Random Variables - STATISTICS -- Lecture no. 3k101.unob.cz/~neubauer/pdf/stat_lecture3.pdf · Random Variable Random Variables STATISTICS – Lecture no. 3 Jiˇr´ı Neubauer Department

Random Variable

Random VariablesSTATISTICS – Lecture no. 3

Jirı Neubauer

Department of Econometrics FEM UO Brnooffice 69a, tel. 973 442029email:[email protected]

13. 10. 2009

Jirı Neubauer Random Variables

Random Variable

Discrete Random VariableContinuous Random VariableMeasures of LocationMeasures of DispersionMeasures of Concentration

Random Variable

Many random experiments have numerical outcomes.

Definition

A random variable is a real-valued function X (ω) defined on thesample space Ω.

The set of possible values of the random variable X is called therange of X .

M = x ;X (ω) = x.


Random Variable


Random Variable

We denote random variables by capital letters X ,Y , . . .(eventually X1,X2, . . . ) and their particular values by smallletters x , y , . . . . Using random variables we can describerandom events, for example X = x ,X ≤ x , x1 < X < x2 etc.

Examples of random variables:

the number of dots when a die is rolled, the range isM = 1, 2, . . . 6the number of rolls of a die until the first 6 appears, the rangeis M = 1, 2, the lifetime of the lightbulb, the range is M = x ; x ≥ 0,


Random Variable


Random Variable

We denote random variables by capital letters X ,Y , . . .(eventually X1,X2, . . . ) and their particular values by smallletters x , y , . . . . Using random variables we can describerandom events, for example X = x ,X ≤ x , x1 < X < x2 etc.

Examples of random variables:

the number of dots when a die is rolled, the range isM = 1, 2, . . . 6the number of rolls of a die until the first 6 appears, the rangeis M = 1, 2, the lifetime of the lightbulb, the range is M = x ; x ≥ 0,


Random Variable


Random Variable

According to the range M we separate random variables to

discrete . . .M is finite or countable,

continuous . . .M is a closed or open interval.


Random Variable


Random Variable

According to the range M we separate random variables to

discrete . . .M is finite or countable,

continuous . . .M is a closed or open interval.


Random Variable


Random Variable

Examples of discrete random variables:

the number of cars sold at a dealership during a given month,M = 0, 1, 2, . . . the number of houses in a certain block, M = 1, 2, . . . the number of fish caught on a fishing trip, M = 0, 1, 2, . . . the number of heads obtained in three tosses of a coin,M = 0, 1, 2, 3

Examples of continuous random variables:

the height of a person, M = (0,∞)

the time taken to complete an examination, M = (0,∞)

the amount of milk in a bottle, M = (0,∞)


Random Variable


Random Variable

Examples of discrete random variables:

the number of cars sold at a dealership during a given month,M = 0, 1, 2, . . . the number of houses in a certain block, M = 1, 2, . . . the number of fish caught on a fishing trip, M = 0, 1, 2, . . . the number of heads obtained in three tosses of a coin,M = 0, 1, 2, 3

Examples of continuous random variables:

the height of a person, M = (0,∞)

the time taken to complete an examination, M = (0,∞)

the amount of milk in a bottle, M = (0,∞)


Random Variable


Random Variable

For the description of random variables we will use some functions:

a (cumulative) distribution function F (x),

a probability function p(x) – only for discrete randomvariables,

a probability density function f (x) – only for continuousrandom variables.

and some measures:

measures of location,

measures of dispersion,

measures of concentration.


Random Variable


Random Variable

For the description of random variables we will use some functions:

a (cumulative) distribution function F (x),

a probability function p(x) – only for discrete randomvariables,

a probability density function f (x) – only for continuousrandom variables.

and some measures:

measures of location,

measures of dispersion,

measures of concentration.


Random Variable


Distribution Function

Definition

Let X be any random variable. The distribution function F (x) ofthe random variable X is defined as

F (x) = P(X ≤ x), x ∈ R.

