Random partition of a set 220 A partition of the set i is a collection of disjoint nonempty sets whose union equals i. e.g. The & partitions of the set i ß#ß$ are ß #ß $ ß # ß $ ß $ ß # #ß$ ß ß # ß $ We refer to each nonempty subset making up a partition of the set i as a block. For any partition of the set i we refer to the number of blocks that have cardinality as the multiplicity of in that partition. It is well known that, the Bell Number, counts the total number of partitions of a set of elements and that Wß, the Stirling Number of the Second Kind, counts the total number of partition of a set of elements into exactly blocks. The article A Review of the Stirling Numbers, Their Generalizations and Statistical Applications, Charalambides, Ch. A.; Singh, Jagbir; Communications in Statistics, Theory and Methods, Vol. 7, No., 9, pages 233--29, is an excellent resource on both Bell and Stirling Numbers. 2 Let i be any set of (distinct) elements. If a partition of the set i is selected uniformly at random from the set of all partitions of i, we will refer to this as a random partition of i. If a partition of i is selected uniformly at random from the set of all Wß partitions of i with blocks, we will refer to this as a random partition of i with blocks. We will need the following definitions. : the infinite product space 0,, á 0,, á â : the set of all vectors s,s, á in such that = #= á # # ß : the set of all vectors s,s, # á in such that = #= # á = = á # ß ß For any T define T T and T T We note that the condition that = #= # á implies that =! for all Þ Hence all vectors in T and Tß are of the form +, á, + ß!ß!ß á Þ
For all T Á!ß!ßá, let be the collection of -dimensional vectors formed by taking each infinite-dimensional vector in and truncating after +ÞSo for example, T +, á, + ß!ß!ß á +, á, + Define ß similarly. For notational consistency it is necessary to separate out the case T!ß!ßá. Let \ equal the multiplicity of in a random partition of a set of elements into blocks. We can construct the set of all partitions of a set of elements into blocks such that \ B ßáß\ B in the following manner. Take any one of the x permutations of the elements and use the first B elements of that permutation to fill the first B blocks, use the next #B# elements of that permutation to fill the next B # blocks, and so on. In total we would use the B #B á B elements to fill the B B á B blocks. # # This yields x set partitions but not all of these set partitions are distinct. In particular this count would assume that rearranging the B blocks of cardinality amongst themselves leads to distinct set partitions - which they do not. Furthermore this count would assume that rearranging elements within a block leads to distinct set partitions - which they do not. Therefore it is necessary to divide the count of x by the number of ways to arrange the B blocks of cardinality amongst themselves ß á ß and by the number of ways to arrange the elements in each block. It follows that there are x â BxâBx x x B B partitions of a set of elements into block s such that \ B ßáß\ B provided B á B, B á B, and B!ßáß a and T \ ß ß\ B ß ßB! x B B B xâbx x â x Wß B áb B áb B!ßßáßa otherwise 2
If we let [ equal the multiplicity of in a random partition of a set of elements then it follows similarly that T [ ß ß[ A ß ßA! x A A A xâax x â x A áa A!ßßáßa otherwise Theorem... E \ ß ß\ E ) - ] ß] # ß xw ß.-.) )! where + ß+ # ßá is any function and + ßáß+ + ßáß+ ß!ß!ßáand for T Á!ß!ßá.. T \ß ß\ ß T ]ß]ß xwß ) - # á T.-.) )! where that ]ß]ßá # is an infinite sequence of independent Poisson random variables such )- )- C x x T ] C C!ßß#ßáand ß#ßá è Cx Theorem 2.. E [ ß ß[ E ] ß] # ß.- - -! where + ß+ # ßá is any function and + ßáß+ + ßáß+ ß!ß!ßáand for T Á!ß!ßá. - T [ ß ß[ T ] ß] # ßá T.- 3
where that ]ß]ßá # is an infinite sequence of independent Poisson random variables such - - C x x T ] C C!ßß#ßá and ß#ßá è Cx Now suppose we randomly distribute distinct objects into distinct urns such that objects are distributed independently and are equally likely to go into any of the urns. If one or more urns are still empty after this distribution we empty all of the urns and start the distribution process over. The distribution process stops as soon as we obtain a distribution of the objects where each of the urns has at least one object. Let Yß ßáß equal the number of objects in the urn after the distribution process has stopped. Then, 2 TYß ßY?ß ß?! x?xâ?xx Wß? á?? ß#ßá a else Let Zß ßáß equal the number of (labeled) urns containing objects after the distribution process has stopped. Then, TZ ß ßZ ß ß! x xâ x x â x Wß á á!ßßáß a otherwise We note that the above probability distribution is based on a model where the urns are distinguishable (labeled) but clearly the latter probability would be the same if the urns were not labeled. However distributing distinguishable objects into like (unlabeled) urns is equivalent to partitioning a set of objects into blocks. That is TZ ß ßZ T \ ß ß\ ß ß where as before the \ ßßáß equal the number of blocks with elements in a random partition of a set of elements into blocks.
