10 Statistical Distributions Solutions

Transcription

1 Communictions Engineeing MSc - Peliminy Reding 1 Sttisticl Distiutions Solutions 1) Pove tht the vince of unifom distiution with minimum vlue nd mximum vlue ( is ) 1. The vince is the men of the sques minus the sque of the men, so fist fo the men of the sques: 1 1 x x p( x) dx= x dx= = nd you might find it ovious tht the men is ( + )/ y symmety, lthough if this isn t ovious you cn wok it out: which gives fo the vince: 1 1 x + xp( x) dx= xdx= = = ( ) σ = = 4 ( ) ( ) ( ) ( + + )( ) 4 = 1 ( ) ( ) = 1 ( ) ( ) ( ) ( ) + = = = ) Wht is the poility tht smple fom Gussin distiution with men vlue of 5 nd stndd devition of is ove 11? Expess you nswe in tems of oth the Q-function nd ef functions. Fist, in tems of the Q-function. The poility tht ny pticul smple x fom this distiution hs vlue gete thn 11 is: so the nswe is just unde.% p( x> 11) = Q = Q() =.8 In tems of the ef function, the sme sum would e witten: 6 Univesity of Yok Pge 1 /4/8

2 Communictions Engineeing MSc - Peliminy Reding p x> 11 = 1 1+ ef = 1 ef =.8 You cn pehps see why I pefe to use the Q-function. ) The vege intelligence quotient (IQ) of people is 1, with stndd devition of 15, nd Gussin distiution. An ognistion clled MENSA is open to the top % of the popultion. Wht is the minimum IQ needed to join MENSA? In tems of the Q-function, this polem cn e expessed s: x 1. = Q 15 Looking up tles of the Q-function, we cn find tht: Q(.575) =., so: x 1 = x = 1.8 4) At school, evey pupil in clss of students ws sked to un 1 metes. The time it took the students fitted Gussin distiution with men of 19 seconds nd stndd devition of seconds. Thee e one million such students in the county ppoximtely how mny of them cn un 1 metes in less thn five seconds? Thee s the sensile nswe, nd the silly nswe. The sensile nswe is none. Congtultions to eveyone who got tht nswe. The silly nswe is to note tht the poility of hving vlue of less thn five seconds fom stndd devition of with men of 19 is: p( x< 5) = 1 Q = 1 Q = 1.5e6 nd theefoe out of one million student, poly one o two cn un the 1 metes in less thn five seconds. This is of couse ludicous. No-one cn un 1 metes in less thn five seconds. The polem is tht this is not elly Gussin distiution. You cn t tke edings fom set of students nd expect them to ccutely pedict wht will hppen lmost five stndd devitions wy fom the men. In ny cse ny distiution of ce times cn t e el Gussin el Gussin hs finite possiility of tking ny vlue, even negtive vlues. Tht would suggest tht it is possile tht child t one exteme of the distiution could finish the ce efoe he o she stted. Tht might seem like silly exmple, ut this is common mistke to mke when woking with poility distiutions. Be vey ceful when using the tils of distiution when you don t hve ny el smple points out thee. 6 Univesity of Yok Pge /4/8

3 Communictions Engineeing MSc - Peliminy Reding 5) Deive the fomuls fo the men, mode, medin nd vince of the Ryleigh distiution. This is good pctice in integtion. Fist, the men: = p() d = exp d χ this cn e done using integtion y pts, whee one pt is, nd the othe is the Ryleigh poility distiution, which cn e integted s: exp d = exp χ (Ty diffeentiting the esult if you e unsue of this. The integtion is done y sustitution of t = /χ.) Then, = p() d = exp d χ = exp exp d χ the emining integtion hee cn e witten in tems of the Gussin poility distiution with zeo men nd vince of χ / : 1 = exp χπ exp d π χ/ 1 = exp + χπ exp d π χ/ now one hlf of Gussin poility distiution (fom zeo to infinity) must e equl to one hlf, so: 1 = exp + χπ exp d π χ/ 1 χπ = + χπ = Right, now the mode. This is the most likely vlue, occus t the mximum vlue of the distiution: 6 Univesity of Yok Pge /4/8

