Expected Value of Statistics on Type-B Permutation Tableaux
Ryan Althoff, Daniel Diethrich, Amanda Lohss, Xin-Dee Low, Emily Wichert
aa r X i v : . [ m a t h . C O ] F e b EXPECTED VALUE OF STATISTICS ON TYPE-BPERMUTATION TABLEAUX
RYAN ALTHOFF, DANIEL DIETHRICH, AMANDA LOHSS, XIN-DEE LOW,AND EMILY WICHERT † Abstract.
Type-B permutation tableaux are combinatorial objects intro-duced by Lam and Williams that have an interesting connection with thepartially asymmetric simple exclusion process (PASEP). In this paper, wecompute the expected value of several statistics on these tableaux. Some ofthese computations are motivated by a similar paper on permutation tableaux.Others are motivated by the PASEP. In particular, we compute the expectednumber of rows, unrestricted rows, diagonal ones, adjacent south steps, andadjacent west steps. Introduction
Permutation tableaux are combinatorial objects introduced in [12] that are in bi-jection with permutations (see [5][Theorem 11]). These tableaux are also connectedto the partially asymmetric simple exclusion process [11], an important model instatistical mechanics. One particular variation of these tableaux are type-B permu-tation tableaux, introduced in [10], which are in bijection with signed permutations(see [3][Theorem 4]).There have been several recent papers on permutation tableaux (see [1], [2], [5],[6], [11], and [12]). In particular, Corteel and Hitczenko calculated the expectedvalue of several statistics on permutation tableaux in [2]. In this paper, we considerthese same statistics on the type-B variation as well as other statistics that areinteresting due to their connection with the PASEP. We will compute the expectedvalue of these statistics using the approach developed in [7] for type-B permutationtableaux which is analogous to the approach used in [2] for permutation tableaux.The paper is organized as follows. In Section 2, we introduce some preliminarydefinitions and terminology. In Section 3, we discuss the relationship between per-mutation tableaux and the partially asymmetric simple exclusion process (PASEP)which is the motivation for several of our results. In Section 4, we will discuss theprobabilistic approach developed in [7] that we will use for our calculations. Theremaining sections include our results on the expected value of statistics on type-Bpermutation tableaux. 2.
Preliminaries A Ferrers diagram is a left aligned sequence of cells with weakly decreasing rows.A Ferrers diagram is said to have half-perimeter n , where n equals the number ofrows plus the number of columns. Each Ferrers diagram has a unique southeastborder with the number of border edges equal to the half-perimeter. We will refer † This research was supported by NSF grant to these edges as steps (south or west) with the first step beginning in the northeastcorner of the diagram. We let S k denote a south step at the k th position and W k denote a west step. ( i ) ( ii ) Figure 1. (i) A Ferrers diagram with half-perimeter 5 and steps W S W W S . (ii) A shifted Ferrers diagram obtained from (i)with the same half-perimeter and border edges.A shifted Ferrers diagram is obtained from a Ferrers diagram by inserting addi-tional rows above. If the Ferrers diagram has k columns then the shifted Ferrersdiagram is the same diagram but with k rows inserted above of lengths k , k − . . . , 1, respectively. The rightmost cells of each added row are called diagonal cells .A shifted Ferrers diagram has the same half-perimeter and border edges as that ofthe original diagram.Permutation tableaux were introduced in [12] as fillings of Ferrers diagrams andtype-B permutation tableaux, introduced in [10], can be defined as fillings of shiftedFerrers diagrams (see [4]). Definition 2.1. A permutation tableau of size n is a Ferrers diagram with half-perimeter n in which each cell contains either a 0 or a 1 and satisfies the followingconditions:(1) Each column must contain at least one 1.(2) A 0 cannot have both a 1 above it in the same column and a 1 to left of itin the same row.1 1 0 11 1 0 10 0 101 ( i ) 10 01 0 10 10 0( ii ) Figure 2. (i) A permutation tableau of size n = 10. (ii) A type-Bpermutation tableau of size n = 6. Definition 2.2. A type-B permutation tableau of size n is a shifted Ferrers diagramwith half-perimeter n , in which each cell contains either a 0 or a 1 and satisfies thefollowing conditions:(1) Each column must contain at least one 1.(2) A 0 cannot have both a 1 above it in the same column and a 1 to left of itin the same row.