The distinction between the two probabilities is that in the case of labeled urns there are Wß possible distributions and in the case of unlabeled urns there are only Wß possible distributions. We will need the following definitions. À the -dimensional product space,#á, â,#á, À the set of all vectors?, áß? in such that? á? ß Define h as that set such that Z ß ßZ Í Y ß ßY h. Theorem 3. d T \ ß ß\ - T ß ], á,] h xwß d- where ]ßáß] are iid zero-truncated Poisson random variables with parameter -. That is, - C - C ß #ß á Cx - T ] C è! else
Applications Problem. (a) Show that the number of partitions of a set of elements into blocks containing exactly blocks of cardinality equals minß x x x W ß (b) Show that the number of partitions of a set of elements containing exactly blocks of cardinality equals x x x provided Ÿ Þ (c) Show that the number of partitions of a set of elements containing at least blocks of cardinality equals x xx provided Ÿ. We note that the special case of (c) where is in agreement with Haigh, J., Random Equivalence Relations, Journal of Combinatorial Theory, Series A, Vol. 3, 972, pages 27-29. Problem 2. 2 (a) Show that the factorial moment of the number of blocks containing elements in a random partition of a set of elements into blocks equals 6
x Wß provided Ÿ and x Wß We note that the special case where is in agreement with Proposition 2. in Recski, A., On Random Permutations, Discrete Mathematics, 6, 976, 73-77. (b) As a check on the result in (a) verify that Wß. Wß 2 (c) Show that the factorial moment of the number of blocks containing elements in a random partition of a set of elements equals x provided Ÿ x We note that the special case where is in agreement with Proposition. in Recski, A., On Random Permutations, Discrete Mathematics, 6, 976, 73-77. Problem 3. (a) Show that the expected number of multiplicities in a random partition of a set of elements into blocks that equal is minß x W ß x x Wß (b) Show that the expected number of multiplicities in a random partition of a set of elements that equal is minß x W ß x x 7
Problem. Show that the expected number of blocks in a random partition of a set with elements equals in agreement with L. H. Harper, Stirling Behavior is Asymptotically Normal, Annals of Mathematical Statistics, Vol 3, 967, pages 0-Þ Problem. (a) Suppose that a partition is picked uniformly at random from the set of Wß partitions of a set of elements into exactly blocks and from that partition a block is then picked uniformly at random. Show that the expected size of this block is (b) Suppose that a partition is picked uniformly at random from the set of partitions of a set of elements and from that partition a block is then picked uniformly at random. Show that the expected size of this block is W ß (c) Suppose that a block is picked uniformly at random from the set of all W ß blocks in the partitions of a set of elements. Show that the expected size of this block is (d) Suppose that a partition is picked uniformly at random from the set of Wß partitions of a set of elements into exactly blocks and from that partition an element is selected uniformly at random. Show that the expected size of the block containing the selected element is
