July 21, 2002, Poisson Randomization (Multinomial Allocation)

Transcription

1 July 21, 2002, Poisson Ranomization (Multinomial Allocation) Theorem 1. Poisson Ranomization Theorem Define to be the prouct space á,, â á,, an let be the set of all vectors =, á, = in such that = á =. Suppose that \ ß, á, \ ß is a multinomial ranom vector. That is, for all =, á,=, x T\ = ßá,\ = : â : = x= xâ= x = = = ß ß # # where : á : an : for, á,. : # Suppose that ]ß áß] is a vector of inepenent Poisson ranom variables an that ] has parameter :,, á,. That is, suppose that for all =, á,=, T ] =ß á,] = T ] = / : 3 = : 3 =x 3 3 = á= / x 3 x = xâ= x = = = # : : â : # Let T an efine T T. Then for ß an T á \ß,,\ ß T / T ],]#, á,] T E1 á \ß,,\ ß / E1],]#, á,] pr9/à[nee to fill in efinition of 1 an 1Ó Theorem 2. Factorial Moments via Poisson Ranomization 1

2 Define the ranom vectors \, á, \ an ], á, ] as in Theorem 1. ß ß Define ranom variables [ B 1 \ ß, á, \ ß an [ C 1 ], á, ] with support on Ö,, á. Define TA to be the event that [ C Aan efine TßA T A so that Then, T \, á,\ T T [ A ß ß ßA B E [ B ÒÓ / T ], á,] A ) A T ) A ) Theorem 3. Expecte Waiting Time to Meet Given Quotas Suppose we istribute balls inepenently into istinguishable boxes such that the probability that a ball is istribute into Box on any given trial is :. When ; 3 balls have been istribute into Box 3 we will say Box 3 has reache its quota. Let [ ÐÀ;ß;ß : ß; Ñ [ ÀU : represent the waiting time until exactly ifferent boxes # á have reache their quota. ÒÓ 2 : À; ß; ßáß; ÀU : Let E[ represent the ascening moment of [. That is, # ÒÓ ÀU : ÀU : ÀU : ÀU : E[ E[ [ â [ Define R ;ß;ß ß; R ÀU to be the number of boxes that have not reache their # á quota after balls have been istribute. We allow for the possibility that : so as to inclue the case where our interest is limite to a particular set of boxes out of a larger set (say ) of possible boxes. 2 Then the ascening moment of [ is given by ÀU : ; ; ÒÓ ÀU : ) E [ Ð â 2

3 : â : á x xâ x á : á: where is the set of all samples of size rawn without replacement from #á,,,, where the orer of sampling is not consiere important. 3

4 Applications Problem 1. Suppose istinguishable balls are istribute inepenently among istinguishable urns an that all urns are equally likely. Suppose that of the urns are marke. Let X equal the number of urns among these marke urns which are occupie by at least one ball. Let [ equal the number of urns among these marke 3 urns which are occupie by exactly 3 balls. We note that without loss of generality we can assume that the marke urns are the first urns when we arrange the urns in a row. (a) Show TX ÒÓ x (b) Show = x E XÒÓ = x = = provie. (c) Show T [ 3 x 3 3x 3 x provie 3. () Show E[ 3 ÒÓ xx 3x 3 x 3

5 provie 3 an Ÿ. A Unifie Derivation of Occupancy an Sequential Occupancy Distributions, Ch. A. Charalambies, Avances in Combinatorial Methos an Applications to Probability an Statistics, pages # 9# 3. Problem 2. Suppose balls are inepenently istribute into istinguishable boxes an that a ball is equally likely to lan in any of the boxes. Let, R the number of boxes containing balls, ßá ß. (a) Show TR ßáßR + ß áß+ x x x x âx + x+ x â+ x where ( i ) + for ŸŸ an + for all ( ii ) + á + the number of boxes, an ( iii) + á + the number of balls. Note that is not necessary to specify whether the balls are istinguishable or not because the require probability woul be the same in either case. (b) Show that the number of ways to istribute istinguishable balls into istinguishable boxes so that exactly + boxes contain exactly balls equals min, x x x provie Ÿ+ Ÿ min,. The special case is consiere in, A recursive solution to an occupancy problem resulting from TDM raio communication application, JyhHorng Wen, JeeWey Wang, Applie Mathematics an Computation, Vol. 101, 1999, no. 1, pages 13. They o not fin a close form solution but rather erive a recurrence relation for the case.

