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3.1
Random
Variables
and
Discrete
DistributionsA
random
variable
is
some
function
thatassigns
a
real
number
X(s)to
each
possiblee
s
S
,
where
S
is
the
sample
space
foran
experiment.Discrete:e.g.Female=1,
Male=0e.g.
Satisfied=1,Average=0,
Unsatisfied=-1Continuous:
height,
score,
priceDiscrete
DistributionsExample:
Coke
vs
PepsiTen
people
are
surveyed
about
whether
theypreference
Coke
or
Pepsi. The
sample
spacecan
be
regarded
as
the
set
of
210
differentsequences.X
could
bethe
number
of
people
who
preferCoke.PPCPPPCPPPCPPPPCP……X(s)=4Example:
Suppose
the
Environment
ProtectionAgency
(EPA)
takes
readings
once
a
month
onthe
amount
of
pesticide
in
the
discharge
water
ofa
chemical
company.
If
the
amount
of
pesticideexceeds
the um
level
set
by
the
EPA,
thecompany
is
forced
to
take
corrective
action
andmay
be
subject
to
penalty.Consider
the
random
variable,
X,
that
is
thenumber
of
months
until
the
company’s
dischargeexceeds
the
EPA’s um
level.What
values
can
X
assume?Solution:The
company’s
discharge
of
pesticide
mayexceed
theum
allowable
level
on
the monthoftesting,the
second
month
of
testing,
and
so
on.It
is
possible
that
the
company’sdischarge
will
neverexceed
the um
level.Thus,the
set
of
possible
values
for
the
number
ofexceeded
is
the
set
ofallmonthsuntil
the
level
ispositive
integers1,2,3,4,…隨量(random
variable):是實(shí)驗(yàn)結(jié)果的數(shù)值描述。試驗(yàn)隨
量隨 量的可能值與5名客戶聯(lián)系訂貨的客戶數(shù)量0,1,2,3,4,5檢查裝運(yùn)的50臺收音機(jī)次品收音機(jī)的數(shù)量0,1,2,…,49,
50飯店?duì)I業(yè)一天客戶的數(shù)量0,1,2,3,…銷售一輛汽車客戶的如果是
則為0;如果是女性則為1。表1:離散型隨 量舉例經(jīng)營一家銀行裝灌一種軟飲料(最大值=12.1盎司)盎司數(shù)x>=00<=x<=12.1試驗(yàn)隨
量
隨客戶在銀行停留的分鐘數(shù)量的可能值表2:連續(xù)型隨量舉例A
complete
description
of
a
discrete
randomvariable
requires
that
we
specifythe
possible
values
the
random
variable
can
assumethe
probability
associated
witheach
value.ProblemSuppose
that
in
some
population,
the
probability
of
preferring
Coke
over
Pepsi
is
50%.Two
people
are
surveyed.Let
X
be
the
number
of
people
who
prefer
Coke.Find
the
probability
associated
with
each
valueof
the
random
variable
X.
Display
these
valuesin
a
table
raph.Solution:P(
X
0)
P(PP)
1
/
4P(
X
1)
P(PC)
P(CP)
1/
2P(
X
2)
P(CC)
1
/
41/41/2f
(x)0
1
2xNow
we
know
the
values
therandom
variable
can
assume(0,1,2)
and
how
the
probability
isdistributed
over
these
values(1/4,1/2,1/4).
This
comple
ydescribes
the
distribution
of
therandom
variable
and
is
referredto
as
the
probability
function,denoted
by
the
notation
f(x).Discrete
DistributionsX
has
a
discrete
distribution
if
X
can
take
only
afinite
number
of
different
values
x1,...,xk
or,atmost,
an
infinite
sequence
of
different
valuesx1,x2,...The
probability
function
(p.f.)
of
X
is
definedtobe
the
function
fsuch
that
for
every
realnumber
x,
f(x)=Pr(X=x).If
x
is
not
one
of
the
possible
values
of
X,f(x)=0.We
haveorki1if
(x
)
1i1If
A
is
any
subset
ofthe
real
line,Pr(
X
A)
f
(xi
)xi
Aif
(x
)
1The
Distribution
FunctionThe
distribution
function
of
a
random
variableX
is
a
function
defined
for
each
real
numberx:
Since
F(x)
is
the
probability
of
the
event
{X
x}At
any
point
x,0
F
(x)
1Thed.f.
ofa
Discrete
DistributionxxIf
a<b
and
if
Pr(a<X<b)=0,
then
F(x)will
beconstantand
horizontal
over
the
interval
a<X<b.At
every
point
x
such
that
Pr(X=x)>0,
the
d.f.
willjump
by
the
amountPr(X=x).f(x)
F(x)The
BinomialDistributionE.g.
