Chpter 1 Continuous probbility distributions 1.1 Introduction We cll x continuous rndom vrible in x b if x cn tke on ny vlue in this intervl. An exmple of rndom vrible is the height of dult humn mle, selected rndomly from popultion. (This tkes on vlues in rnge.5 x 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 1.7-1.8 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. This leds us to redefine our ide of the probbility, using continuous function in plce of the discrete br-grph seen in the previous chpter. 1.2 Bsic definitions nd properties Definition A function p(x) is probbility density provided it stisfies the following conditions: 1. p(x) for ll x. 2. p(x) dx = 1 where the possible rnge of vlues of x is x b. s The probbility tht rndom vrible x tkes on vlues in the intervl 1 x 2 is defined 2 1 p(x) dx. v.25.1 - Jnury 5, 29 1
Mth 13 Notes Chpter 1 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. (Remrk: the probbility tht x is exctly equl to b is the integrl b p(x) dx = ; the vlue is zero, by properties of the definite integrl.) Definition The cumultive distribution function F(x) represents the probbility tht the rndom vrible tkes on ny vlue up to x, i.e. F(x) = x p(s) ds. The cumultive distribution is simply the re under the probbility density. The bove definition hs severl implictions: Properties of continuous probbility 1. Since p(x), the cumultive distribution is n incresing function. 2. The connection between the probbility density nd its cumultive distribution cn be written (using the Fundmentl Theorem of Clculus) s 3. F() =. This follows from the fct tht p(x) = F (x). F() = By property of the definite integrl, this is zero. 4. F(b) = 1. This follows from the fct tht p(s) ds. by property 2 of p(x). F(b) = p(s) ds = 1 5. The probbility tht x tkes on vlue in the intervl 1 x 2 is the sme s F( 2 ) F( 1 ). This follows from the dditive property of integrls: 2 p(s) ds 1 p(s) ds = 2 1 p(s) ds v.25.1 - Jnury 5, 29 2
Mth 13 Notes Chpter 1 Finding the normliztion constnt Not every 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. However, in mny cses we cn convert function to probbility density by simply multiplying it by constnt, C, equivlent to the reciprocl of the totl re under its grph over the intervl of interest. This process is clled normliztion, nd the constnt C is clled the normliztion constnt. 1.2.1 Exmple Consider the function f(x) = sin(πx/6) for x 6. () Normlize the function so tht it describes probbility density. (b) Find the cumultive distribution function, F(x). Solution The function is positive in the intervl x 6, so we cn define the desired probbility density. Let p(x) = C sin( π 6 x). () We must find the normliztion constnt, C, such tht Here is how we find the constnt: 1 = 6 6 p(x) dx = 1. C sin( π 6 x) dx = C 6 π ( cos( π 6 x) ) 6 1 = C 6 12 ( cos(π) + cos()) = C π π Thus, we find tht the desired constnt of normliztion is C = π 12. Once we rescle our function by this constnt, we get the desired probbility density, p(x) = π 12 sin(π 6 x). This density hs the property tht the totl re under its grph over the intervl x 6 is 1. A grph of this probbility density function is shown in Figure 1.1(). (b) F(x) = x p(s) ds = π 12 x sin( π s) ds 6 v.25.1 - Jnury 5, 29 3
Mth 13 Notes Chpter 1 F(x) = π ( 6π ) x 12 cos(π6 s) = 1 (1 cos( π ) 2 6 x). This cumultive distribution function is shown in Figure 1.1(b). 1.25.2.8.15.6.1.4.5.2 1 2 3 4 5 6 1 2 3 4 5 6 x x Figure 1.1: () The probbility density p(x), (left) nd (b) the cumultive distribution F(x) (right) for exmple 1. 1.3 Men nd medin Recll tht we hve defined the men of distribution of grdes or mss in previous chpter. For mss density ρ(x), the ide of the men coincides with the center of mss of the distribution, x = xρ(x) dx. ρ(x) dx This definition cn lso be pplied to probbility density, but in this cse the integrl in the b denomintor is simply 1 (by property 2), i.e. p(x) dx = 1. (The simplifiction is nlogous to n observtion we mde for expected vlue in discrete probbility distribution.) We define the men of probbility density s follows: Definition For rndom vrible in x 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. 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. v.25.1 - Jnury 5, 29 4