(3) If a diagonal cell contains a 0, then all of cells in that row must also be 0.An important statistic for our calculations later is that of an unrestricted row .A restricted k th step by U k . For example,consider the type-B permutation tableau in Figure 2(ii). This tableau has threeunrestricted rows on the last step ( U = 3). Prior to that last step, the number ofunrestricted rows are U = 1, U = 2, U = 3, U = 2, and U = 2.In this paper, we will calculate the following statistics on type-B permutationtableaux of size n ,(1) The expected number of rows (Theorem 3).(2) The expected number of unrestricted rows (Theorem 4).(3) The expected number of 1’s on the diagonal (Theorem 5).These calculations are motivated by analogous work for permutation tableaux in [2].In that paper, all three of those statistics were calculated for permutation tableauxas well as a fourth statistic which we were unable to calculate here for type-Bpermutation tableaux (see [2][Theorem 1]).We will also calculate the following statistics on type-B permutation tableaux ofsize n ,(1) The expected number of two adjacent south steps (Theorem 6).(2) The expected number of two adjacent west steps (Theorem 7).These statistics are interesting due to the connection between permutation tableauxand the PASEP. This will be discussed in detail in the next section, but these twostatistics, along with a result from [7][Theorem 6], provide the expected value of allpossible configurations of two adjacent steps.3. The PASEP
An important motivation for studying permutation tableaux is due to the con-nection between certain types of tableaux with the partially asymmetric simpleexclusion process (PASEP). The PASEP is an important particle model in statis-tical mechanics that can be described as Markov chain, a stochastic process thattransitions from one state to another with probabilities dependent only on the pre-vious state.In particular, the PASEP is a Markov Chain on n sites where each site is eitherempty or occupied by one particle. Empty sites are denoted by a ◦ and occupiedsites by a • . Particles enter and exit the system in one direction through the firstand last sites. A particle enters from the left with rate α and exits the system fromthe right with rate β . Within the system, particles hop one site at a time with rate ALTHOFF, DIETHRICH, LOHSS, LOW, AND WICHERT q for a left hop. See Figure 3 below for an example ofparticular state of the PASEP when n = 10. α q β Figure 3.
An example of the PASEP as defined by a Markovchain of size 10.Various types of tableaux have been used to give a combinatorial formula forthe steady state distribution of the PASEP including permutation tableaux butalso tree–like tableaux [8], alternative tableaux [13],and stammering tableaux [9].Each particular tableau is associated with a state of the PASEP which is called thetableau’s type .The type of each permutation and type-B permutation tableau corresponds tothe shape of its southeast border. In particular, each south step corresponds toa filled site and each west step corresponds to an empty site. For permutationtableaux, the first step is ignored since this step is always south. For type-B per-mutation tableaux, the Markov chain is doubled, preserving the site structure.Therefore, type-B permutation tableaux are connected with symmetric states ofthe PASEP. See Figure 4 for an example of the type associated with a particularpermutation tableau and a type-B permutation tableau.1 1 0 10 0 11 ◦ • ◦ ◦ • ◦ • ( i ) 10 00 1 • ◦ ◦ • • ◦ ◦ • ( ii ) Figure 4. (i) A permutation tableau of size 8 and its type, aMarkov chain on 7 sites. (ii) A type-B permutation tableau of size4 and its type, a symmetric Markov chain on 8 sites.In [7], the expected number of corners in permutation and type-B permutationtableaux were computed (see Theorems 5 and 10). The motivation for these calcu-lations was because a corner (south step followed by a west step) corresponds to aposition in the PASEP where a filled site is followed by an empty site. From thenumber of corners, it is immediate to determine the number of inner corners whichis a west step followed by a south step.In this paper, we extend these results to provide calculations for all possibleconfigurations of adjacent sites in the case of type-B permutation tableaux. Wecalculate the number of adjacent south steps in Section 8 which correspond toadjacent filled sites in the PASEP, and in Section 9, we calculate the number ofadjacent west steps which correspond to adjacent empty sites. This connectionwith the PASEP is the motivation for these two calculations as mentioned in the introduction and is also further motivation for calculating the number of rows whichcorresponds to the number of filled sites in the PASEP.4.
