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Statistics and Mathematics for Economics

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Statistics and Mathematics for Economics

Statistics Component: Lecture Six

- 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 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

- 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

- 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 â€“ 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

Mathematical presentation of the function:

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

f(x) = 0,otherwise.

- 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 â€“ 1) = 1/5.

f(x) = 1/5

0

1 2 3 4 5 6 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).

When X is a discrete random variable:

P(X = 1 or 2) = ïƒ¥P(X = x).

When X is a continuous random variable:

2

P(1 < X < 2) = ïƒ²f(x).dx,

1

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

Upon substitution:

2

P(1 < X < 2) = ïƒ²(1/5)dx

1

2

P(1 < X < 2) = ïƒ²(1/5)dx

1

2

= [x/5]

1

= (2/5) â€“ (1/5)

= 1/5

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] = ïƒ¥x.P(X = x)

When X is a continuous random variable:

ï‚¥

E[X] = ïƒ²x.f(x).dx

-ï‚¥

In the current example:

6

E[X] = ïƒ²x.(1/5).dx

1

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

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

ï‚¥

E[X2] = ïƒ²x2.f(x).dx

-ï‚¥

Upon substitution:

6

E[X2] = ïƒ²x2(1/5)dx

1

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

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

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 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.

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:

ï‚¥

ïƒ²f(x).dx = 1.

-ï‚¥

In the case of the current probability density function:

6

ïƒ²f(x).dx = 1.

0

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

integrals:

6 2 4 6

ïƒ²f(x).dx = ïƒ²f(x).dx + ïƒ²f(x).dx + ïƒ²f(x).dx = 1

0 0 2 4

Upon substitution:

6 2 4 6

ïƒ²f(x).dx = ïƒ²(cx/2)dx + ïƒ²c.dx + ïƒ²(3c â€“ cx/2)dx = 1

0 0 2 4

2 4 6

= [cx2/4] + [cx] + [3cx â€“ cx2/4] = 1

0 2 4

(c â€“ 0) + (4c â€“ 2c) + (18c â€“ 9c) â€“ (12c â€“ 4c) = 1

So, 4c = 1, such that c = Â¼.

Mathematical form of the probability density function:

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

f(x) = Â¼,2 < x < 4;

f(x) = Â¾ - x/8,4 < x < 6.

6

E[X] = ïƒ²x.f(x).dx

0

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

integrals:

2 4 6

E[X] = ïƒ²x.f(x).dx + ïƒ²x.f(x).dx + ïƒ²x.f(x).dx

0 2 4

Upon substitution:

2 4 6

E[X] = ïƒ²x(x/8)dx + ïƒ²x(1/4)dx + ïƒ²x(3/4 â€“ x/8)dx

0 2 4

2 4 6

E[X] = ïƒ²(x2/8)dx + ïƒ²(x/4)dx + ïƒ²(3x/4 â€“ 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 â€“ 0) + (16/8 - 4/8) + (108/8 â€“ 216/24) - (48/8 â€“ 64/24)

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

= -144/24 + 72/8

= 3

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

6

where E[X2] = ïƒ²x2.f(x).dx.

0

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

integrals:

2 4 6

E[X2] = ïƒ²(x3/8)dx + ïƒ²(x2/4)dx + ïƒ²(3x2/4 - x3/8)dx.

0 2 4

2 4 6

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

0 2 4

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

- (192/12 â€“ 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