Download Pr obability Distributions

ility
/3 of
,
also
I
that
ge or
broner
b and
; how
f 3 of
l
green
ch ball
ilraw is
3 defecpurchase
he defec-
Random Variables and
Pr obability Distributions
J."/ Concept af a Random Variable
The field of statistids is concerned with making inferences about populations
and pop-ulation- qhar.actffistics. Experiments are'ioiiducied with results that
'are
subject to chance. The testing of a number of electronic components is an
example of a statistical experiment, a term that is used to describe any process
by which several chance observations are generated. It is often very important
*) tg allocata".a nuqerical description !9 the outcome. For example, the sample
space giving a detailed description of each possible outcome when three elect1o-9jc_.qppponents are tested may be written
5:
{NNN, NND, NDN, DNN, NDD, DND' DDN,,DDD\,
-.
'
I,r
where N denotes "nondefective" and D denotes "defecti#.n'One is naturally
concerned with the number of defectives that occur. Thus each point in the
sample space will be assigned a numericsl value of 0, I, 2, or 3. These values
are, of course, random quantities determined by the outcome of iei*priiilinf.
fn*y *"y be viewed as values assumed by the random uariable' X, the number
gi-Ce,{ectivg items when three electronic components are
tested.
4
j
Ch.
44
Definition
2.1
.4 random variable
a func
2'
Rsndom Variables and Probability Distributions
tion tAq-
qs ssCt:q! e,s q-r gql luy1b-91-Vtt!-e
rch:JSmep in
the.s*mple space.
We shall use a capital letter, say X, to denote a random variable and its
corresponding small letter, x in this case, for one of its values. In the electronic
component testing illustration above, we notice that the random variable X
assumes the value 2 for all elements in the subset
5:
{DDN, DND, NDD}
of the sample space S. That is, each possible value of X_represents an
that is a subset of the sample space for the given experiment.
Example 2.1
Two balls are drawn in succession without replacement from an urn containing 4 red balls and 3 black balls. Th_e pA-s-lible outcomes and the valueg-;r of
the random variable
,n/ \*j
't..t
t,).,a'v i *^
.,:.r
:,-!,..r .J1.1
Example 2.2
eveg-t
Lol"", t*r*qh
",
*$j:I
it
ll1".::ryEt-q!9{ b{ls'
Sample Space
v"
RR
2
RB
1
BR
1
BB
0
are
A stockroom clerk returns three safety helmets at random to three steel mill
employees, who had previously checked them. If Smith, Jones, and Brown, in
thai oider, receive one of the three hats, list the sample points for the possible
orders of returning the helmets and find the values m 9'-f lhe random variable
M that represents the number of lneclSatches'
Salution. If S, J, and B stancl for Smith's, Jones', and Brown's helmets, respectively, then the possible arrangements in which the helmets may be returned
and the number of correct matches are
Sample Space
fl
SJB
SBJ
JSB
JBS
BSJ
BJS
3
1
1
0
0
1
tns
,,
iec. 2.1
'
45
Concept of a Rantlom Vaiable
each of the two preceding examples the sample space containp a finite
number of elements. On tne other hand,-yhg! a d-i9 i1lhJown unti! a 5 occurs,
In
in
..*a'afmna-sempii slace m-rmannnena_ing
s:
its
nic
sequence of elements,
{F, NF, NNF, NNNF,...},
N represent, respectively, the occurrence and nonoccurrence Of a
can be equated to the
5. But even in this experiment, the number of elements
where F and
tX
second element, a
number of whole numbers so that tllere is a first element, a
and so on, and in this sense can be counted'
ihird
"taro"nt,
&qt
tleffnitiOn
possibilities or an unending sequence
2.2 If a sample space contoins a fnite number ofnumbers,
it is called a discrete sample
with as
m-any elements as there are whole
space.
rin-
tof
Theoutcomesofsonrestatisticalexperimentsmaybeneitherfinitenor
an investigation
countable. Such is the case, for example,-when one conducts
travel over a
will
automobile
oi
certain
measuring the distances that a
-aiii
to be a varidistance
Assuming
gasoline.
of
Of*;i;i test course on 5 litersof accuracy, then clearly we have an infinite
able measured to any degree
be equated to
number of possible distanies in the sample space that cannot
length of time
the
record
to
were
one
if
the num-beF-o$*whsle*numher$, Also,
time intervals
possible
the
again
place,
once
i;;;il;cal reaction to take
We see
uncountable'
and
in
number
sample ,puc" utt infinite
making up our
now that all sample spaces need not be discrete'
mill
illdrution 2.3 If a sample space contains
on infnite numbey of possibilities equal to the number of
pointson.alinesegment,itiscalledacontinuoussamplespace.
n, in
cible
iable
ryec-med
@
possible
A random variable is called a discrete ranilom variable if its set of
0, 1,
are
in Example 2'1
outcomes is countable. Since the possible values of Y
it
follows
uii z, and the possible values of M in Example 2.2 are 0, 1, and 3,
can
that y and M are discrete random variables. When a random variable
vsriable'
random
take on values on a-c-qn-tiDuqEs- scgle, it is called a continuous
precisely the
are
variable
random
of
i
continuous
nutu"r
oil"o ttr" possible
space._ Such is the
same values that are contained in the continuous sample
random variable represents the measured distance that a certain
gasoline'
oiuotomobile will travel over a test course on 5 liters of
case when the
2 |
--'zF
.^..