Probabilistic approach
In [7], probabilistic techniques were developed in order to compute the numberof corners in type-B permutation tableaux. This approach is similar to the onedeveloped in [2] for permutation tableaux. We will describe this approach here anduse it to obtain our results in the subsequent sections.Let B n denote the set of all type-B permutation tableaux of size n . Note that |B n | = 2 n · n ! as these tableaux are in bijection with signed permutations. Define P n to be the probability distribution defined on B n such that for all T ∈ B n , P n ( T ) = 12 n · n ! , and let E n denote the expected value with respect to this measure.There is a relationship between the measures P n − and P n through an extensionprocedure. Any tableau of size n − n byinserting a new row or a new column. If a new column is inserted, there are severalpossible ways to fill the new column and these possibilities depend on the numberof unrestricted rows.Recall that an unrestricted row does not contain a 0 with a 1 above it andtherefore, this new column could place a 0 or a 1 in that row without contradictingthe definition of type-B permutation tableaux. So as we extend tableaux of size n −
1, the unrestricted rows are where choices can be made to determine the particulartableau of the extension. As mentioned in the preliminaries, we denote this keystatistic by U k where U k = U k ( T ) is the number of unrestricted rows on the k thstep of a fixed tableau T ∈ B n .One tableau in B n − extends to a group of tableaux in B n depending on thechoices made in the extension–whether a row or a column is inserted and if a columnis inserted, whether 0’s or 1’s are inserted in the unrestricted rows. Let F n − denotethe σ -subalgebra on B n that groups together all the tableaux obtained by extendingthe same tableau in B n − (see Figure 4).01 101 1 10 01 1 1 10 00 1 1 00 01 1 1 Figure 5.
A type-B permutation tableau of size 2 with U = 1and the collection of tableaux obtained by extending this tableauto size 3.Using the terminology and notation described above, several key relationshipsbetween B n and B n − were established in [7] such as the following relationship ALTHOFF, DIETHRICH, LOHSS, LOW, AND WICHERT between the measures,(1) E n X n − = 1 n E n − (cid:0) U n − X n − (cid:1) where X n − is a random variable defined on B n − .The conditional distribution of U n was also derived,(2) L ( U n |F n − ) = 1 + Bin( U n − )where Bin( U n − ) is a binomial random variable with U n − trials and 1 / S k denotes a south step and W k denotes a west step at the k thposition. The probability of a south step occurring during the extension procedurewas derived in [7][Equation 14],(3) P ( S k |F k − ) = 12 U k − +1 . The following calculations can be found in the proofs of [7][Proposition 5] and[7][Theorem 7] and will be established here in order to simplify our work whichutilize these results frequently.
Lemma 1.
For any random variable X on B m − and a ∈ R , E m Xa U m = am E m − X ( a + 1) U m − . Proof.
Applying Equation 1 and the law of total expectation respectively, E m Xa U m = 1 m E m − X · U m − a U m = 1 m E m − X · U m − E ( a U m |F m − ) . Now we will apply Equation 2 and use the fact that E a Bin( n ) = (cid:18) a + 12 (cid:19) n to obtainthe desired result, E m ( a U m ) = 1 m E m − X · U m − E ( a U m − ) |F m − )= 1 m E m − X · U m − a (cid:18) a + 12 (cid:19) U m − = am E m − X ( a + 1) U m − . (cid:3) Lemma 2.
For any a ∈ R , E m ( a U m ) = Γ( m + a − m ! · Γ( a − . Proof.
