Continuous probability distributions

Transcription

1 Chpter 8 Continuous probbility distributions 8.1 Introduction We begin by extending the ide of discrete rndom vrible 28 to the continuous cse. We cll x continuous rndom vrible in pple x pple b if x cn tke on ny vlue in this intervl. An exmple of rndom vrible is the height of person, sy n dult mle, selected rndomly from popultion. (This height typiclly tkes on vlues in the rnge.5 pple x pple 3 meters, sy, so =.5 nd b =3.) If we select mle subject t rndom from lrge popultion, nd mesure his height, we might expect to get result in the proximity of meters most often - thus, such heights will be ssocited with lrger vlue of probbility thn heights in some other intervl of equl length, e.g. heights in the rnge 2.7 <x<2.8 meters, sy. Unlike the cse of discrete probbility, however, the mesured height cn tke on ny rel number within the intervl of interest. 8.2 Bsic definitions nd properties Compred to discrete probbility, one of the most importnt differences is tht we now consider probbility density, rther thn vlue of the probbility per se. First nd foremost, we observe tht now p(x) will no longer be probbility, but rther probbility per unit x. This ide is nlogous to the connection between the mss of discrete beds nd continuous mss density, encountered previously in Chpter 5. Definition A function p(x) is probbility density provided it stisfies the following properties: 1. p(x) for ll x. 2. R b p(x) dx =1where the possible rnge of vlues of x is pple x pple b. 28 see Mth

2 28 Chpter 8. Continuous probbility distributions The probbility tht rndom vrible x tkes on vlues in the intervl 1 pple x pple 2 is defined s Z 2 1 p(x) dx. The trnsition to probbility density mens tht the quntity p(x) does not crry the sme mening s our previous nottion for probbility of n outcome x i, nmely p(x i ) in the discrete cse. In fct, p(x)dx, or its pproximtion p(x) x is now ssocited with the probbility of n outcome whose vlues is close to x. Unlike our previous discrete probbility, we will not sk wht is the probbility tht x tkes on some exct vlue? Rther, we sk for the probbility tht x is within some rnge of vlues, nd this is computed by performing n integrl 29. Hving generlized the ide of probbility, we will now find tht mny of the ssocited concepts hve nturl nd stright-forwrd generliztion s well. We first define the cumultive function, nd then show how the men, medin, nd vrince of continuous probbility density cn be computed. Here we will hve the opportunity to prctice integrtion skills, s integrls replce the sums in such clcultions. Definition For experiments whose outcome tkes on vlues on some intervl pple x pple b, we define cumultive function, F (x), s follows: F (x) = Z x p(s) ds. Then F (x) represents the probbility tht the rndom vrible tkes on vlue in the rnge (, x) 3. The cumultive function is simply the re under the probbility density (between the left endpoint of the intervl,, nd the point x). The bove definition hs severl implictions: Properties of continuous probbility 1. Since p(x), the cumultive function is n incresing function. 2. The connection between the probbility density nd its cumultive function cn be written (using the Fundmentl Theorem of Clculus) s p(x) =F (x). 29 Remrk: the probbility tht x is exctly equl to b is the integrl p(x) dx. But this integrl hs vlue b zero, by properties of the definite integrl. 3 By now, the reder should be comfortble with the use of s s the dummy vrible in this formul, where x plys the role of right endpoint of the intervl of integrtion.

3 8.2. Bsic definitions nd properties F () =. This follows from the fct tht F () = Z p(s) ds. By property of the definite integrl, this is zero. 4. F (b) =1. This follows from the fct tht F (b) = p(s) ds =1 by Property 2 of the definition of the probbility density, p(x). 5. The probbility tht x tkes on vlue in the intervl 1 pple x pple 2 is the sme s F ( 2 ) F ( 1 ). This follows from the dditive property of integrls nd the Fundmentl Theorem of Clculus: Z 2 p(s) ds Z 1 p(s) ds = Finding the normliztion constnt Z 2 1 p(s) ds = Z 2 1 F (s) ds = F ( 2 ) F ( 1 ) Not every rel-vlued function cn represent probbility density. For one thing, the function must be positive everywhere. Further, the totl re under its grph should be 1, by Property 2 of probbility density. Given n rbitrry positive function, f(x), on some intervl pple x pple b such tht f(x)dx = A>, we cn lwys define corresponding probbility density, p(x) s p(x) = 1 f(x), pple x pple b. A It is esy to check tht p(x) nd tht R b p(x)dx =1. Thus we hve converted the originl function to probbility density. This process is clled normliztion, nd the constnt C =1/A is clled the normliztion constnt Exmple: probbility density nd the cumultive function Consider the function f(x) =sin( x/6) for pple x pple 6. () Normlize the function so tht it describes probbility density. (b) Find the cumultive distribution function, F (x). 31 The reder should recognize tht we hve essentilly rescled the originl function by dividing it by the re A. This is relly wht normliztion is ll bout.

4 21 Chpter 8. Continuous probbility distributions Solution The function is positive in the intervl pple x pple 6, so we cn define the desired probbility density. Let p(x) =C sin 6 x. () We must find the normliztion constnt, C, such tht Property 2 of continuous probbility is stisfied, i.e. such tht 1= Crrying out this computtion leds to Z 6 C sin 6 x dx = C 6 Z 6 p(x) dx. 6 cos 6 x = C 6 12 (1 cos( )) = C (We hve used the fct tht cos() = 1 in step here.) But by Property 2, for p(x) to be probbility density, it must be true tht C(12/ ) =1. Solving for C leds to the desired normliztion constnt, C = 12. Note tht this clcultion is identicl to finding the re A = Z 6 sin 6 x dx, nd setting the normliztion constnt to C =1/A. Once we rescle our function by this constnt, we get the probbility density, p(x) = 12 sin 6 x. This density hs the property tht the totl re under its grph over the intervl pple x pple 6 is 1. A grph of this probbility density function is shown s the blck curve in Figure 8.1. (b) We now compute the cumultive function, F (x) = Z x Crrying out the clcultion 32 leds to F (x) = 12 6 p(s) ds = 12 Z x sin 6 s ds x cos 6 s = 1 1 cos 2 6 x. This cumultive function is shown s red curve in Figure Notice tht the integrtion involved in finding F (x) is the sme s the one done to find the normliztion constnt. The only difference is the ultimte step of evluting the integrl t the vrible endpoint x rther thn the fixed endpoint b =6.