*utd
i;
most practical problems,
99!-4_Luo-qs- random- YaqAllgS -Lep!9!9-nt-"msa-
or life
telqpqrallules, distaqce$,
w-94-ngj9,cua3se!!9uoUt"-frtigltlt*trehtt,
k
/
-*,'
such as lhe
."tta"",*' ,frf
wt*"* airdt"
ffi:- ,--t''
* I ffi-'
drn
""ri"ut"-*reBL?,seatJpunL-da-ta,
qltlhe luslbpr of highway-,iel.a-lities
{, ;ffi; sf dged="_"t a_sample of k itp$s
variables Y an{.M of E11m-f-ryrt;-iry4-gvel -stata 59i*1har-the.-rand-om
rdi
pd;.i;;a fi-ioth repreienicount data, f the number of red balls and r14
:tr*
I
*r,"--14-
v
-
;;;U"t
of corrgct hat rnatches'
,
t
'r"*, @
]#
I'li"li.{ r.
TT
Ff
6.
pt(
'1-
jr' "'. ++r
I J'
,r' !'
t lTr
Ll
46
T
KY
"4
H
t_plf
,?- 1 rrl-l
Ch.
2'
Random Variables and probability Distributions
2.2 Discrete Probability Distributions
,{ dis.c-r91e randgm variable ?s-suTes each of its values with a cer1ain-probabil", ity"Jn the case of tossing a coiiTfi-ree-iiffi&;'tr6-14;it6l" x,-iepresenting the
number of heads, assumes the value 2 with probability 3/g, since 3 of the g
equally likely sample points result in two heads and one tail. If one assumes
equal weights for the simple events in Example 2.2, the probability that no
employee gets back his right helmet, that is, the probability that M assumes
the value zero, is l/3. The possible values m of u and theii probabilities are
given by
i
-.ir')'
!1ri":!r.
t:
t\
!
.r
i
i't/:{i
P(M
:r,ii
:
m)
111
326
L-l
le(tL
..,:.t . ,.-.,.i.. i
i:
,\ -'s 1 ,'
Note that the values of m exhaust all possible
add to l.
cases and hence the probabilities
Frequently, it is convenient to represent all the probabilities of a random
variable X by a formula. such a formula would necessarily be a function of the
numerical values x that we shall denote by f (x),g(x), r(x), and so forth. Therefore, we write /(x) : P(X : x); that is, /(3) : p(X: 3). The set of ordered
pairs (x,/(x)) is called the probability function or probability distribution of rhe
discrete random variable X.
Definition
2.4
The set of ordered pairs (x,f (x));s a p1qlghiljlx.fg4plior-r, probability mass funcrion,
or ppla-bllly ,Qqt1ilrrlion of the discrete random uaiiable X rf,for each possible
outcome
x'
lr'r t>' o
2it"':"t
{"
l./(x)>0.
2.lf1x1: t.
Z.
Example
23
PtX:;r) :/(x).
A shipment of 8 similar microcomputers to a retail outlet contains 3 that are
defective. If a school makes a random purchase of 2 of these computers, find
the probability distribution for the number of defectives.
sslntion.
Let
X
be a random variable whose values
x
are the possible
numbers of defective computers purchasec by the school. Then x can be any of
.
j -
-1
:- ':
.]'
Sec.
2.2
,4ri*rl.:,*{
'*l
i. ,.
" i
!''|,r"r /i t li { ' (
47
Discrete Probability Distributions
the numbers 0, 1, and 2.
'
"
.
/:\
r;10
/to1:P(X:0):7f:X'
-t
a
Now'
,0
\,1
.:,."'.: ...
e
8
s
o
:
t."
e\ ty'
:s
f (2) : P{X
:2) :
'ii", ;il;."i"uii',v Ji',ribuiion ol X is
0
)m
he
te-
ed
the
Example2.4If50%oftheautomobilessoldbyanagencyforacertainforeigncarare
equrppedwitlldiesele4g!ne$n.a"r"'-"r"foitheprobabilitydistributionof
the next 4 cars sold bv this agency'
dl
r,
tt. n,i,.,'U",
_
-\:..--ales66"olJilnons
"'^'"-'
'
j
,.*. 'i-
-SD,
ible
---
gasoline model is
probability of selling a diesel model or a
Therein the sample space are equally iikely to occur'
function' will be
;11 prouuuititi"t, and.also for our
^Solulion,$iace..the
tO'Uoints
0.5,
;;;';i:
,or","iil"'o*;t'";;;;
i6..fg-qbt-aa$9-lu-loberofw'ays.ofselling3diesel..models"weneedtocon.
two cells with 3
of *uy, oi partitioning 4 -outcomes into
*<--rla"i-t[-ndUei
\, diesel models assigned to one Lu ana a gasoline model assigned to the other'
general, the event of seliing x diesel
This can be done - (i) - 4 ways. In
.
-....
r
models and 4
.:
-
/a\
where x can be 0'
x gasoline models can occur trr (;) ways'
2,3,ot4' Thus the probabilnt uttT:itton/(x)
/(x)==
t are
find
sible
ay of
''/
\r/
i?
for
1'
: P(x: x) is
x:0. l, 2, 3,4'
Therearemanyproblemsinwhichwewishtocomputetheprobabilitythat
theobservedvalueofu.u,'do-variableXwillbelessthanorequaltosome
w-a-def,ne
gpJ'
-aumber"-v'real number x. writigs fqi :-l-tljl -"1 :l'"-ry f
f thE random variable'x'
r(');;e"ii";ffi"dffiaisi'q
--Wililil1
-
Ch.
