Download Chapter5-1

Chapter 5. Continuous Random
Variables
Continuous Random Variables
• Discrete random variables
– Random variables whose set of possible values is
either finite or countably infinite.
• Continuous random variables
– Random variables whose set of possible values is
uncountable.
– The lifetime of a light bulb.
– The amount of precipitation in a year.
2
Definition of Continuous Random
Variable
• Probability of continuous random variable
– If there exists a nonnegative function f, defined for
all real x ϵ (-∞, ∞), having the property that for any
set B of real numbers
P( X  B)   f ( x)dx
B
• The function f is called the probability density
function of the random variable X.
3
• The probability of the whole sample space is 1.

1  P( X  (, ))   f ( x)dx

• The probability of a certain region B = [a,b] is
b
P(a  X  b)   f ( x)dx
a
• The probability at a certain point is zero
a
P( X  a)   f ( x)dx  0
a
a
P( X  a)  P( X  a)  F (a)   f ( x)dx

4
• Ex 1a. Suppose that X is a continuous random
variable whose probability density function is
given by
 C (4 x  2 x 2 ) 0  x  2
f ( x)  
otherwise
0
(a) What is the value of C?
(b) Find P(X > 1)
5
(a) Because


-
f ( x)dx  1
2
C  (4 x  2 x 2 )dx  1
0
 2 2 x3 
C 2 x 

3


C  3/8
(b) P( X  1)  

1
x 2
1
x 0
3 2
1
2
f ( x)dx   (4 x  2 x )dx 
8 1
2
6
• Ex 1b. The amount of time, in hours, that a computer
functions before breaking down is a continuous
random variable with probability density function
given by
e  x /100 x  0
f ( x)  
x0
0
What is the probability that
(a) a computer will function between 50 and 150
hours before breaking down;
(b) it will function less than 100 hours?
7

(a ) Since 1   λe -x /100dx

1   (100)e  x /100 |0  100
  1 / 100
The probabilit y that a computer w ill function between
50 and 150 hours before breaking down is
150 1
P(50  X  150)  
e  x /100dx  e  x /100 |150
50
50 100
 e 1/ 2  e 3 / 2  .384
(b) The probabilit y that the computer w ill function
less than 100 hours is
100
P( X  100)  
0
1  x /100
1
e
dx   e  x /100 |100

1

e
 .633
0
100
8
• Ex 1c. The lifetime in hours of a certain kind of radio
tube is a random variable having a probability density
function given by
0
x  100

100
f ( x)  
x  100
2

x
What is the probability that exactly 2 of 5 such tubes
in a radio set will have to be replaced within the first
150 hours of operation? Assume that the events Ei, i =
1, 2, 3, 4, 5, that the i-th such tube will have to be
replaced within this time, are independent.
9
150
P ( Ei )  
0
150
 100 
100
f ( x)dx
2
x dx
 1/ 3
 5  1   2 
80
     
 2  3   3  243
2
3
10
Distribution Function and PDF
F (a )  P ( X  (, a ])  
a

f ( x)dx
d
F (a)  f (a)
da
P(a 


a  / 2
 X a ) 
f ( x)dx  f (a)
a


/
2
2
2
• Probability density function can be considered
as the measure of how likely that the random
variable will be near a.
11
• Ex 1d. If X is continuous with distribution
function FX and density function fx, find the
density function of Y = 2X.
FY (a)  P(Y  a)
 P(2 X  a)
 P( X  a / 2)
 FX (a / 2)
Differenti ation gives
1
fY (a)  f X (a / 2)
2
12
Another way to determine fY
 fY (a)  P(a   / 2  Y  a   / 2)
 P(a   / 2  2 X  a   / 2)
 P(a / 2   / 4  X  a / 2   / 4)
  / 2 f X (a / 2)
13
Expectation and Variance of
Continuous Random Variables
• Discrete random variable
E[ X ]   xp( x)
x
• Continuous random variable

E[ X ]   xf ( x)dx

14
• Ex 2a. Find E[X] when the density function of
X is
2 x if 0  x  1
f ( x)  
 0 otherwise
E[ X ]   xf ( x)dx
1
  2 x 2 dx
0
2x

3
3 1
0
 2/3
15
• Ex 2b. The density function of X is given by
1 if 0  x  1
f ( x)  
0 otherwise
Find E[eX]
Let Y  e X . Determine FY first. For 1  y  e,
FY ( y )  P(Y  y )
 P (e X  y )
 P( X  log( y ))

log( y )
0
f ( x)dx
 log( y )
16
Differentiating FY ( y ),
fY ( y )  1/ y
1 y  e
E[e x ]  E[Y ]

  yfY ( y )dy

e
  dy  e  1
1
17
• Proposition 2.1
• If X is a continuous random variable with
probability density function f(x), then for any
real-valued function g,

E[ g ( x)]   g ( x) f ( x)dx

Ex. 2b
1
E[e ]   e x dx  e  1
X
0
18
• Ex 2c. A stick of length 1 is split at a point U
that is uniformly distributed over (0,1).
Determine the expected length of the piece that
contains the point p, 0 ≤ p ≤ 1.
21
Let L p (U ) denote the length of the substick t hat contains p,
1  U U  p
L p (U )  
Up
U
U is uniformly distribute d in [0,1], so pdf of U is f (u )  1
From Propositio n 2.1 we have
1
E[ L p (U )]   L p (u ) f (u )du
0
p
1
  (1  u )du   udu
0
p
1 (1  p ) 2 1 p 2
 
 
2
2
2 2
1
  p (1  p )
2
22
• Corollary 2.1
If a and b are constants, then
E[aX + b] = aE[X] + b
• Variance of continuous random variable
Var(X) = E[(X-μ)2]
• Var(X) = E[X2] - (E[X])2
23
• Ex 2e. Find Var(X) if X has the pdf
2 x if 0  x  1
f ( x)  
 0 otherwise

E[ X ]   x 2 f ( x)dx
2

1
  2 x dx
3
0
 1/ 2
Because E[ X ]  2 / 3,
Var ( x)  1 / 2  (2 / 3) 2  1 / 18
24