Wß Wß (e) Suppose that a partition is picked uniformly at random from the set of partitions of a set of elements and from that partition an element is selected uniformly at random. Show that the expected size of the block containing the selected element is Problem 6. Show that the number of partitions of a set of elements into blocks where exactly < of the blocks have an even number of elements is #< < #< < # x < #!! The special case < (all blocks have an even number of elements) simplifies to # # x #! and the special case <! (all blocks have an odd number of elements) to # x #! in agreement with (2.2 and 2.2) of Set Partitions, L. Carlitz, Fibonacci Quarterly, Nov. 976, pages 327 32, apart from an obvious typo in equation 2.2. Problem 7. Suppose that a partition is picked uniformly at random from the set of 9
partitions of a set of elements and from that partition a block size is picked uniformly at random from the set of all block sizes in that partition. Show that the expected size of the selected block size is 0
Proof (Theorem ) Consider an infinite sequence ] ß] ßá # - of independent Poisson random variables where - C x x T ] C C!ß ß #ß á and ß #ß á Cx Let C be a nonnegative integer for ß #ß âþ Then T] C ß] C ßá # # x C - C - - - C x C x C x x C and T ] ât] C ß] # C# ßá C ßC# ßá C C - x - âx x C ßC# ßá C x C - - x It follows that for T Á!ß!ßá
T] ß] # ßá T T] ß] # ßá T ± ] T ] Therefore!! T - - C x - x - x C - x C C# C - C xc xâc x x #x â x # - x - - - [ [ T ß ß x x - # T] ß] ßá T T[ ß ß[ - x and <..- < - T] ß] # ßá T T[ ß ß[ <xiö< x Therefore, for T Á!ß!ßá. T[ ß ß[ T] ß] # ßá T.- - 2
Proof (Theorem 2) Consider an infinite sequence ]ß]ßá # )- of independent Poisson random variables where )- C x x T ] C C!ßß#ßá and ß#ßá Cx Let C be a nonnegative integer for ß#ßâÞ Then T ] Cß] Cßá # # C C )- ) - x C C x C C - ) ) - C x x x C and T ] and ] ât ] Cß] # Cßá # C ßC# ßá C C ) - )- x C x âx C ßC# ßá C x C C - )- Wß x ) It follows that 3
T ] ß] # ßá T T ]ß]ß # á T ± ] ß ] T ] ß ]!!!! T ß ß ) - )- )- C x Wß Cx ) - ) - x # x C C CxCxâCx x #x â C x # Wß x ) - )- Wß ) - e )- T \ ß ß\ ß Wß x x )- Wß Therefore ) - Wß T ] ß] # ßá T T \ ß ß\ ß ) - x and < =...-.) < = ) - T ] ß] # ßá T Wß T \ ß ß\ ß <x=xi < I = x )! Therefore, for T Á!ß!ßá.. T \ß ß\ ß T ]ß]ß xwß ) - # á T.-.) )!
Proof (Theorem 3) T \ ß ß\ ß TZ ß ßZ ß TYß ßY h But by the Zero-Truncated Poisson Randomization Theorem, d TY ß ßY h T ] ] -, á, h xwß d- where ]ßáß] are iid zero-truncated Poisson random variables with parameter -. That is, - C - C ß #ß á Cx - T ] C! else
Solutions Problem (a) This problem can be handled by a direct application of the inclusion-exclusion principle but we will apply Theorem to illustrate its use. Clearly the solution equals WßT\. Now define ß +, + #, á, + + á + ß + á + and + Tß +, + #, á, + ß!ß!ßá + á + ß+ á + and + T +, +, á, + ß+ ß+ ßá + # Then by Theorem, # T \ T \ß ß\ ß.. á ) - T ] ß] ß xw ß.. # T - ) where T ] ß] # ßá T T ] 3 0,, á T ] 3Á T ] )- x )- )! Therefore, T \.. )- )- ) xwß.-.) - x.. ) xw ß x. -.) )- - - )!. Wß xx.- - - - )! 6
. - - Wß xx.- -!. x W 3ß 3 - Wß xx.- x 3x - -! 3. x W 3ß 3 Wß xx.- x 3x -- -! 3. x W 3ß 3 Wß xx.- x 3x -! 3 x W 3ß Wß xx x 3x x I3! 3 x W 3ß Wß xx x 3x x I 3! 3 x Wß Wß x x x x x Ißá! minß x Wß x Wß x x x x! Therefore the number of partitions of the set of elements into blocks which contain exactly blocks of cardinality is minß x x x W ß Problem (b). -.- # T [ ß ß[ T ] ß] ßá T 7
Clearly the solution equals T [. Now define +, + #, á, + + á + and + T +, + #, á, + ß!ß!ßá + á + and + T +, +, á, + ß+ ß+ ßá + # Then by Theorem 2, # T [ T [ß ß[. - T ] ß] ßá. # T - where T ] ß] # ßá T T ] 3 0,, á T ] 3Á T ] - x - Therefore,. - T [.- x. x.- - - - - -. 3 3 - - x. 3xx 3!! 3 x I3 x 3xx 3!! 3 x I 3 x 3xx! 3!