Note: distribution function = cumulative distribution function


Random Variable



We mention some important properties of F (x):

for every real x : 0 ≤ F (x) ≤ 1,

F (x) is a non-decreasing, right-continuous function,

it has limits

limx→−∞

F (x) = 0, limx→∞

F (x) = 1,

if range of X is M = x ; x ∈ (a, b〉 then F (a) = 0 aF (b) = 1,

for every real numbers x1 and x2:P(x1 < X ≤ x2) = F (x2)− F (x1).


Random Variable






it has limits

limx→−∞

F (x) = 0, limx→∞

F (x) = 1,




Random Variable






it has limits

limx→−∞

F (x) = 0, limx→∞

F (x) = 1,




Random Variable






it has limits

limx→−∞

F (x) = 0, limx→∞

F (x) = 1,




Random Variable


Discrete Random Variable

For a discrete random variable X , we are interested in computingprobabilities of the type P(X = xk) for various values xk in rangeof X .

Definition

Let X be a discrete random variable with range x1, x2, . . . (finiteor countably infinite). The function

p(x) = P(X = x)

is called the probability function of X .

Note: probability function = probability mass function


Random Variable



We mention some important properties of p(x)

for every real number x , 0 ≤ p(x) ≤ 1,

∑x∈M

p(x) = 1

for every two real numbers xk and xl (xk ≤ xl):

P(xk ≤ X ≤ xl) =

xl∑xi=xk

p(xi ).


Random Variable




for every real number x , 0 ≤ p(x) ≤ 1,∑x∈M

p(x) = 1


P(xk ≤ X ≤ xl) =

xl∑xi=xk

p(xi ).


Random Variable




for every real number x , 0 ≤ p(x) ≤ 1,∑x∈M

p(x) = 1


P(xk ≤ X ≤ xl) =

xl∑xi=xk

p(xi ).


Random Variable


Probability Function

The probability function p(x) can be described by

the table,

X x1 x2 . . . xi . . .∑

p(x) p(x1) p(x2) . . . p(xi ) . . . 1


Random Variable



the graph [x , p(x)],


Random Variable



the formula, for example

p(x) =

π(1− π)x x = 0, 1, 2, . . . ,

0 otherwise,

where π is a given probability.


Random Variable


Example

The shooter has 3 bullets and shoots at the target until the firsthit or until the last bullet. The probability that the shooter hits thetarget after one shot is 0.6. The random variable X is the numberof the fired bullets. Find the probability and the distributionfunction of the given random variable. What is the probability thatthe number of the fired bullets will not be larger then 2?


Random Variable


Example

Random variable X is discrete with the range M = 1, 2, 3. Theprobability function is:

p(1) = P(X = 1) = 0.6,

p(2) = P(X = 2) = 0.4 · 0.6 = 0.24,

p(3) = P(X = 3) = 0.4·0.4·0.6+0.4·0.4·0.4 = 0.4·0.4 = 0.16.


Random Variable


Example


p(1) = P(X = 1) = 0.6,

p(2) = P(X = 2) = 0.4 · 0.6 = 0.24,

p(3) = P(X = 3) = 0.4·0.4·0.6+0.4·0.4·0.4 = 0.4·0.4 = 0.16.


Random Variable


Example


p(1) = P(X = 1) = 0.6,

p(2) = P(X = 2) = 0.4 · 0.6 = 0.24,

p(3) = P(X = 3) = 0.4·0.4·0.6+0.4·0.4·0.4 = 0.4·0.4 = 0.16.


Random Variable


Example

All results are summarized in the table

x 1 2 3∑

p(x) 0.6 0.24 0.16 1

The probability function can be described by the formula

p(x) =

0.6 · 0.4x−1 x = 1, 2,

0.42 x = 3,0 otherwise.