Applying Lemma 1 m − E m ( a U m ) = a ( a + 1) · · · ( a + m − m · · · E ( a + m − U . Since U = 1 for all tableau of size 1, E m ( a U m ) = a ( a + 1) · · · ( a + m − m · · · m + a − m ! · Γ( a − . (cid:3) Expected number of rows
In this section, we will determine the expected number of rows in type-B per-mutation tableaux of size n . Note that we are only considering the number of rowsin the original Ferrers diagram. This number is equivalent to the number of southsteps in the diagram and therefore interesting in terms of the PASEP. Besides,the total number of rows in the shifted Ferrers diagram of a type-B permutationtableaux is simply n . This is because the number of rows added in the shiftedFerrer’s diagram is equal to the number of columns in the original.We will first calculate the probability of a south step at the k th position in atype-B permutation tableaux of size n and then use that result to determine theexpected number of rows. Proposition 1.
For ≤ k ≤ n , P n ( I S k ) = 12 (cid:18) − k − n (cid:19) . Proof.
Since we are working with an indicator random variable, P n ( I S k ) = E n ( I S k ) . Applying Lemma 1 ( n − k ) times, E n ( I S k ) = E n ( I S k U n )= ( n − k )! n · ( n − · · · ( k + 1) E k ( I S k ( n − k + 1) U k ) . Using the law of total expectation to condition on F k − , E n ( I S k ) = ( n − k )! n · ( n − · · · ( k + 1) E k E ( I S k ( n − k + 1) U k | F k − )Since the kth step is south, U k = U k − + 1 and U k − is constant under the condi-tional expectation. Thus, E n ( I S k ) = ( n − k )! n · ( n − · · · ( k + 1) E k E ( I S k ( n − k + 1) U k − +1 | F k − )= ( n − k + 1)! n · ( n − · · · ( k + 1) E k ( n − k + 1) U k − E ( I S k | F k − )= ( n − k + 1)! n · ( n − · · · ( k + 1) E k ( n − k + 1) U k − P ( I S k | F k − )where the last equality follows since we are working with an indicator randomvariable. Applying Equation 3 and Equation 1 respectively, E n ( I S k ) = ( n − k + 1)! n · ( n − · · · ( k + 1) E k ( n − k + 1) U k − U k − +1 = ( n − k + 1)! n · ( n − · · · ( k + 1) k E k − U k − ( n − k + 1) U k − U k − +1 = ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − ( n − k + 1) U k − . (4) ALTHOFF, DIETHRICH, LOHSS, LOW, AND WICHERT
Applying Lemma 2 with m = k − a = n − k + 1, E n ( I S k ) = 12 · ( n − k − n · ( n − · · · k (cid:18) ( n − k − n − k )! (cid:19) = 12 (cid:18) − k + 1 n (cid:19) as desired. (cid:3) Theorem 3.
The expected number of rows in type-B permutation tableaux of size n is ( n + 1) / .Proof. Let R n denote the number of rows in type-B permutation tableaux of size n . Notice that E n R n = E n n X k =1 I S k ! = n X k =1 E n I S k . Applying Proposition 1, n X k =1 E n I S k = n X k =1 (cid:18) − k − n (cid:19) = n − n (cid:18) n ( n + 1)2 (cid:19) + 12= n + 14 . (cid:3) Expected number of unrestricted rows
In this section, we have an immediate result on the expected number of unre-stricted rows in type-B permutation tableaux of size n . An analogous calculationwas performed in [2] for permutation tableaux and each equation used for thatcalculation apply in the same manner to type-B permutation tableaux. Therefore,we omit the proof as there is no difference from that of the proof of [2][Lemma 4]. Theorem 4.
