Upload
others
View
25
Download
0
Embed Size (px)
Citation preview
Chapter 6Continuous Random Variables
Wen-Guey Tzeng
Computer Science Department
National Chiao Tung University
Probability density functions
Definition.• Let X be a random variable that takes values on all real
numbers • There exists a non-negative real-valued function
f: R[0, )• For an interval I=[a b], let
b
aIdxxfdxxfIXP )()()(
2
3
• X is called a continuous random variable
• The function f is called the probability density function
(pdf), or the density function of X.
• Density is not probability• f(x): we cannot say that the probability that x occurs is f(x)
• Area is probability• P(XA): the probability that X occurs in A
4
5
6
7
Distribution function for continuous random variable
• If X is a continuous random variable, then the function
𝐹 𝑡 = −∞𝑡𝑓 𝑡 𝑑𝑡
is called the (cumulative) distribution function of X.
2014 Fall 8
9
2014 Fall 10
11
F(b)-F(a)
dxxf
bXaPbXaP
bXaPbXaP
dxxfbXaP
xfxF
dxxf
dxxftF
b
a
b
a
t
)(
)()(
)()( (e)
.)()( b,a realsFor (d)
).()(' ,continuous is f If (c)
.1)( (b)
.)()( a)(
Properties
• The distribution function of a random variable X is given by
31
322/112/
212/1
104/
00
)(
x
xx
x
xx
x
xF
2014 Fall 12
• Compute the following quantities:• P(X<2)
• P(X=2)
• P(1<=X<3)
• P(X>3/2)
• P(X=5/2)
• P(2<X<=7)
2014 Fall 13
• Suppose that a bus arrives at a station every day between 10:00 A.M. and 10:30 A.M., at random.
• Let X be the arrival time; find the distribution function of X and sketch its graph.
Sol:
5.101
5.1010)10(2
00
)(
t
tt
t
tF
2014 Fall 14
• Experience has shown that while walking in a certain park, the time X, in minutes, between seeing 2 people smoking has a density function of the form
.0
0)(
otherwise
xxexf
x
15
(a) Calculate the value of (b) Find the probability distribution function of X.(c) What is the probability that Jeff, who has just seen a
person smoking, will see another person smoking in 2 to 5 minutes? In at least 7 minutes?
16
Sol:
1.or 1(-1)]-[0 Therefore
.01
lim1
lim)1(limget werule sHopital'l'By
.1])1([ Thus
.)1( parts,by n integratioBy
.)(1 )(
0
00
xxxx
x
x
x
xx
xx
ee
xex
ex
exdxxe
dxxedxxedxxfa
17
.007.0)7(1)7(
37.0)31()61()2()5()52()(
.1)1(])1([)()(
0,For t 0. tif 0)()(
25
0
FXP
eeFFXPc
etexdxxftF
tFb
ttt
x
18
• Sketch the graph of the function f(x)• Show that it is the probability density function of a
random variable X.• Find F, the distribution function of X, and show that it
is continuous.• Sketch the graph of F.
,0
51|3|4
1
2
1
)(
otherwise
xxxf
19
Density function of a function of a random variable
• f(x) is the density function of a random variable X
• What is the density function g(y) of Y=h(X)?
Two methods
• The method of distributions• Compute the distribution G(t) of Y• g(y) = G(y)’
• The method of transformations• Compute g(y) directly
20
By the method of distributions:
• Let X be a continuous random variable with the probability density function
• Find the distribution and density functions of Y=X2.
.0
21 /2)(
2
elsewhere
xifxxf
21
Sol: Let G and g be the distribution function and the density function of Y, respectively.
.2
2]2
[2
)1(
Now
.41
41)1(
10
)()()()(
11 2
2
txdx
xtXP
t
ttXP
t
tXtPtXPtYPtG
tt
22
.0
41 1/t(t)G'g(t)
:G ofation differenti
by found is g, Y, offunction density The
.41
412
2
10
)(
Therefore,
elsewhere
tift
t
tt
t
tG
23
By the method of transformations
• X is continuous with density function fX and the possible value set A.
• If Y=h(X) is invertible on B=h(A)={ h(a): a in A}.
• Then, the density function fY of Y is
B |,)()'(|))(()( 11 yyhyhfyf XY
24
Expectations and variances
Definition. If X is a continuous r. v. with the density function f(x), the expected value of X is defined by
.)()( dxxxfXE
25
• In a group of adult males, the difference between the uric acid value and 6, the standard value, is a random variable X with the following density function:
• Calculate the mean of these differences for the group.
.0
11 )32(2
1
)(2
elsewhere
xifxxxf
26
Sol:
• Remark: If X is a continuous r. v. with density function f, X is said to have a finite expected value if
.3
2]
4
3
3
2[
2
1-
)32(2
1)()(
1
1
43
1
1
32
xx
dxxxdxxxfXE
.)(|| dxxfx
27
• A r. v. X with density function f(x) is called a Cauchy random variable
(a) Find c.(b) Show that E(X) does not exist.
,x- ,1
)(2
x
cxf
28
Sol: (a)
(b)
./1 So,
.)]2
(2
[] [arctan1
1 -- 2
c
ccxcx
dxc
.)]1[ln(1
)1(2
)1(
||)(||
02
0 2- 2-
x
x
xdx
x
dxxdxxfx
29
• Let X be a continuous r. v. with density function f
• For any function h: R R
.)()()]([ dxxfxhXhE
30
• Let X be a continuous r. v. with density function f(x). Let h1, h2, …, hn be real-valued functions, and let a1, a2, …, an be real numbers. Then
)]([)]([)]([
)]()()([
2211
2211
XhEaXhEaXhEa
XhaXhaXhaE
nn
nn
31
• A point X is selected from the interval (0, /4) randomly
• Calculate E(cos2X) and E(cos2X)
Sol:
.1
2
1.
2
2
1
2
1)2cos
2
1
2
1()(cos
.2
]2sin2
[2cos4
)2(cos
6.3, Theoremby Now
.0
4/0/4)(
2
4/0
4/
0
XEXE
xxdxXE
otherwise
ttf
32
Definition. X is a continuous r. v. with density function f
• Var(X) = E(X2) - (EX)2
.)())((]))([()( 22 dxxfXExXEXEXVar
33
• The time elapsed, in minutes, between the placement of an order of pizza and its delivery is random with the density function
(a) Determine the mean and standard deviation of the time it takes for the pizza shop to deliver pizza.
(b) Suppose that it takes 12 minutes for the pizza shop to bake pizza. Determine the mean and standard deviation of the time it takes for the delivery person to deliver pizza.
.0
4025 15/1)(
otherwise
xifxf
34
• Sol:
33.4
.5.2012)()12( )(
.33.4)( hence and
,75.18)5.32(1075)( So
.107515
1)(
,5.3215
1)( )(
12-XX
X
2
40
25
22
40
25
XEXEb
XVar
XVar
dxxXE
dxxXEa
35