Deeper Understanding, Faster Calculation --Exam P Insights & Shortcuts. 19th Edition. by Yufeng Guo. For SOA Exam P/CAS Exam 1. Spring, 2012 Edition

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1 Deeper Understnding, Fster Clcultion --Exm P Insights & Shortcuts 9th Edition by Yufeng Guo For SOA Exm P/CAS Exm Spring, 0 Edition This electronic book is intended for individul buyer use for the sole purpose of prepring for Exm P. This book cn NOT be resold to others or shred with others. No prt of this publiction my be reproduced for resle or multiple copy distribution without the express written permission of the uthor. 00, 0, 0 By Yufeng Guo of 447

2 Tble of Contents Chpter Exm-tking nd study strtegy... 5 Top Horse, Middle Horse, Wek Horse... 5 Truths bout Exm P... 6 Why good cndidtes fil... 8 Recommended study method... 0 CBT (computer-bsed testing) nd its implictions... Chpter Doing clcultions 00% correct 00% of the time... Wht clcultors to use for Exm P... Criticl clcultor tips... 7 Comprison of best clcultors... 6 Chpter Set, smple spce, probbility models... 7 Chpter 4 Multipliction/ddition rule, counting problems... 4 Chpter 5 Probbility lws nd whodunit Chpter 6 Conditionl Probbility Chpter 7 Byes theorem nd posterior probbilities Chpter 8 Rndom vribles Discrete rndom vrible vs. continuous rndom vrible Probbility mss function Cumultive probbility function (CDF) PDF nd CDF for continuous rndom vribles Properties of CDF... 8 Men nd vrince of rndom vrible... 8 Men of function Chpter 9 Independence nd vrince Chpter 0 Percentile, men, medin, mode, moment... 9 Chpter Find E( X), Vr( X), E( X Y), Vr( X Y) Chpter Bernoulli distribution... Chpter Binomil distribution... Chpter 4 Geometric distribution... Chpter 5 Negtive binomil... 9 Chpter 6 Hypergeometric distribution Chpter 7 Uniform distribution... 4 Chpter 8 Exponentil distribution Chpter 9 Poisson distribution Chpter 0 Gmm distribution... 7 Chpter Bet distribution... 8 Chpter Weibull distribution... 9 Chpter Preto distribution Chpter 4 Norml distribution Chpter 5 Lognorml distribution... Yufeng Guo, Deeper Understnding: Exm P of 447

3 Chpter 6 Chi-squre distribution... Chpter 7 Bivrite norml distribution... 6 Chpter 8 Joint density nd double integrtion... Chpter 9 Mrginl/conditionl density... 5 Chpter 0 Trnsformtion: CDF, PDF, nd Jcobin Method... 6 Chpter Univrite & joint order sttistics Chpter Double expecttion Chpter Moment generting function Key MGF formuls you must memorize... 0 Chpter 4 Joint moment generting function... 6 Chpter 5 Mrkov s inequlity, Chebyshev inequlity... Chpter 6 Study Note Risk nd Insurnce explined Deductible, benefit limit Coinsurnce... 5 The effect of infltion on loss nd clim pyment Mixture of distributions Coefficient of vrition... 6 Norml pproximtion Security loding Chpter 7 On becoming n ctury Guo s Mock Exm Solution to Guo s Mock Exm Finl tips on tking Exm P... 4 About the uthor Vlue of this PDF study mnul User review of Mr. Guo s P Mnul Bonus Problems Yufeng Guo, Deeper Understnding: Exm P of 447