The expected number of unrestricted rows in type-B permutationtableaux of size n is H n , where H n is the n th harmonic number. Expected number of ones on the diagonal
In this section, we will determine the expected number of ones along the diagonalin type-B permutation tableaux of size n . For a fixed tableau of size n , let D n denotethe number of ones on the diagonal, and let G k denote the position of the topmostone on the k th step, numbering only unrestricted rows. For a south step, we leave G k undefined.Notice that when G k = 1, the kth diagonal cell contains a 1 and otherwise, doesnot. Therefore, E n D n = E n n X k =1 I G k =1 = n X k =1 E n I G k =1 . To calculate the expectation on the right–hand side, we will use the following prop-erty of a binomial random variable as was used in [2],(5) E I G =1 a G +Bin( m − G ) = aa + 1 (cid:18) a + 12 (cid:19) m . In this equation, G represents the position of the first success in the binomialrandom variable Bin( m ). Proposition 2.
For ≤ k ≤ n E n I G k =1 = 12 Proof.
Consider E n I G k =1 U n and apply Lemma 1 ( n − k ) times, E n I G k =1 = ( n − k )! n · ( n − · · · ( k + 1) E k ( I G k =1 ( n − k + 1) U k ) . Now apply the law of total probability to condition on F k − , E n I G k =1 = ( n − k )! n · ( n − · · · ( k + 1) E k E ( I G k =1 ( n − k + 1) U k | F k − ) . Since we are working with the indicator random variable I G k = 1, we may assumethere is a 1 on the diagonal at the k th step. This implies that the k th step is W and therefore, a column is created. For type-B permutation tableaux, there are U k − + 1 unrestricted rows in a column created on the k th step (see Section 4for more details). In our case, one of those rows (the first one) is occupied bya 1. Therefore, there are U k − unrestricted rows remaining. Therefore, U k =Bin( U k − ) + 1 in the conditional expectation, E n I G k =1 = ( n − k )! n · ( n − · · · ( k + 1) E k E ( I G k =1 ( n − k + 1) Bin( U k − )+1 | F k − )Applying Equation 5 with m = U k − + 1, E n I G k =1 = ( n − k )! n · ( n − · · · ( k + 1) E k n − k + 1 n − k + 2 (cid:18) n − k + 22 (cid:19) U k − +1 ! . Applying Equation 1, E n I G k =1 = ( n − k + 1)( n − k )!2 n · ( n − · · · ( k + 1) k E k − U k − (cid:18) n − k + 22 (cid:19) U k − ! = ( n − k + 1)( n − k )!2 n · ( n − · · · ( k + 1) k E k − ( n − k + 2) U k − Finally, apply Lemma 2 to the expectation with m = k − a = n − k + 2 toobtain the result, E n I G k =1 = ( n − k + 1)( n − k )!2 n · ( n − · · · ( k + 1) k (cid:18) n !( k − n − k + 1)! (cid:19) = 12 . (cid:3) Theorem 5.
The expected number of ones on the diagonal of a type-B permutationtableau of size n is n/ . Proof.
Applying Proposition 2, E n D n = n X k =1 E n I G k =1 = n X k =1
12 = n . (cid:3) Adjacent south steps
In this section, we will determine the expected number of two adjacent southsteps in type-B permutation tableaux of size n . This statistic corresponds to adja-cent filled sites in the PASEP.First, we will calculate the probability of a south step at both the kth and the(k+1)th position of a type-B permutaiton tableaux of size n and then use thatresult to compute the expected number of adjacent south steps. Proposition 3.
For ≤ k ≤ n , P n ( I S k ,S k − ) = ( n − k + 1) n ( n − Proof.