x! xx x I!ßßá xx x x I!ßßá! provided!! x x x x Therefore the number of partitions of the set of elements which contain exactly blocks of cardinality is x x x provided! Problem (c) By Theorem 3, the number of partitions of a set of elements into blocks containing at least blocks of cardinality equals where x d d- - T ], á,] h h? ßáß? ± at least of the values? ßáß? equal and ] ßáß] are iid zero-truncated Poisson random variables with parameter -. That is, T ] C! - C - - Cx C ß #ß á else 9
Summing this result over all yields the total number of partitions of a set of elements containing at least blocks of cardinality. However as an application of the Zero-Truncated Poisson Randomization Theorem we showed that the number of ways that distinguishable balls can be distributed among distinguishable urns such that at least urns will contain balls and no urn is left empty equals d d - - T ], á,] h minß x x Wß xx Dividing this result by x and summing this result over all possible values of gives minß x Wß xx x Wß xx x Wß xx x xx Problem 2(a) It is well known and easy to verify that the factorial moment of a Poisson random variable with parameter 7 is just 7. Therefore it follows from Theorem that if we define ] µ Poisson )- then 2 20
E\ \ â \.. ) xwß.-.) E] ] â ] ) )-.. xwß.-.) - )! - )! xwß x.- ßßá x. - - I Wß.- x. - - I ßßá. - Wß - Ißßá Wß. x xwß x I Ißßá Wß x x Wß Wß ßßá ßßá I I x Wß provided Ÿ and. x Wß Problem 2(b) As a check on our work we note that \ á \ and hence E\ á\ E\ which can serve as a check on the above result for E Wß Wß Wß Wß \. We note that 2
Wß! W ß Wß W ß Wß W ß Wß Wß W ß W ß Wß W ß Þ In this proof we used two well known recurrence relations for the Stirling Numbers of the Second Kind. Namely, and Wß W <ß < < W ß Wß Wß Problem 2(c) By Theorem 2, if we define ] µ Poisson - then E[ [ â [..- ] ] â ] - E. - -.-. - -. x! x x I! 22
Ÿ x x provided Problem 3(a) We have defined the random variable \ as the multiplicity of in a random partition of a set of elements into blocks. It follows that I \ is the number of multiplicities that equal in a random partition of a set of elements into blocks. The problem asks for the expected value of this sum. However and from Problem (a) E I \ T \ T \ minß x W ß x x Wß Hence the expected number of multiplicities equaling in a random partition of a set of elements into blocks is x W ß x x Wß minß Problem 3(b) We have defined the random variable [ as the multiplicity of in a random partition of a set of elements. It follows that I [ is the number of multiplicities that equal in a random partition of a set of elements. The problem asks for the expected value of this sum. However, if we let X equal the number of blocks in a random partition, then 23
EI [ EEI [ X E I \ TX minß x x x Wß Problem Let X equal the number of blocks in a random partition of a set with elements. EX TX Wß Wß Wß Wß Wß Wß Wß Problem (a) The given two step procedure for selecting a block is equivalent to selecting a block uniformly at random from the set of all Wß blocks making up the Wß partitions of a set of elements into exactly block s. As before, we let \ equal the multiplicity of in a random partition of a set of elements into blocks. Then, 2
TP number of blocks of size among the set of all Wß blocks Wß min ß T\Wß! Wß. E\ Wß Wß It follows that EP TP Wß Wß Wß Wß Wß w W ß! w w W ß Wß. Problem (b) Let X represent the number of blocks in a randomly selected partition. Then EP EEPX EP TX Wß Wß 2
Problem 6 From Theorem 3 the solution will equal where x d d- - T ], á,] h h? ßáß? ± exactly < of the values? ßáß? are even numbers and ] ßáß] are iid zero-truncated Poisson random variables with parameter -. That is, T ] C! - C - - Cx C ß #ß á else The zero-truncated Poisson distribution is a member of the Power Series Family with + C Cx,, - -, and 7. H ence from Theorem PSMM, taking 7#, we find $ and T ] #ß%ßá, -, - #, - # - - - - - - # - - # # # - # and T ] ß$ßá, -, - #, - 26
# - - - - - - # Therefore, and x T ] ] - -, á, h - - # # < - - # # d d- -, á, h T ] ] - - # # # # - - d x d - - < - # - # #< d - - < x < # d- # # - - #< < < d #< < # x < # d - - #< <!! #< < d # - x < #!! d- #< < x < #!! #< < # < < < Problem 7 27
Notes: For using Derive, define W +ß, -23!ß +ß ß! -23!ß,ß ß! W3<63# +ß, because Derive mistakenly defines W3<63#!ß!!Þ d db +B x B + Ißßá x B! and and - Wß- x x - - x! Also read Arratia, R. and Tavare, S. (992). Independent process approximations for random combinatorial structures. Advances in Mathematics.??? 2
2 : Descending Factorial Moment of \ E\ \â\. Suppose \µ Poisson -. Then â T \.! â T \ â x - - - - - - x 29