48
2'
Random Variables and Probability Distributions
Definition2.5,,'oW)ofadiscreterandomuariableXwithprobabiIity
r'(x): P(X<x):
I "f(0
for
,<I
-a <x< cc.
For the random variable M, the number of correct matches in Example
2.2,
we have
F(2.4)
:
P(M < 2.4)
The cumulative distribution of
:f
(0\
+/(1)
:
:
(*) + (i)
*.
M is given by
|.o form<0
m<l
r1'l:lil+ foro<
rorl< m<3,'
. U form>3.
a-i;)
{ Examnle
^
\
2,5" tF;na the cumulative
-''
-{' ,
lt
ti, t/,
'/t
,. j. "'
il}
i
'''
.1 .,'
\G
fL?;:'
Itt I
:
{
lr )f
tl
*
_.7
IF
-iti= /"
2.4.
of the probability distribution of Example
2.4
A.t."ti
"--;"'
ir-
(2)
:
r(0):/(0):
t
tlt6,f (r) :
give/(0) :
d
*
distribution of the random variable X in Example
,r^t-- F(x),
r1--\ .,^-:f,,
(2\:- lle
318.
verifY ;L^+
thatf/'/,)\
Using
Solation. Direct calculations
lhr
I}
f iu 1-
One should pay particular notice to the fact that the cumulative distribution is
defined not only for the values assumed by the given random variable bui for
all real numbers.
8
Il4,f
318,f (3)
:
U4, andf (4)
r(r):/(0) +/(l): *
F(2) : f (o) + f(t) + f (2) : {+
r(3) :/(o) + f(t) + f(2) +/(3)
r(4) :/(0) + f(t) + f(2) +/(3)
:
+*
+
f(4)
Hence
, i,,)= 'lr
|.t
r('):
Now
f(2)
lf
{
forx<0
for0<x<1
forl<x<2
lit
for2<x<3
for3<x<4
Li'
forx>4.
l_i
: F(2) -F(1) : 1+ - * :
t.
:
:
1.
Ur6' Therefore,
l
ihutions
Discr ete Pr obabilit
y
49
Distributions
fiabilitY
6l t6
s
lt6
4l16
t6
3l
nrple 2.2,
2lt6
I
Figure
2.1
lt6
Bar chart.
is often helpful to look at a probability distribution in graphic form. one
might plot the points (x,,f(x)) of Example 2.4 to obtain Figure 2.l.By joining
the points to the x axis either with a dashed or solid line, we obtain what is
commonly called a bar chgrt. Figure 2.1 makes it very easy to see what values
of X are most likely to occur, and it also indicates a perfectly symmetric situation in this case.
Instead of plotting the points (x,"f(x)), we more frequently construct rectangles, as in Figure 2.2.Herc the rectangles are bonstructed so that their bases of
equal width are centered at each valus x and their heights are equal to the
corresponding probabilities given by"f(t).The bases are constructed so as to
leave no space between the rectangles. Figure 2.2 is called a probability histo-
It
ibution,is-ile
but for
rmple 2.4.
gram.
nmple 2.4
herefore,
I
since each base in Figure 2.2 has'.nit width, the P(x : x) is equal to the
area of the rectangle centered at x. Even if the bases were not of unit width' we
could adjust the heights of the rectangles to give areas that would still equal
the probabilities of X assuming any of its values x. This concept of using areas
to represent probabilities is necessary for our consideration of the probability
distribution of a continuous random variable'
The graph of the cumulative distribution of Example 2.5, which appears as a
step function in Figure 2.3, is obtained by plotting the points (x, F(x))'
6l 16
s
lt6
4lt6
3116
z.l
t6
I lr6
0l
Flgure
2.2
Frobability histogram.
50
Ch.
2'
Random Variables and Probability Distributions
r--
I
r___________l
314
I
r------J
t12
I
I
t14
?------------J
I
I
Figure
2.3
Discrete cumulative distribution.
Certain probability distributions are applicable to more than one physical
situation. The probability distribution of Example 2.4, for example, also
applies to the random variable Y, where Y is the number of heads when a coin
is tossed 4 times, or to the random variable W, where W is the number of red
cards that occur when 4 cards are drawn at random from a deck in succession
with each card replaced and the deck shuflled before the next drawing. Special
discrete distributions that can be applied to many different experimental situations will be considered in Chapter 4.
2.3 Continuous Probability Distributions
>1
'.(' ,J (.\, A continuous random
a probability of zero of assuming exactlv
.._ variable has
.',,,,:--'-':-----=- cannot be-givet.
distribution
m €-it-61-m-vamECTonsequentlylitFprobability
-- in
q
,'ti\.!
,l
ibfrlnt
first this may seem startling, but it becomes
more plausible
ffid,
-''''t -..:^'
-=:--.r
E
.'; -f'
r- when
.\
L
we consider a particular example. Let us discuss a random variable
.... 1i {i.. -, r,, !
.-
whose values are the heights of all people over 2l years of age. Between any
.ll,
.*t.o ,,u }f,..
-"
\" {. t lr.
. two values, say 163.5 and 164.5 centimeters, or even 163.99 and 164.01 centit -ndL+r"\^.',' meters- there are an infinite number of heishts. one of which is 164 centimeters.