4 Chpter 8 Exponentil distribution Key Points Gin deeper understnding of exponentil distribution: Why does exponentil distribution model the time elpsed before the first or next rndom event occurs? Exponentil distribution lcks memory. Wht does this men? Understnd nd use the following integrtion shortcuts: For ny > 0 nd 0 : x e dx = e x x e dx = ( + ) e x x e dx = ( + ) + e You will need to understnd nd memorize these shortcuts to quickly solve integrtions in the het of the exm. Do not ttempt to do integrtion by prts during the exm. Explntions Exponentil distribution is the continuous version of geometric distribution. While geometric distribution describes the probbility of hving N trils before the first or next success (success is rndom event), exponentil distribution describes the probbility of hving time T elpse before the first or next success. Let s use simple exmple to derive the probbility density function of exponentil distribution. Clims for mny insurnce policies seem to occur rndomly. Assume tht on verge, clims for certin type of insurnce policy occur once every months. We wnt to find out the probbility tht T months will elpse before the next clim occurs. To find the pdf of exponentil distribution, we tke dvntge of geometric distribution, whose probbility mss function we lredy know. We divide ech month into n intervls, ech intervl being one minute. Since there re 0*4*60=4,00 minutes in month (ssuming there re 0 dys in month), we convert ech month into 4,00 46 of 447

5 intervls. Becuse on verge one clim occurs every months, we ssume tht the chnce of clim occurring in one minute is 4, 00 How mny minutes must elpse before the next clim occurs? We cn think of one minute s one tril. Then the probbility of hving m trils (i.e. m minutes) before the first success is geometric distribution with p = 4, 00 Insted of finding the probbility tht it tkes exctly m minutes to hve the first clim, we ll find the probbility tht it tkes m minutes or less to hve the first clim. The ltter is the cumultive distribution function which is more useful. P (it tkes m minutes or less to hve the first clim) = P (it tkes more thn m minutes to hve the first clim) The probbility tht it tkes more thn m trils before the first clim is ( p). To see why, you cn reson tht to hve the first success only fter m trils, the first m trils m p. must ll end with filures. The probbility of hving m filures in m trils is Therefore, the probbility tht it tkes m trils or less before the first success is m p. Now we re redy to find the pdf of T : PT ( t) = P(4,00t trils or fewer before the st success) m t 4,00t 4,00t = = e 4, 00 4, 00 Of course, we do not need to limit ourselves by dividing one month into intervls of one minute. We cn divide, for exmple, one month into n intervls, with ech intervl of /,000,000 of minute. Essentilly, we wnt n. ( t) PT = P ( nt trils or fewer before the st success) t 47 of 447

6 t nt n = = = e n n t (s n ) If you understnd the bove, you should hve no trouble understnding why exponentil distribution is often used to model time elpsed until the next rndom event hppens. Here re some exmples where exponentil distribution cn be used: Time until the next clim rrives in the clims office. Time until you hve your next cr ccident. Time until the next customer rrives t supermrket. Time until the next phone cll rrives t the switchbord. Generl formul: Let T=time elpsed (in yers, months, dys, etc.) before the next rndom event occurs. Alterntively, t = ( ) =, f ( t) = e, where = E( T) t ( > ) = = Ft PT t e PT t Ft e t = ( ) = t, f ( t) λe λ Ft PT t e λ t PT> t = Ft = e λ t λ =, where Men nd vrince: E( T) = =, Vr ( T ) = = λ λ = E( T) Like geometric distribution, exponentil distribution lcks memory: PT> + bt> = PT> b We cn esily derive this: P T > + b T > P T > + b e PT> PT> e ( b) + b PT ( > + bt> ) = = = = e = PT> b In plin English, this lck of memory mens tht if component s time to filure follows exponentil distribution, then the component does not remember how long it hs been working (i.e. does not remember wer nd ter). At ny moment when it is working, the component strts fresh s if it were completely new. At ny moment while the component 48 of 447

