aa r X i v : . [ m a t h . S T ] J un On absolute central moments of Poisson distribution
Pavel S. Ruzankin ∗ Sobolev Institute of Mathematics, Novosibirsk, Russia Novosibirsk State University, Novosibirsk, Russia
Abstract
A recurrence formula for absolute central moments of Poisson distribution is suggested.
Keywords:
Poisson distribution, absolute central moment.
Let X be a Poisson random variable with mean m . In this paper, we study absolute centralmoments E | X − a | r about a for naturals r . Explicit representations for such moments may beuseful in situations where we want to test whether observations in a large sample are indepen-dent Poisson with given means, which is the case, for instance, for some image reconstructiontechniques in emission tomography (e.g., see Ben Bouall`egue et al., 2013 and Hebert, 1990).An explicit representation for the mean deviation was obtained independently by Crow (1958)and Ramasubban (1958): E | X − m | = 2 e − m m ⌊ m ⌋ +1 ⌊ m ⌋ ! , where ⌊·⌋ denotes the floor function.Kendall (1943, relations (5.21) and (5.22) on p. 121) has showed that, for all integers r ≥ E ( X − m ) r = m r − X k =0 (cid:18) r − k (cid:19) E ( X − m ) k , (1) E ( X − m ) r +1 = rm E ( X − m ) r − + m ddm E ( X − m ) r . (2)Denote by F the confluent hypergeometric function of the first kind F ( α, β, z ) = ∞ X n =0 z n n ! n − Y j =0 α + jβ + j , ∗ email: [email protected] Q − j =0 = 1. Katti (1960) has derived the following representation for absolute centralmoments of X about a for all odd r ≥ E | X − a | r = − E ( X − a ) r + e − m m ⌊ a ⌋ +1 ( ⌊ a ⌋ + 1)! G ( r ) (0 , , (3)where G ( β, t ) = exp { t ( ⌊ a ⌋ − a + β + 1) } F ( β + 1 , β + ⌊ a ⌋ + 2 , me t ) ,G ( r ) ( β, t ) is its r -th partial derivative with respect to t , which, for t = 0, can be computed bythe recurrence formula G ( s +1) ( β,
0) = ( ⌊ a ⌋ − a + β + 1) G ( s ) ( β,
0) + m ( β + 1) β + ⌊ a ⌋ + 2 G ( s ) ( β + 1 , s = 1 , , . . . , where G (0) ( β,
0) := G ( β,
0) = F ( β + 1 , β + ⌊ a ⌋ + 2 , m ) . The main goal of this paper is to suggest a simpler recurrence formula for absolute centralmoments of X about a .Define the sign function sign( y ) = ( − y ≤ , y > F ( b ) := P ( X ≤ b ) ,C ( r, a ) := E ( X − a ) r ,D ( r, a, b ) := E ( X − a ) r sign( X − b ) ,B ( r, a, f ) := E ( X − a ) r f ( X ) , where 0 = 1 by definition, f is a real-valued function. Here F ( b ) is the cumulative distributionfunction of X , and C ( r, a ) is the r -th central moment about a . The above definition of the signfunction is not common for y = 0, but is chosen here for the sake of convenience, to provide theequality D (0 , a, b ) = 1 − F ( b ) . (4)We have E | X − a | r = ( C ( r, a ) if r is even, D ( r, a, a ) if r is odd . heorem 1. For all integers r ≥ , reals b ≥ and a , and functions f such that E X r | f ( X ) | < ∞ , the following recurrence relations are valid: C ( r, a ) = ( m − a ) C ( r − , a ) + m r − X k =0 (cid:18) r − k (cid:19) C ( k, a ) (5)= mC ( r − , a − − aC ( r − , a ) , (6) D ( r, a, b ) = ( m − a ) D ( r − , a, b ) + m r − X k =0 (cid:18) r − k (cid:19) D ( k, a, b )+2( ⌊ b ⌋ + 1 − a ) r − e − m m ⌊ b ⌋ +1 ⌊ b ⌋ ! (7)= mD ( r − , a − , b − − aD ( r − , a, b ) , (8) B ( r, a, f ) = ( m − a ) B ( r − , a, f ) + m r − X k =0 (cid:18) r − k (cid:19) B ( k, a, f )+ m r − X k =0 (cid:18) r − k (cid:19) B ( k, a, ∆ f ) , (9) where = 1 by definition, ∆ f ( j ) := f ( j + 1) − f ( j ) , and all the expectations B are finite. Remark 1.