6 (c) Show that TR Wß provie max, ŸŸ an where Wß is the Stirling Number of the Secon Kin. Problem 3. Special cases illustrating Theorem 3. (1) Coupon Collector's Problems (nee references) ; ;# á ; ÒÓ E ÐÑ [ x Ð : ÀU : ) á : : : á : : an ; ; á ; # # ÒÓ E[ x ÐÑ ÀU : : which agrees with Charalambiesß page #, with : ajustments. Furthermore, in this case after appropriate notational 6

7 ÐÑ E[ ÀU : : : (nees proof. relate to sum of inep. but not ientically ist. geometric ranom vars.) : : : : an ; ; ; # á # á # Newman, Donal an Shepp, Lawrence; Double Dixie Cup Problem, American Mathematical Monthly, Vol. 6, 960, pages 6. Holst, Lars; On Birthay, Collector's, Occupancy an Other Classical Urn Problems, International Statistical Review, 96,,, pp. #. (Two chilren collecting cars in cooperation.) ÒÓ E ÐÑ [ À# : â Ð ) : : á á x á : ÐÑ 3 Ð) x (Birthay problem. Waiting for the first person to meet their quota.) ÒÓ : â : á x E[ À; : â xâ x á : á : ; ; ; ; á â x xâ x : : â : á á : : : # á : Îß ; ; # á ; ; an

8 ; ; ÒÓ E[ á x À; : â xâ x á : : # á : Îß ; ; # á ; # an á ÒÓ E[ À# : â á x 3 3 x 3 3 (2) Knock'm Down Benjamin, Arthur an Fluet, Matthew, The Best Way to Knock'm Down, The UMAP Journal, # 0., 999, pages # 0. Mauer, Stephen, The King Chicken Theorems, Mathematics Magazine, Vol 3, 90, pages 60. The game of Knock'm Down is playe by two players, each of whom is given a 6sie ie, tokens, an a car with the numbers # through #. Each player allocates tokens among the eleven numbers on their car. It is permissible to allocate multiple tokens to the same number of a car. The players roll the ice together an each removes one token from their boar on the value equal to the sum of the ice. If a player oes not have any tokens on their boar on the value equal to the sum of ice, that player's boar is unchange uring that turn. Play continues until a player has remove all tokens from their boar. The first player to remove all tokens from their boar is the winner. If both players remove their last token on the same roll, then the game is a raw. How shoul a player allocate their tokens? Strategy. Determine the allocation which has the smallest expecte number of turns to remove all tokens. Strategy. Determine the allocation (if it exists) which wins at least as frequently # as it loses no matter what allocation your opponent has chosen.

9 As it will be seen these two strategies can lea to ifferent conclusions. First consier Strategy. We will illustrate our notation for the allocation (quotas) through two examples. If $ an we allocate one of the three tokens to the number 6 an the remaining two tokens to the number, then we will efine #, ; an ;# #, an & 6 : Tsum of two ice 6, an : Tsum of two ice. $' # If $ an we allocate one of the three tokens to the number 6, one to the number an one to the number, then we will efine $, ;, ; #, an ; $, an & 6 : Tsum of two ice 6 $', :# Tsum of two ice $', : $ Tsum of two ice. In this way for any allocation U we have $' E[ : expecte number of turns to remove all tokens (satisfy all quotas) ÀU ; ; â Ð ) á x xâ x : : â : á á : Strategy reuces to the task of etermining an allocation Uwhich minimizes the above expectation. Exercises..) Write a computer program which accepts the number of tokens you choose to allocate to each number # through # as input an returns E as output. $' [ ÀU : #.) Verify that on average it takes.6 turns to remove four tokens if one token is allocate to each of the numbers five through eight. Verify that this is optimal in the sense of Strategy. Be sure to exploit symmetry an intuitive conitions given in Benjamin an Fluet's paper to reuce the search space. 9

