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Block 5 Stochastic &
Dynamic Systems
There is no
longer a need
now to be
discrete!
Lesson 17 – Stochastic Models
Continuous Random
Variables and their
Probability Distributions
Charles Ebeling
University of Dayton
1
Continuous Random Variables Examples






T = a continuous random variable, time to failure
X = a continuous random variable, the distance in
miles between cable defects
Z = a continuous random variable, the monthly
consumption of power in watts
T = a continuous random variable, the repair time of
a failed machine
Y = a continuous random variable, the time between
arrivals of customers at City National Bank
X = a continuous random variable, the procurement
lead-time for a critical part
2
The Probability Density Function
(PDF)
Let X = a continuous random variable
f ( x) is a probability density function if :
is never negative
1. f ( x)  0

2.

f ( x) dx  1
area under the curve = 1

The graph of f(x) provides the shape of the probability distribution
3
A real, honest-to-goodness
PDF
f ( x)  .1e.1x ; x  0
1. f ( x)  0
.1 x 

.1e
.1x
2.  .1e dx 
.1
0
0
.1
.1e0
 lim

 0 1  1
.1
x
x  .1e
.1
0.12
0.1
0.08
0.06
0.04
0.02
area = 1
0
-5
5
15
25
35
45
55
4
Yet another one…
0.2  0.02t 0  t  10
f (t )  
0 elsewhere
1. f (t )  0
10
2.   0.2  0.02t  dt  .2t  .01t 2   2  1  1
0
10
0
0.25
1
area  10  f (0)  5 .2   1
2
0.2
0.15
0.1
0.05
0
-1
1
3
5
7
9
11
5
Finding
Probabilities
b
P(a  X  b)   f ( x) dx
a
given : f ( x)  .1e.1x ; x  0
15
P(5  X  15)   .1e
.1 x
dx  e
.1 x 15
5
 e1.5  e.5  .3834
5
0.12
0.1
0.08
0.06
0.04
0.02
0
-5
5
15
25
35
45
55
6
Finding More Probabilities
b
P(a  X  b)   f ( x) dx
a
f (t )  0.2  0.02t
3
P(1  X  3) 
  0.2  0.02t  dt  .2t  .01t
2 3
  .51  .19  .32
1
1
0.25
0.2
0.15
0.1
0.05
0
-1
1
3
5
7
9
11
7
The Cumulative Distribution
Function (CDF)
x
Define: F ( x)  P( X  x) 

f (t )dt
t is the variable
of integration

Then: P(a  X  b)  F (b)  F (a)
and: P( X  x)  1  F ( x)
f(x)
F(b)
complementary
cumulative
F(a)
a
b
x
8
Our real, honest-to-goodness
PDF revisited
f ( x)  .1e.1x ; x  0
x
.1t x
.1e
F ( x)   .1e dt 
.1
0
.1t
 e
0.1 x
1  1 e
0.1 x
0
P(5  X  15)  F (15)  F (5)  1  e 1.5   1  e .5   .3834
P( X  10)  1  F (10)  e1  .367879
9
Let’s do it one more time…
0.2  0.02t 0  t  10
f (t )  
0 elsewhere
t
F (t ) 
  0.2  0.02 y  dy  .2 y  .01y
2 t
  .2t  .01t 2
0
0
2
2



P(1  X  3)  F (3)  F (1)  .2  3  .01 3  .2 1  .011 

 

 .51  .19  .32
2

P( X  4)  1  F (4)  1  .2  4   .01 4    1  .64  .36


10
The Expected Value
  E[ X ] 

 xf ( x)dx

f ( x)  .1e .1x ; x  0

E  X    .1te
0
.1t

dt .1 te
0
.1t
dt 
.1
 .1
2
 10
From our little
table of improper
definite integrals:

 xe
0
 ax
1
dx  2
a
11
Even More Expectation
0.2  0.02t 0  t  10
f (t )  
0 elsewhere
10
10
0
0
E[T ]   t  0.2  0.02t  dt    0.2t  0.02t 2  dt
10
 .2t .02t 
20 30 20 10
1


 .1100  
 
 3

3 0
3
3
3
3
3
 2
2
3
  E[ X ] 

 xf ( x)dx

12
A Bonus Moment
The Median Defined
The median of a probability distribution is
that value of the random variable X such that
P(X< xmed) = F(xmed) = .5
0.12
0.1
0.08
area = .5
0.06
0.04
0.02
0
-5
5
15
xmed
25
35
45
55
13
Let’s find a median…
f ( x)  .1e
.1 x
F ( x)  1  e
e
0.1 x
xmed
; x0
0.1 x
 .5
 1  .5  .5
1

ln .5  6.9315
.1
recall E[X] = 10
14
Another median…
0.2  0.02t 0  t  10
f (t )  
0 elsewhere
1
and E[T ]  3
3
F (t )  .2t  .01t 2  .5
.01t 2  .2t  .5  0
t 2  20t  50  0
b  b 2  4ac 20  400  4  50 
t

 2.9289, 17.071
2a
2
tmed  2.9289
15
The Mode
The mode of a probability distribution is its most likely
value. Specifically, xmode is the mode of the distribution if
f ( xmod e )  max f ( x)
0.12
0.25
0.1
xmode = 0
0.08
0.2
xmode = 0
0.15
0.06
0.1
0.04
0.05
0.02
0
0
-5
5
15
25
35
45
55
-1
1
3
5
7
9
11
16
Much More Mode Methods
Professor I. M. Brite has determined that the driving time
in minutes to the University from her home has the
following probability density function:
I shall be
f ( x)  .025ln( x)  .0011881x , 10  x  30
.025
f '( x) 
 .0011881  0
x
.025
xmode 
 21.042
.0011881
.025
f ''( x)   2  0
x
arriving in
21.042
minutes.
17
Let’s explore this very interesting
density function
f ( x)  .025ln( x)  .0011881x , 10  x  30
2 x
x
t
F ( x)   .025ln(t )  .0011881t  dt  .025  t ln t  t   .0011881
2
10
10
x2
 .025  x ln x  x   .0011881  0.266239773
2
30
E[ X ] 
 .025 x ln( x)  .0011881x  dx
2
10
x
x
 .025  ln x 
4
 2
2
2
 .0011881x

3

3
30
 21.94557  1.8572  20.09
10
18
What’s this strange PDF look
like?
That is a strange
one all right.
median = 20.12 minutes
0.052
f(x)
0.051
0.05
0.049
0.048
0.047
0.046
0.045
0
5
10
15
20
25
30
x 35
19
This has been
Adventures with Continuous
Random Variables
Presented by the Department of
Engineering Management & Systems
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