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 Component: Lecture Six
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
A Feature of the Graph
Area underneath the
graph = 1
0
1 2 3 4 5 6 x
Mathematical presentation of the function:
f(x) = 1/5,1 < x < 6,
f(x) = 0,otherwise.
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
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
2
P(1 < X < 2) = (1/5)dx
1
2
= [x/5]
1
= (2/5) – (1/5)
= 1/5
The probability density function of X is symmetrical. Hence, the
expected value of the variable corresponds to the midpoint 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
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
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.
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 upwardsloping 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 downwardsloping 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:
f(x).dx = 1.

In the case of the current probability density function:
6
f(x).dx = 1.
0
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
(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.
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
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
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
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