Random Variable


Example

We can calculate some values of the distribution function F (x):

F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example


F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example


F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example


F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example


F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example


F (0) = P(X ≤ 0) = 0,

F (1) = P(X ≤ 1) = p(1) = 0.6,

F (1.5) = P(X ≤ 1.5) = P(X ≤ 1) = p(1) = 0.6,

F (2) = P(X ≤ 2) = p(1) + p(2) = 0.84,

F (3) = P(X ≤ 3) = p(1) + p(2) + p(3) = 1,

F (4) = P(X ≤ 4) = p(1) + p(2) + p(3) = 1.


Random Variable


Example

We can write

F (x) =

0 x < 1,

0.6 1 ≤ x < 2,0.84 2 ≤ x < 3,1 x ≥ 3.


Random Variable


Example

Figure: The probability and the distribution function


Random Variable


Example

What is the probability that the number of the fired bullets will notbe larger then 2?

P(X ≤ 2) = P(X = 1) + P(X = 2) = p(1) + p(2) = F (2) == 0.6 + 0.24 = 0.84.


Random Variable


Probability Density Function

If the cumulative distribution function is a continuous function,then X is said to be a continuous random variable.

Definition

The probability density function of the random variable X is anon-negative function f (x) such that

F (x) =

x∫−∞

f (t)dt, x ∈ R.


Random Variable



Some properties of f (x):∞∫−∞

f (x)dx =∫M

f (x)dx = 1,

f (x) = dF (x)dx = F ′(x), where the derivative exists,

P(x1 ≤ X ≤ x2) = P(x1 < X < x2) = P(x1 < X ≤ x2) =

P(x1 ≤ X < x2) = F (x2)− F (x1) =x2∫x1

f (x)dx

If X is a continuous random variable, then P(X = x) = 0.


Random Variable




f (x)dx =∫M

f (x)dx = 1,


P(x1 ≤ X ≤ x2) = P(x1 < X < x2) = P(x1 < X ≤ x2) =

P(x1 ≤ X < x2) = F (x2)− F (x1) =x2∫x1

f (x)dx



Random Variable




f (x)dx =∫M

f (x)dx = 1,


P(x1 ≤ X ≤ x2) = P(x1 < X < x2) = P(x1 < X ≤ x2) =

P(x1 ≤ X < x2) = F (x2)− F (x1) =x2∫x1

f (x)dx



Random Variable




f (x)dx =∫M

f (x)dx = 1,


P(x1 ≤ X ≤ x2) = P(x1 < X < x2) = P(x1 < X ≤ x2) =

P(x1 ≤ X < x2) = F (x2)− F (x1) =x2∫x1

f (x)dx



Random Variable



The function f (x) we can describe by a formula or a graph, forexample

f (x) =

15e−

x−25 pro x > 2,

0 pro x ≤ 2.


Random Variable


Example

The random variable X has the probability density function

f (x) =

cx2(1− x) 0 < x < 1,

0 otherwise.

Determine a constant c in order that f (x) is a probability densityfunction. Find a distribution function of the random variable X .Calculate the probability P(0.2 < X < 0.8).


Random Variable


Example

The probability density function has to fulfil∫M

f (x)dx = 1.

1∫0

cx2(1− x)dx = c1∫0

(x2 − x3)dx = c[

x3

3 −x4

4

]1

0=

= c[

13 −

14

]= c

12 = 1,

we get c = 12.


Random Variable


Example

The distribution function can be calculated by the definition ofthe probability density function. We can write for 0 < x < 1

F (x) =x∫0

12t2(1− t)dt = 12x∫0

(t2 − t3)dt = 12[

t3

3 −t4

4

]x

0=

= 12[

x3

3 −x4

4

]= 4x3 − 3x4.

F (x) =

0 x ≤ 0,

x3(4− 3x) 0 < x < 1,1 otherwise.


Random Variable


Figure: The probability density function and the distribution function


Random Variable


Example

Using the probability density function we can calculate

P(0.2 < X < 0.8) =

0.8∫0.2

12x2(1− x)dx =[4x3 − 3x4

]0.8

0.2= 0.792.

