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Statistics and Mathematics for Economics. Statistics Component: Lecture Six. Objectives of the Lecture. To supply a distinction between a discrete and a continuous random variable To provide examples of the probability density function of a continuous random variable

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Statistics and mathematics for economics

Statistics and Mathematics for Economics

Statistics Component: Lecture Six


Objectives of the lecture
Objectives of the Lecture

  • To supply a distinction between a discrete and a continuous random variable

  • To provide examples of the probability density function of a continuous random variable

  • To outline an approach towards calculating the probability that the value of a continuous random variable lies within a specified range of values

  • To indicate how to calculate the expected value and the value of the variance of a continuous random variable


A discrete random variable
A Discrete Random Variable

  • A distinguishing feature of a discrete random variable is that it can be equal to only a finite or countably infinite number of values

  • The earliest example which was given of a discrete random variable was the number which is obtained following a single throw of a dice




Continuous random variable
Continuous Random Variable X

  • X is now fulfilling the role of a continuous random variable

  • A distinguishing feature of a continuous random variable is that, between any two specified values, the variable can be equal to an infinite number of values

  • Examples are the height, weight, width and length of an object, the speed at which an object is travelling, and the distance of an object from a specified location


Statistics and mathematics for economics

A Feature of the Graph X

Area underneath the

graph = 1

0

1 2 3 4 5 6 x


The probability density
The Probability Density X

  • If the area underneath the graph is equal to 1 then it is possible to work out the height of the rectangle

  • The area of a rectangle = height x width

  • Therefore, 1 = height x (6 – 1)

  • So, height = 1/5

  • The height of the rectangle is not the probability, but, rather, the probability density, f(x)

  • A continuous random variable does not have a probability distribution but, rather, a probability density function


The probability density function of x
The Probability Density Function of X X

Mathematical presentation of the function:

f(x) = 1/5, 1 < x < 6,

f(x) = 0, otherwise.


The probability of interest
The Probability of Interest X

  • In the case of a continuous random variable, the probability that the variable is equal to any individual value is (approximately) equal to zero

  • Hence, the concern tends to be with the probability that the value of the variable lies within a specified range of values

  • This probability is equal to the area underneath the graph between the corresponding points on the horizontal axis


Statistics and mathematics for economics

The Probability that the Value of X Falls between 1 and 2 X

The probability that the value of X lies between 1 and 2 is equal to the area

of the shaded region. The height x the width = 1/5 x (2 – 1) = 1/5.

f(x) = 1/5

0

1 2 3 4 5 6 x


Mathematical approach
Mathematical Approach X

  • This probability can also be obtained mathematically,

  • using the technique of integration.

  • Two fundamental rules of integration:

  • The consequence of integrating a constant (k) with

  • respect to X is kX;

  • (ii) The consequence of integrating kXn with respect to X

  • is kXn+1/(n+1).


Mathematical calculation
Mathematical Calculation X

When X is a discrete random variable:

P(X = 1 or 2) = P(X = x).

When X is a continuous random variable:

2

P(1 < X < 2) = f(x).dx,

1

where dx denotes a small change in the value of X.

Upon substitution:

2

P(1 < X < 2) = (1/5)dx

1


Evaluation of the integral
Evaluation of the Integral X

2

P(1 < X < 2) = (1/5)dx

1

2

= [x/5]

1

= (2/5) – (1/5)

= 1/5


Calculation of the expected value of x
Calculation of the Expected Value of X X

The probability density function of X is symmetrical. Hence, the

expected value of the variable corresponds to the mid-point along the

horizontal axis (= (6 + 1)/2 = 3.5)). This result can be confirmed

mathematically using integration.

When X is a discrete random variable: E[X] = x.P(X = x)

When X is a continuous random variable:

E[X] = x.f(x).dx

-

In the current example:

6

E[X] = x.(1/5).dx

1


Result for the expected value
Result for the Expected Value X

Upon integrating x/5, with respect to x, we achieve x2/10.

Hence:

6

E[X] = [x2/10]

1

= (36/10) - (1/10)

= 35/10 units


Calculation of the value of the variance of x
Calculation of the value of the variance of X X

var.(X) = E[X2] - (E[X])2

E[X2] = x2.f(x).dx

-

Upon substitution:

6

E[X2] = x2(1/5)dx

1

Upon integrating x2/5, with respect to x, we achieve x3/15.