A
machine
produces
a
defective
item
withprobability
p
(0<p<1)
and
produces
anondefectiveitem
with
probability
q=1-p.n
independent
itemsare
examined. Let
X
denote
thenumber
of
these
items
that
are
defective.
X
can
takevalues
0,1,2,...,n.The
BinomialDistributionThis
discrete
distribution
is
called
Binomialdistribution
with
parameters
n
and
p.
x
nfor
x
0,1,...notherwisex n
x0f
(x)
p
qfor x
0for j
x
j
1,
j
0,1,,
(n
1)jpiqnii0for x
n
n
F
(x)
i0
1
Continuous
DistributionsXhas
a
continuous
distribution
ifthere
exists
anonnegative
function
f
defined
on
the
realline,
s.t.
forany
subset
A
of
the
real
line,Pr(
X
A)
f
(x)dxAf
is
called
the
probability
density
function(p.d.f.)
of
X. Every
p.d.f.
must
satisfy
tworequirements:f
(x)
0
f
(x)dx
1The
probability
density
function
(pdf)
for
acontinuous
random
variable
is
afunction
which
canbe
integrated
toobtain
the
probability
that
the
randomvariable
takes
a
value
ina
given
interval.P(a<=X<=b)=P(X=a)=0f
(x)dxbaContinuous
DistributionsProbability
for
IndividualValuesSoPr(a
X
b)
Pr(a
X
b)
Pr(a
X
b)
Pr(a
X
b)For
every
individual
value
x,xxf
(t)dt
0Pr(
X
x)
The
d.f.
of
a
Continuous
DistributionSuppose
X
has
a
continuous
distribution
with
p.d.f.f(x). Since
the
probability
of
each
individual
point
xis
0,
the
d.f.
F(x)
willhave
no
jumps.Furthermore,xF
( )
f
(t)dtdxat
each
point
x
at
which
f(x)
is
continuous,F
'(x)
dF
(x)
f
(x)Uniform
Distribution
on
An
Intervala
and
b
are
two
givenreal
numbers
suchthata<b.A
point
X
is
selected
from
the
intervalS
{x
:
a
x
b}The
probability
that
X
will
belong
to
anysubinterval
of
S
is
proportional
to
the
lengthof
that
subinterval.This
distribution
is
called
the
uniformdistribution
on
the
interval
(a,b).The
p.d.f.f(x)
of
Xmust
be
0
outside
S.f(x)
must
be
constant
throughout
S.Also,The
p.d.f.
of
Xmust
beba
Sf
(x)dx
f
(x)dx
1for
a
x
botherwise1f
(x)
b
a
0b
afor x
afor
a
x
bfor x
b
10F
(x)
x
aThe
value
of
the
constantis
the
reciprocal
of
thelength
oftheinterval.It
is
not
possible
to
define
a
uniform
distributionoverthe
interval
x
a because
the
length
of
this
intervalis
infinite.Since
the
probability
is
0
at points
a
or
b,
itisirrelevant
whetherthe
distribution
is
regarded
as
auniform
distribution
on
[a,b],
(a,b],
[a,b)
or
(a,b).Example Calculating
Probabilities
from
ap.d.f.0
otherwiseThep.d.f.
of
a
X
has
the
followingform:for
0
x
4f
(x)
cxc
?Pr(1
X
2)
?Pr(X
2)
?Example Calculating
Probabilities
from
ap.d.f.Solution:0
otherwiseThep.d.f.
of
a
X
has
the
followingform:for
0
x
4f
(x)
cxc
?Pr(1
X
2)
?Pr(X
2)
?14042
8Pr(X
2)
1
xdx
34163xdx
Pr(1
X
2)
cxdx
8c
1
c
182
18Note:
c
is
often
called
the
normalizing
constant.Its
value
is
unique.102
116for x
0x for
0
x
4for x
4F
(x)
Example:
Unbounded
Random
Variables(1
x)2for
x
01
0In
an
electrical
system,
the
voltage
X
is
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