If the distribution function is known, we can do simpler calculation

P(0.2 < X < 0.8) = F (0.8)− F (0.2) == 0.83(4− 3 · 0.8)− 0.23(4− 3 · 0.2) = 0.792.


Random Variable


Example

Using the probability density function we can calculate

P(0.2 < X < 0.8) =

0.8∫0.2

12x2(1− x)dx =[4x3 − 3x4

]0.8

0.2= 0.792.

If the distribution function is known, we can do simpler calculation

P(0.2 < X < 0.8) = F (0.8)− F (0.2) == 0.83(4− 3 · 0.8)− 0.23(4− 3 · 0.2) = 0.792.


Random Variable


Example

A random variable X is described by the distribution function

F (x) =

0 x ≤ 0,

1− e−x x > 0.

Find a probability density function.


Random Variable


Example

Using mentioned formula

f (x) =dF (x)

dx

and the fact that ddx (1− e−x) = e−x we get

f (x) =

0 x ≤ 0,

e−x x > 0.


Random Variable


Measures of Location

The distribution function (the probability function or theprobability density function) gives us the complete informationabout the random variable. Sometimes is useful to know somesimpler and concentrated formulation of this information such asmeasures of location, dispersion and concentration.

The best known measures of location are a mean (an expectedvalue), quantiles (a median, an upper and a lower quartile, . . . )and a mode.


Random Variable


Expected Value

Definition

The mean (the expected value) E (X ) of the random variable X(sometimes denoted as µ) is the value that is expected to occurper repetition, if an experiment is repeated a large number oftimes. For the discrete random variable is defined as

E (X ) =∑M

xip(xi ),

for the continuous random variable as

E (X ) =

∫M

xf (x)dx

if the given sequence or integral absolutely converges.Jirı Neubauer Random Variables

Random Variable


Expected Value

We mention some properties of the mean:

the mean of the constant c is equal to this constant

E (c) = c ,

the mean of the product of the constant c and the randomvariable X is equal to the product of the given constant c andthe mean of X

E (cX ) = cE (X ),


Random Variable


Expected Value

We mention some properties of the mean:

the mean of the constant c is equal to this constant

E (c) = c ,

the mean of the product of the constant c and the randomvariable X is equal to the product of the given constant c andthe mean of X

E (cX ) = cE (X ),


Random Variable


Expected Value

the mean of the sum of random variables X1,X2, . . . ,Xn isequal to the sum of the mean of the given random variables,

E (X1 + X2 + · · ·+ Xn) = E (X1) + E (X2) + · · ·+ E (Xn),

if X1,X2, . . . ,Xn are independent, then the mean of theirproduct is equal to the product of their means

E (X1X2 . . .Xn) = E (X1)E (X2) . . .E (Xn).


Random Variable


Expected Value

the mean of the sum of random variables X1,X2, . . . ,Xn isequal to the sum of the mean of the given random variables,

E (X1 + X2 + · · ·+ Xn) = E (X1) + E (X2) + · · ·+ E (Xn),

if X1,X2, . . . ,Xn are independent, then the mean of theirproduct is equal to the product of their means

E (X1X2 . . .Xn) = E (X1)E (X2) . . .E (Xn).


Random Variable


Random Variables and Independency

The random variables X1,X2, . . . ,Xn are independent if and only iffor any numbers x1, x2, . . . , xn ∈ R is

P(X1 ≤ x1,X2 ≤ x2, . . . ,Xn ≤ xn) =

= P(X1 ≤ x1) · P(X2 ≤ x2) · · ·P(Xn ≤ xn).

Let us have the random vector X = (X1,X2, . . . ,Xn), whosecomponents X1,X2, . . . ,Xn are the random variables.F (x) = F (x1, x2, . . . , xn) = P(X1 ≤ x1,X2 ≤ x2, . . . ,Xn ≤ xn) isthe distribution function of the vector X andF (x1),F (x2), . . . ,F (xn) are the distribution functions of therandom variables X1,X2, . . . ,Xn. The random variablesX1,X2, . . . ,Xn are independent if and only if

F (x1, x2, . . . , xn) = F (x1) · F (x2) · · ·F (xn).