- q The prdba-.fiIi1! of selecting a person at random w-ho is qx4ctly 164 centim_eters
i'
, A '' ., ^ / tall and not one of the infinitely large set of heights so close to 16a centimeieii
you cannot humanly measure the difference ltg:fr,ole, and thus we assign
, s-"Fr* "ntr'' ,nN, ] thatprobability
p I"-t
of zero to the event. This is not the case, however, if we,f-alk
[a
M
about the probability of selecting a person who is at least 163 centimeters'6u1@ \t ,^5 I-1;d'
W .,id}" \u
not more than 165 centimeters tall. Now we are dealing with an interval rather
'$''
,
tttun u poini uutu" of our *nao* variable.
We shall concern ourselves with computing probabilities for various intervals of continuous random variables such as P(a < X < b), P(W > c), and so
forth. Note that when X is continuous
P(a
< X < b) : P{a < X < b) + P(X
:
P(s< X <b).
:
b
I
I
Scc. 2-3
C
ontinuous Pr ob abilit y Distributions
5I
ical
also
:oin
'red
sion
rial
itu-
(c)
Figure
9e
),
.That
2.4
is..
it
Typical density functions.
does not matter whether we include an end point of the interval or
X is discrete.
| - not. This is not true. though. when
Although the probability distribution of a continuous random variable
it can have a formula. Such a formula
would necessarily be a function of the numerical values of the continuous
vaiiable X and as such will be represented by the functional notation/(x). In
, ^,\. / dealing with continuous variables,/(x) is usually called the probability density
[*
function, or simply the density function of X. Since X is defined over a continuous sample space, it is possible for/(x) to have a finite number of discontin-.:,',t5
.;.. f
uities. However, most density functions that have practical applications in the
.. ,, analysis of statistical data are continuous and their graphs may take any of
' ,, ,
several forms, some of which are shown in Figure 2.4. Because areas will be
used to represent probabilities and probabilities are positive numerical values,
.' ,-,
the density _fu,Irgli_on must lie entirely above the x axis.
itv density
densitv lunction
function is constructed so that the area under its curve
A probabilitv
ualtolw
bounded bv the x axi
qyer,ltlqlg-lgr el X_!gI
x) is defined. Should this range of X be a finite interval, it is always
possible to exten
interval to include the entire set of real numbers by
defining/(x) to be zero at all points in the extended portions of the interval. In
Figure 2.5, the probability that X assumes a value between a and b is equal to
the shaded area under the density function between the ordinatss at x : o and
x : b, and from integral calculus is given by
cannot be presented in tabular form,
actly
Eg
ln
lsible
iable
r any
Entiqlers.
rcters
aeteid"
ssign
;
talk
rs but
rather
inter-
nd
so
tl,
1
IL-P@. x <b): J"
I /tx) a_r.[
--l
52
Ch.
2'
Random Yariables and probability Distributions
Figure25 P(a<X<b).
Definition
2.6
The functionl(x) is c probability density function /or the continuous ranilom aariable X, defined ouer the set of real numbers R, if
\
:i.f(,t)>o
2.
ri
^
r@
forauxeR. \
\\
dx:
r.
fg)
\
I
J---
I'
i:ti
i
I
i-fDi
, 3. p(a<x<b)= Jo
| /(x)dx.
-..,-. ..-
\
,....',
Example
2.6
Suppose that the error in the reaction temperature, in oC, for a controlled
laboratory experiment is a continuous random variable X having the probability density function
It
-l<x<2
rr,): J I .
t0,
(a) Yerify condition 2 of Definiti on
(b) FindP(0<X<l).
elsewhere-
2.6.
Solution
t*t
8r
9'9
J' rt,l o,: I_,t o.:il'_, --+-:l
(b)P(o<x<l):,[
,*:ul,
I
9'
I
Sec.
2.3
.
Continuoas Probability Distributians
Definition
2.7
53
The cumulative rlistribution F(x) of a continuous ranilom Dariable
function[(x\ k giaen by
F(x):
P(X <
ft
x): I f(t) dt
J--
X with density
for-m<x<@.
As an immediate consequence of Definition 2-'l one can write the two results
P(a<X<b):F(b)-F(a)
and
f(x):;
dFkl
if the derivative exists.
Exarnple
2J
For the density function of Example 2.6 find F(x) and
P(0<x<l).
Solution For-L<x<2,
t' n',,
a
Therefore,
j-
I
-"1'
9l-, -
x<
-l
-l<x<2
h+f
x>
=a,!X
n*l
2.
tt
a
L
) yl rtv = 'i b
,z
,l 7
-I
use
--
gve la
:r,t t =
{t-tf
(-l-'|_=$
g - '91 I
z
-l
Figure
2.6
r,o
08
06
o'4
o
I
Continuouscumulative distribution.
x3+1
it to evaluate
Ch.
2'
Random Variables and probability Distibutions
The cumulative distribution F(x) is expressed graphically in Figure
2.6. Now,
P(0 <
X<
1)
:
r(1I_
F(0)
:
4
_
t
: *,
which agrees with the result obtained by using the density function in Example
2.6.
Exercises
" l.
Classify the following random variables as discrete 06. From a box containing 4 dimes
and 2 nickels, 3
or continuouscoins are selected at random without replacement.
Find the probability distribution for the total T of
the number of automobile accidents per year
the 3 coins. Express the probability distribution
in Virginia.
graphically as a probability histogram.
Y: the length of time to play 18 holes of golf.