The same approach can be used to obtain similar relations for binomial andhypergeometric distributions.The following statement is a direct consequence of relations (7) and (4).
Corollary 1.E | X − m | = m (1 − F ( m )) + 2 (cid:0) ( m − ⌊ m ⌋ ) + 2 ⌊ m ⌋ + 1 (cid:1) e − m m ⌊ m ⌋ +1 ⌊ m ⌋ ! , E | X − m | = (cid:0) m + m (cid:1) (1 − F ( m )) + 2 (cid:16) ( ⌊ m ⌋ + 1 − m ) +2 m (cid:0) m − ⌊ m ⌋ ) + 7 ⌊ m ⌋ + 7 − m (cid:1)(cid:17) e − m m ⌊ m ⌋ +1 ⌊ m ⌋ ! . Notice that the cumulative distribution function F can be calculated, e.g., by the built-infunction ppois() in R or by the function scipy.stats.poisson.cdf() in Python.3 roof of Theorem 1. First, let us prove (7). By Proposition 1 in Borisov and Ruzankin(2002, p. 1660), we have the equivalences E | ( X − a ) r f ( X ) | < ∞ ⇔ E X r | f ( X ) | < ∞ ⇔ E | ∆ r f ( X ) | < ∞⇔ E X r − | ∆ f ( X ) | < ∞ ⇔ E | ( X − a ) r − ∆ f ( X ) | < ∞ , where the ⇔ sign means “if and only if”, ∆ r means applying the ∆ operator r times. Hence,all the expectations B in (7) are finite. For r ≥
1, we have e m B ( r, a, f ) = ∞ X j =0 ( j − a ) r f ( j ) m j j ! = ∞ X j =0 r X k =0 (cid:18) rk (cid:19) j k ( − a ) r − k f ( j ) m j j != ∞ X j =0 r X k =1 (cid:18) r − k − (cid:19) j k ( − a ) r − k f ( j ) m j j ! + ∞ X j =0 r − X k =0 (cid:18) r − k (cid:19) j k ( − a ) r − k f ( j ) m j j != ∞ X j =0 j ( j − a ) r − f ( j ) m j j ! − ∞ X j =0 a ( j − a ) r − f ( j ) m j j != ∞ X j =0 ( j + 1 − a ) r − f ( j + 1) m j +1 j ! − ae m B ( r − , a, f ) (10)= ∞ X j =0 ( j + 1 − a ) r − ( f ( j ) + ∆ f ( j )) m j +1 j ! − ae m B ( r − , a, f )= m r − X k =0 (cid:18) r − k (cid:19) ∞ X j =0 ( j − a ) k ( f ( j ) + ∆ f ( j )) m j j ! − ae m B ( r − , a, b )= m r − X k =0 (cid:18) r − k (cid:19) e m ( B ( k, a, f ) + B ( k, a, ∆ f )) − ae m B ( r − , a, b ) , which proves (7).Relation (5) follows immediately from (9) with f ( j ) ≡
1. Analogously, (10) implies (6).Let us now prove (7). By (10), for r ≥
1, we have e m D ( r, a, b ) = ∞ X j =0 ( j + 1 − a ) r − sign( j + 1 − b ) m j +1 j ! − ae m D ( r − , a, b ) (11)4hich proves (8). Notice that sign( j + 1 − b ) = sign( j − b ) for j = ⌊ b ⌋ only. Hence, (11) equals= ∞ X j =0 ( j + 1 − a ) r − sign( j − b ) m j +1 j ! + 2( ⌊ b ⌋ + 1 − a ) r − m ⌊ b ⌋ +1 ⌊ b ⌋ ! − ae m D ( r − , a, b )= m r − X k =0 (cid:18) r − k (cid:19) ∞ X j =0 ( j − a ) k sign( j − b ) m j j ! + 2( ⌊ b ⌋ + 1 − a ) r − m ⌊ b ⌋ +1 ⌊ b ⌋ ! − ae m D ( r − , a, b )= m r − X k =0 (cid:18) r − k (cid:19) e m D ( k, a, b ) + 2( ⌊ b ⌋ + 1 − a ) r − m ⌊ b ⌋ +1 ⌊ b ⌋ ! − ae m D ( r − , a, b ) , which proves (7).The theorem is proved. References
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