Random Variable


Random Variables and Independency

If X is the random vector whose components are the discreterandom variables, the functionp(x) = p(x1, x2, . . . , xn) = P(X1 = x1,X2 = x2, . . . ,Xn = xn) is theprobability function of the vector X, p(x1), p(x2), . . . , p(xn) are theprobability functions of X1,X2, . . . ,Xn, then:X1,X2, . . . ,Xn are independent if and only if

p(x1, x2, . . . , xn) = p(x1) · p(x2) · · · p(xn).

If X is the random vector whose components are the continuousrandom variables, the function f (x) = f (x1, x2, . . . , xn) is theprobability density function of the vector X, f (x1), f (x2), . . . , f (xn)are the probability density functions of X1,X2, . . . ,Xn, then:X1,X2, . . . ,Xn are independent if and only if

f (x1, x2, . . . , xn) = f (x1) · f (x2) · · · f (xn).


Random Variable


Quantile

Definition

100P% quantile xP of the random variable with the increasingdistribution function is such value of the random variable that

P(X ≤ xP) = F (xP) = P, 0 < P < 1.


Random Variable


Quantile

The quantile x0.50 we call median Me(X ), it fulfilsP(X ≤ Me(X )) = P(X ≥ Me(X )) = 0.50. The quantile x0.25 iscalled the lower quartile, the quantile x0.75 is called the upperquartile. The selected quantiles of some important distributionsare tabulated.


Random Variable


Mode

Definition

The mode Mo(X ) is the value of the random variable with thehighest probability (for the discrete random variable), or the value,where the function f (x) has the maximum (for the continuousrandom variable ).


Random Variable


Example

Find the mean (the expected value) and the mode of the randomvariable defined as the number of fired bullets (see the examplebefore). The probability function is

p(x) =

0.6 · 0.4x−1 x = 1, 2,

0.42 x = 3,0 otherwise.


Random Variable


Example

The mean (the expected value) we get using the formula from thedefinition of E (X )

E (x) =3∑

i=1

xip(xi ) = 1 · 0.6 · 0.40 + 2 · 0.6 · 0.41 + 3 · 0.42 = 1.56.

The mode is the value of the given random variable with thehighest probability which is Mo(X ) = 1, because p(1) = 0.6.


Random Variable


Example

The mean (the expected value) we get using the formula from thedefinition of E (X )

E (x) =3∑

i=1

xip(xi ) = 1 · 0.6 · 0.40 + 2 · 0.6 · 0.41 + 3 · 0.42 = 1.56.

The mode is the value of the given random variable with thehighest probability which is Mo(X ) = 1, because p(1) = 0.6.


Random Variable


Example

The random variable X is described by the probability densityfunction

f (x) =

12x2(1− x) 0 < x < 1,

0 otherwise.

Find the mean (the expected value) and the mode.


Random Variable


Example

We calculate the mean using the definition formula

E (X ) =

1∫0

xf (x)dx =

1∫0

x ·12x2(1−x) = 12

[x4

4− x5

5

]1

0

=3

5= 0.6.

The mode is the maximum of the probability density function. Wehave to find the maximum of f (x) on the interval 0 < x < 1,ddx

[12x2(1− x)

]= 12(2x − 3x2) = 0, x(2− 3x) = 0, we get x = 0

or x = 2/3. The maximum of f (x) is in x = 2/3 ⇒ Mo(X ) = 2/3.


Random Variable


Example

We calculate the mean using the definition formula

E (X ) =

1∫0

xf (x)dx =

1∫0

x ·12x2(1−x) = 12

[x4

4− x5

5

]1

0

=3

5= 0.6.