10 Proof (Theorem 1) T] á ] T ] =,] =, á,] = Thus, # # / = á= x : : # â : x =x=xâ=x x / = : : x =x=xâ=x / x T ] ß á, ] Ñ T] á ] T ] ß á, ] T an ] á ] T ] á] T ] ß á, ] T T ] á] # # = = = # = = # â: # T / = á = x # = # = x = = x= # xâ= x : : â : e x T / x # = # = x = = x= # xâ= x : : â : e x T x =x=xâ=x # = = = # : : â : # T\, á, \ T ß ß Therefore, T ],], á,] T # 10

11 T ], á,] T ] TY T\, á, \ / ß ß T x 3 3 an / T ],]#, á,] T T\ ß, á, \ ß T x It follows that / T ],]#, á,] T T\ ß, á, \ ß T x \ \ T, á, T ß ß x T\, á, \ T I ß ß T\, á, \ T ß ß Thus for ß T á \ß,,\ ß T / T ],]#, á,] T 11

12 Proof (Theorem 2) E [ E[ [ â[ = B ÒÓ B B B AA âa T [ A A A T [ A A ÒÓ B A T \, á,\ T ÒÓ =ßA A. A.) ) T \, á,\ T=ßA A ). A )., á, T=ßA ) A ) T \ \ B Therefore, = E = [ B ÒÓ =. A ) T \, á,\ TßA =x.) =x = = A ) =. A ) T \, á,\ T=ßA.) =x A =. A ) T ], á,] TA/.) A ) ) It follows that,.. A E B ÒÓ ), á, T A..) A [ T ] ] /.. / T ], á,] T )..) A A A ) ) 12

13 Proof (Theorem 3) Within the proof of the Poisson Ranomization Theorem, we showe It follows that T ] ] ] /,, á, T \,\, á, \. # # TR TR U T ÀU nx / nx. However, [ ÀÀU Í RÀU Therefore, T [ TR ÀÀU ÀU an TR T [ U T ÀÀU / x. Thus, T T R U / T [ ÀÀU. T [ / ÀÀU x x T [ ÀÀU x T [ Ð Ñ ÀÀU x ÒÓ 13

14 T [ ÀÀU ÒÓ E [ ÀÀU ÒÓ Ðsee Problem??? Ñ. But [ [ ÀÀU ÒÓ ÒÓ ÀU : Thus T ÒÓ T R U [ ÀU E : an ÒÓ T ÀU : U E[ TR Now let E 3 be the event that ] 3 ; 3. From the General Probability Theorem, we have TR U T Tat least of the events E, E, á, E occurs # ÐÑ Ð Ñ Ð ) ÐÑ ÐÑ ÐÑ Ð ) where Ð ß á, Ñ TE â E Ÿ Ÿ 1

15 an where is the set of all samples of size rawn without replacement from Ö á,,, when the orer of sampling is consiere unimportant. In the Poisson moel, Also, as the / TE T ] ; 's are inepenently, ] 3 TE â E TE TE â TE # / â / â ; : : x ; ; : : : / : â x x ; ; : á: á : â : xâ x Therefore, ; ; : á: á / : â : â Ð ß Ñ xâ á, & x ; / â ; : á: á : â : xâ x an TR U T â ; ; : á: á / : â : xâ x ; ; : á: á / : â : â xâ x Ð) ÐÑ 1

16 an ÒÓ T ÀU : U E[ TR Ð ) ÐÑ ; ; : á: á : â : / â ; ; ÐÑ j âð ) : â : xâ x ; ; / xâ ÐÑ â Ð ) : â : xâ x : x : á: á á x á á : 16

17 Solutions Problem 1(a) 2 Let \ equal the number of balls that go into the urn. Then \, á,\ is a multinomial ranom vector. That is, for all ( =, á, = ), T \ =, á, \ = x =xâ= = = â x x =xâ= x Therefore we can use Theorem 1 with T efine as that subset of where exactly of ], á, ] are greater than an the ranom variables ], á, ] can be anything. Therefore, / / T (], á,] ) T x x an / / / / / / 1

18 /T(], á,] ) T / / Therefore, / / TX T( \, á,\ ) T / T( ], á,] ) T / / ÒÓ x which agrees with Charlambies, page # 6, an with Barton an Davi, Contagious Occupancy, Journal of the Royal Statistical Society, Series B, #, # Problem 1(b). From Theorem 2 1