M:
the amount of milk produced yearly
N:
particular cow.
the number of eggs laid each month by
P:
Q:
by a-./7. From a box containing 4 black balls and 2 green
balls, 3 balls are drawn in succession, each ball
a
hen.
the number of building permits issued each
month in a certain city.
the weight of grain produced per
,2. An
acre.
being replaced in the box before the next draw is
made. Find the probability distribution for the
number of green balls.
'/8' Find the probability distribution of the random
variable 17 in Exercise 3, assuming that the
coin is
overseas shipment of 5 foreign automobiles
biased so that a head is twice as likely to occur as
contains 2 that have slight paint blemishes. If an
a tail.
agency receives 3 of these automobiles at random,
list the elements of the sample space S using the '/9. Find the probability distribution for the number
of jarz records when 4 records are selected at
letters B and N for ..blemished - and
random from a collection consisting of 5 jezz
" nonblemished," respectively; then to each sample
records, 2 classical records, and 3 polka records.
point assign a vdlue x of the random variable X
Express your results by means of a formula.
representing the number of automobiles purchased by the agency with paint blemishes. ./10- Find
a formula for the probability distribution of
t 3. Let W be a random variable giving the number of
the random variable X representing the outcome
heads minus the number of tails in three tosses of
when a single die is rolled once.
a coin. List the elements of the sample space S for
/ll.
A shipment of 7 television sets contains 2 defective
the three tosses of the coin and to each sample --'
sets. A hotel makes a random purchase of 3 of the
point assign a value w ofW.
sets. If X is the number o[ defective sets purchased
/4. A cain is flipped until 3 heads in succession occur.
by the hotel, find the probability distribution of X.
List only those elements of the sample space that
Express the results graphically as a probability
require 6 or less tosses. Is this a discrete sample
histogram.
space? Explain.
lL Three cards are drawn in succession from a deck
v'5. Determine the value c so that each of the followwithout replacement. Find the probability dising functions can serve as a probability distributribution for the number of spades.
tion of the discrete random variable X:
13' Find the cumulative distribution ol the random
t(a) f(x) : c(x2 + 4) for x : 0, 1,2,3;
variable l/ in Exercise 8. Using F(w), find
(b)
f(xt:
"(:)(,I.)
for
x:0,
r,2
(a) P(W > o);
(b) P(-t<w <3).
;ec.
63
2.5 Joint Probability Distibutions
Thus a
coded by the letter b; and so forth'
lb and
of
value
stem
a
has
1.29
as
such
number
a leaf equal to 29'
(b) Set up i relative frequency distribution'
decimal point, each repeated five times such
that the double-digit leaves 00 through 19 are
a;
associated with stems coded by the letter
stems
with
associated
are
39
through
20
leaves
1.5 Joint Probability Distributigr'!
(e)ourstudyofrandomvariablesandtheirprobabilitydistributionsinthepre.
ced i n g sec ti o ns * ;';.;;;;t'd tggl,':q iT'l'i " " il' l++"-:p:"="-::
:l:: H
i:
".
f*.";0
;I;i,"li3:.!
r'v
rrrv4Jurv the
might measure
we
w€ trrrBrrl
.ri-ancinnal
'ving rise
-:-^ .^
^ +rr,^
two-dimensional
to a
from a controlled chemical experiment
be-interested in
(p'
sample space consistd^;ilil;;tcomes
:l'
"-t "t"::lT
T of cold-drawn copper resulting in the
the hardness H and ,"i'if"
'tt""gth
io J"i".-ine the likelihood of success in college'
ourcomes (h, 4. In ;;;
"
use a three-dimensional sample space
based on high school J*", on" might
andrecordforeachindividualhisorheraptitudetestscore,highschoolrank
year in college'
gr
endof the freshman
in class, and grade-poi;; ;;;;;g" at the
the probability distribution
If X and y ur" t*o-dlr"."tJ."rroom variables,
represented by a function with values
lor their simultaneous occurrence can be
variables X
puit or *ru"t 1t' v) within the range of the random
I;;;;";;"v
distribuprobability
the
as
ioint
and y, It is customary ,o-i"r"t ,. this function
case'
tion of X and Y' Hence, in the discrete
f{x,Y}:P(x:x'Y:Y);
that outcomes x and y occur at
that is, the values /(x, y) give the probability
set is to be serviced and X repthe same time' For "*L'u:ptin, if aielevision
of the set and Y represents the number of
resents the age to ttre iea're'i y"u'
that the television set is
in trr" r"t, it"n i6,z)is the probability
defective tubes
5 years old and needs 3 new tubes'
2"8
distribution or probability
f (x, y)is a ioint probability
Y
if
uariables X and
the discrete
The function
mass function o/
'lon''lo^
7. f (x, y) >
,
0 for all (x, Y)'
y): 1.
\\.f(x,
3. P(X:x,Y:Y):f(x,Y)'
For
any region
A in the xy plane' Pl(X' Y) e Af
: I I 16' il'
64
Ch.
Example 2.8
2 - Random Yariables
and Probability Distributions
Two refills for a ballpoint pen are selected at random from a box that contains
3 blue refills, 2 red refills, and 3 green refills. If X is the numbcr of blue refills
and Y is the number of red refills selected, find (a) the joint probability func{oni(x, d, ana 0)"1(X,Yf e Af,whdre.4 is the region {(t, y)lx + y < 1}.