7 is working, if you reset the clock to zero nd count the time elpsed from then until the component breks down, the time elpsed before brekdown is lwys exponentilly distributed with the identicl men. This is clerly n idelized sitution, for in rel life wer nd ter does reduce the longevity of component. However, in mny rel world situtions, exponentil distribution cn be used to pproximte the ctul distribution of time until filure nd still give resonbly ccurte result. A simple wy to see why component cn, t lest by theory, forget how long it hs worked so fr is to think bout geometric distribution, the discrete counterprt of exponentil distribution. For exmple, in tossing coin, you cn clerly see why coin doesn t remember its pst success history. Since getting heds or tils is purely rndom event, how mny times you hve tossed coin so fr before getting heds relly should NOT hve ny bering on how mny more times you need to toss the coin to get heds the second time. The clcultion shortcuts re explined in the following smple problems. Smple Problems nd Solutions Problem The lifetime of light bulb follows exponentil distribution with men of 00 dys. Find the probbility tht the light bulb s life () Exceeds 00 dys () Exceeds 400 dys () Exceeds 400 dys given it exceeds 00 dys. Solution Let T = # of dys before the light bulb burns out. t = e, where = E( T) = 00 t ( > ) = = F t PT t Ft e ( 00) ( 00) ( 400) ( 400) PT> = F = e = e =0.679 PT> = F = e = e =0.08 ( > 400) ( > 00) PT e PT ( > 400 T> 00) = = = e PT e = of 447

8 Or use the lck of memory property of exponentil distribution: ( ) ( ) PT> T> = PT> = e = e = Problem The length of telephone converstions follows exponentil distribution. If the verge telephone converstion is.5 minutes, wht is the probbility tht telephone converstion lsts between minutes nd 5 minutes? Solution /.5 = F t e t P < T < = e e = e e = 5/.5 /.5 /.5 5/ Problem f t = e t. The rndom vrible T hs n exponentil distribution with pdf / Find E( TT> ), Vr ( T T > ), E( TT ), ( ) Vr T T. Solution First, let s understnd the conceptul thinking behind the symbol E( TT> ). Here we re only interested in T >. So we reduce the originl smple spce T [ 0, ] to T [, ]. The pdf in the originl smple spce T [ 0, ] is f ( t ) ; the pdf in the f ( t) reduced smple spce t [, ] is. Here the fctor is PT> PT> ( ) normlizing constnt to mke the totl probbility in the reduced smple spce dd up to one: f ( t) dt = f ( t) dt = P ( T > ) = PT> PT> PT> 50 of 447

9 Next, we need to clculte E( TT> ) spce T [, ]: [, ] f ( t), the expected vlue of T in the reduced smple E ( T T > ) = t dt = t f ( t) dt = t f ( t) dt PT> PT> F / e F = / = 5e Reduced Smple spce T tf t dt (integrtion by prts) / 5e E ( T T > ) = tf ( t) dt = = 5 / F e Here is nother pproch. Becuse T does not remember wer nd ter nd lwys strts new t ny working moment, the time elpsed since T = until the next rndom event (i.e. T ) hs exponentil distribution with n identicl men of. In other words, T T > is exponentilly distributed with n identicl men of. So E( T T ) > =. E T T > = E T T > + = + = 5 Next, we will find Vr ( T T > ). ( ) ( T ) t / E T T > = t f ( t) dt = t e dt Pr > Pr ( T > ) t / / t e dt = 9e (integrtion by prts) ( T ) E T / 9e > = = 9 / e VrTT> = E T T> E TT> = 9 5 = 4 = 5 of 447

10 Vr T. To see why, we know It is no coincidence tht Vr ( T T > ) is the sme s Vr ( T T > ) = Vr ( T T > ). This is becuse Vr ( X c) Vr ( X ) constnt c. Since ( T T ) + = stnds for ny > is exponentilly distributed with n identicl men of, then Vr T T > = = = 4. Next, we need to find E( TT ). ( ) f ( t) ( T < ) F 0 0 E T T < = t dt = tf t dt Pr F / ( ) = e tf t dt tf t dt tf t dt = E ( T ) tf t dt = = / = 5e tf t dt (we lredy clculted this) / 5e E ( T T < ) = tf ( t) dt F = e 0 / Here is nother wy to find E( TT< ). = ( < ) ( < ) + ( > ) ( > ) E T E T T P T E T T P T The bove eqution sys tht if we brek down T into two groups, T > nd T <, then the overll men of these two groups s whole is equl to the weighted verge men of these groups. Also note tht PT= ( ) is not included in the right-hnd side becuse the probbility of continuous rndom vrible t ny single point is zero. This is similr to the concept tht the mss of single point is zero. 5 of 447

