (E(X))2)= E

(X2)2E(X)E(X)+

(E(X)

)2 = E(X2) (E(X))2.Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Theorem.

Let UA,B be a random variable that is uniformly

distributed over the interval [A,B]. Then V(UA,B) =(BA)2

12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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Theorem. Let UA,B be a random variable that is uniformly

distributed over the interval [A,B].

Then V(UA,B) =(BA)2

12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B)

= E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2

=

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2

= B

Ax2

1BA

dx (B+A)2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4

=1

3(BA)x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4

=(BA)

(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4

=(BA)2

12



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12.

Proof.

V(UA,B) = E(

U2A,B)(E(UA,B)

)2 =

x2fA,B(x) dx(

B+A2

)2=

BA

x21

BAdx (B+A)

2

4=

13(BA)

x3BA (B+A)

2

4

=B3A3

3(BA) (B+A)

2

4=

(BA)(B2 +AB+A2

)3(BA)

(B+A)2

4

=B2 +AB+A2

3 B

2 +2AB+A2

4=

(BA)2

12



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Theorem.

Let W be a random variable that is exponentiallydistributed with parameter . Then

V(W) = 2.

Proof. Good exercise in integration by parts.



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Theorem. Let W be a random variable that is exponentiallydistributed with parameter .

Then

V(W) = 2.




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Theorem. Let W be a random variable that is exponentiallydistributed with parameter . Then

V(W) = 2.




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V(W) = 2.

Proof.

Good exercise in integration by parts.



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V(W) = 2.




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Theorem.

Let N, be a random variable that is normallydistributed with parameters and . Then V(N, ) = 2.


leads to

dzdx

=1

or dx = dz.

V(N, ) =

(x)2 1

2e

(x)2

22 dx =

(z)212

ez22 dz

= 2

z z 12

ez22 dz (integration by parts)

= 2[z 1

2e

z22

zz

1

2e

z22 dz

]= 2 [0+1] = 2.



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Theorem. Let N, be a random variable that is normallydistributed with parameters and .

Then V(N, ) = 2.


leads to

dzdx

=1

or dx = dz.

V(N, ) =

(x)2 1

2e

(x)2

22 dx =

(z)212

ez22 dz

= 2

z z 12


= 2[z 1

2e

z22

zz

1

2e

z22 dz

]= 2 [0+1] = 2.



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Theorem. Let N, be a random variable that is normallydistributed with parameters and . Then V(N, ) = 2.


leads to

dzdx

=1

or dx = dz.

V(N, ) =

(x)2 1

2e

(x)2

22 dx =

(z)212

ez22 dz

= 2

z z 12


= 2[z 1

2e

z22

zz

1

2e

z22 dz

]= 2 [0+1] = 2.



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Proof.

Substitution z :=x

leads to

dzdx

=1

or dx = dz.

V(N, ) =

(x)2 1

2e

(x)2

22 dx =

(z)212

ez22 dz

= 2

z z 12


= 2[z 1

2e

z22

zz

1

2e

z22 dz

]= 2 [0+1] = 2.



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leads to

dzdx

=1

or dx = dz.

V(N, ) =

(x)2 1

2e

(x)2

22 dx =

(z)212

ez22 dz

= 2

z z 12


= 2[z 1

2e

z22

zz

1

2e

z22 dz

]= 2 [0+1] = 2.



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Proof.

Documents

expected_values_continuous.pdf