Solation
i.,.
j
(a) The possible pairs of values (x, y) are (0, 0), (0, 1), (1, 0), (1, 1), (0, 2), and
(2,0). Now,/(0, 1), for example, represents the probability that a red and a
green refill are selected. The total number of equally likely ways of selec-
:
ting any 2 refills from the
t-
red from 3 red refills and
I green from
0
28. The number of ways of selecting I
3 green refills is
(?Xi) :
6. Hence
1) :6128 -- 3114. Similar calculations yield the probabilities for the
other cases, which are presented in Table 2.6. Note that the probabilities
sum to 1. In Chapter 3 it will become clear that the joint probability distribution of Table 2.6 can be represented by the
,!
,f(0,
formula
'
., \-00(,-1-,)
Jtx,y):T__,,
\r/
for x : A, \2; | : 0, 1,2;0 < x + Y <
(b) P[(X, Y] e A]: P(X + Y < 1)
:"f(0,
0)
+/(0'
:rr3+*+*
1)
2.
+/(l' 0)
:&
Table 2.6 Joint Probability
Distribution for Example 2.8
x
f(x,
yl
v)
0
ii
o
i^.\
Row
Totals
2
'_tle
-L
l28l28i
28
i -r- I -r-
ll'
2A
rli
i 2R:i
-L
2A
3
7
,1411+
2
Column
Totals
,r-r i
-r
La. 2g 2a
i
I
l-.r'
t,
-o;
I
uions
Sec.
2.5
Joint
Pr ob
density function
and
P[(X,
Y\
Af,
where,4 is any
e
f(x,y)
region in the xy olane- is eoual to tht uolg-e of th" !!ght t
the base .4 and the surfiG]-
a@,
is
refills
func-
r).
Definition
2.9 ,**;1t?;f(*,y\
'selec-
*ing
65
X and Y are continuous random variables, the joint
When
ntains
l), and
and a
abiltt y Distributions
l.f(x,v)>
1
r.
Hence
fa
.J__ .l__
3.
or the
fo
O
isc jointdensityfunction of thecontinuousrsndomoariablesX
for all (x, y).
f(x, y) dx dy :
Pt(x, Y) e
Af:
bilities
1.
iitO, fl
dx dy
ty disfor any region A in the xy plane.
Erample Z.g
'
) A candy company distributes boxes of chocolates with a mixture
toffees, and nuts coated
in both light and dark
of creams,
chocolate. For a randomly
let X and Y, respectively, be the proportions of the light and
dark chocolates that are creams and suppose that the joint density function is
selected box,
given by
f(x,
v): {::t
+3y), 0<x<1,0<y<l
elsewhere.
l.
condition 2 of Definition 2.9. *)
(b) Find P[(X, Y) e A], where.4 is the region
(a) Verify
Solution
(a)
fo
-,
fo
l- l* 1,",
J-*J-*
y) dx
dy:
ft rr;-*\\
.l. .J, itz,
*t,u'
!e:t'/t'
r
"
i'"'
* 3y) d) dy
6xvr'=r
:l ,rr*r
.) +=l dv
) l'-o
Jo
2v 3v'l'
: Jof' (?*
a,
9)
'':T*
t l.
\) J/
23
---!--l- t - -
))
t.
-'
J'r"-{'s
66
Ch.
(b) P[(x, Y) e Af
2'
Random Variebles anil Probability Distributians
:P(o<x<t,+<Y<+)
f Ltz f Ltz 1
:
+ 3Y) ttx dY
J,,. Jo i{z'
6xyl'= rz
:l | ''' 2r'
+-:l
) l'=o dy
Jrn )r
:(* . +) *: fi * #l:,,',
:
l,:
:*
(1.;)- (i. *)l:,60.11
Given the joint probability distribution/(x, y) of the discrete random variables X and Y, the probability distribution g(x) of X alone is obtained by
summing f (x, y) over the values of Y. Similarly, the probability distribution
h$) of Y alone is obtained by summing f (x, y) over the values of X. We define
g(x) and h(y) to be the marginal distributions of X and Y, respectively. When X
and Y are continuous random variables, summations are replaced by infegrals.
t
We can now make the following general definition.
,
Definition
2.10
The marginal distributions of
s(x):L.f
for
x
alone and of
t*,
yl
alone are giuen by
and h(y):L f$,
y)
the discrete case and by
l€
e(x):1 f8'Y)dY
J-for
y
and h(y): Ifo f(x. y) dx
J--
the continuous csse.
The term marginal is used here because, in the discrete case, the values of
g(x) and h(y) arejust the marginal totals of the respective columns and rows
when the values off(x, y) are displayed in a rectangular table.
Example
2.10
Show that the column and row totals of Table 2.6 give the marginal distribution of X alone and of Y alone.
Sotation. For the random variable X, we see that
p(x :0)
2
: s(0) : I "r(0, y) :,f(0, 0) +/(0, 1) + f (0,2)
v=o
:zt+*+*:*,
\
t
67
J oint ProbsbilitY Distributions
2
P(X:1):9(1):lf0,t)
v=0
:/(1,0) +/(1,
+f(r,2)
1)
--&+rnr+o:4;'
2
L f(2, v) :f(2,0) + f(2' r) + f(2'2)
a P(X : 2) : g(2) : )=0
:*+0+0:*,
lu:
2'6' In a similar manner we could
which are just the column totals of Table
In tabular fornr' these
given by
show that the values
ll9 to't" totals'
"*
"i-lttyl
as follows:
margittut distributions may be written
Lxample 2.11
function of Example 2'9'
Find g(x) and h(y) for the joint density
Solution. BY definition'
ra
s$):
v) dY
.J__r,*,
4xv 6v'lt=
:
f
(
1,
and g(x)
:
("
;
1 4x * 3
.|.