As in Proposition 1, we will consider the expected value of I S k ,S k − andwork to reduce the measure. The beginning of the proof is almost identical toProposition 1 until we have reduced to the ( k − E n ( I S k I S k − ) = ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − I S k − ( n − k + 1) U k − . Using the law of total expectation to condition on F k − , E n ( I S k I S k − ) = ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − E ( I S k − ( n − k + 1) U k − | F k − )Since the ( k − U k − = U k − + 1 and U k − is constant under theconditional expectation. Thus, E n ( I S k I S k − ) = ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − E ( I S k − ( n − k + 1) U k − +1 | F k − )= ( n − k + 1) · ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − ( n − k + 1) U k − E ( I S k − | F k − )= ( n − k + 1) · ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − ( n − k + 1) U k − P ( I S k − | F k − )where the last equality follows since we are working with an indicator randomvariable. Applying Equation 3 and Equation 1 respectively, E n ( I S k I S k − ) = ( n − k + 1) · ( n − k + 1)! n · ( n − · · · ( k + 1) k · E k − ( n − k + 1) U k − U k − +1 = ( n − k + 1) · ( n − k + 1)! n · ( n − · · · k ( k − · E k − U k − ( n − k + 1) U k − U k − +1 = ( n − k + 1) · ( n − k + 1)! n · ( n − · · · k ( k − · E k − ( n − k + 1) U k − . Applying Lemma 2 with m = k − a = n − k + 1, E n ( I S k I S k − ) = ( n − k + 1) · ( n − k + 1)! n · ( n − · · · k ( k − · (cid:18) ( n − k − n − k )! (cid:19) = ( n − k + 1) n ( n − (cid:3) Theorem 6.
The expected number of two adjacent south steps in type-B permuta-tion tableaux of size n is n − .Proof. Notice that E n n X k =2 I S k ,S k − ! = n X k =2 E n I S k ,S k − . Applying Proposition 3, E n n X k =2 I S k ,S k − ! = n X k =2 ( n − k + 1) n ( n − n ( n − n ( n − − n n ( n − n X k =2 ( k −
1) + 14 n ( n − n X k =2 ( k − = n − n − n − X j =1 j + 14 n ( n − n − X j =1 j = n − n − · n ( n − n ( n − · n ( n − n − n − . (cid:3) Adjacent west steps
In this section, we will determine the expected number of two adjacent west stepsin type-B permutation tableaux of size n . This statistic corresponds to adjacentempty sites in the PASEP.First, we will calculate the probability of a west step at both the kth and the(k+1)th position of a type-B permutation tableaux of size n and then use thatresult to compute the expected number of adjacent west steps. Proposition 4.
For ≤ k ≤ n , P n (cid:0) I W k ,W k − (cid:1) = kn − n + ( n − k + 1) n ( n − . Proof.
Since we are working with an indicator random variable, P n ( I W k ,W k − ) = E n ( I W k I W k − ) . Applying Lemma 1 ( n − k ) times, E n ( I W k I W k − ) = E n ( I W k I W k − U n )= ( n − k )! n · ( n − · · · ( k + 1) E k ( I W k I W k − ( n − k + 1) U k ) . Using the law of total expectation to condition on F k − , E n ( I W k I W k − ) = ( n − k )! n · ( n − · · · ( k + 1) E k E ( I W k I W k − ( n − k + 1) U k | F k − ) . Since I W k − is constant under the conditional expectation, E n ( I W k I W k − ) = ( n − k )! n · ( n − · · · ( k + 1) E k I W k − E ( I W k ( n − k + 1) U k | F k − ) . (6)Now use the complement to rewrite E ( I W k ( n − k + 1) U k | F k − ) = P ( I W k ( n − k + 1) U k | F k − )= P (( n − k + 1) U k | F k − ) − P ( I S k ( n − k + 1) U k | F k − )= E (( n − k + 1) U k | F k − ) − E ( I S k ( n − k + 1) U