11 Of course, you cn lso write: = ( ) ( ) + ( > ) ( > ) E T E T T P T E T T P T Or E( T) = E( T T < ) P( T < ) + E( T T ) P( T ) You should get the sme result no mtter which formul you use. = ( < ) ( < ) + ( > ) ( > ) E T E T T P T E T T P T ( ) E TT< = ( > ) ( > ) E T E T T P T ( < ) PT / / ( ) e 5e + E( TT< ) = = e e Next, we will find ( ) E T T < : ( ) PT / / t / E T T < = t f t dt = t e dt < PT< 0 0 t/ x/ / 0 Alterntively, t e dt = t e = 8 9e 0 ( ) PT / t / 8 9e < = = / > e 0 E T T t e dt = ( < ) ( < ) + ( > ) ( > ) E T E T T P T E T T P T ( T ) E T < = ( > ) ( > ) E T E T T P T ( < ) PT PT 9 > 8 9e = = PT< e / / 5 of 447

12 / / 8 9e 5e VrTT ( < ) = E( T T< ) E ( TT< ) = / / e e In generl, for ny exponentilly distributed rndom vrible T with men > 0 nd for ny 0 : T T> is lso exponentilly distributed with men E T T> =, VrT T> = E( TT ), Vr( TT ) > = + > = E ( T T > ) = ( t ) f ( t) dt PT> ( ) ( ) = ( > ) ( > ) = t f t d te T T P T e E ( T T > ) = tf ( t) dt PT> ( ) = ( > ) ( > ) = ( + ) tf t d te T T P T e E ( T T < ) = tf ( t) dt PT< ( ) 0 tf ( t) d = te( T T < ) P( T < ) = E( T T < )( e ) 0 E( T) = E( T T < ) P( T < ) + E( T T > ) P( T > ) ( ) = E TT< e + E TT> e ( ) = ( < ) ( < ) + ( > ) ( > ) E T E T T P T E T T P T You do not need to memorize the bove formuls. However, mke sure you understnd the logic behind these formuls. Before we move on to more smple problems, I will give you some integrtion-by-prts formuls for you to memorize. These formuls re criticl to you when solving 54 of 447

13 exponentil distribution-relted problems in minutes. You should memorize these formuls to void doing integrtion by prts during the exm. Formuls you need to memorize: For ny > 0 nd 0 x/ / e dx = e () x/ / x e dx = ( + ) e () x/ / x e dx = ( + ) + e () You cn lwys prove the bove formuls using integrtion by prts. However, let me give n intuitive explntion to help you memorize them. Let X represent n exponentilly rndom vrible with men of, nd f( x ) is the probbility distribution function, then for ny 0, Eqution () represents x / P( X > ) = F( ), where F( x) = e is the cumultive distribution function of X. You should hve no trouble memorizing Eqution (). For Eqution (), from Smple Problem, we know / = ( > ) ( > ) = ( + ) xf x d xe X X P X e To understnd Eqution (), note tht / ( > ) = P X e = ( > ) ( > ) x f x d xe X X P X ( > ) = ( > ) + ( > ) E X X E X X V rx X ( > ) = ( + ), Vr ( X X > ) = E X X Then x f x dx = + + e / 55 of 447

14 We cn modify Eqution (),(),() into the following equtions: For ny > 0 nd b 0 b x/ / b/ e dx = e e (4) x e dx ( ) e ( b ) e b x e x/ dx / b/ ( ) e ( b ) e = (6) b x/ / b/ = + + (5) We cn esily prove the bove eqution. For exmple, for Eqution (5): b x/ x/ x/ x e dx x e dx x e dx = b = ( + ) ( + ) We cn modify Eqution (),(),() into the following equtions: For ny > 0 nd 0 x/ x/ e dx = e + c (7) x/ x/ x e dx = ( x + ) e + c (8) x/ x/ x e dx = x + + e + c (9) e b e / b/ Set λ =. For ny λ > 0 nd 0 λx λx λe dx = e + c (0) λx λx x ( λe ) dx = x + e + c λ () λx λx x ( λe ) dx = x + + e + c λ λ () So you hve four sets of formuls. Just remember one set (ny one is fine). Equtions (4),(5),(6) re most useful (becuse you cn directly pply the formuls), but the formuls re long. 56 of 447

