Download Chapter 6 Continuous Random Variables and Distributions

Chapter 6 Continuous Random Variables and Distributions
Many economic and business measures such as sales, investment, consumption
and cost can have the continuous numerical values so that they can not be
represented by discrete random variables. In addition measures of time,
distance and temperature fit into the same category and they are all
represented by continuous random variables. Probability statements for
continuous variables can be explained by continuous distribution functions.
Cumulative Distribution function
Let X be a discrete or continuous random variable and the cumulative
distribution function F (X ) gives the probability that X does not exceed the
value of x and given by
F (X ) = P (X ≤ x )
Remember that for each discrete random variable X, we had a probability
distribution function f whose histogram representation consists of percentages.
x1
x2
x3
x4
points marked on horizontal axis shows the values that X can take and heights of
rectangles gives the corresponding probabilities. If you arrange intervals around
x k ’s in such a way that all they are equal and has length 1, then we can measure
probabilities by the areas of rectangles.
e.g. P (X = x 3 ) is the area of the shaded region below.
x1
x2
x3
x4
and P (x 2 ≤ X ≤ x 3 ) is the area of the shaded region below.
x1
x2
x3
x4
Similarly to find the probability of continuous random variables we need a
function f, called a Probability Density Function which should satisfy:
1) f (x ) ≥ 0 for all values of x.
2) The total area under the f probability density function is equal to 1.
3) The cumulative distribution function F (x 0 ) is the area under the probability
density function f (x ) up to x 0
x0
F (x 0 ) = ∫f (x )dx
xm
Where x m is the minimum value of the random variable x .
4) P (a ≤ X ≤ b ) = F (b ) − F (a ) is the area of the region bounded by the by the
graph of f (x ) and the horizontal axis from a to b.
0
a
b
d
Remark: If X is a continuous random variable then P (X = 0) = 0 for all x (i.e.
the probability of a single value is always equal to zero.)
So, the following statement is correct for all continuous random variables.
P (a < X < b ) = P (a ≤ x ≤ b ) = P (a < X ≤ b ) = P (a ≤ X < b )
The Uniform Distribution
If a continuous random variable X has a constant probability for all values of x
and the probability distribution function is given as follows

1
,
0≤x ≤1

0
,
otherwise
f (x ) = 
X is said to be Uniformly distributed over the interval (0, 1).
In general X is a uniform random variable on the interval (a, b) if its probability
density function is given by
1

a <x <b
,
 b −a
f (x ) = 

0
,
otherwise

And the distribution function of a uniform random variable on the interval (a,b)
is given by

0

 x − a
F (x ) = 
 b −a

1

Example : If a continuous random variable
function.
,
x ≤a
,
a <x <b
,
x ≥b
has the following probability density

1/5
,
−2 ≤ x ≤3

0
,
otherwise
f (x ) = 
a) Find P ( −2 < X < 3) ?
b) Find the probability that X is greater than 1?
c) Find the probability that X is greater than or equsl to 1?
So the cumulative distribution function is as follows.

x ≤ −2
0
,

 x + 2
,
−2 < x < 3
F (x ) = 
5


1
,
x ≥3

Now we can calculate the probabilities.
P ( −2 < X < 3) = F (3) − F ( −2) = 1 − 0 = 1
P (X > 1) = F (3) − F (1) = 1 −
3 2
=
5 5
P (X ≥ 1) = P (X > 1) = F (3) − F (1) = 1 −
3 2
=
5 5
Example (6.7): The incomes of all families in a particular suburb can be
represented by a continuous random variable. It is known that the median
income for all families in this suburb is $60,000 and that 40% of all families in
the suburb have incomes above $72,000.
a) For a randomly chosen family what is the probability that its income will
be between $60000 and $72000.
b) If the distribution function for the income is known to be uniform what is
the probability that a random chosen family has an income below $65000.
We know that the total area below a probability density function for a
continuous random variable is equal to 1. Moreover, we know that the median is
the value which determines the middle (50%) of the probability density
function. So, we can write that, P (X ≤ 60000) = 0,5 and P (X ≥ 72000) = 0,4
and P (60000 ≤ X ≤ 72000) = 1 − P (X ≤ 60000) − P (X ≥ 72000) = 0,1
If the distribution function of the income is Uniform the shape of the
probability density function for this continuous random variable X should be as
follows,
f(x)
0,10
$60,000
$72,000
From this chart we can say that f (x ).(72,000 − 60,000) = 0,1 so we can find the
value of probability density function f (x ) = 0,1 / 12,000 . Now we can calculate
the probability that a random chosen family has an income below $65000.
P (X ≤ 65,000) = P (X ≤ 60,000) + P (60,000 ≤ X ≤ 65,000) = 0,5 + 5,000.(0,1 / 12,000)
= 0,5 + 0,0416 = 0,5416
Expectations for Continuous Random Variables
Suppose that a random experiment gives results that can be represented by a
random continuous variable. And if g (x ) is any function of the random variable
X then the expected value of X is denoted by E ( g (X )) and calculated by as
follows.
E [g (x )] = ∫ g (x )f (x )dx
x
For continuous random variables the mean of X is denoted by µ x
µ x = E (X ).
and
2
2
2
2
2
The variance is denoted by σ x and σ x = E ((X − µ x ) ) = E (X ) − µ x . And the
standard deviation is σ x = σ x2 .
Example: Suppose that X is a continuous random variable with the probability
density function

0.25
,
0≤x ≤4

0
,
otherwise
f (x ) = 
A)
B)
C)
D)
Find
Find
Find
Find
the probability X is more then 3.
the expected value of X.
the expected value of Y=2X+3.
the variance and the standard deviation of X.
4
4
P (X > 3) = ∫ 0.25dx = 0.25x ∫ = 0.25( 4) − 0.25(3) = 0.25
3
3
4
4
0
0
µ x = E (X ) = ∫f (x )xdx = ∫ 0.25xdx = 0.25
x24
2
∫
=2−0 =2
0
E (Y ) = E (2X + 3) = 2E (X ) + 3 = 2.2 + 3 = 7
4
σ x2 = E (X 2 ) − µ x2 = ∫ x 2f (x )dx − (2) 2 =0,25
0
σ x = σ x2 = 1,33 = 1,53
x3
3
4
∫
0
− 4 = 5,33 − 4 = 1,33