5 8.3. Men nd medin F(x) p(x).. 6. Figure 8.1. The probbility density p(x) (blck), nd the cumultive function F (x) (red) for Exmple Note tht the re under the blck curve is 1 (by normliztion), nd thus the vlue of F (x), which is the cumultive re function is 1 t the right endpoint of the intervl. 8.3 Men nd medin When we re given distribution, we often wnt to describe it with simpler numericl vlues tht chrcterize its center : the men nd the medin both give this type of informtion. We lso wnt to describe whether the distribution is nrrow or ft - i.e. how clustered it is bout its center. The vrince nd higher moments will provide tht type of informtion. Recll tht in Chpter 5 for mss density (x), we defined center of mss, x = R b R b x (x) dx (x) dx. (8.1) The men of probbility density is defined similrly, but the definition simplifies by virtue of the fct tht R b p(x) dx =1. Since probbility distributions re normlized, the denomintor in Eqn. (8.1) is simply 1.Consequently, the men of probbility density is given s follows: Definition For rndom vrible in pple x pple b nd probbility density p(x) defined on this intervl, the men or verge vlue of x (lso clled the expected vlue), denoted x is given by x = xp(x) dx.

6 212 Chpter 8. Continuous probbility distributions To void confusion note the distinction between the men s n verge vlue of x versus the verge vlue of the function p over the given intervl. Reviewing Exmple my help to dispel such confusion. The ide of medin encountered previously in grde distributions lso hs prllel here. Simply put, the medin is the vlue of x tht splits the probbility distribution into two portions whose res re identicl. Definition The medin x med of probbility distribution is vlue of x in the intervl pple x med pple b such tht Z xmed p(x) dx = x med p(x) dx = 1 2. It follows from this definition tht the medin is the vlue of x for which the cumultive function stisfies F (x med )= Exmple: Men nd medin Find the men nd the medin of the probbility density found in Exmple Solution To find the men we compute x = 12 Z 6 x sin 6 x dx. Integrtion by prts is required here 33. Let u = x, dv =sin 6 x dx. Then du = dx, v = 6 cos 6 x. The clcultion is then s follows: x = x cos 6 x + 6 Z! 6 cos 6 x dx! = 1 6 x cos 2 6 x sin 6 x = 1 6 cos( )+ 6 2 sin( ) 6 sin() = 6 =3. (8.2) 2 33 Recll from Chpter 6 tht R R udv = vu vdu. Clcultions of the men in continuous probbility often involve Integrtion by Prts (IBP), since the integrnd consists of n expression xp(x)dx. The ide of IBP is to reduce the integrtion to something involving only p(x)dx, which is done essentilly by differentiting the term u = x, s we show here.

7 8.3. Men nd medin 213 (We hve used cos( ) = 1, sin() = sin( ) =in the bove.) To find the medin, x med, we look for the vlue of x for which F (x med )= 1 2. Using the form of the cumultive function from Exmple 8.2.1, we find tht 1. F(x).5.. x med 6. Figure 8.2. The cumultive function F (x) (red) for Exmple in reltion to the medin, s computed in Exmple The medin is the vlue of x t which F (x) =.5, s shown in green. Z xmed sin 6 s ds = 1 2 ) cos 6 x med = 1 2. Here we must solve for the unknown vlue of x med. 1 cos 6 x med =1, ) cos 6 x med =. The ngles whose cosine is zero re ± /2, ±3 /2 etc. We select the ngle so tht the resulting vlue of x med will be inside the relevnt intervl ( pple x pple 6 for this exmple), i.e. /2. This leds to 6 x med = 2 so the medin is x med =3. In other words, we hve found tht the point x med subdivides the intervl pple x pple 6 into two subintervls whose probbility is the sme. The reltionship of the medin nd the cumultive function F (x) is illustrted in Fig 8.2.

8 214 Chpter 8. Continuous probbility distributions Remrk A glnce t the originl probbility distribution should convince us tht it is symmetric bout the vlue x =3. Thus we should hve nticipted tht the men nd medin of this distribution would both occur t the sme plce, i.e. t the midpoint of the intervl. This will be true in generl for symmetric probbility distributions, just s it ws for symmetric mss or grde distributions How is the men different from the medin? p(x) p(x) x x Figure 8.3. In symmetric probbility distribution (left) the men nd medin re the sme. If the distribution is chnged slightly so tht it is no longer symmetric (s shown on the right) then the medin my still be the sme, which the men will hve shifted to the new center of mss of the probbility density. We hve seen in Exmple tht for symmetric distributions, the men nd the medin re the sme. Is this lwys the cse? When re the two different, nd how cn we understnd the distinction? Recll tht the men is closely ssocited with the ide of center of mss, concept from physics tht describes the loction of pivot point t which the entire mss would exctly blnce. It is worth remembering tht men of p(x) = expected vlue of x = verge vlue of x. This concept is not to be confused with the verge vlue of function, which is n verge vlue of the y coordinte, i.e., the verge height of the function on the given intervl. The medin simply indictes plce t which the totl mss is subdivided into two equl portions. (In the cse of probbility density, ech of those portions represents n equl re, A 1 = A 2 =1/2 since the totl re under the grph is 1 by definition.) Figure 8.3 shows how the two concepts of medin (indicted by verticl line) nd men (indicted by tringulr pivot point ) differ. At the left, for symmetric probbility density, the men nd the medin coincide, just s they did in Exmple To the right, smll portion of the distribution ws moved off to the fr right. This chnge did not ffect the loction of the medin, since the totl res to the right nd to the left of the verticl line re still equl. However, the fct tht prt of the mss is frther wy to the right leds to shift in the men of the distribution, to compenste for the chnge.