: s *lol,=o
for 0 < x
r')
+
3Y) dY
5
0 elsewhere' Similarly'
ro
h(y):
.|_to,
r)
dx:
lr )
J.
i,t.
+
3) dx
-4r t_3v\
)
:0 elsewhere'
for 0 < Y S 1, and h(Y)
prob*
g(x) and hlyYt: inlef the
The fact that the marginal distributions
be
easily
can
alone
Y
variiL. X "nA
abitity distributions ,iiir-iiai"idual
are
2'6
Definition
or
of Definition 2.4
verified by showing ;1";;;""ditions
case
continuous
..iitn"a. io, examplc, in the
[*
j--
n(r)
ar:
v\ dv dx: I
,,',
l*
i'
J-o J-o
Ch.
2'
X<
b)
68
Random Yariables anil Probabtlity Distributions
and
P(a <
:
X 1b, -co < Y <
P(a <
:l lb Il*
JoJ-
co)
f@,v)dvdx
: Ifbg$\ dx'
Jo
2.1 we stated that the value x of the random variable X represents an event that is a subset of the sample space. If we use the definition of
conditional probability as given in Chapter 1,
In Section
P(Bl A)
where
/
:
P(A
o
B)
P(.4)
P(A)
and B are now the events defined by
>
0,
X : x and Y: y, respectively,
then
P(Y
: ylX: x): P(X--x,Y:l)
P(X : x\
:
IJ+!,
s@)
s(x) > o,
*Itlr X and Y are disglgte'raadam-raua-bles*."
Aiffilt t; show that the function f(x,y\ls!), which is strictly a
function""i
of y with x fixed, satisfies all the conditions of a probability distribuwhen
tion. This is also true when/(x, y) and g(x) arc the joint density and marginal
distribution of continuous random variables. Expressing such a probability
distribution by the symbol/(y lx), we have the following definition.
Definition
2.11
Let X and Y be two random variables, discrete or continuous. Ihe*cgndjtional
distribution of the random uariable Y, giun thot X : ,, it gir.4
f ulx)
-<
:
s(x) > o.
W,
Shnilarly, the conditional distribution of the random variable
Y : y, is giuen by
f(xli:W,
X,
giuen that
h(v)>0.
If one wished to find the probability that the discrete random variable X
falls between a and b when it is known that the discrete variable f - y, wE
evaluate
P(a
< X < blY
:
y)
: I.f(rly),
lttl
]j
,tl
69
J oint Pt obabilitY Distributions
over all values of
where the summation extends
and Y are continuous, we evaluate
P(a <
Solurton.
x < blY
: !) :
:
w" needl find/(xly), where y
= x=O
i .f(x, 1):
h(l)
Now
between a and b' When
lb
X
til
a*'
J"ft*li
1' First we find that
* + * +o :+'
f(x, l) t .(r,
/(xl1):7f=3r ,),
Therefore,
X
lii
irir
x
:
o,
l,
Z.
: +f(a,l): (3x*): *
: (3x*) : I
:
/(11 1) !f(r,l)
: (3x0) : 0
:
/(21 1) zf {2,1)
/(0t1)
of
and the conditional distribution
X' given that Y
: l' is
f(x I l)
FinallY,
/-' ^
*fu
O
r€t
Y: 1):.r(ol1):
*'
is red' we have
a
that 1 of-the 2 pen refills selected
G)Therefore, if it is known
blue'
to-r/z tt'ut the other refill is not
-;;;;;;uiiv
"qutt
Y of female
r-^^+r^n Y nf male runners and the fraction
"qlp1s-z,reJw.*Jn'ff JfiT'*,f,*?:::,':#TFilil#;ii":"*'densitv
o!\(t{r
function
x
bnra
r(x, y):
til'
o<
I,i"J,i;.
/<1
1/8 of the
that f:Y-":
and determine the probability
x)'
'1"'n
that
h(v),/(y
I
known
g(x),
is
it
Find
i"tshed if
women entered
;xafi; ip;i;"
i'
u pJrti"oru, marathon
d;lly
the race'
male runners completed
70
Ch.
2 . Random variables
snd probability Dis*ibutions
Solution. By definition,
s@):
fo
lx
I f6, y) dy : Jo| 8xy dy
J-lY=t
:4xy2l :4xt,
o<x<1,
lY=o
and
h(y):
fo
fr
I .f&, t) dx: l$ry a,
J-C
r
lx=
:
- 4x'yl
lx=y
4il1
-
o<ycl.
yz),
Now,
o<y<x,
Example 2.14
find9(x), h(y),f(xly),andevaluate P(+
<X < +ly:
+).
Solution. By definition,
s',): Il*tr*, v) dv: 'o
[o'
xv rur [r= t
:i+Tl,:o:i,
x
+-tY')
o,
o<x<2,
and
h(v):
f€
r)
dx:
'T
f2 xe +
J"
x2 3xryr1'=, l+3y,
:T+--1,=r:
J_*I@,
2-,
3y2)
o*
o<v<1.
listributions
J
7I
oint Probability Distributions
Therefore,
f
(x, y) _ x(l + 3y2)14
(xlt): f h(y)
-tl
- (1 +3y\2 -/".
t
and
(t
Pl
- < x <)1,
\4
:n"ru mi,stic
0<x<2.