Mth 13 Notes Chpter 1 Definition The medin x m of probbility distribution is vlue of x in the intervl x m b such tht xm p(x) dx = x m p(x) dx = 1 2. It follows from this definition tht the medin is the vlue of x for which the cumultive distribution stisfies F(x m ) = 1 2. 1.3.1 Exmple Find the men nd the medin of the probbility distribution found in Exmple 1.2.1. Solution To find the men we compute x = π 12 6 x sin( π x) dx. 6 Integrtion by prts is required here. Let u = x, dv = sin( πx) dx. Then du = dx, v = 6 6 π cos(πx) 6 The result is ( x = π x 6 6 12 π cos(π 6 x) + 6 ) 6 cos( π π 6 x)dx ( x = 1 x cos( π 6 2 6 x) + 6 ) 6 π sin(π 6 x) x = 1 ( 6 cos(π) + 6π 2 sin(π) 6π ) sin() = 6 2 = 3 To find the medin, x m, we set F(x m ) = 1 2. Using the form of the cumultive distribution from exmple 1, we find tht 1 ( 1 cos( π ) 2 6 x m) = 1 2. 1 cos( π 6 x m) = 1 v.25.1 - Jnury 5, 29 5
Mth 13 Notes Chpter 1 cos( π 6 x m) = The ngles whose cosine is zero re ±π/2, ±3π/2 etc. We select the ngle in the relevnt intervl, i.e. π/2. This leds to π 6 x m = π 2 so the medin is x m = 3. 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. 1.3.2 How is the men different from the medin? p(x) p(x) x x Figure 1.2: We hve seen in Exmple 2 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. 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.) v.25.1 - Jnury 5, 29 6
Mth 13 Notes Chpter 1 Figure 1.2 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 2. 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 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. 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. 1.4 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 tht n tom would decy t time t. This function represents the frction of the toms tht decy per unit time. It turns out tht good cndidte for such function is p(t) = Ce kt, where k is constnt tht represents the rte of decy of the specific rdioctive mteril. In principle, this function is defined over the intervl t ; 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, introducing compliction tht we will lern how to hndle in the context of this exmple. Using this function we cn chrcterize the men nd medin decy time for the mteril. Normliztion We first find the constnt of normliztion, i.e. set We need to ensure tht p(t) dt = 1. Ce kt dt = 1. An integrl of this sort in which one of the endpoints is t infinity is clled n improper integrl. We must see if it mkes sense by computing it s limit, i.e. by clculting I T = T Ce kt dt v.25.1 - Jnury 5, 29 7
Mth 13 Notes Chpter 1 nd computing limit: This is shown below I = lim T I T. I T = C T [ ] e e kt kt T dt = C k = 1 k C(1 e kt ). Now recll tht the exponentil function decys to zero so tht lim T e kt =. Thus, the second term in brces will vnish s T so tht the vlues of the improper integrl will be I = lim T I T = 1 k C. (We will discuss improper integrls more fully in lter chpter.) To find the constnt of normliztion C we set this equl to 1, i.e. C = 1, which mens tht 1 k Thus the probbility density for the decy is C = k. p(t) = ke kt. This mens tht the frction of toms tht decy between time t 1 nd t 2 is t2 k e kt dt. t 1 Cumultive decys The frction of the toms tht decy between time nd time t (i.e. by time t ) is F(t) = We cn simplify this expression: t p(s) ds = k t e ks ds. [ ] e ks t F(t) = k = [ e kt e ] = 1 e kt. k 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, s expected for cumultive distribution function. v.25.1 - Jnury 5, 29 8
Mth 13 Notes Chpter 1 Medin decy time We cn use the cumultive distribution function to help determine the medin decy time, t m. To determine t m, the time t which hlf of the toms hve decyed, we set F(t m ) =.5, giving us we get F(t m ) = 1 e ktm = 1 2 e ktm = 1 2 e ktm = 2 kt m = ln 2 So we find tht 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 = 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 = T T tp(t) dt, t = lim T I T. T tke kt dt = k te kt dt. Let u = t, dv = e kt dt. Then du = dt, v = e kt /( k), so tht I T = ] I T = k [t e kt e kt T ( k) ( k) dt [ te kt + ] T e kt dt ] = [ te kt e kt T k v.25.1 - Jnury 5, 29 9.