9 8.3. Men nd medin 215 Simply put, the men contins more informtion bout the wy tht the distribution is rrnged sptilly. This stems from the fct tht the men of the distribution is sum - i.e. integrl - of terms of the form xp(x) x. Thus the loction long the x xis, x, not just the mss, p(x) x, ffects the contribution of prts of the distribution to the vlue of the men Exmple: nonsymmetric distribution 1. F(x).5 p(x).. x med 6. Figure 8.4. As in Figures 8.1 nd 8.2, but for the probbility density p(x) = ( /36)x sin( x/6). This function is not symmetric, so the men nd medin re not the sme. From this figure, we see tht the medin is pproximtely x med =3.6. We do not show the men (which is close but not identicl). We cn compute both the men nd the medin for this distribution using numericl integrtion with the spredsheet. We find tht the men is x = Note tht the most probble vlue, i.e. the point t which p(x) is mximl is t x =3.9, which is gin different from both the men nd the medin. We slightly modify the function used in Exmple to the new expression f(x) =x sin ( x/6) for pple x pple 6. This results in nonsymmetric probbility density, shown in blck in Figure 8.4. Steps in obtining p(x) would be similr 34, but we hve to crry out n integrtion by prts to find the normliztion constnt nd/or to clculte the cumultive function, F (x). Further, to compute the men of the distribution we hve to integrte by prts twice. Alterntively, we cn crry out ll such computtions (pproximtely) using the spredsheet, s shown in Figure 8.4. We cn plot f(x) using sufficiently fine increments x long the x xis nd compute the pproximtion for its integrl by dding up the quntities f(x) x. The re under the curve A, nd hence the normliztion constnt (C =1/A) 34 This is good prctice, nd the reder is encourged to do this clcultion.

10 216 Chpter 8. Continuous probbility distributions will be thereby determined (t the point corresponding to the end of the intervl, x =6). It is then n esy mtter to replot the revised function f(x)/a, which corresponds to the normlized probbility density. This is the curve shown in blck in Figure 8.4. In the problem sets, we leve s n exercise for the reder how to determine the medin nd the men using the sme spredsheet tool for relted (simpler) exmple. 8.4 Applictions of continuous probbility In the next few sections, we explore pplictions of the ides developed in this chpter to vriety of problems. We tret the decy of rdioctive toms, consider distribution of heights in popultion, nd explore how the distribution of rdii is relted to the distribution of volumes in rindrop drop sizes. The interprettion of the probbility density nd the cumultive function, s well s the mens nd medins in these cses will form the min focus of our discussion Rdioctive decy Rdioctive decy is probbilistic phenomenon: n tom spontneously emits prticle nd chnges into new form. We cnnot predict exctly when given tom will undergo this event, but we cn study lrge collection of toms nd drw some interesting conclusions. We cn define probbility density function tht represents the probbility per unit time tht n tom would decy t time t. It turns out tht good cndidte for such function is p(t) =Ce kt, where k is constnt tht represents the rte of decy (in units of 1/time) of the specific rdioctive mteril. In principle, this function is defined over the intervl pple t pple 1; tht is, it is possible tht we would hve to wit very long time to hve ll of the toms decy. This mens tht these integrls hve to be evluted t infinity, leding to n improper integrl. Using this probbility density for tom decy, we cn chrcterize the men nd medin decy time for the mteril. Normliztion We first find the constnt of normliztion, i.e. find the constnt C such tht Z 1 p(t) dt = Z 1 Ce kt dt =1. Recll tht n integrl of this sort, in which one of the endpoints is t infinity is clled n improper integrl 35. Some cre is needed in understnding how to hndle such integrls, nd in prticulr when they exist (in the sense of producing finite vlue, despite the 35 We hve lredy encountered such integrls in Sections nd 7.3. See lso, Chpter 7 for more detiled discussion of improper integrls.

11 8.4. Applictions of continuous probbility 217 infinitely long domin of integrtion). We will dely full discussion to Chpter 7, nd stte here the definition: Z 1 Z T I = Ce kt dt lim I T where I T = Ce kt dt. T!1 The ide is to compute n integrl over finite intervl pple t pple T nd then tke limit s the upper endpoint, T goes to infinity (T!1). We compute: Now we tke the limit: Z T pple e I T = C e kt kt dt = C k T = 1 k C(1 e kt ). I = lim I 1 T = lim T!1 T!1 k C(1 e kt )= 1 C(1 lim k e kt ). (8.3) T!1 To compute this limit, recll tht for k>,t >, the exponentil term in Eqn. 8.3 decys to zero s T increses, so tht lim T!1 e kt =. Thus, the second term in brces in the integrl I in Eqn. 8.3 will vnish s T!1so tht the vlue of the improper integrl will be I = lim T!1 I T = 1 k C. To find the constnt of normliztion C we require tht I =1, i.e. 1 C =1, which mens k tht C = k. Thus the (normlized) probbility density for the decy is p(t) =ke kt. This mens tht the frction of toms tht decy between time t 1 nd t 2 is Cumultive decys Z t2 k e kt dt. t 1 The frction of the toms tht decy between time nd time t (i.e. ny time up to time t or by time t - note subtle wording 36 ) is F (t) = Z t p(s) ds = k Z t e ks ds. 36 Note tht the precise English wording is subtle, but very importnt here. By time t mens tht the event could hve hppened t ny time right up to time t.

12 218 Chpter 8. Continuous probbility distributions We cn simplify this expression by integrting: pple e ks F (t) =k k t = e kt e =1 e kt. Thus, the probbility of the toms decying by time t (which mens nytime up to time t) is F (t) =1 e kt. We note tht F () = nd F (1) =1, s expected for the cumultive function. Medin decy time As before, to determine the medin decy time, t m (the time t which hlf of the toms hve decyed), we set F (t m )=1/2. Then so we get 1 2 = F (t m)=1 e ktm, e ktm = 1 2, ) ektm =2, ) kt m =ln2, ) t m = ln 2 k. Thus hlf of the toms hve decyed by this time. (Remrk: this is esily recognized s the hlf life of the rdioctive process from previous fmilirity with exponentilly decying functions.) Men decy time The men time of decy t is given by t = Z 1 tp(t) dt. We compute this integrl gin s n improper integrl by tking limit s the top endpoint increses to infinity, i.e. we first find nd then set I T = To compute I T we use integrtion by prts: I T = Z T Z T tp(t) dt, t = lim T!1 I T. Z T tke kt dt = k te kt dt.