15 If you cn memorize ny one set, you cn void doing integrtion by prts during the exm. You definitely do not wnt to clculte messy integrtions from scrtch during the exm. Now we re redy to tckle more problems. Problem 4 After n old mchine ws instlled in fctory, Worker John is on cll 4-hours dy to repir the mchine if it breks down. If the mchine breks down, John will receive service cll right wy, in which cse he immeditely rrives t the fctory nd strts repiring the mchine. The mchine s time to filure is exponentilly distributed with men of hours. Let T represent the time elpsed between when the mchine ws first instlled nd when John strts repiring the mchine. Find ET nd Vr( T ). Solution / T is exponentilly distributed with men =. F( t) = e t. We simply pply the men nd vrince formul: Vr ( T ) E T = =, = = = 9 Problem 5 After n old mchine ws instlled in fctory, Worker John is on cll 4-hours dy to repir the mchine if it breks down. If the mchine breks down, John will receive service cll right wy, in which cse he immeditely rrives t the fctory nd strts repiring the mchine. The mchine ws found to be working tody t 0:00.m.. The mchine s time to filure is exponentilly distributed with men of hours. Let T represent the time elpsed between 0:00.m. nd when John strts repiring the mchine. Find E( T ) nd Vr ( T ). Solution 57 of 447

16 Exponentil distribution lcks memory. At ny moment when the mchine is working, it forgets its pst wer nd ter nd strts fresh. If we reset the clock t 0:00 nd observe T, the time elpsed until brekdown, T is exponentilly distributed with men of. = =, Vr ( T ) E T = = = 9 Problem 6 After n old mchine ws instlled in fctory, Worker John is on cll 4-hours dy to repir the mchine if it breks down. If the mchine breks down, John will receive service cll right wy, in which cse he immeditely rrives t the fctory nd strts repiring the mchine. Tody, John hppens to hve n ppointment from 0:00.m. to :00 noon. During the ppointment, he won t be ble to repir the mchine if it breks down. The mchine ws found working tody t 0:00.m.. The mchine s time to filure is exponentilly distributed with men of hours. Let X represent the time elpsed between 0:00.m. tody nd when John strts repiring the mchine. Find E( T ) nd Vr ( T ). Solution Let T =time elpsed between 0:00.m. tody nd brekdown. T is exponentilly X = mx, T. distributed with men of. X, if T = T, if T > You cn lso write X, if T < = T, if T As sid before, it doesn t mtter where you include the point T = becuse the probbility density function of continuous vrible t ny single point is lwys zero. 58 of 447

17 Pdf is lwys / f t = e t no mtter T or T >. t/ t/ E ( X ) = x( t) f ( t) dt = e dt t e dt e dt = e t / / t e dt = ( + ) e = 5e t / / / E X = e + 5e = + e / / / t = = + / t/ E X x f t dt e dt t e dt e dt = e = 4 e t / / / t e dt = ( 5 + ) e = 4e t / / / E X = 4 e + 4e = 4 + 0e / / / / / 4 0 = = + ( + ) Vr X E X E X e e We cn quickly check tht / e EX = + is correct: X, if T = T, if T > 0, if T X = T, if T > 0 ( ) / = ET ( T> ) PT ( > ) = e E X = ETT< PT< + ET T> PT> E X = E X + = + e / You cn use this pproch to find E( X ) too, but this pproch isn t ny quicker thn using the integrtion s we did bove 59 of 447