The mode is the maximum of the probability density function. Wehave to find the maximum of f (x) on the interval 0 < x < 1,ddx

[12x2(1− x)

]= 12(2x − 3x2) = 0, x(2− 3x) = 0, we get x = 0

or x = 2/3. The maximum of f (x) is in x = 2/3 ⇒ Mo(X ) = 2/3.


Random Variable


Example

Find the median,the upper and the lower quartile of the randomvariable X with the distribution function

F (x) =

1− 1

x3 x > 1,0 x ≤ 1.


Random Variable


Example

The quantile is defined by the formula F (xP) = P.

1− 1

x3P

= P,

xP =1

3√

1− P.

median x0.50 = 13√1−0.50

= 1.260,

lower quartile x0.25 = 13√1−0.25

= 1.101,

upper quartile x0.75 = 13√1−0.75

= 1.587.


Random Variable


Example

The quantile is defined by the formula F (xP) = P.

1− 1

x3P

= P,

xP =1

3√

1− P.

median x0.50 = 13√1−0.50

= 1.260,

lower quartile x0.25 = 13√1−0.25

= 1.101,

upper quartile x0.75 = 13√1−0.75

= 1.587.


Random Variable


Measures of Dispersion

The elementary and widely-used measures of dispersion are thevariance and the standard deviation.


Random Variable


Variance

Definition

The variance D(X ) of the random variable X (sometimes denotedas σ2) is defined by the formula

D(X ) = E[X − E (X )]2

.

The variance of the discrete random variable is given by

D(X ) =∑M

[xi − E (X )]2p(xi ),

the variance of the continuous random variable is

D(X ) =

∫M

[x − E (X )]2f (x)dx .


Random Variable


Variance

Some properties of the variance:

D(k) = 0, where k is a constant,

D(kX ) = k2D(X ),

D(X + Y ) = D(X ) + D(Y ), if X and Y are independent,

D(X ) ≥ 0 for every random variable,


Random Variable


Variance



D(kX ) = k2D(X ),




Random Variable


Variance



D(kX ) = k2D(X ),




Random Variable


Variance



D(kX ) = k2D(X ),




Random Variable


Variance

D(X ) = E (X 2)− E (X )2,D(X ) = E [X − E (X )]2 = E [X 2 − 2XE (X ) + E (X )2] =

= E (X 2)− E [2XE (X )] + E [E (X )2] == E (X 2)− 2E (X )E (X ) + E (X )2 = E (X 2)− E (X )2

For the discrete random variable

D(X ) =∑M

x2i p(xi )− E (X )2,

for the continuous random variable

D(X ) =

∫M

x2f (x)dx − E (X )2.


Random Variable


Standard Deviation

Definition

The standard deviation σ(X ) of the random variable X isdefined as the square root of the variance

σ(X ) =√

D(X ).

The standard deviation has the same unit as the random variableX .


Random Variable


Figure: Relation between the mean and the standard deviation


Random Variable


Example

Find the variance and the standard deviation of the randomvariable defined as the number of fired bullets (see the examplebefore).


Random Variable


Example

The mean is E (X ) = 1.56 (see the previous example). For thepurpose of calculation of the variance we use the formula

D(X ) = E (X 2)− E (X )2

E (X 2) =∑M

x2i p(xi ) =

3∑i=1

x2i p(xi ) = 12·0.6+22·0.24+32·0.16 = 3,

thenD(X ) = 3− 1.562 = 0.566.

The standard deviation is the square root of the variance

σ(X ) =√

D(X ) = 0.753.


Random Variable


Example


D(X ) = E (X 2)− E (X )2

E (X 2) =∑M

x2i p(xi ) =

3∑i=1

x2i p(xi ) = 12·0.6+22·0.24+32·0.16 = 3,

thenD(X ) = 3− 1.562 = 0.566.


σ(X ) =√

D(X ) = 0.753.