13 8.4. Applictions of continuous probbility 219 Let u = t, dv = e kt dt. Then du = dt, v = e kt /( I T = k pplet e kt ( k) pple = te kt Z e kt k e kt T ( k) dt T pple = pple = Te kt k), so tht Z te kt + e kt k + 1 k e kt dt T Now s T!1, we hve e kt! so tht Thus the men or expected decy time is t = lim T!1 I T = 1 k. t = 1 k Discrete versus continuous probbility In Chpter 5.3, we compred the tretment of two types of mss distributions. We first explored set of discrete msses strung long thin wire. Lter, we considered single br with continuous distribution of density long its length. In the first cse, there ws n unmbiguous mening to the concept of mss t point. In the second cse, we could ssign mss to some section of the br between, sy x = nd x = b. (To do so we hd to integrte the mss density on the intervl pple x pple b.) In the first cse, we tlked bout the mss of the objects, wheres in the ltter cse, we were interested in the ide of density (mss per unit distnce: Note tht the units of mss density re not the sme s the units of mss.) As we hve seen so fr in this chpter, the sme dichotomy exists in the topic of probbility. The exmple below provides some further insight to the connection between continuous nd discrete probbility. In prticulr, we will see tht one cn rrive t the ide of probbility density by refining set of mesurements nd mking the pproprite scling. We explore this connection in more detil below Exmple: Student heights Suppose we mesure the heights of ll UBC students. This would produce bout 3, dt vlues 37. We could mke grph nd show how these heights re distributed. For exmple, we could subdivide the student body into those students between nd 1.5m, nd those between 1.5 nd 3 meters. Our br grph would contin two brs, with the number of students in ech height ctegory represented by the heights of the brs, s shown in Figure 8.5(). Suppose we wnt to record this distribution in more detil. We could divide the popultion into smller groups by shrinking the size of the intervl or bin into which height is subdivided. (An exmple is shown in Figure 8.5(b)). Here, by bin we men 37 I m grteful to Dvid Austin for developing this exmple.

14 22 Chpter 8. Continuous probbility distributions p(h) p(h) p(h) h h h h h Figure 8.5. Refining histogrm by incresing the number of bins leds (eventully) to the ide of continuous probbility density. little intervl of width h where h is height, i.e. height intervl. For exmple, we could keep trck of the heights in increments of 5 cm. If we were to plot the number of students in ech height ctegory, then s the size of the bins gets smller, so would the height of the br: there would be fewer students in ech ctegory if we increse the number of ctegories. To keep the br height from shrinking, we might reorgnize the dt slightly. Insted of plotting the number of students in ech bin, we might plot number of students in the bin. h If we do this, then both numertor nd denomintor decrese s the size of the bins is mde smller, so tht the shpe of the distribution is preserved (i.e. it does not get fltter). We observe tht in this cse, the number of students in given height ctegory is represented by the re of the br corresponding to tht ctegory: number of students in the bin Are of bin = h = number of students in the bin. h The importnt point to consider is tht the height of ech br in the plot represents the number of students per unit height. This type of plot is precisely wht leds us to the ide of density distribution. As h shrinks, we get continuous grph. If we normlize, i.e. divide by the totl re under the grph, we get probbility density, p(h) for the height of the popultion. As noted, p(h) represents the frction of students per unit height 38 whose height is h. It is thus density, nd hs the pproprite units. In this cse, p(h) h represents the frction of individuls whose height is in the rnge h pple height pple h + h Exmple: Age dependent mortlity In this exmple, we consider n ge distribution nd interpret the menings of the probbility density nd of the cumultive function. Understnding the connection between the 38 Note in prticulr the units of h 1 ttched to this probbility density, nd contrst this with discrete probbility tht is pure number crrying no such units.

15 8.4. Applictions of continuous probbility 221 verbl description nd the symbols we use to represent these concepts requires prctice nd experience. Relted problems re presented in the homework. Let p() be probbility density for the probbility of mortlity of femle Cndin non-smoker t ge, where pple pple 12. (We hve chosen n upper endpoint of ge 12 since prcticlly no Cndin femle lives pst this ge t present.) Let F () be the cumultive distribution corresponding to this probbility density. We would like to nswer the following questions: () Wht is the probbility of dying by ge? (b) Wht is the probbility of surviving to ge? (c) Suppose tht we re told tht F (75) =.8 nd tht F (8) differs from F (75) by.11. Interpret this informtion in plin English. Wht is the probbility of surviving to ge 8? Which is lrger, F (75) or F (8)? (d) Use the informtion in prt (c) to estimte the probbility of dying between the ges of 75 nd 8 yers old. Further, estimte p(8) from this informtion. Solution () The probbility of dying by ge is the sme s the probbility of dying ny time up to ge. Restted, this is the probbility tht the ge of deth is in the intervl pple ge of deth pple. The pproprite quntity is the cumultive function, for this probbility density F () = Z p(x) dx. Remrk: note tht, s customry, x is plying the role of dummy vrible. We re integrting over ll ges between nd, so we do not wnt to confuse the nottion for vrible of integrtion, x nd endpoint of the intervl. Hence the symbol x rther thn inside the integrl. (b) The probbility of surviving to ge is the sme s the probbility of not dying before ge. By the elementry properties of probbility discussed in the previous chpter, this is 1 F (). (c) F (75) =.8 mens tht the probbility of dying some time up to ge 75 is.8. (This lso mens tht the probbility of surviving pst this ge would be 1-.8=.2.) From the properties of probbility, we know tht the cumultive distribution is n incresing function, nd thus it must be true tht F (8) >F(75). Then F (8) = F (75)+.11 = =.91. Thus the probbility of surviving to ge 8 is 1-.91=.9. This mens tht 9% of the popultion will mke it to their 8 th birthdy. (d) The probbility of dying between the ges of 75 nd 8 yers old is exctly Z 8 75 p(x) dx.