18 Problem 7 After n old mchine ws instlled in fctory, Worker John is on cll 4-hours dy to repir the mchine if it breks down. If the mchine breks down, John will receive service cll right wy, in which cse he immeditely rrives t the fctory nd strts repiring the mchine. Tody is John s lst dy of work becuse he got n offer from nother compny, but he ll continue his current job of repiring the mchine until :00 noon if there s brekdown. However, if the mchine does not brek by noon :00, John will hve finl check of the mchine t :00. After :00 noon John will permnently leve his current job nd tke new job t nother compny. The mchine ws found working tody t 0:00.m.. The mchine s time to filure is exponentilly distributed with men of hours. Let X represent the time elpsed between 0:00.m. tody nd John s visit to the mchine. Find E( X ) nd Vr ( X ). Solution Let T =time elpsed between 0:00.m. tody nd brekdown. T is exponentilly X = min, T. distributed with men of. X t, if T =, if T > Pdf is lwys / f t = e t no mtter T or T >. t/ t/ E X = t e dt + e dt 0 0 t e dt = ( + ) e t / / e dt = e t / / / / / = 5 + = E X e e e 60 of 447

19 Vr X, we need to clculte To find E X. = = + E X x t f t dt t f t dt f t dt t f t dt = e + + e = 8 4e 0/ / / f t dt = 4e / E X = 8 4e + 4e = 8 0e / / / / / ( 8 0 = = ) ( ) Vr X E X E X e e E X We cn esily verify tht / We know tht = e is correct. Notice: T + = min ( T, ) + mx ( T, ) E( T + ) = E min ( T, ) + E mx ( T, ) min (, ) = / mx (, ) = + / ( + ) = E( T) + = + E T e (from this problem) E T e (from the previous problem) E T So the eqution E( T + ) = E min ( T, ) + E mx ( T, ) We cn lso check tht E( X ) T + = min T, + mx T, holds. = is correct. / 8 0e ( T + ) = min ( T, ) + mx ( T, ) ( T ) ( T ) ( T ) ( T ) = min, + mx, + min, mx, 6 of 447

20 ( T ) min, t if t = if t >, mx ( T, ) if t = t if t > min T, mx T, = t ( T + ) = min ( T, ) + mx ( T, ) + ( t) E T + = E min T, + mx T, = E min T, + E mx T, + E t E T + = E T + 4t+ 4 = E T + 4E t + 4 = = 4 / = E min T, 8 0e (from this problem) / = + = = = E mx T, 4 0e (from previous problem) E t 4E t E min T, E mx T, E t / / = 8 0e e + = 4 So the eqution Problem 8 E T + = E min T, + mx T, holds. An insurnce compny sells n uto insurnce policy tht covers losses incurred by policyholder, subject to deductible of $00 nd mximum pyment of $00. Losses incurred by the policyholder re exponentilly distributed with men of $00. Find the expected pyment mde by the insurnce compny to the policyholder. Solution Let X =losses incurred by the policyholder. X is exponentilly distributed with men / 00 of 00, f ( x) = e x. 00 Let Y =clim pyment by the insurnce compny. 0, if X 00 Y = X 00, if 00 X , if X of 447

21 E( Y ) = y x f x dx = 0 f x dx + x 00 f x dx + 00 f x dx f x dx = 0 ( 00) = x f x dx xf x dx f x dx xf x dx = e e = 00e 600e 00/ / 00 / 00/ / 00 / = ( ) = ( ) 00 f x dx 00 e e 00 e 400/ 00 = = 00 f x dx 00 e 00 e Then we hve E X = 00e 600e 00 e e + 00e = 00 e e / ( / / ) ( ) Alterntively, we cn use the shortcut developed in Chpter : d+ L x 00 / 00 x / 00 Pr 00 00( / = > = = = ) 400 E X X x dx e dx e e e d 00 Problem 9 An insurnce policy hs deductible of. Losses re exponentilly distributed with men 0. Find the expected non-zero pyment by the insurer. Solution Let X represent the losses nd Y the pyment by the insurer. Then Y = 0 if X ; Y = X if E YY> 0. X >. We re sked to find ( > 0) = ( > ) E YY E X X X X > is n exponentil rndom vrible with the identicl men of 0. So E X X > = E X = 0. 6 of 447

