Expected Value and Variance for Continuous Random Variables€¦ · expected value gives the long term averages of sample values. Bernd Schroder¨ Louisiana Tech University, College

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




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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)





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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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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) := µX :=∫

∞

−∞

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 =∫

∞

−∞

xρ(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) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X)

:= µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX

:=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(x) dx.





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E(X) := µX :=∫

∞

−∞

xfX(x) dx.


Cx =∫

∞

−∞

xρ(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 average”would be more accurate, but “expected value” is customary.)



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qE(X)




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qE(X)




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qE(X)




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q

E(X)




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qE(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 average”would 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 average”would 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

B−A ; 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

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

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


0; otherwise.

The expected


2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B)

=∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞

xf (x;A,B) dx

=∫ B

Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


A

x1

B−Adx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)

=B+A

2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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




2.

Proof.

E(UA,B) =∫

∞

−∞


Ax

1B−A

dx

=1

2(B−A)x2∣∣∣∣BA

=B2−A2

2(B−A)=

B+A2



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Visualization

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2

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Visualization

A BuB−A

2

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Visualization

A BuB−A

2

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Visualization

A BuB−A

2

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Visualization

A

BuB−A

2

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BuB−A

2

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A B

uB−A

2

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A B

uB−A

2

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Visualization

A Bu

B−A2

AAA�

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A BuB−A

2

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Visualization

A BuB−A

2

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Theorem.

The probability density function of an exponentiallydistributed random variable WΘ is

f (x;Θ) ={ 1

Θe−

xΘ ; for x≥ 0,0; otherwise.

The expected value is

E(WΘ) = Θ.

Proof. Good exercise for integration by parts.

Warning. Exponential distributions are also often given using

the parameter λ =1Θ

.



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

f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.




.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.




.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.

Proof.

Good exercise for integration by parts.



.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.




.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.




.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.


Warning.

Exponential distributions are also often given using


.



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f (x;Θ) ={ 1

Θe−



E(WΘ) = Θ.




.



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Visualization

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ΘLLL�

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Theorem.

The probability density function of a normallydistributed random variable Nµ,σ with parameters µ and σ is

f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .

The expected value is E(Nµ,σ ) = µ.

Proof. Substitution z :=x−µ

σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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

f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .


Proof.

Substitution z :=x−µ

σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σ

leads todzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ )

=∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx

=∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz

= 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ

= µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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f (x; µ,σ) =1

σ√

2πe−

(x−µ)2

2σ2 .



σleads to

dzdx

=1σ

or dx = σ dz.

E(Nµ,σ ) =∫

∞

−∞

x1

σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

x1

σ√

2πe−

( x−µσ )2

2 dx

=∫

∞

−∞

(zσ + µ)1

σ√

2πe−

z22 σ dz

=∫

∞

−∞

zσ1√2π

e−z22 dz+

∫∞

−∞

µ1√2π

e−z22 dz = 0+ µ = µ.



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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

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-

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average

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Definition.

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

V(X) := E((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.

The standard deviation of X is σX :=√

V(X).

Theorem. V(X) = E(X2)− (E(X)

)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).

Theorem.

V(X) = E(X2)− (E(X)

)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.

V(X) = E((

X−E(X))2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X)

= E((

X−E(X))2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(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((

X−E(X))2)

=∫

∞

−∞

(x−E(X)

)2fX(x) dx.


V(X).


)2.

Proof.V(X) = E

((X−E(X)

)2)

= E(

X2−2XE(X)+(E(X)

)2)

= E(X2)−2E(X)E(X)+

(E(X)

)2 = E(X2)− (E(X)

)2.



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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) =(B−A)2

12.

Proof.

V(UA,B) = E(

U2A,B

)−(E(UA,B)

)2 =∫

∞

−∞

x2fA,B(x) dx−(

B+A2

)2

=∫ B

Ax2 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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) =(B−A)2

12.

Proof.

V(UA,B) = E(

U2A,B

)−(E(UA,B)

)2 =∫

∞

−∞

x2fA,B(x) dx−(

B+A2

)2

=∫ B

Ax2 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4

=1

3(B−A)x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4

=(B−A)

(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4

=(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)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 1

B−Adx− (B+A)2

4=

13(B−A)

x3∣∣∣∣BA− (B+A)2

4

=B3−A3

3(B−A)− (B+A)2

4=

(B−A)(B2 +AB+A2)

3(B−A)− (B+A)2

4

=B2 +AB+A2

3− B2 +2AB+A2

4=

(B−A)2

12



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Theorem.

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

V(WΘ) = Θ2.

Proof. Good exercise in integration by parts.



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

Then

V(WΘ) = Θ2.




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

V(WΘ) = Θ2.




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V(WΘ) = Θ2.

Proof.

Good exercise in integration by parts.



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V(WΘ) = Θ2.




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V(WΘ) = Θ2.




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Theorem.

Let Nµ,σ be a random variable that is normallydistributed with parameters µ and σ . Then V(Nµ,σ ) = σ

2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π

e−z22 dz (integration by parts)

= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



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Theorem. Let Nµ,σ be a random variable that is normallydistributed with parameters µ and σ .

Then V(Nµ,σ ) = σ2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



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Theorem. Let Nµ,σ be a random variable that is normallydistributed with parameters µ and σ . Then V(Nµ,σ ) = σ

2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



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2.

Proof.

Substitution z :=x−µ

σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



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2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



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2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ )

=∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx

=∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π

e−z22 dz

(integration by parts)

= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2

[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]

= σ2 [0+1] = σ

2.



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2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2

[0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0

+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1]

= σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



logo1



2.


σleads to

dzdx

=1σ

or dx = σdz.

V(Nµ,σ ) =∫

∞

−∞

(x−µ)2 1σ√

2πe−

(x−µ)2

2σ2 dx =∫

∞

−∞

(σz)2 1√2π

e−z22 dz

= σ2∫

∞

−∞

z · z 1√2π


= σ2[−z

1√2π

e−z22

∣∣∣∣z→∞

z→−∞

−∫

∞

−∞

− 1√2π

e−z22 dz

]= σ

2 [0+1] = σ2.



Expected Value and Variance for Continuous Random Variables€¦ · expected value gives the long term averages of sample values. Bernd Schroder¨ Louisiana Tech University, College

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