16 222 Chpter 8. Continuous probbility distributions However, we cn lso stte this in terms of the cumultive function, since Z 8 75 p(x) dx = Z 8 p(x) dx Z 75 p(x) dx = F (8) F (75) =.11 Thus the probbility of deth between the ges of 75 nd 8 is.11. To estimte p(8), we use the connection between the probbility density nd the cumultive distribution 39 : Then it is pproximtely true tht p(x) p(x) =F (x). (8.4) F (x + x) F (x). (8.5) x (Recll the definition of the derivtive, nd note tht we re pproximting the derivtive by the slope of secnt line.) Here we hve informtion t ges 75 nd 8, so x = 8 75 = 5, nd the pproximtion is rther crude, leding to p(8) F (8) F (75) 5 =.11 5 =.22 per yer. Severl importnt points merit ttention in the bove exmple. First, informtion contined in the cumultive function is useful. Differences in vlues of F between x = nd x = b re, fter ll, equivlent to n integrl of the function R b p(x)dx, nd re the probbility of result in the given intervl, pple x pple b. Second, p(x) is the derivtive of F (x). In the expression (8.5), we pproximted tht derivtive by smll finite difference. Here we see t ply mny of the themes tht hve ppered in studying clculus: the connection between derivtives nd integrls, the Fundmentl Theorem of Clculus, nd the reltionship between tngent nd secnt lines Exmple: Rindrop size distribution In this exmple, we find rther non-intuitive result, linking the distribution of rindrops of vrious rdii with the distribution of their volumes. This reinforces the cution needed in interpreting nd hndling probbilities. During Vncouver rinstorm, the distribution of rindrop rdii is uniform for rdii pple r pple 4 (where r is mesured in mm) nd zero for lrger r. By uniform distribution we men function tht hs constnt vlue in the given intervl. Thus, we re sying tht the distribution looks like f(r) = C for pple r pple 4. () Determine wht is the probbility density for rindrop rdii, p(r)? Interpret the mening of tht function. 39 In Eqn. (8.4) there is no longer confusion between vrible of integrtion nd n endpoint, so we could revert to the nottion p() =F (), helping us to identify the independent vrible s ge. However, we hve voided doing so simply so tht the formul in Eqn. (8.5) would be very recognizble s n pproximtion for derivtive.

17 8.4. Applictions of continuous probbility 223 (b) Wht is the ssocited cumultive function F (r) for this probbility density? Interpret the mening of tht function. (c) In terms of the volume, wht is the cumultive distribution F (V )? (d) In terms of the volume, wht is the probbility density p(v )? (e) Wht is the verge volume of rindrop? Solution This problem is chllenging becuse one my be tempted to think tht the uniform distribution of drop rdii should give uniform distribution of drop volumes. This is not the cse, s the following rgument shows! The sequence of steps is illustrted in Figure 8.6. p(r) () F(r) (b) r 4 4 r p(v) (c) F(V) (d) V V Figure 8.6. Probbility densities for rindrop rdius nd rindrop volume (left pnels) nd for the cumultive distributions (right) of ech for Exmple () The probbility density function is p(r) = 1/4 for pple r pple 4. This mens tht the probbility per unit rdius of finding drop of size r is the sme for ll rdii in pple r pple 4, s shown in Fig. 8.6(). Some of these drops will correspond to smll volumes, nd others to very lrge volumes. We will see tht the probbility per unit volume of finding drop of given volume will be quite different. (b) The cumultive function is F (r) = Z r A sketch of this function is shown in Fig. 8.6(b). 1 4 ds = r, pple r pple 4. (8.6) 4

18 224 Chpter 8. Continuous probbility distributions (c) The cumultive function F (r) is proportionl to the rdius of the drop. We use the connection between rdii nd volume of spheres to rewrite tht function in terms of the volume of the drop: Since we hve r = V = 4 3 r3 (8.7) 1/3 3 V 1/3. 4 Substituting this expression into the formul (8.6), we get F (V )= 1 4 1/3 3 V 1/3. 4 We find the rnge of vlues of V by substituting r =, 4 into Eqn. (8.7) to get 4 V =, Therefore the intervl is pple V pple or pple V pple (256/3). The function F (V ) is sketched in pnel (d) of Fig (d) We now use the connection between the probbility density nd the cumultive distribution, nmely tht p is the derivtive of F. Now tht the vrible hs been converted to volume, tht derivtive is little more interesting : Therefore, p(v )= 1 4 p(v )=F (V ) 1/ V 2/3. Thus the probbility per unit volume of finding drop of volume V in pple V pple is not t ll uniform. This probbility density is shown in Fig. 8.6(c) This results from the fct tht the differentil quntity dr behves very differently from dv, nd reinforces the fct tht we re deling with density, not with probbility per se. We note tht this distribution hs smller vlues t lrger vlues of V. (e) The rnge of vlues of V is nd therefore the men volume is pple V pple 256 3, V = Z 256 /3 Vp(V )dv = 1 1/3 Z /3 V V /3 dv = 1 1/3 Z /3 V 1/3 dv = = / /3 256 /3 3 4 V 4/3 4/3 = mm3.

19 8.5. Moments of probbility density Moments of probbility density We re now fmilir with some of the properties of probbility distributions. On this pge we will introduce set of numbers tht describe vrious properties of such distributions. Some of these hve lredy been encountered in our previous discussion, but now we will see tht these fit into pttern of quntities clled moments of the distribution Definition of moments Let f(x) be ny function which is defined nd positive on n intervl [, b]. We might refer to the function s distribution, whether or not we consider it to be probbility density. Then we will define the following moments of this function: zero th moment M = first moment M 1 = second moment M 2 = f(x) dx xf(x) dx x 2 f(x) dx n th moment M n =. x n f(x) dx. Observe tht moments of ny order re defined by integrting the distribution f(x) with suitble power of x over the intervl [, b]. However, in prctice we will see tht usully moments up to the second re usefully employed to describe common ttributes of distribution Reltionship of moments to men nd vrince of probbility density In the prticulr cse tht the distribution is probbility density, p(x), defined on the intervl pple x pple b, we hve lredy estblished the following : M = p(x) dx =1. (This follows from the bsic property of probbility density.) Thus The zero th moment of ny probbility density is 1. Further M 1 = xp(x) dx = x = µ.