22 Generlly, if X is n exponentil loss rndom vrible with men, then for ny positive deductible d E( X d X > d) = E( X) =, Problem 0 E X X > d = E X d X > d + d = + d Clims re exponentilly distributed with men of $8,000. Any clim exceeding $0,000 is clssified s big clim. Any clim exceeding $60,000 is clssified s super clim. Find the expected size of big clims nd the expected size of super clims. Solution This problem tests your understnding tht the exponentil distribution lcks memory. Let X represents clims. X is exponentilly distributed with men of =8,000. Let Y =big clims, Z =super clims. = E( X) + 0, 000 = + 0, 000 = 8, 000 E Y = E X X > 0, 000 = E X 0, 000 X > 0, , 000 = E( X) + 60, 000 = + 60, 000 = 68, 000 E Z = E X X > 60, 000 = E X 60, 000 X > 60, , 000 Problem x /5 x x e dx. Evlute ( + ) Solution /5 /5 /5 /5 x x x x x + x e dx = 5 x + x e dx = 5 x e dx 5 x e dx x /5 /5 x e dx = e, x = ( 5+ ) 5 x x e dx e 5 /5 /5 x + x e dx = ( e = 405e /5 /5 /5 64 of 447

23 Problem (two exponentil distributions competing) You hve cr nd vn. The time-to-filure of the cr nd the time-to-filure the vn re two independent exponentil rndom vribles with men of 8 yers nd 4 yers respectively. Clculte the probbility tht the cr dies before the vn. Solution Let X nd Y represent the time-to-filure of the cr nd the time-to-filure of the vn P X < Y. respectively. We re sked to find X nd Y re independent exponentil rndom vribles with men of 8 nd 4 respectively. Their pdf is: fx x e 8 fy y e 4 x 8 x 8 =, F ( x) e y 4 y 4 =, F ( y) e X Y =, where x 0 =, where y 0 Method # X nd Y hve the following joint pdf: x 8 y 4 fxy, ( xy, ) = fx( x) fy( y) = e e of 447

24 The shded re is x 0, y 0, nd x< y. x 8 y 4 P( X < Y ) = f ( x, y) dxdy = f ( x, y) dydx = e e dydx 8 4 shded Are x 8 x 4 x 8 e ( e ) dx e dx x 0 = = = 8 8 x Method P X Y P x X x dx P Y x dx ( < ) = ( < + ) ( > + ) 0 The bove eqution sys tht to tht find P( X Y) ( x, x + dx]. Next, we set Y x dx flls in the intervl ( x, x + dx]. To find P( X < Y) when X flls [ ] integrte P ( x < X x + dx) P ( Y > x + dx) over the intervl [ 0, ]. x 8 ( < + ) = = 8 PY ( x dx) PY ( x) <, we first fix X t tiny intervl > +. This wy, we re gurnteed tht X < Y when X P x X x dx f x dx e dx > + = > becuse dx is tiny x = F x = e = e Y 4 x 4 0,, we simply 66 of 447

25 8 4 x + x x 8 4 P ( X < Y ) = f 8 X x P Y > x = e dx e = e dx 8 8 = = This is the intuitive mening behind the formul 8. In this problem, we hve cr nd vn. The time-to-filure of the cr nd the time-to-filure the vn re two independent exponentil rndom vribles with men of 8 yers nd 4 yers respectively. So on verge cr filure rrives t the speed of /8 per yer; vn filure rrives t the speed of ¼ per yer; nd totl filure (for crs nd vns) rrives speed of per yer. Of the totl filure, cr filure ccounts for 8 = of the totl filure With this intuitive explntion, you should esily memorize the following shortcut: In generl, if X nd Y re two independent exponentil rndom vribles with prmeters of λ nd λ respectively: X x = λ nd f ( y) = λ e λ f x e λ Then Y y λx λx ( λ+ λ) x λ X λ λ λ λ P X < Y = f x P Y > x dx = e e dx = e dx = Similrly, PY< X = λ. λ + λ λ λ Now you see tht P( X < Y) + P( X > Y) = + =. This mens tht λ+ λ λ+ λ P( X = Y) = 0. To see why P( X = Y) = 0, plese note tht X = Y is line in the -D plne. A line doesn t hve ny re (i.e. the re is zero). If you integrte the joint pdf over line, the result is zero. If you hve trouble understnding why P( X Y) 0 = =, you cn think of probbility in -D plne s volume. You cn think of the joint pdf in -D plne s the height function. In order to hve volume, you must integrte the height function over n re. A line doesn t hve ny re. Consequently, it doesn t hve ny volume. 67 of 447