Result for the value of the variance
Result for the value of the variance X

6

E[X2] = [x3/15]

1

= (216/15) - (1/15)

= 215/15 units squared

Consequently,

var.(X) = 215/15 - (7/2)2

= 43/3 - 49/4 = 25/12 units squared


Statistics and mathematics for economics

Second Example of the Probability Density Function of a Continuous Random Variable

f(x)

0 2 4 6 X


Attempting to achieve a mathematical presentation
Attempting to achieve a mathematical presentation Continuous Random Variable

Assign to the maximum probability density the arbitrary constant, c.

Consequently,

f(x) = c, 2 < x < 4.

The first element of the probability density function has the diagrammatic

appearance of an upward-sloping straight line. Thus, mathematically, it

complies with the general equation of a straight line:

f(x) = a + bx, 0 < x < 2.

It can be shown that, more specifically:

f(x) = cx/2, 0 < x < 2.


The third element of the probability density function
The third element of the probability density function Continuous Random Variable

The third element of the probability density function has the diagrammatic

appearance of a downward-sloping straight line. Hence, mathematically,

it complies with the general equation of a straight line:

f(x) = a + bx, 4 < x < 6.

It can be shown that, more specifically:

f(x) = 3c - cx/2, 4 < x < 6.


Determining the value of c
Determining the value of c Continuous Random Variable

The value of the constant, c, can be determined using one of the

properties of the probability density function of a continuous

random variable.

Over the possible values of the random variable, the sum of the

probabilities is equal to one:

f(x).dx = 1.

-

In the case of the current probability density function:

6

f(x).dx = 1.

0


Decomposition of the integral
Decomposition of the integral Continuous Random Variable

The single integral may be expressed as the sum of three separate

integrals:

6 2 4 6

f(x).dx = f(x).dx + f(x).dx + f(x).dx = 1

0 0 2 4

Upon substitution:

6 2 4 6

f(x).dx = (cx/2)dx + c.dx + (3c – cx/2)dx = 1

0 0 2 4

2 4 6

= [cx2/4] + [cx] + [3cx – cx2/4] = 1

0 2 4


Evaluating the integrals
Evaluating the integrals Continuous Random Variable

(c – 0) + (4c – 2c) + (18c – 9c) – (12c – 4c) = 1

So, 4c = 1, such that c = ¼.

Mathematical form of the probability density function:

f(x) = x/8, 0 < x < 2;

f(x) = ¼, 2 < x < 4;

f(x) = ¾ - x/8, 4 < x < 6.


Calculation of the expected value of the random variable
Calculation of the Expected Value of the Random Variable Continuous Random Variable

6

E[X] = x.f(x).dx

0

The single integral may be expressed as the sum of three separate

integrals:

2 4 6

E[X] = x.f(x).dx + x.f(x).dx + x.f(x).dx

0 2 4

Upon substitution:

2 4 6

E[X] = x(x/8)dx + x(1/4)dx + x(3/4 – x/8)dx

0 2 4


Evaluating the integrals1
Evaluating the integrals Continuous Random Variable

2 4 6

E[X] = (x2/8)dx + (x/4)dx + (3x/4 – x2/8)dx

0 2 4

2 4 6

E[X] = [x3/24] + [x2/8] + [3x2/8 - x3/24]

0 2 4

E[X] = (8/24 – 0) + (16/8 - 4/8) + (108/8 – 216/24) - (48/8 – 64/24)

= 8/24 + 12/8 + 60/8 - 152/24

= -144/24 + 72/8

= 3


Calculating the value of the variance of the random variable
Calculating the value of the variance of the random variable Continuous Random Variable

var.(X) = E[X2] - (E[X])2,

6

where E[X2] = x2.f(x).dx.

0

The single integral may be presented as the sum of three separate

integrals:

2 4 6

E[X2] = (x3/8)dx + (x2/4)dx + (3x2/4 - x3/8)dx.

0 2 4


Evaluating the integrals2
Evaluating the integrals Continuous Random Variable

2 4 6

E[X2] = [x4/32] + [x3/12] + [3x3/12 - x4/32]

0 2 4

E[X2] = (16/32 – 0) + (64/12 – 8/12) + (648/12 – 1296/32)

- (192/12 – 256/32)

E[X2] = 16/32 + 56/12 + 456/12 - 1040/32

= -1024/32 + 512/12

= -128/4 + 128/3 = 128/12 units squared

Thus, var.(X) = 128/12 - 32 = 20/12 = 5/3 units squared