Download Continuous Random Variables and Probability Distributions

Systems Engineering Program
Department of Engineering Management, Information and Systems
EMIS 7370/5370 STAT 5340 :
PROBABILITY AND STATISTICS FOR SCIENTISTS AND ENGINEERS
Continuous Probability
Distributions
Continuous Random Variables &
Probability Distributions
Dr. Jerrell T. Stracener, SAE Fellow
Leadership in Engineering
1
Random Variable
•Definition - A random variable is a mathematical
function that associates a number with every
possible outcome in the sample space S.
• Definition - If a sample space contains an infinite number
of possibilities equal to the number of points on a line
segment, it is called a continuous sample space and a
random variable defined over this space is called a
continuous random variable.
• Notation - Capital letters, usually X or Y, are
used to denote random variables. Corresponding
lower case letters, x or y, are used to denote
particular values of the random variables X or Y.
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Continuous Random Variable
For many continuous random variables or (probability
functions) there exists a function f, defined for all
real numbers x, from which P(A) can for any event
A  S, be obtained by integration:
PA    f x dx
A
Given a probability function P() which may be
represented in the form of
PA    f x dx  area
A
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Continuous Random Variable
in terms of some function f, the function f is called
the probability density function of the probability
function P or of the random variable X, and the
probability function P is specified by the
probability density function f.
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Continuous Random Variable
Probabilities of various events may be obtained
from the probability density function as follows:
Let A = {x|a < x < b}
Then
P(A) = P(a < X < b)
  f x dx
A
b
  f x dx
a
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Continuous Random Variable
Therefore
P( A) = area under the density function curve
between x = a and x = b.
f(x)
Area = P(a < x <b)
0
0
a
b
x
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Probability Density Function
The function f(x) is a probability density function for
the continuous random variable X, defined over the
set of real numbers R, if
1. f(x)  0 for all x  R.

2.
 f (x )dx  1.

b
3. P(a < X < b) =
 f (x)dx.
a
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Probability Distribution Function
The cumulative probability distribution function,
F(x), of a continuous random variable X with
density function f(x) is given by
x
Note:
F( x )  P(X  x )   f ( t )dt.

d
f(x) 
Fx 
dx
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Probability Density and Distribution Functions
f(x) = Probability Density Function
Area = P(x1 < X<x2)
x
x1
x2
F(x) = Probability Distribution Function
1
cumulative area
F(x2)
P(x1 < X<x2) = F(x2) - F(x1)
F(x1)
x1
x2
x
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Mean & Standard Deviation
of a Continuous Random Variable X
• Mean or Expected Value

μ  EX    x f x dx
• Remark

Interpretation of the mean or expected value:
The average value of X in the long run.
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Mean & Standard Deviation
of a Continuous Random Variable X
•Variance of X:

Var X   σ   (x - μ) f(x) dx
2
2

•
 
Var X   E X  μ
2
2

  x f x dx  μ
2
2

•Standard Deviation of
X : σ  Var X 
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Rules
If a and b are constants and if  = E (X ) is the mean
and 2 = Var ( X ) is the variance of the random
variable X, respectively, then
EaX  b  aμ  b
and
Var aX  b  a Var X 
2
12
Rules
If Y = g(X) is a function of a continuous random variable
X, then

μ Y  Egx    gx f x dx

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Example
If the probability density function of X is
f (x) 
2(1  x)
for 0 < x < 1
0
elsewhere
then find
(a)  and 
(b) P(X>0.4)
(c) the value of x* for which P(X<x*)=0.90
14
Example
First, plot f(x):
2
f(x)
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
x
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Example Solution
Find the mean and standard deviation of X,
1
  E ( X )   xf ( x)dx
0
1
1
0
0
  x  2(1  x)dx  2 [ x  x 2 ]dx
3 1
x
x 
 1 1
 2    2  
 2 3
 2 3 0
2 1
 1 
3 3
2
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Example Solution
 2  Var( X )  E ( X 2 )   2
1
  x f ( x)dx   
 3
0
1
2
2
4 1
x x  1
1
  x  2(1  x)dx   2   
9
 3 4 0 9
0
1
3
2
1 1 1 2 1
 2    2  
 3 4  3 12 9
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Example Solution
1 2 1 1 2  1
     
3  4 3  3  12  18
and the standard deviation is
1

 0.236
18
18
Example Solution
(b)
x
x
0
0
F ( x)  P( X  x)   f ( x) dx   2  2 x dx
 2x  x2
for 0<x<1
P(X  0.4)  1  P(X  0.4)
 1  2 * 0.4  0.4
2

 1  0.64  0.36
P( X  x*)  P( x*)  2( x*)  ( x*)2  0.9
(c)
therefore x*  0.68 or 1.32
Since 1.32>1, so x*  0.68
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Uniform Distribution
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Uniform Distribution
Probability Density Function
 1

, for a  x  b, for a  0
f ( x)   b  a

0
, elsewhere
f(x)
1/(b-a)
0
a
b
x
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Uniform Distribution
Probability Distribution Function
0 for x  a
x a
F ( x)  P( X  x) 
for a  x  b
ba
1 for x  b
F(x)
1
0
a
b
x
22
Uniform Distribution
• Mean
 = (a+b)/2
• Standard Deviation
ba

12
23
Example – Uniform Distribution
24
Example Solution – Uniform Distribution
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