4 Communictions Engineeing MSc - Peliminy Reding dp() = = exp = exp exp d χ χ χ χ χ = χ χ χ = χ nd since thee is only one tuning point, this must e the mode. Next: the medin. The integtion of the poility distiution fom zeo to the medin must e one-hlf, so: medin exp =.5 χ medin medin.5 = p d= exp d χ χ medin medin.5 = exp = 1exp χ χ medin = χln.5 = ln nd finlly, the vince. To clculte the vince, we ll need the men of the sques: = p() d = exp d χ χ gin, we ll need to poceed y integtion y pts, this time with one pt eing : exp d = exp + exp d χ χ = exp + χ exp d χ χ = exp χ exp = χ nd finlly, the vince is the men of the sque minus the sque of the men : 6 Univesity of Yok Pge 4 /4/8

5 Communictions Engineeing MSc - Peliminy Reding χπ χπ π σ = χ = χ = χ ) When my telephone ings, I pick up the phone. Duing ny given second in the susequent phonecll, the chnces of my putting the phone ck down e.1. Wht pecentge of my phoneclls lst moe thn ten minutes? If the poility of n event hppening duing ny vey smll intevl of time is constnt, then the time until the next event hs negtive exponentil distiution, nd the nume of events in ny given fixed peiod hs Poisson distiution. ( Vey smll intevl in this sense indictes n intevl much shote thn the vege time etween events.) The sitution hee cn e ppoximted y such negtive exponentil solution, since the poility of n event hppening (the end of the phonecll) in one second is so smll. This mkes the vege length of phonecll much longe thn one second, nd one second cn e egded s smll intevl of time. The men vlue of negtive exponentil distiution is 1/λ, whee λdt is the poility tht the event hppens in smll intevl dt. Put dt = one second, nd λ =.1, so the men time etween events is 1 /.1 = 1 seconds. Note tht fte I pick up the phone, the men time until the end of the phonecll is lwys 1 seconds, it doesn t mtte how long I ve een on the phone ledy. It s the supising esult, ut this distiution hs een used to model the length of phoneclls, (If we didn t wnt to mke the ssumption tht one second could e teted s smll time with espect to the length of phonecll, we d hve to use the geometic distiution. Howeve, we d get exctly the sme nswe.) 7) Evey week, I uy one lottey ticket, nd I hve done this ll my life. If the chnce of winning pize in ny month is 1/4, wht is the poility tht I'll win thee pizes in one 7-ye lifetime? The poility of the event hppening in one month is vey smll, so fo these puposes, one month cn e consideed to e vey shot time intevl. The poility of winning is constnt evey month, so this is Poisson distiution polem. Thee e 7 * 1 = 84 months in my lifetime, so the poility of winning thee times is: ( T ) qx ( x) = exp T x! x ( 84/ 4) (.5) qx () = exp 84/ 4! qx () = exp.5 = 6.9e6! 6 Univesity of Yok Pge 5 /4/8

6 Communictions Engineeing MSc - Peliminy Reding so thee is only one chnce in 145, tht I will win thee times. The chnces of winning once is only one in thity. (I hven t won nything yet.) + 1 8) Conside the Peto distiution: p( x) = / x, whee < x < (the distiution hs vlue of zeo when x is less thn ). Sketch the distiution, nd detemine the men nd vince s function of nd. Notice nything unusul? Fo wht vlues of nd does this distiution hve n infinite vince, nd n infinite men? Is this just mthemticl cuiosity, o is it possile distiution with n infinite vince o men might e useful in engineeing? The shpe of the Peto distiution vies with the pmetes nd, ut few epesenttive distiutions look like this: =.5, =1 =1, =1 =, = Fist, woking out the men: + 1 x x = xp x dx= x dx= x dx 1 + x ( 1) ( 1 = ) = ( ) when > 1, this gives: ( 1) x= ( ) = 1 1 ut if < 1 then the men is infinite. To wok out the vince, we ll need the men sque vlue s well: 6 Univesity of Yok Pge 6 /4/8

7 Communictions Engineeing MSc - Peliminy Reding nd this time, if >, then this gives: x 1 x = x p x dx= x dx= x dx + + ( ) + x = = + + ( ) x= = ut if < then the men sque vlue, nd hence the vince, e infinite. Is this sensile? Well, yes. Thee s nothing wong with hving poility distiution with n infinite men, nd in fct this distiution is commonly used fo modelling tffic on communictions netwoks. 6 Univesity of Yok Pge 7 /4/8