19 .. E XÒÓ / T ] ].., á, T) ).. /..) ).. ).... ) ). / ).) / / )..) 6 6 ) ) ) ).. 6 ) / )..) ) / )..).. /..) ) ) ) ) ) We have simplifie the initial expression in such a way so as to simplify the process of taking the erivatives with respect to ) an an evaluating the resulting expression at ) an Þ In particular, 19

20 EX / ÒÓ ) ). 3.) 3.)3 3 ). x /. xiö Ð3Ñ IÖ3ß á ÐÑ 3 3x 3. x IÖß IÖ. á x. x. I Öß á IÖ x x x IÖß á ÐÑIÖ x x x IÖ x = = x = = x IÖ = = x = x = IÖ x = = / x ÐÑ / x ÐÑ = We note that there is a misprint in Charlambies, page # 6, where he has omitte a constant factor of x (in his notation) in his statement of the above result. Problem 1(c). Define T to be that subset of where exactly of ], á, ] equal 3 an the ranom variables ],,] can be anything. Hence, á / / T (], á,] ) T 3x 3x / 33 3x 3x / / an x 33

21 /T(], á,] ) T / 3 3x 3 Therefore, T [ 3 T( \, á,\ ) T / T( ], á,] ) T / 3 3x / 3 3x 3 3 However for integers + an, +, / + x + Ößß á + + Öß ß á ß +, +x It follows that, T [ / x 3 x 3 I 3x 3x 3 x I 3x 3 x 3 x 3x 3 x provie 3. Ö 3 Ö 3 21

22 This results agrees with Charlambies, page # 6. Problem 1()... E [ 3 ÒÓ / T ], á,] T)..) ).. / 3 )..) 3 3x ).. / 3 )..) 3 3x ).. 3 / )..) 3 3x ).. / 3 )..) 3 3x ).. 3 /..) 3 ) 3x.. 3 / )..) 3 3x ). 3. / 3 ). 3x.) ). 3 3 / x Ößß ß. I á 3x 3 x x 3 Ö 3 Ößß ß 3x I I á 3 x 3 xx 3x 3 x provie 3 an Ÿ. 22 )

23 This results agrees with Charlambies, page 26 an with Buoncristiani, Cerosoli (???) who consiere the special case. 23

24 Solution Ð2aÑ Let \ equal the number of balls that lan in box, #á,,,. Define to be the prouct space á,, â á,, an let be the set of all vectors,, á, in such that á. # # Let be that set of all,, á, such that Then, # () i + of the values in the vector,#, á, equal, #á,,,, an ( ii ),, á,. # TR, R, á, R +, +, á, + T\,\, á, \ # where \,\#, á,\ is a multinomial ranom vector with equal cell probabilities. That is, x # T (\,\# #, á, \ ) x xâ x â. # Now efine be that set of all,#, á, such that () i + of the values in the vector,#, á, equal, #á,,,, an ( ii ),, á,. # By the Poisson Ranomization Theorem we have, T \ \ \,, á, / T],], á,] # # where ],]#, á,] are inepenent, ientically istribute Poisson ranom variables with parameter. Now efine, an T R the number of ] 3's that equal for #á,,, T 3 R the number of ] 's. Then, T T],]#, á,] T T T T R, R, á, R, R +, +, á, +, 2

25 where the vector T T T T R,,,, R á R R follows a multinomial istribution with parameters ),),,), an ), with á / ) x ) T ] an T ]. Thus, T T T T T R R R R +,, á,,, +, á, +, x x+ xâ xx ) ) â ) + ) + x / + x+ xâ+ xx x + x + x + x + â x + x+ xâ+ x á á / x + x x â x + x+ xâ+ x + + /. Therefore, T \,\, á,\ / T ],], á,] # # x x x â x + x+ xâ+ x / + + / + x + + x + x â x + x+ xâ+ x x x + x + + x â x x xâ x Solution Ð2bÑ We can procee exactly as in ÐÑ a except now efine to be the set of all,#, á, such that ( i ) + of the values in the vector,#, á, equal (ii),, á, # 2