:):
x3
l,': td*: a'
al Independence
If /(x ly) lg9qlo!_{:e91{g_l_1, as was the case in Example 2.14, then
f (x, y) : f (xly)h(y)
into the marginal distribution of X. That
is,
fa
fo
s$): J-I .f(", y) dy : J-I f Vlilh(v) dv,
If/(x ly)
does not depend on y, we may write
fo
s(x):f(xlil |
Now
|
-
h(y) dy.
'--.
,
,(ot
-' dv:
r,
J--
since h(y) is the probability density function of Y. Therefore,
s{x):f(xlv}
and then
f(x,
y)
:
s(x)h(y).
It should make sense to the reader that ifl(x ly) does not depend on y, then
of course the outcome of the random variable Y has no impact on the
outcome of the random variable X. In other words, we say that X and Y are
independent
random variables. We now offer the following formal definition of
L.
-'--
Statistical independence.
Mdmimtqmn
2.12
Let X anil Y be two random uariables, discrete or continuous,with joint probability distribution f (x, y) anil marginal ilistributions g(x) and h(y), respectiuely' The
rsndom uariables {, and Y pre said ro be statistically independent d anil only if
Yk'-rl:
9(x)h{t\
for all (x, y) within their range.
7
2
Ch.
2 - RandoTn
Variables and Probability Distributions
Sec. 2.5
The continuous random variables of Example 2.L4 are statistically independent, since the product of the two marginal distributions gives the joint density
function. This is obviously not the case, however, for the continuous variables
of Example 2.13. Checking for statistical independence of discrete random
variables requires a more thorough investigation, since it is possible to have
the product of the marginal distributions equal to the joint probability distribution for some but not all combinations of (x, y). If you can find any point
(x, y) for which/(x, y) is defined such that/(x, y) + S6)h(y), the discrete variables X and Y are not statistically independent.
Example
2.15 Show that the random variables of Example
2.8 are not statistically indepen-
dent.
Solution.'Let us consider the point (0, 1). From Table 2.6 we find the three
probabilities,f(0, 1), g(0), and i(1) to be
/(0'
Definitr
l): t
)
I
s(o):
f$, v):
*+
I
3
14rr 28-t4
-l-
*
*i
'=O
h(1):
2
I
x=O
,f(x,
1):
+
+
o:;.
Exi
Clearly,
/(0, l) # g9)h(t),
and therefore
X and Y are not statistically independent-
All the preceding definitions concerning two random variables can be generalized to the case of n random variables. Let f(xr, x2, ..., x,) be the joint
probability function of the random variables Xy Xr, ..., X,. The marginal
distribution of Xr, for example' is given by
s(x):I I f(xyxz,...,xJ
for the discrete case and by
s(xr):
I:
*
f
.rtrr, xz,
...,
xi;,)
dx, dx3 '' '
dxn
for the continuous case. We can now obtain ioint marginal distributions such
as @(x1, xr), where
6(xt, xr)
:
It
l-t
Lxn f@r,
xz,
-.',
(discrete case)
Xo)
ll- "J-'-tt'"x2r"'r
x^) dx3 dxa
"' dx,
(continuous case).
t
istributions
Sec,
2.5
Joint Probability
y indepen-
73
one could consider numerous conditional distributions. For
example, the joint
conditional distribution of Xr, X.r, and X., given that Xnl
int density
s variables
Xn
te random
le to have
rbility disany point
:
x,,
rn, X,
is written
f (xt, xz, xtlx+, xs, ...,
where g(xa,
crete vari-
:
x5, ...,
f lxt xz' ' ", x)
"' : g(x+,
x5, .. ., xJ'
xn)
xs,..., x,) is the joint marginal distribution of the random vari..., X n.
ables Xn, X s,
A generalization of Definition 2.12 leads to the following definition
for the
mutually statistical independence of the variables X X . .-. Xn.
,, r,
,
r indepen-
the three
Distributions
mk'finition 2.13
Let Xr,Xr,.", xnbenrandomuariobles,discrete or continuous,withjointprobability distribution f (x,., xr, . . ., xo) and marginal distributions
fr1xr1,, yrlxrj, ...,
f,(x,), respectiuely. The random uariqbles X ,, X r, ..., Xn or, ,iti'ti b:i*iiually
if
f (x,., xr, .. ., x,) : fr$)fz(xz)
statistically independent if and only
for all (xr,
frample 2.16
x,...,
Xn)
the joint
marginal
.
f ,(x,)
within their range.
Suppose that the shelf life, in years, of a certain perishable food product
packaged in cardboard containers is a random variable whose proUaUitity
density
function is given by
x>o
tvt:{'^"
[0,
be gener-
..
Let X r, Xr, and
X,
elsewhere.
represent the shelf lives
selected independently and find
p(X, <
2, 1
for three of these containers
< X, 1 3, X, > 2).
Solution. Since the containers were selected independently, we can assume
that the random variables X r X z, and X. are statistically independent, having
the joint probability density
f (xy xz, xr) -,f(xr) f $rlf(xr\
: g- x\ e- xze- xl
:
ons such
as€)
rs case).
for
x, > 0, xz 2
0,
xr
)
0,
e-tt
aqd/(xr, xz,
-J2-r3
xs):
P(J' < 2, 1 < x2 < 3, xt > 2) :l-,[t
:
(1
0elsewhere. Hence
lr', r-',-xz-x1
- "-2)'1r-r -
:0.0376.
d,x1 d.x2
e-3)e-2
dx,