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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

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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

• 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 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

Diagrammatic Presentation of the Probability Distribution of X

P(X)

1/6

0

1 2 3 4 5 6 X

Modification of the Probability Distribution

0

1 2 3 4 5 6 X

### Continuous Random Variable

• 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

A Feature of the Graph

Area underneath the

graph = 1

0

1 2 3 4 5 6 x

### The Probability Density

• 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

Mathematical presentation of the function:

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

f(x) = 0,otherwise.

### The Probability of Interest

• 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

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

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

• 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

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

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

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

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

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

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

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

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 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

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

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

(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

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 integrals

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

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 integrals

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