20 226 Chpter 8. Continuous probbility distributions Tht is, The first moment of probbility density is the sme s the men (i.e. expected vlue) of tht probbility density. So fr, we hve used the symbol x to represent the men or verge vlue of x but often the symbol µ is lso used to denote the men. The second moment, of probbility density lso hs useful interprettion. From bove definitions, the second moment of p(x) over the intervl pple x pple b is M 2 = x 2 p(x) dx. We will shortly see tht the second moment helps describe the wy tht density is distributed bout the men. For this purpose, we must describe the notion of vrince or stndrd devition. Vrince nd stndrd devition Two children of pproximtely the sme size cn blnce on teeter-totter by sitting very close to the point t which the bem pivots. They cn lso chieve blnce by sitting t the very ends of the bem, eqully fr wy. In both cses, the center of mss of the distribution is t the sme plce: precisely t the pivot point. However, the mss is distributed very differently in these two cses. In the first cse, the mss is clustered close to the center, wheres in the second, it is distributed further wy. We my wnt to be ble to describe this distinction, nd we could do so by considering higher moments of the mss distribution. Similrly, if we wnt to describe how probbility density distribution is distributed bout its men, we consider moments higher thn the first. We use the ide of the vrince to describe whether the distribution is clustered close to its men, or spred out over gret distnce from the men. Vrince The vrince is defined s the verge vlue of the quntity (distnce from men) 2, where the verge is tken over the whole distribution. (The reson for the squre is tht we would not like vlues to the left nd right of the men to cncel out.) For discrete probbility with men, µ we define vrince by V = X (x i µ) 2 p i. For continuous probbility density, with men µ, we define the vrince by V = (x µ) 2 p(x) dx. The stndrd devition The stndrd devition is defined s

21 8.5. Moments of probbility density 227 = p V. Let us see wht this implies bout the connection between the vrince nd the moments of the distribution. Reltionship of vrince to second moment From the eqution for vrince we clculte tht V = Expnding the integrl leds to: (x µ) 2 p(x) dx = (x 2 2µx + µ 2 ) p(x) dx. V = = x 2 p(x)dx x 2 p(x)dx 2µ 2µx p(x) dx + xp(x) dx + µ 2 µ 2 p(x) dx p(x) dx. We recognize the integrls in the bove expression, since they re simply moments of the probbility distribution. Using the definitions, we rrive t Thus V = M 2 2µµ+ µ 2. V = M 2 µ 2. Observe tht the vrince is relted to the second moment, M 2 nd to the men, µ of the distribution. Reltionship of vrince to second moment Using the bove definitions, the stndrd devition, cn be expressed s = p V = p M 2 µ Exmple: computing moments Consider probbility density such tht p(x) =C is constnt for vlues of x in the intervl [, b] nd zero for vlues outside this intervl 4. The re under the grph of this function 4 As noted before, this is uniform distribution. It hs the shpe of rectngulr bnd of height C nd bse (b ).

22 228 Chpter 8. Continuous probbility distributions for pple x pple b is A = C (b ) 1 (enforced by the usul property of probbility density), so it is esy to see tht the vlue of the constnt C should be C =1/(b ). Thus p(x) = 1, pple x pple b. b We compute some of the moments of this probbility density M = p(x)dx = 1 b 1 dx =1. (This ws lredy known, since we hve determined tht the zeroth moment of ny probbility density is 1.) We lso find tht M 1 = xp(x) dx = 1 b xdx= 1 b This lst expression cn be simplified by fctoring, leding to µ = M 1 = (b )(b + ) 2(b ) x 2 2 = b + 2. b = b2 2 2(b ). The vlue (b+)/2 is midpoint of the intervl [, b]. Thus we hve found tht the men µ is in the center of the intervl, s expected for symmetric distribution. The medin would be t the sme plce by simple symmetry rgument: hlf the re is to the left nd hlf the re is to the right of this point. To find the vrince we clculte the second moment, M 2 = Fctoring simplifies this to x 2 p(x) dx = 1 b 1 x x 2 3 dx = b 3 b = b3 3 3(b ). M 2 = (b )(b2 + b + 2 ) 3(b ) = b2 + b The vrince is then V = M 2 µ 2 = b2 + b The stndrd devition is = (b + ) 2 4 (b ) 2 p 3. = b2 2b = (b )2. 12

23 8.6. Summry Summry In this chpter, we lerned tht the continuous probbility density is function specifying the probbility per unit vlue (of the vrible of interest), so tht p(x)dx = probbility tht x tkes vlue in the intervl (, b). We lso defined nd studied the cumultive function F (x) = Z x p(s)ds = probbility of vlue in the intervl (, x). We noted tht by the Fundmentl Theorem of Clculus, F (x) is n ntiderivtive of p(x) (or synonymously, p (x) =F (x).) The men nd medin re two descriptors for some fetures of probbility densities. For p(x) defined on n intervl pple x pple b nd zero outside, the men, ( x, or sometimes clled µ) is x = wheres the medin, x med is the vlue for which xp(x)dx F (x med )= 1 2. Both men nd medin correspond to the center of symmetric distribution. If the distribution is non-symmetric, long til in one direction will shift the men towrd tht direction more strongly thn the medin. The vrince of probbility density is V = nd the stndrd devition is (x µ) 2 p(x) dx, = p V. This quntity describes the width of the distribution, i.e. how spred out (lrge clumped (smll ) it is. We defined the n th moment of probbility density s ) or M n = x n p(x)dx, nd showed tht the first few moments re relted to men nd vrince of the probbility. Most of these concepts re directly linked to the nlogous ides in discrete probbility, but in this chpter, we used integrtion in plce of summtion, to del with the continuous, rther thn the discrete cse.

24 23 Chpter 8. Continuous probbility distributions 8.7 Exercises Exercise 8.1 In Figures 8.7 () nd (b), you re sked to indicte the following by sketching directly on the digrms. () The probbility tht the vlue of x is in the rnge [2, 4] for the probbility density p(x). See Figure 8.7() (b) The loction of the medin for the cumultive distribution F (x) in Figure 8.7(b) P(X) 1 F(X) 1/ X X Figure 8.7. Figures for problem 1.1 () nd (b). Exercise 8.2 Figure 8.8 depicts the grphs of probbility density function, p(x) nd of cumultive distribution function, F (x) corresponding to it. Wht is the vlue of the constnt K? Wht is the specil feture of the point mrked * (one word nswer)? Is the men the sme s the medin in this cse? (Justify!) K F(X) K/2 P(X) * Figure 8.8. Figure for problem 1.2 Exercise 8.3 Use one sentence to explin clerly wht cumultive probbility distribution represents.

