Math 135, Spring 2012: HW 7

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1 Mth 3, Spring : HW 7 Problem (p. 34 #). SOLUTION. Let N the number of risins per cookie. If N is Poisson rndom vrible with prmeter λ, then nd for this to be t lest.99, we need P (N ) P (N ) ep( λ) λ ln(.) Recll tht λ E[N] is the verge number of risins per cookie. Problem (p. 34 #). SOLUTION. If we ssume tht no two microbes lie on top of one nother nd they re evenly smered cross the entire plte, then we my ssume tht the microbes re distributed cross the plte ccording to Poisson rndom sctter. The number N of microbes in the viewing field is Poisson rndom vrible with men µ 4.. So the probbility of t lest one microbe is P (N ) P (N ) e. Problem 3 (p. 34 #). SOLUTION. Since X, Y, Z re independent Poisson rndom vribles, we know tht X + Y is Poisson rndom vrible with men E[X] + E[Y ], nd X + Y + Z is Poisson rndom vrible with men 3. This gives ) P (X + Y 4) e 4 4! ; b) E [ (X + Y ) ] V [(X + Y )] + (E [X + Y ]) c) P (X + Y + Z 4) e ! Problem 4 (p. 3 #6). SOLUTION. For prt (), if we ssume the number of chips is distribution ccording to Poisson sctter through given volume of cookie dough, nd if we define N s the number of chips in 3 cubic inches of dough, then N is Poisson rndom vrible with men µ 6 chips. So 4 e 6 6 k P (N 4) k! For prt (b), let M, M, M 3 be the number of mrshmllows in cookies, nd 3; nd let C, C, nd C 3 be the number of chips in cookies,, nd 3. By ssumption M i nd C i, where i, or 3, re independent Poisson rndom vribles, nd the prmeter for M is ; the prmeter for M nd M 3 is 3; the prmeter for C is 4; the prmeter for C nd C 3 is 6. Now, since the sums of independent Poisson rndom vribles is lso Poisson, we know tht N i M i + C i re independent Poisson rndom vribles with prmeters tht re simply the sums of the prmeters for M i nd C i. k P (t most one cookie hs no mrshmllows nd no chips) P (N i > for ll i) + P (N, N >, N 3 > ) + P (N >, N, N 3 > ) + P (N >, N >, N 3 ) ( e 6 )( e 9 ) + e 6 ( e 9 ) + ( e 6 )e 9 ( e 9 ) + ( e 6 )( e 9 )e 9 ( e 9 ) + e 9 ( e 6 )( e 9 )

2 Problem (p.76 #4). SOLUTION. Note tht c must be the vlue tht gurntees tht the integrl of the density over the whole rel line is : c c ( ) d ( ) d to evlute this integrl, epnd the integrnd: ( ) ( + ) the finl evlution is left to you; you will find tht c 3. Now compute E(X) nd E(X ) E(X) E(X ) [ c 3 ( ) d 3 ] ] [ c 4 ( ) d so the vrince is V (X) E(X ) [E(X)] Problem 6 (p.76 #). SOLUTION. The grph of f is left to you. To find P ( X ) we write ( ) d + ( + ) d ech of these integrls cn be evluted by substitution: put u, for instnce, nd then the first integrl reduces to u du.[.].. So P ( X ) To find P ( X > ), note tht P ( X > ) [P ( X ) + P ( < X )] [ ] ( ) d + ( + ) d Finlly, E(X) is not defined becuse the epecttion of X is given by n integrl which does not converge: E( X ) ( ) d + ( + ) d ( + ) d To see tht the bove integrl diverges either use integrtion by prts or substitute u +.

3 Problem 7 (p.76 #8). SOLUTION. For prt (), let X i be the weight of single lump of metl ( ).8. P (.8 X i.) P X i.. ( ) ( )..8 Φ Φ.. For prt (b), we pply the Centrl Limit Theorem to the smple verge X P n i Xi. Here, we do not hve to ssume tht the individul mesurements re normlly distributed, becuse the CLT gurntees tht the smple men will be symptoticlly norml for lrge n regrdless of the distribution of the individul rndom vribles X i, provided the vribles X i re independent, identiclly distributed, nd hve finite vrince. ( P (.8 X.8.) P./ X ) µ./../ ( ) ( )..8 Φ./ Φ./ Problem 8 (p. 76 #). SOLUTION. Let X be the coordinte, rndomly selected from the region in prts (), (b), (c), nd (d). Let lso A denote the re of the region. () First note tht if < b, then If, on the other hnd, < b P [ X b] A A P [ X b] A Therefore, X hs density f X () given by f X () A A ( + ) ( ) d ( + 4) d. ( + ) ( ) d ( + 4) d. + 4 if + 4 if <. 4 + if 4 + if <. (b) In similr fshion, for < b we see tht P [ X b] A + d 3

4 nd if < b P [ X b] A Therefore the density f X () is precisely f X () A + d + if + if < if if < (c) For < b for < b, nd for < b Hence P [ X b] A A P [ X b] A A P [ X b] A A ( + ) ( ) d + d, ( + ) ( ) d d, ( + ) ( 3) d + d, f X () + if < A if < + if + if < if < + if Problem 9 (p.93 #). SOLUTION: Let I i (t) be the indictor of the event tht the i th tom is still present t time t. So for prt () we hve P (I (t) ) e λt, where λ is the rte t which n tom dies. Since the hlf-life is one yer, e λ nd λ log. Therefore P (I () ) e (log ) (/). 4