k | F k − )and combine with Equation 6, E n ( I W k I W k − ) = ( n − k )! n · ( n − · · · ( k + 1) (cid:16) E k I W k − E (( n − k + 1) U k | F k − ) − E k I W k − E ( I S k ( n − k + 1) U k | F k − ) (cid:17) . (7)Now let’s consider each part in the parenthesis of Equation 7 separately. For thefirst conditional expectation on the right-hand side, we will apply Equation 2 anduse the fact that E a Bin( n ) = (cid:18) a + 12 (cid:19) n to obtain E k I W k − E (( n − k + 1) U k | F k − ) = E k I W k − E (( n − k + 1) U k − ) | F k − )= E k I W k − ( n − k + 1) · ( n − k + 2) U k − U k − . Applying Equation 1, E k I W k − E (( n − k + 1) U k | F k − ) = n − k + 1 k E k − U k − I W k − · ( n − k + 2) U k − U k − = n − k + 1 k E k − I W k − · ( n − k + 2) U k − . Next, we apply the law of total probability to condition on F k − and use thefact that I S k − is the complement of I W k − , E k I W k − E (( n − k + 1) U k | F k − ) = n − k + 1 k E k − E (cid:0) I W k − · ( n − k + 2) U k − | F k − (cid:1) = n − k + 1 k E k − " E (cid:0) ( n − k + 2) U k − | F k − (cid:1) − E (cid:0) I S k − ( n − k + 2) U k − | F k − (cid:1) . Next, we will apply Equation 2 and the fact that E a Bin( n ) = (cid:18) a + 12 (cid:19) n to thefirst conditional expectation on the right-hand side. For the second conditionalexpectation, since the ( k − U k − = U k − + 1 and then U k − isconstant under the conditional expectation. All together, we obtain E k I W k − E (( n − k + 1) U k | F k − )= n − k + 1 k E k − " ( n − k + 2) ( n − k + 3) U k − U k − − ( n − k + 2)( n − k + 2) U k − E (cid:0) I S k − | F k − (cid:1) . By Equation 3, E ( I S k − | F k − ) = P ( I S k − | F k − ) = 12 U k − +1 and therefore, E k I W k − E (( n − k + 1) U k | F k − )= n − k + 1 k E k − " ( n − k + 2) ( n − k + 3) U k − U k − − ( n − k + 2)( n − k + 2) U k − U k − +1 . Applying Equation 1 and simplifying, E k I W k − E (( n − k + 1) U k | F k − )= n − k + 1 k ( k − E k − " U k − ( n − k + 2) ( n − k + 3) U k − U k − − U k − ( n − k + 2)( n − k + 2) U k − U k − +1 = ( n − k + 1)( n − k + 2) k ( k − E k − " ( n − k + 3) U k − −
12 ( n − k + 2) U k − . The last step for this part is to apply Lemma 2 to each part of the expected valuewith m = k − a = n − k + 3 and m = k − a = n − k + 2 respectively, E k I W k − E (( n − k + 1) U k | F k − )= ( n − k + 1)( n − k + 2) k ( k − (cid:20)(cid:18) n !( k − n − k + 2)! (cid:19) − (cid:18) ( n − k − n − k + 1)! (cid:19)(cid:21) = 1 k ! (cid:20)(cid:18) n !( n − k )! (cid:19) − (cid:18) ( n − n − k + 2)( n − k )! (cid:19)(cid:21) (8)Now returning to Equation 7, let’s compute the remaining term. E k I W k − E ( I S k ( n − k + 1) U k | F k − ) . Since the k th step is south, U k = U k − + 1 and then U k − is constant under theconditional expectation. Therefore, E k I W k − E ( I S k ( n − k + 1) U k | F k − ) = E k I W k − ( n − k + 1) U k − +1 E ( I S k | F k − ) . By Equation 3, E k I W k − E ( I S k ( n − k + 1) U k | F k − ) = E ( I S k | F k − ) = P ( I S k | F k − ) = 12 U k − +1 and therefore, E k I W k − E ( I S k ( n − k + 1) U k | F k − ) = E k I W k − ( n − k + 1) U k − +1 U k − +1 . Applying Equation 1, E k I W k − E ( I S k ( n − k + 1) U k | F k − ) = 1 k E k − U k − I W k − ( n − k + 1) U k − +1 U k − +1 = n − k + 12 k E k − I W k − ( n − k + 1) U k − Applying the law of total probability and the fact that I S k − is the