Mth 13 Notes Chpter 1 Now s T, we hve I T = [ Te kt e kt + 1 ]. k k Te kt, e kt so tht t = lim I T = 1 T k. Thus the men or expected decy time is t = 1 k. 1.5 Discrete versus continuous probbility In n erlier chpter, 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 x 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.) The sme dichotomy exists in the topic of probbility. In n erlier chpter, we were concerned with the probbility of discrete events whose outcome belongs to some finite set of possibilities (e.g. Hed or Til for coin toss, llele A or in genetics). But mny rndom processes led to n infinite, continuous set of possible outcomes. We need the notion of continuous probbility to del with such cses. In continuous probbility, we consider the probbility density - nlogous to mss density. We cn ssign probbility to some rnge of vlues of the outcome between x = nd x = b. (To do so we hve to integrte the probbility density on the intervl x b.) The exmples below provide 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. 1.5.1 Exmple: Student heights Suppose we mesure the heights of ll UBC students. This would produce bout 3, dt vlues. 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 1.3(). Suppose we wnt to keep 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 v.25.1 - Jnury 5, 29 1
Mth 13 Notes Chpter 1 p(h) p(h) p(h) h h h h h Figure 1.3: Refining histogrm by incresing the number of bins leds (eventully) to the ide of continuous probbility density. in Figure 1.3(b)). Here, by bin we men 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 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 bin Are of bin = h = number of students in 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 whose height is h. It is thus density, nd hs the pproprite units. More generlly, p(x) x represents the frction of individuls whose height is in the rnge x h x + x. 1.5.2 Exmples: Age dependent mortlity Let p() be probbility density for the probbility of mortlity of femle Cndin non-smoker t ge, where 12. (We hve chosen n upper endpoint of ge 12 since prcticlly no Cnv.25.1 - Jnury 5, 29 11
Mth 13 Notes Chpter 1 din femle lives pst this ge t present.) Let F() be the cumultive distribution corresponding to this probbility density. () 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. 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, i.e. it is F() = p(s) ds, i.e. it is the cumultive distribution for this probbility density. Tht, precisely, is the interprettion of the cumultive function. (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) 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 =.8 +.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, ccording to this nlysis. (d) The probbility of dying between the ges of 75 nd 8 yers old is exctly 8 75 p(x) dx. However, we cn lso stte this in terms of the cumultive function, since 8 75 p(x) dx = 8 p(x) dx 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: p(x) = F (x). Then it is pproximtely true tht p(x) F(x) F(x x). x v.25.1 - Jnury 5, 29 12
Mth 13 Notes Chpter 1 (Recll the definition of the derivtive - the limit of the slope of the secnt line s the width increments x pproch.) Thus p(8) F(8) F(75) 5 =.11 5 =.22 per yer 1.5.3 Exmple: Rindrop size distribution During Vncouver rinstorm, the density function which describes the rdii of rindrops is constnt over the rnge r 4 (where r is mesured in mm) nd zero for lrger r. () Wht is the density function p(r)? (b) Wht is the cumultive distribution F(r)? (c) In terms of the volume, wht is the cumultive distribution F(V )? (d) In terms of the volume, wht is the density function 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 1.4. p(r) F(r) r 4 4 r p(v) F(V) V V Figure 1.4: Probbility densities for rindrop rdius nd rindrop volume (left pnels) nd for the cumultive distributions (right) of ech. v.25.1 - Jnury 5, 29 13