26 an efine to be the set of all,#, á, such that ( i ) + of the values in the vector,#, á, equal (ii),, á,. # Then, TR + T \,\, á,\ # an by the Poisson Ranomization Theorem we have, # # T \,\, á,\ T ],], á,]. However T T T],], á,] TR + # where R the number of ] 3's that equal an R follows a binomial istribution with parameter, with ) T Thus, T T R + / ) x T ]. + + x / + x + x x / x + x / + + / + x + x x x / x Hence,. T T \,\, á,\ TR + # 26

27 / x + + x x + x + By a change of variable, this solution can also be expresse in the form : x x x. Note : / x. x Solution Ð2cÑ For Ÿ+ Ÿ TR Ð+ Ñ + x + + x x x + 2

28 W, +. + Taking +, we have, TR Wß. 2

29 From the above it is clear that the two probability moels are relate. However the critical ifference shoul be stresse. \,\#, á,\ is a vector of epenent ranom variables while ],]#, á,] is a vector of inepenent ranom variables. It is the purpose of the Poisson Ranomization Theorem given below to restructure problems involving the multinomial moel into equivalent problems involving the Poisson moel. By so oing, we will be able to exploit the inepenence of the 's. ] 3 29

30 Theorem Suppose an experiment consists of inepenent trials an that every trial has istinct outcomes. Let : equal the probability of outcome on any trial an let G equal the number of times outcome occurs in these (multinomial) trials. Then E G G, á, G / EG^, á, ^ where ^ßáß^ are inepenent Poisson ranom variables such that ^ :,, á,. That is Theorem T^ D / 3 3 : 3 : D x 3 3 D 3 D3 ß ß á. Let H equal the number of G's which equal, ßßáß. Then has parameter H H á H an H H á H an EG HßHß.. EG..) / )/ á ß H ß ß ß á ^ß^ß á ) x where ^ß^ßáare inepenent an ^ µ Poisson ßßá ) Proof ) x Let ^ µ Poisson ß ß á T^ Dß^ Dßá ) ) / á x x ) D ) D x â x D xd xâ D D â )/ D D á D D á DxDxâ x x / ) T^ Dß^ Dßá ±D D á an D D á 30

31 D D x â x D xd xâ D D x â x â D xd xâ D ßD ßá D D á D D á D D â x x D D á D xd xâ D D â x x â D ßD ßá D D á D xd xâ D D á D D á D D â x x D xd xâ D D â x x â D xd xâ D ßD ßá D D á D D á x D x D âd x D xâ â x D x D âd x D xâ D ßD ßá D D á D D á provie D D á an D D á x x x D x D âd xd xâ x x â x D x D âd xd xâ D ßD ßá D D á D D á provie D D á an D D á x x D D x x âd xd xâ provie D D á an D D á 31

32 E G ^ß^ßá EG^ß^ß á ± T ^^á EGH ßH ßá ß H ß ß ß á T ^ ^ á ^ ^ á ^ ^ á ^ ^ á ^ ^ á )/ H ßH ß / E G á ß H ß ß ß á ) x x....) )/ / EG ^ ß^ ß á ) H ßH ß.. E G á ß H ß ß ß á ) x x..) ) EG H ßH ßá ß H ß ß ß á EG HßHß.. EG..) / )/ á ß H ß ß ß á ^ß^ß á ) T ^^á ^ ^ á )/ / â x x ) D D â ) / ) x x x x D xd xâ / x x D ßD ßá D D á D D á 32

33 )/ / EG ^ ß^ ß á H ßH ß E G á ß H ß ß ß á ) x x 33

34 Suppose ientical balls are inepenently istribute into equally likely boxes ( i.e. a multinomial moel or classical allocation scheme) until any of the boxes have at least balls each. Let Q enote the number of boxes containing exactly balls when the istribution of balls into boxes stops. Let ^µpoisson ). For TQ = x xx=x=x T^ Ÿ = = ) T^ T^ ) /.) = ß á ß an equals else. In the case EQ x xxx T^ Ÿ ) T^ T^ ) /.). To illustrate we note the special case simplifies to = = TQ = 3 3 =3 x = 3x x=x= x = an EQ xx x x. Holst [0], [0] consiers the special case an aitionally assumes but the form of the solutions given in these papers is complicate. Proof 3