complementof I W k − , E k I W k − E ( I S k ( n − k + 1) U k | F k − ) = n − k + 12 k E k − E (cid:0) I W k − ( n − k + 1) U k − | F k − (cid:1) = n − k + 12 k E k − (cid:20) E (cid:0) ( n − k + 1) U k − | F k − (cid:1) − E (cid:0) I S k − ( n − k + 1) U k − | F k − (cid:1) (cid:21) . Next, we will apply Equation 2 and the fact that E a Bin( n ) = (cid:18) a + 12 (cid:19) n to thefirst conditional expectation on the right-hand side. For the second conditionalexpectation, since the ( k − U k − = U k − + 1 and then U k − isconstant under the conditional expectation. All together, we obtain E k I W k − E ( I S k ( n − k + 1) U k | F k − )= n − k + 12 k E k − (cid:20) ( n − k + 1) ( n − k + 2) U k − U k − − ( n − k + 1) U k − +1 E (cid:0) I S k − | F k − (cid:1) (cid:21) = ( n − k + 1) k E k − (cid:20) ( n − k + 2) U k − U k − − ( n − k + 1) U k − E (cid:0) I S k − | F k − (cid:1) (cid:21) . By Equation 3, E k I W k − E ( I S k ( n − k + 1) U k | F k − )= ( n − k + 1) k E k − (cid:20) ( n − k + 2) U k − U k − − ( n − k + 1) U k − U k − +1 (cid:21) . Applying Equation 1, E k I W k − E ( I S k ( n − k + 1) U k | F k − )= ( n − k + 1) k ( k − E k − (cid:20) U k − ( n − k + 2) U k − U k − − U k − ( n − k + 1) U k − U k − +1 (cid:21) = ( n − k + 1) k ( k − E k − (cid:18) ( n − k + 2) U k − −
12 ( n − k + 1) U k − (cid:19) . The last step for this part is to apply Lemma 2 to each part of the expectedvalue with m = k − a = n − k + 2 and m = k − a = n − k + 1 respectively, E k I W k − E ( I S k ( n − k + 1) U k | F k − )= ( n − k + 1) k ( k − (cid:20) ( n − k − n − k + 1)! −
12 ( n − k − n − k )! (cid:21) = 12 k ! (cid:20) ( n − n − k + 1)( n − k )! −
12 ( n − n − k + 1) ( n − k )! (cid:21) . (9)Now returning to Equation 7 and plugging in Equation 8 and Equation 9, E n ( I W k I W k − ) = ( n − k )! n · · · ( k + 1) (cid:20) k ! (cid:18) n !( n − k )! −
12 ( n − n − k + 2)( n − k )! (cid:19) − k ! (cid:18) ( n − n − k + 1)( n − k )! −
12 ( n − n − k + 1) ( n − k )! (cid:19) (cid:21) = 1 − ( n − k + 2)2 n − ( n − k + 1)2 n + ( n − k + 1) n ( n − kn − n + ( n − k + 1) n ( n − (cid:3) Theorem 7.
The expected number of two adjacent west steps in type-B permutationtableaux of size n is given by, n − n . Proof.
The expected value can be found by summing the result from Proposition 4, n X k =2 P n (cid:0) I W k ,W k +1 (cid:1) = n X k =2 (cid:18) kn − n + ( n − k + 1) n ( n − (cid:19) = 1 n n − X j =1 ( j + 1) − n − n + 124 n ( n − n − X j =1 j = ( n − n n + n − n − n − n + ( n − n (2 n − n ( n − n − n . (cid:3) Conclusion
In this paper, we calculated the expected value of five statistics on type-B permu-tation tableaux, specifically rows, unrestricted rows, diagonal ones, adjacent southsteps and adjacent west steps. The first three statistics were previously calculatedfor permutation tableaux in [2] and the other two calculations are motivated bythe PASEP. It would be interesting to compute the expected number of superflu-ous ones in a type-B permutation tableaux as this was done in [2] for permutationtableaux but we were unable to do so here for type-B permutation tableaux.
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Isaac Newton institute. , 2007. Department of Computing, Mathematics and Statistics, Messiah University, One Uni-versity Avenue, Mechanicsburg, PA 17055, USA
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