26 Problem (Smple P #90, lso My 000 #0) An insurnce compny sells two types of uto insurnce policies: Bsic nd Deluxe. The time until the next Bsic Policy clim is n exponentil rndom vrible with men two dys. The time until the next Deluxe Policy clim is n independent exponentil rndom vrible with men three dys. Wht is the probbility tht the next clim will be Deluxe Policy clim? (A) 0.7 (B) 0. (C) (D) (E) Solution B T = time until the next Bsic policy is sold. Let B λ = =. B D Let T = time until the next Deluxe policy is sold. D with λ = =. D B T is exponentil rndom vrible with D T is exponentil rndom vrible The next clim is Deluxe policy mens tht T D B < T. D < = T 0.4 D B T + T = = 5 = + D B ( T ) PT Homework for you: # My 000; #9, #4, #4 Nov 000; #0 My 00; #5 Nov 00; #4 My of 447

27 About the uthor Yufeng Guo ws born in centrl Chin. After receiving his Bchelor s degree in physics t Zhengzhou University, he ttended Beijing Lw School nd received his Msters of lw. He ws n ttorney nd lw school lecturer in Chin before immigrting to the United Sttes. He received his Msters of ccounting t Indin University. He hs pursued life cturil creer nd pssed exms,,, 4, 5, 6, nd 7 in rpid succession fter discovering successful study strtegy. Mr. Guo s exm records re s follows: Fll 00 Pssed Course Spring 00 Pssed Courses, Fll 00 Pssed Course 4 Spring 004 Pssed Course 6 Fll 004 Pssed Course 5 Spring 005 Pssed Course 7 Mr. Guo currently teches n online prep course for Exm P, FM, MFE, nd MLC. For more informtion, visit If you hve ny comments or suggestions, you cn contct Mr. Guo t yufeng.guo.ctury@gmil.com. 44 of 447

28 Vlue of this PDF study mnul. Don t py the shipping fee (cn cost $5 to $0 for U.S. shipping nd over $0 for interntionl shipping). Big sving for Cndin cndidtes nd other interntionl exm tkers.. Don t wit week for the mnul to rrive. You downlod the study mnul instntly from the web nd begin studying right wy.. Lod the PDF in your lptop. Study s you go. Or if you prefer printed copy, you cn print the mnul yourself. 4. Use the study mnul s flsh crds. Click on bookmrks to choose chpter nd quiz yourself. 5. Serch ny topic by keywords. From the Adobe Acrobt reder toolbr, click Edit ->Serch or Edit ->Find. Then type in key word. User review of Mr. Guo s P Mnul Mr. Guo s P mnul hs been used extensively by mny Exm P cndidtes. For user reviews of Mr. Guo s P mnul t click here Review of the mnul by Guo. Testimonies: Second time I used the Guo mnul nd ws ble to do some of the similr questions in less thn 5% of the time becuse of knowing the shortcut. Testimony # of the mnul by Guo I just took the exm for the second time nd feel confident tht I pssed. I used Guo the second time round. It ws very helpful nd gives lot of shortcuts tht I found very vluble. I thought the mnul ws kind of expensive for n e-file, but if it helped me pss it ws well worth the cost. Testimony # of the mnul by Guo I took the lst exm in Feb 006, nd I rn out of time nd I ended up with five. I needed to do the questions quicker nd more efficiently. The Guo's study guide relly did the job. Testimony # of the mnul by Guo Yufeng Guo, Deeper Understnding: Exm P 45 of 447