35 For TQ = T 2 Q = an box is last box to receive a ball TQ = an = box is last box to receive a ball TT = box is last box to receive a ball where event T is T : exactly of boxes 2 through have at least balls an exactly = of boxes 2 through have exactly balls. State in this way we can view the problem as asking for the probability of event T = where the stopping rule is to stop when the box has exactly balls. The purpose of casting the problem in this way is to make Theorem 0.0. appropriate. Let G equal the number of balls istribute into box ß ß á ß, before the trials stop. Then efine GG G, á, exactly of G, á, G are an exactly = of G, á, G equal else. It follows from Theorem 0.0. that TQ = exactly of ^, á, ^ are ) T ) /.) x an exactly = of ^, á, ^ equal where ^ßáß^ are ii Poisson ). The final form follows on recognizing that exactly of ^, á, ^ are at least an exactly = of ^, á, ^ equal if an only if exactly of ^, á, ^ are less than, exactly = of ^, á, ^ equal, an exactly =, á, ^ excee. of ^ The factorial moment result follows from efinition. We have EQ = TQ = = 3

36 x x x T ^ Ÿ / ) ) T ^ T ^. = x= x = = ) = an the final form follows on applying the binomial theorem to remove the sum. x xxx T^ Ÿ ) T^ T^ ) /.). x = x x=x= x ) T^ Ÿ T^ x T^ = ) /.) = xx=x= xt^ Ÿ T^ T^ = ) ) /.) an EQ = TQ = = = x xx=x= xt^ Ÿ = = = ) T^ T^ ) /.) 36

37 x T^ Ÿ = x x = x = x = = ) T^ T^ ) /.) x ) xxt^ Ÿ ) /.) T ^ T ^ = x= x = = = x ) xxt^ Ÿ ) /.) x T^ T^ x = x=x = = = x ) xxt^ Ÿ ) /.) T^ T ^ x x xxx T^ Ÿ T^ ) T^ ) /.) For an = ßßáß 3

38 TQ = x T ^ Ÿ x = x=x x T ^ = ) T^ T^ ) /.) = an EQ x x xx T^ Ÿ ) T^ T^ ) /.). For an = ßáß TQ = x xx=x=x T^ Ÿ = = ) T^ T^ T^ ) /.) an EQ x xxx T^ Ÿ ) T^ T^ ) /.). Case 1. TQ = T 2 Q = an box is last box to receive a ball before stopping TQ = an = box is last box to receive a ball before stopping exactly of boxes 2 through have balls exactly = of boxes 2 through have exactly balls T an exactly of ^, á, ^ x T # exactly = of of ^ ^ ), á, equal /. ) x # 3 = ) xx=x=x T^ Ÿ T^ box is last box to receive a ball before stopping =

39 = ) T^ T^ ) /.) an EQ = TQ = = = x x= x=x xt^ Ÿ T^ = = = ) T^ T^ ) /.) x T^ Ÿ T^ = x = x = x x = = ) T^ T^ ) /.) x ) xx T^ ) /.) x = = = = T^ Ÿ T^ T^ x ) xx T^ ) /.) x T^ T^ Ÿ x x x xt^ Ÿ T^ T^ ) ) /.) Case 3. TQ = T 2 Q = an box is last box to receive a ball before stopping TQ = an = box is last box to receive a ball before stopping 39

40 exactly of boxes 2 through have balls exactly = of boxes 2 through have exactly balls T an = box is last box to receive a ball before stopping x = xx=x= xt^ Ÿ T^ T^ T^ = ) ) /.) an EQ = TQ = = = x xx=x= xt^ Ÿ = = = ) T^ T^ T^ ) /.) x T^ Ÿ = x x = x = x = = ) T^ T^ T^ ) /.) x ) xxt^ Ÿ ) /.) x T^ T^ T^ x = x=x = = = x ) xxt^ Ÿ ) /.) T^ T ^ x x xxx T^ Ÿ T^ ) T^ ) /.) 0