Random Variable


Example


D(X ) = E (X 2)− E (X )2

E (X 2) =∑M

x2i p(xi ) =

3∑i=1

x2i p(xi ) = 12·0.6+22·0.24+32·0.16 = 3,

thenD(X ) = 3− 1.562 = 0.566.


σ(X ) =√

D(X ) = 0.753.


Random Variable


Example

Find the variance and the standard deviation of the randomvariable X with the probability density function

f (x) =

12x2(1− x) 0 < x < 1,

0 otherwise.


Random Variable


Example

The mean is E (X ) = 3/5 (see the previous example).

D(X ) = E (X 2)− E (X )2

E (X 2) =∫M

x2f (x)dx =1∫0

x2 · 12x2(1− x)dx = 12[

x5

5 −x6

6

]1

0

= 25 = 0.4,

D(X ) =2

5−

(3

5

)2

=1

25= 0.04.

The standard deviation is

σ(X ) =√

D(X ) =1

5= 0.2.


Random Variable


Example

The mean is E (X ) = 3/5 (see the previous example).

D(X ) = E (X 2)− E (X )2

E (X 2) =∫M

x2f (x)dx =1∫0

x2 · 12x2(1− x)dx = 12[

x5

5 −x6

6

]1

0

= 25 = 0.4,

D(X ) =2

5−

(3

5

)2

=1

25= 0.04.

The standard deviation is

σ(X ) =√

D(X ) =1

5= 0.2.


Random Variable


Measures of Concentration

We will focus on the measures describing the shape of randomvariables distribution (skewness and kurtosis). These measures aredefined by moments.


Random Variable



Definition

The r th moment µ′r of the random variable X is defined by theformula

µ′r (X ) = E (X r ) for r = 1, 2, . . . .

The r th moment of the discrete random variable is given by

µ′r (X ) =∑M

x ri p(xi ),

r th moment of the continuous random variable is

µ′r (X ) =

∫M

x r f (x)dx .


Random Variable



Definition

The r th central moment µr of the random variable X is definedby the formula

µr (X ) = E [X − E (X )]r for r = 2, 3, . . . .

The r th central moment of the discrete random variable is given by

µr (X ) =∑M

[xi − E (X )]rp(xi ),

the r th central moment of the continuous random variable is

µr (X ) =

∫M

[x − E (X )]r f (x)dx .


Random Variable


Skewness

Definition

The skewness α3(X ) is defined by the formula

α3(X ) =µ3(X )

σ(X )3.

According to the values of the skewness we can tell whetherdistribution is symmetric or asymmetric.

α3 = 0, distribution is symmetric,

α3 < 0, distribution is skewed to the right,

α3 > 0, distribution is skewed to the left.


Random Variable


Skewness

Definition

The skewness α3(X ) is defined by the formula

α3(X ) =µ3(X )

σ(X )3.

According to the values of the skewness we can tell whetherdistribution is symmetric or asymmetric.

α3 = 0, distribution is symmetric,

α3 < 0, distribution is skewed to the right,

α3 > 0, distribution is skewed to the left.


Random Variable


Skewness

Figure: The skewnessJirı Neubauer Random Variables

Random Variable


Kurtosis

Definition

The kurtosis α4(X ) is defined by the formula

α4(X ) =µ4(X )

σ(X )4− 3.

Kurtosis is a measure of how outlier-prone a distribution is. Thekurtosis of the normal distribution is 0. Distributions that are moreoutlier-prone than the normal distribution have kurtosis greaterthan 0; distributions that are less outlier-prone have kurtosis lessthan 0.


Random Variable


Kurtosis

Definition

The kurtosis α4(X ) is defined by the formula

α4(X ) =µ4(X )

σ(X )4− 3.

Kurtosis is a measure of how outlier-prone a distribution is. Thekurtosis of the normal distribution is 0. Distributions that are moreoutlier-prone than the normal distribution have kurtosis greaterthan 0; distributions that are less outlier-prone have kurtosis lessthan 0.