25 8.7. Exercises 231 Exercise 8.4 The probbility density distribution of grdes on mid-term test (out of totl score of 1) ws found to be p(x) =Ce x 1 pple x pple 1 () Clculte C. (b) Ws the test esy or hrd? Explin your nswer. Exercise 8.5 For the following probbility density functions, find the constnt C nd the cumultive distribution function, F (x), for b pple x pple b. () p(x) =C (b) p(x) = C(x 2 b 2 ) Exercise 8.6 Consider function y = f(x) > defined on some intervl [, b]. The medin of f is defined to be the vlue of the independent vrible, x, sy x = m, which splits the re under f(x) between nd b into two equl portions, i.e. such tht Z m f(x) dx = m f(x) dx = 1 2 f(x) dx Use this definition to find the medin of the following functions on the indicted intervl. () f(x) =1 x 2 for 1 <x<1 (b) f(x) = 1 x for 1 <x<1 (c) f(x) =5 x for <x<5 (d) f(x) =sin(2x) for <x< Note: it my help to sketch the given function nd intervl nd consider symmetry for some of these exmples. Exercise 8.7 Figure 8.9 shows the result of n experiment with three groups of people. In ech group, every person ws sked to throw bll. The distnce x ws recorded. The probbility p i (x) of throwing given distnce is shown for ech of the groups i =1, 2, 3. p (x) 1 p (x) 2 p (x) 3 d x d x d x Figure 8.9. For problem 1.7 () Compre the verge (men) distnce thrown in the three groups.

26 232 Chpter 8. Continuous probbility distributions (b) Compre the medin distnces thrown in the three cses. Exercise 8.8 Shown below is the probbility density function which describes how much of the Erth s surfce is t given elevtion (negtive elevtion mens below se level). Figure 8.1. For problem 1.8 () Is more of the Erth s surfce bove or below se level? Use the concept of medin to explin. (b) Estimte wht frction of the Erth s surfce is below se level. Exercise 8.9 Suppose p(x) is the probbility tht student in Mth 13 gets grde x. The medin of p(x) is tht vlue of x for which typicl student is eqully likely to score bove s below. () Suppose the rnge of scores is pple x pple b nd denote the medin by m. Explin why Z m p(x) dx = m p(x) dx = 1 2 (b) Suppose tht the students mrks re described by the constnt probbility density function p(x) = 1 1 on the intervl pple x pple 1. Find the verge mrk nd the medin mrk. Wht is the chnce tht typicl student scores better thn the verge mrk? better thn the medin? Ws the course esy or hrd for these students? (c) Now suppose tht the students mrks re described by p(x) =Cx on the intervl pple x pple 1. Wht is the constnt C? Find the verge mrk nd the medin. Wht is the chnce tht typicl student scores better thn the verge mrk? better thn the medin? In this cse ws the course esier or hrder thn in prt (b). Explin why. Exercise 8.1 Suppose F (x) is the cumultive distribution function for the height, x, of North Americn dult mles (in cm), nd suppose p(x) is the probbility density. () Wht is the mening of F (2) =.9? (b) Interpret R 16 p(x) dx =.2. 15

27 8.7. Exercises 233 (c) Suppose tht F (2) nd F (22) differ by.13. Which one is bigger? Use this informtion to estimte the vlue of p(2). Exercise 8.11 Consider the two probbility density functions p 1 (x) =C 1 (1 x ) nd p 2 (x) =C 2 (1 x 2 ) on the intervl [ 1, 1]. (See Figure 8.11 for sketch.) p1(x) p2(x) 1 1 x 1 1 x Figure For problem 1.11 () Find the constnts C 1,C 2 in ech cse so tht the probbility distribution is normlized (i.e. the totl probbility is 1). (b) Find the verge vlue (men) for both of these functions. (Note: creful use of symmetry will simplify your tsk.) (c) Find the medin for ech distribution. Exercise 8.12 Consider the uniform probbility distribution p(x) =C for pple x pple b. () Find the vlue of the constnt C. (b) Compute the men nd the medin of the distribution. Exercise 8.13 Consider the function p(x), b pple x pple b, which is mde up of the two stright line segments which connect ( b, ) to (,c) nd (,c) to (b, ), where b, c re both positive.

28 234 Chpter 8. Continuous probbility distributions c b Figure For problem 1.12 () Grph p(x) nd determine formul for p(x). (b) Determine c so tht p(x) is probbility distribution. (c) Find the men nd medin for p(x). Exercise 8.14 The probbility tht the height of bsketbll plyer is x meters is given by the probbility density distribution p(x) =Cxe x/2 where <x<3. () Wht is the constnt C? (b) Wht is the most probble height? (c) Wht is the men height? Exercise 8.15 Ants re lwys found within 1 mile of their hill. The frction of nts found within distnce x of their hill (where pple x pple 1 mile) is given by the function D(x) = 4 rctn(x). () Wht is the medin distnce tht nts re found from their hill? (b) Wht is the most likely position to find n nt? (c) Wht is the verge or men distnce? Exercise 8.16 Mortlity is often highly ge dependent. Suppose tht the probbility of dying per yer t ge x (in yers) is p(x) =C(1 + (x 2) 2 ) where pple x pple 1. () Find the constnt C. (b) Find the frction of the popultion tht hs died by ge 2. (c) Find the frction of the popultion tht hs died by ge 8.