41 Example 0.0. A company hires people. It eclares a holiay on the birthay of any employee. What value of maximizes the expecte work force per year? This question was pose in Chance, Vol. 13, No., Fall 2000, page. of open Assume that people's birthays are inepenently etermine an that every ay the year is equally likely to occur as a birthay. Let a year consist of ays where equals the number of potential work ays (ays the company will be if no employee has a birthay on that ay) an equals the number of nonpotential work ays (ays the company woul not be open even if no employee has a birthay on that ay, e.g. weekens, July th, etc.). Let X equal the number of potential work ays where the company eclares a holiay because one or more of the employees has a birthay on that ay. Let [ equal the work force per year if the company has employees. We will efine work force per year as the prouct of the work force per ay (number of employees) an the number of ays the company is open per year. That is, [ X Show that E[ It an that E[ is maximize at, the total number of ays in a year. is interesting that the solution epens on an only through their sum. Proof Clearly, E[ EX EX Furthermore, from part (b) of this problem with (see Poisson Ranomization problem 1(b)) we have 1

42 x = EX = x = = Thus, E[ A little calculus shows that E[, the expecte work force per year, achieves its maximum at both an. Can also easily solve this problem by letting some employee has a birthay on Day I else Also let Then : probability a person is born on Day for any a potential workay E E X I : 2

43 Suppose we perform multinomial trials with istinct possible outcomes. We will assume that of the outcomes have been classifie as type E outcomes an that the remaining outcomes as type F outcomes. We will assume that all type E outcomes occur with constant probability : an that all type F outcomes occur with constant probability : so that : :. # # Let G equal the number of type E outcomes which occur exactly times in the trials an let H equal the number of type F outcomes which occur exactly times in the trials, ß ß á ß. Then, an x TG 3 : : 3x 3 x 3 3 EG âg H âh α α x x x E E E E αx xe E x x â x # # α α : : : α : α α α where α α á α á Eα α á α E á The first result is given in Charalambies [0]. Davi an Barton [0] give the first result for the special case 3. Charalambies [0] gives the secon result for the special case α, α for 3Áan for all. We note that \ by efinition. 3 ÐÑ Proof For the first result let equal the number of balls in urn an let Z 3

44 GZ Z exactly of Z ßáßZ equal 3, á, else. By Theorem TG EG Z, á, Z 3 / Texactly of ^ ßáß^ equal 3 / T^ 3 T^ 3 where ^µ Poisson :. The final form follows on substituting for T^3 an extracting the coefficient of. For the joint factorial moment result, we have by Theorem 0.0. EG âg H âh α α... )/ / : / : # E[ â [ ^ â^..). α α ).....). : : )/ / # / E[ â E[ E^ âe^ α α using the inepenence of all the ranom variables [ß[ßâß^ß^ßâ where. 3 3 ) : : 3 3x 3 3x # [ µ Poisson an ^ µ Poisson. ) + However E\ + $ for \ µ Poisson $ an the final form follows on substituting an extracting the coefficient of )

45 ... : : )/ / ) : α ) : α : :..). x x x x / # â # â # α á α á : : x â x # α α ).....). : : )/ / α áα á α á α # / ) ) α á α á : : x â x # α α : :.. / / # ) x..) ) α α α α á á á á x / ) α á α á : : x x â x á x # α α.. : / ) # /..) / : ) α á α α á α á ) α á α á : : x x â x á x # α α : α α : / x. α α á á á α áα x. # / α á α á : : x x x â x á x α áα x # α α. :# / : /. α á α á α áα α á α á : : x x x â x á x α áα x # α α

46 . : : # α á α / α á α á. α á α á : : x x x â x á x α áα x # α α α á α : α áα : # á x α á α x x x x : : x â x α α x α α x # á á x á α α : α á α : á # α á α á α á α +, +x + x +, + Ößß + / á + Öß ß á ß ) : : x x # where [ µ Poisson, ^ µ Poisson #, : : an where all ranom variables are inepenent. Proof Let G equal the number of balls in urn an let G G exactly of G ßáßG equal á 3 else. G,, an G, á, G IG 3áI G 3. G # By Theorem

47 T [ EG G, á, G 3 / Texactly of ^ ßáß^ equal 3 / T^ 3 T^ : : : : / / / 3x 3x : 3 / 3 : 3x x : : 3 3 3x 3x an E[ 3 EG # G, á, G / EI^ 3 á I ^ 3 / x T^ 3 where ^µ Poisson. The final form for both problems follows on substituting for T^ 3 an extracting the coefficient of.