Random Variable


Kurtosis

Figure: The kurtosisJirı Neubauer Random Variables

Random Variable


Example

Calculate the skewness and the kurtosis of the random variabledefined as the number of fired bullets (see the previous examples).


Random Variable


Example

First of all we have to calculate 3rd and 4th central moment. Themean of the given random variable is E (X ) = 1.56, the standarddeviation is σ(X ) = 0.753.

µ3 =3∑

i=1[xi − E (X )]3p(xi ) = (1− 1.56)3 · 0.6+

+(2− 1.56)3 · 0.24 + (3− 1.56)3 · 0.16 = 0.393

µ4 =3∑

i=1[xi − E (X )]4p(xi ) = (1− 1.56)4 · 0.6+

+(2− 1.56)4 · 0.24 + (3− 1.56)4 · 0.16 = 0.756


Random Variable


Example


µ3 =3∑

i=1[xi − E (X )]3p(xi ) = (1− 1.56)3 · 0.6+

+(2− 1.56)3 · 0.24 + (3− 1.56)3 · 0.16 = 0.393

µ4 =3∑

i=1[xi − E (X )]4p(xi ) = (1− 1.56)4 · 0.6+

+(2− 1.56)4 · 0.24 + (3− 1.56)4 · 0.16 = 0.756


Random Variable


Example


µ3 =3∑

i=1[xi − E (X )]3p(xi ) = (1− 1.56)3 · 0.6+

+(2− 1.56)3 · 0.24 + (3− 1.56)3 · 0.16 = 0.393

µ4 =3∑

i=1[xi − E (X )]4p(xi ) = (1− 1.56)4 · 0.6+

+(2− 1.56)4 · 0.24 + (3− 1.56)4 · 0.16 = 0.756


Random Variable


Example

The skewness is equal to

α3 =µ3

σ3= 0.922,

the kurtosis isα4 =

µ4

σ4− 3 = −0.644.


Random Variable


Example


α3 =µ3

σ3= 0.922,

the kurtosis isα4 =

µ4

σ4− 3 = −0.644.


Random Variable


Example

Calculate the skewness and the kurtosis of the random X with theprobability density function

f (x) =

12x2(1− x) 0 < x < 1,

0 otherwise.


Random Variable


Example

The mean of the given random variable is E (X ) = 3/5. thestandard deviation is 1/5.

First of all we calculate 3rd and 4th

central moment.

µ3 =

1∫0

[x − 0.6]312x2(1− x)dx = · · · = − 2

875= −0.00229,

µ4 =

1∫0

[x − 0.6]412x2(1− x)dx = · · · = 33

8750= 0.00377.


Random Variable


Example

The mean of the given random variable is E (X ) = 3/5. thestandard deviation is 1/5. First of all we calculate 3rd and 4th

central moment.

µ3 =

1∫0

[x − 0.6]312x2(1− x)dx = · · · = − 2

875= −0.00229,

µ4 =

1∫0

[x − 0.6]412x2(1− x)dx = · · · = 33

8750= 0.00377.


Random Variable


Example

The mean of the given random variable is E (X ) = 3/5. thestandard deviation is 1/5. First of all we calculate 3rd and 4th

central moment.

µ3 =

1∫0

[x − 0.6]312x2(1− x)dx = · · · = − 2

875= −0.00229,

µ4 =

1∫0

[x − 0.6]412x2(1− x)dx = · · · = 33

8750= 0.00377.


Random Variable


Example


α3 =µ3

σ3= −2

7= −0.286,

the kurtosis is

α4 =µ4

σ4− 3 = − 9

14= −0.643.


Random Variable


Example


α3 =µ3

σ3= −2

7= −0.286,

the kurtosis is

α4 =µ4

σ4− 3 = − 9

14= −0.643.


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Random Variables - STATISTICS -- Lecture no. 3k101.unob.cz/~neubauer/pdf/stat_lecture3.pdf · Random Variable Random Variables STATISTICS – Lecture no. 3 Jiˇr´ı Neubauer Department