29 8.7. Exercises 235 (d) Wht is the men lifetime in this popultion? Exercise 8.17 Nonsymmetric distribution Consider the function f(x) = cos x for pple x pple 3. 6 This function is symmetric on the given intervl. () Find the corresponding (normlized) probbility density. (b) Find the most probble vlue of x. (c) Compute the men nd the medin. (d) Use spredsheet to determine the men for this distribution by numericl integrtion. (e) Use the spredsheet to plot p(x) nd the corresponding cumultive function F (x) on the sme grph. Using F (x), determine the vlue of the medin. Exercise 8.18 Suppose tht x is vrible tht tkes on rndom non-negtive vlues only. () Show tht the function below is probbility density distribution. f(x) = 2, pple x<1. (1 + x) 3 Note: the vlue of x here is not bounded, so you will hve to compute n integrl whose limits re from to infinity. Recll tht this is clled n improper integrl. To compute its vlue, integrte from to b nd then consider b!1. (b) Find the probbility tht x tkes on vlues in the intervl [1, 4] Exercise 8.19 The probbility density function of light bulb filing t time t, pple t< 1, is given by p(t) =Ce (.1)t, where t is in dys. () Determine the constnt C so tht p(t) is probbility density function. In other words determine C so tht R 1 Ce t/1 dt =1holds. This problem requires clculting n improper integrl. (b) Find the function tht describes the frction of light bulbs filing by time t. (c) Wht frction of light bulbs re working fter 2 dys? (d) Determine the men nd medin times of filure. Exercise 8.2 According to L. Glss nd M. Mckey, the uthors of recent book From clocks to chos, the rhythms of life p. 44: The probbility tht cell drwn t rndom from lrge popultion hs divided t time t 2 (minutes) is p(t) =Re R(t 2),t 2, where R>is some constnt. (They ssume tht cells never divide before time t = 2 min.)

30 236 Chpter 8. Continuous probbility distributions () Find the men time for division. (b) Wht frction of the cells hve not divided by time t = 4 min? Exercise 8.21 Given the probbility distribution p(x) =C 1 (1+x) 2, defined on pple x< 1. () Find the probbility tht x tkes on vlues in the intervl [1, 4]. (You will first hve to find the vlue of the constnt, C). (b) Find the cumultive distribution function, F (x), for this probbility density. Note tht this problem involves clculting improper integrls. (c) Sketch both p(x) nd F (x), Exercise 8.22 The probbility tht seeds will be dispersed distnce x (meters) wy from prent tree is found to be p(x) =Ce x/1. () Determine the vlue of the constnt C so tht p(x) is probbility density function. (b) Determine the men nd medin distnces of dispersl. (c) Wht is the vrince V in the dispersl distnce? (d) Seeds tht fll right under the tree (for x<1 meter) fil to thrive becuse of competition with the prent tree nd shding tht stunts the growth of the seedling. Wht frction of the seeds will fil to thrive? (Assume 1-dimensionl rrngement.) Note tht this problem involves clculting improper integrls. Exercise 8.23 Survivl Cells in tissue culture re treted with n experimentl medicine. Let t be the survivl time of the cells in hours fter receiving the tretment. The probbility density function which describes this sitution is p(t) =Ce Ct where pple t pple1.if cell s 7% chnce of surviving two or more hours, find the vlue of the constnt C. Exercise 8.24 A mixture of cubic crystls of vrious sizes is mde in the lbortory, nd their sizes re mesured. The probbility density function, p(x), for the size of crystl, x, (in millimeters), shown in Figure 8.13, is constnt for 1 pple x pple 4. () Write down the functionl form of the probbility density p(x) for crystl size nd the corresponding cumultive distribution F (x). (b) Using the fct tht the volume of cube is V = x 3, express this cumultive distribution in terms of the volume, i.e. find F in terms of V. (We ll cll this F (V )). (c) Using wht you hve found for F (V ), determine the corresponding probbility density function p(v ) for crystl volume. (Hint: wht is the connection between F (V ) nd p(v )?) (d) Use your result in (c) to clculte the men volume, V, of the crystls.

31 8.7. Exercises 237 P(X) 1 4 X Figure For problem Exercise 8.25 A rndom bulb from btch of light bulbs hs the probbility D(t) = 1 e rt of filing before t hours. If the men lifetime of such bulbs is 1 hours, wht is the vlue of r? Exercise 8.26 Let M(t) be the mount of certin rdioctive substnce left t time t. Then M(t) =M e kt for some positive constnt k, where M is the initil mount. The cumultive mount tht hs decyed in the first t yers is F (t) =M (1 e kt ). () If k = ln(2)/1, show tht F (1) = M /2. (b) Show tht the function p(t) = F (t) F (1) stisfies ll the requirements of probbility density distribution for pple t pple 1. (c) Sketch the functions M(t), F (t), F (t). (d) Use integrtion by prts to clculte the men time t of decy. Exercise 8.27 If p(x) is probbility density distribution on [, b] then its first moment M 1, is just its men, M 1 = x = R b xp(x) dx, nd its second moment M 2 is defined by M 2 = R b x2 p(x) dx. We lso define the vrince, V, nd stndrd devition,, s follows: V = M 2 (M 1 ) 2 nd = p V. Use these definitions to compute the men x, vrince V, nd stndrd devition of the following probbility density functions: C, () p(x) = pple x pple b, otherwise. (b) p(x) =ke kx for x. (c) p(x) =2(1 x) for pple x pple 1.

32 238 Chpter 8. Continuous probbility distributions Exercise 8.28 Consider the uniform probbility distribution p(x) =C for pple x pple b. () Find the vrince nd the stndrd devition. (b) Find the cumultive distribution function, F (x) for this probbility density. Exercise 8.29 Consider the function p(x) for b pple x pple b, which is given by 8 c >< x + c if b pple x pple p(x) = b c >: x + c if <xppleb b Find the vrince V nd the stndrd devition of p(x). Exercise 8.3 The function p(x) =C sin 6 x, pple x pple 6, is the probbility density function for rndom jump of x units t some lrge trck meeting. () Determine the constnt C. (b) Find the zeroth, the first, nd the second moments of this probbility distribution, i.e. clculte I = Z 6 p(x) dx, I 1 = Z 6 xp(x) dx, I 2 = Z 6 x 2 p(x) dx. (c) Use the results of (b) to determine the men, the vrince, nd the stndrd devition of the jump distnce. Exercise 8.31 Consider the Norml (or Gussin) distribution function given by f(x) =Ce x2 2, 1 <x<1, nd suppose you re told (since it is not esy to clculte) tht Z 1 1 e x2 2 dx = p 2 () Wht vlue of the constnt C should be used to mke this function represent probbility density distribution? This is clled the Stndrd Norml distribution. (b) Show tht the men nd medin re both. (c) Show tht the vrince nd the stndrd devition re both 1. Note tht this problem involves clculting improper integrls.