Mth 13 Notes Chpter 1 () The density function is p(r) = 1 r 4. This mens tht the probbility per unit rdius 4 of finding drop of size r is the sme for ll rdii in r 4. 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 distribution function is F(r) = r 1 4 ds = r 4 r 4. (c) The cumultive distribution function 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 V = 4 3 πr3 we hve nd so r = F(V ) = r 4 = 1 4 ( ) 1/3 3 V 1/3 4π ( ) 1/3 3 V 1/3. 4π We find the rnge of vlues of V by substituting r =, 4 into the eqution V = 4 3 πr3, to get V =, 4 3 π43. Therefore the intervl is V 4 3 π43 = (256/3)π. (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 : p(v ) = F (V ) Therefore, p(v ) = 1 4 ( ) 1/3 3 1 4π 3 V 2/3. Thus the probbility per unit volume of finding drop of volume V in V 4 3 π43 is not t ll uniform. 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 V 4 3 π43 = 256π 3 v.25.1 - Jnury 5, 29 14
Mth 13 Notes Chpter 1 nd therefore the men volume is V = 256π/3 = 1 12 = 1 12 = 1 12 = 1 16 V p(v )dv ( ) 1/3 3 256π/3 V V 2/3 dv 4π ( ) 1/3 3 256π/3 V 1/3 dv 4π ( ) 1/3 3 3 4π 4 V 256π/3 4/3 ( ) 1/3 ( ) 4/3 3 256π 4π 3 = 64π 3 67mm3. 1.6 Moments of probbility distribution 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. 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 distribution. 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 x f(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 v.25.1 - Jnury 5, 29 15
Mth 13 Notes Chpter 1 up to the second re usefully employed to describe common ttributes of distribution. 1.6.1 Moments of probbility density distribution In the prticulr cse tht the distribution is probbility density, p(x), defined on the intervl x 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 = x p(x) dx = x = µ. 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 x 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 kids 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. v.25.1 - Jnury 5, 29 16
Mth 13 Notes Chpter 1 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 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 σ = 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 + x p(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. Plugging in these fcts, we rrive t v.25.1 - Jnury 5, 29 17
Mth 13 Notes Chpter 1 Thus V = M 2 2µ µ + µ 2 V = M 2 µ 2 Thus the vrince is clerly relted to the second moment nd to the men of the distribution. Reltionship of vrince to second moment From the definitions given bove, we find tht σ = V = M 2 µ 2 1.6.2 Exmple Consider the continuous distribution, in which the probbility is constnt, p(x) = C, for vlues of x in the intervl [, b] nd zero for vlues outside this intervl. Such distribution is clled uniform distribution. (It hs the shpe of rectngulr bnd of height C nd bse (b ).) It is esy to see tht the vlue of the constnt C should be 1/(b ) so tht the re under this rectngulr bnd will be 1, in keeping with the property of probbility distribution. Thus the eqution of this probbility is p(x) = 1. We compute the first moments of this probbility density b 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 = x p(x) dx = 1 x dx b = 1 x 2 b b 2 = b2 2 2(b ). This lst expression cn be simplified by fctoring, leding to µ = M 1 = (b )(b + ) 2(b ) = b + 2 Thus we hve found tht the men µ is in the center of the intervl [, b], s expected. 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. v.25.1 - Jnury 5, 29 18
Mth 13 Notes Chpter 1 To find the vrince we might first clculte the second moment, M 2 = x 2 p(x) dx = 1 x 2 dx b It cn be shown by simple integrtion tht this yields the result which cn be simplified to M 2 = b3 3 3(b ), M 2 = (b )(b2 + b + 2 ) 3(b ) = b2 + b + 2. 3 We would then compute the vrince After simplifiction, we get The stndrd devition is then V = M 2 µ 2 = b2 + b + 2 3 V = b2 2b + 2 12 σ = = (b ) 2 3. (b )2. 12 (b + )2. 4 v.25.1 - Jnury 5, 29 19