5 For prt (b), suppose we strt with n toms. We wnt to find t so tht: n EI i (t).n i Since EI i (t) e (log )t for every i, we wish to solve For prt (c) we solve for t in: ne (log )t.n t log log 3.3 yers. 4 EI i (t) i 4 e (log )t t. For prt (d) we ssume tht the toms survive independently, so: P (I () I () I 4 () ) 4 i P (I i () ) [ e (log ) ] 4 [ /4] 4 e.3679 Problem (p.93 #). SOLUTION. Let T 4 nd T 3 be the times of the fourth nd third clls, respectively. We know tht the interrrivl times in Poisson rrivl process hve n eponentil distribution, so T 4 T 3 is n eponentil rndom vribles with prmeter λ. Thus P (T 4 T 3 < ) e Net, let N t the number of clls tht hve occurred by time t. This is Poisson rndom vrible with prmeter µ t, so P (N 4) P (N ) P (N ) P (N ) P (N 3) nd the probbility P (N k) e k k!. The time of the fourth cll is sum of the interrrivl cll times between the first, second, third, nd fourth clls, so this hs gmm distribution with shpe prmeter r 4 nd λ, so E[T 4 ] 4 4. Problem (p. 93 #6). SOLUTION. The hits recorded by Geiger counter cn be viewed s Poisson rrivl process with rte λ. In this cse, the times between hits re lso eponentil rndom vribles with prmeter λ, so T 3 hs gmm distribution with r 3, λ. You cn integrte by prts twice in the following integrl to get numericl nswer: P ( T 3 4) 4 t e t dt

6 Alterntively (nd perhps more esily), you cn write the probbility in terms of N t Poisson(t), the number of rrivls by time t P ( T 3 4) P (T 3 4) P (T 3 < ) P (N 4 3) P (N 3) e 4 [ /] ( e [ + + /]) e 3e 4 Problem (p. 94 #9). Consider the gmm function Γ(r) r e d, where r >. ) Use integrtion by prts to show tht Γ(r + ) rγ(r). SOLUTION. Γ(r + ) r e d. Let U() r, du/d r r. Let V () e. Then integrtion-by-prts yields Γ(r + ) r e d [ lim b r e b + ] + b + r rγ(r) b) Deduce from () tht Γ(r) (r )! r e d r r e d SOLUTION. This follows immeditely from () by induction nd from the fct tht Γ(). Note tht Γ(3) Γ() [Γ()]. So Γ(4) 3Γ(3) 3, etc. c) If T hs eponentil distribution with rte, then show tht E[T n ] n! nd SD(T ). SOLUTION. Observe tht if f(t) is the eponentil density with rte, then f(t) e t, nd so E[T n ] o t n f(t)dt t n e t dt Γ(n + ) n! by using the results from prts () nd (b). The vrince clcultion is strightforwrd V (T ) E[T ] (E[T ]) d) If T hs eponentil distribution with rte λ, then show λt hs eponentil distribution with rte, hence E(T n ) n! λ, nd SD(T ) n λ. SOLUTION. Observe tht if T λt, then P ( T > t) P (λt > t) P (T > t/λ) e ( λ t λ) e t 6

7 Therefore, λ n E[T n ] E[(λT ) n ] E[ T n ] n! by the previous prt, becuse T is eponentil with rte. The stndrd devition gin follows esily. Problem 3 (p. 94, #). SOLUTION. Let Z int(t ). Then P (Z k) P (int(t ) k) P (T k) e λk [ e λ] k q k where q e λ is the probbility of filure on ech tril; therefore p e λ is the probbility of success on ech tril. Net, suppose T hs eponentil distribution with prmeter λ. Let T m int(mt )/m so P (mt m k) P (int(mt ) k) P (mt k) P (T k m ) [e λ/m] k q k m where q m e λ/m, so p m e λ/m. Conversely, suppose mt m hs geometric p m distribution for ll nonnegtive integers m. Let q m p m be the filure probbility on ny given tril. The key point is to notice the following chin: q m m P (mt m m) P (int(mt ) m) P (int(t ) ) P (T ) q nd s result, log(q ) m log(q m ) nd therefore for ll m, m log(q m ) is independent of m. Consequently, we cn set λ m log(q m ). For ny rtionl number, t Q, we cn write t n m for some integers n nd m. P (T t) P (T n/m) P (int(mt ) n) q n m [e log(qm)] n [e m log(qm)] n/m [ e λ] t e λt To complete the proof for ll nonnegtive rel numbers t observe tht we cn pproimte ny irrtionl number by rtionl numbers. Tht is, tke n incresing sequence of rtionls r n nd decresing sequence of rtionls t n so tht lim r n lim t n t. So r n < t < t n for every n nd P (T t n ) < P (T t) < P (T r n ). The probbilities on the left nd right converge to e λt, so by the Squeeze Theorem from clculus, P (T t) e λt. 7