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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,