A characterization of Exponential Distribution and the Sukhatme-Renyi Decomposition of Exponential Maxima
aa r X i v : . [ m a t h . P R ] D ec A Characterization of Exponential Distribution and theSukhatme-R´enyi Decomposition of ExponentialMaxima
George P. Yanev and Santanu Chakraborty
School of Mathematical and Statistical SciencesThe University of Texas Rio Grande Valley
Abstract
A new characterization of the exponential distribution is established. It isproven that the well-known Sukhatme-Renyi necessary condition is also suf-ficient for exponentiality. A method of proof due to Arnold and Villasenorbased on the Maclaurin series expansion of the density is utilized.
Keywords: characterization, exponential distribution, Sukhatme – Renyidecomposition, maxima, random shifts
1. Introduction and Main Results
A number of characterizations of the exponential distribution are based onthe distributional equation X + T d = Y involving a pair of random variables( X, Y ) and a random translator (shift) variable T , independent of X . Char-acterizations making use of this equation when X , Y , and T are either order statistics or record values were obtained in Wesolowski and Ahsanullah (2004),Castano-Martinez et al. (2012), and Shah et al. (2014) among others. In allstudies so far the translator T was assumed to follow a certain distribution.This restriction is removed in our theorem below.Suppose X , X , . . . , X n is a random sample of size n ≥ X with absolutely continuous cdf F , such that F (0) = 0. Denote the maximumorder statistic by X n : n . Preprint submitted to Journal of L A TEX Templates August 20, 2018 rnold and Villasenor (2013) obtained a series of characterizations of theexponential distribution based on a random sample of size two. In particular,they proved that, under some additional conditions on the cdf F , X + 12 X d = X characterizes the exponential distribution with some positive parameter. Theyalso made conjectures for extensions to larger sample sizes. In Chakrabortyand Yanev (2013) and Yanev and Chakraborty (2013) some of the results fromArnold and Villasenor (2013) were generalized to random samples of size n ≥ assumptions on the cdf F as in the case n = 2, that for a fixed n ≥ X n − n − + 1 n X n d = X n : n (1)characterizes the exponential distribution.The contribution of the present paper is twofold. (i) The characterizationequation (1) is extended to the case of maxima of n and n − s random variablesfor 1 ≤ s ≤ n . (ii) The technique of proof from Arnold and Villasenor (2013) for a random sample of size two is expanded to the case of sample size n ≥ n . The proof of the main result makes use of combinatorial identities,which might be of independent interest. We believe that this technique will beuseful in obtaining other characterization results in the future. Theorem.
Let X be a non-negative random variable with pdf f ( x ). Assume that f ( x ) is complex analytic for every x and f (0) >
0. Let n and s be fixedintegers such that 1 ≤ s ≤ n −
1. If X n − s : n − s + 1 n − s + 1 X n − s +1 + . . . + 1 n X n d = X n : n , (2)then X is exponential with some positive parameter.It is well-known (cf. Conway, (1978), p.35) that every complex analytic func-tion is infinitely differentiable and, furthermore, has a power series expansion about each point of its domain.Note that the Theorem has been applied in constructing goodness-of-fit testsfor exponential distribution in Jovanovic et al. (2015) and Volkova (2015).2he following direct corollary of the Theorem is of its own interest. Corollary.
Let X be a non-negative random variable with pdf f ( x ). Assume that f ( x ) is complex analytic for every x and f (0) >
0. If for fixed nX + 12 X + 13 X + . . . + 1 n X n d = X n : n , (3)then X is exponential with some positive parameter.Equation (3) is a particular case (for maxima) of the well-known Sukhatme-R´enyi decomposition (cf. Arnold et al., 2008, p.73) of the k th order statisticin a random sample X , X , . . . , X n from an exponential distribution. It is known (cf. Arnold and Villasenor, 2013) that if (3) holds for every n , thennecessarily X , X , . . . , X n have a common exponential distribution. Under theassumptions of the Corollary, for X , X , . . . , X n to be exponential it is sufficientthat (3) holds for one fixed n only.In the next section we state three lemmas, to be used in the proof of the Theorem. The main steps in the proof of the Theorem are outlined in Section 3.Details of the proof of the Theorem are given in Section 4. Section 5 containsthe proofs of Lemmas 1 and 2. Concluding remarks are given in the last section.
2. Preliminaries
Introduce, for all non-negative integers n and i , and any real number x , H n,i ( x ) := n X j =0 ( − j (cid:18) nj (cid:19) ( x − j ) i . (4)We start with identities involving H n,i ( x ), which may be of independent interest. Lemma 1
Let s and r be positive integers. Then( i ) r − X j =0 (cid:18) rj (cid:19) H s − ,j ( s ) = H s,r ( s + 1) . (5)( ii ) r − X j =0 (cid:18) rj + 1 (cid:19) H s,j ( s + 1) = 1 s + 1 H s +1 ,r ( s + 2) . (6)( iii ) r − X j =0 ( s + 2) r − − j H s,j ( s + 1) = 1 s + 1 H s +1 ,r ( s + 2) . (7)3efine G m ( x ) := F m ( x ) f ( x ) for m ≥ G ( x ) := F ( x ). Assuming (8), we calculate the derivatives of G m ( x ) at 0 for m ≥ Lemma 2
Let m ≥ d be integers, such that d ≥ − m . Assume F (0) = 0. In case d is positive, also assume, f ( k ) (0) = (cid:20) f ′ (0) f (0) (cid:21) k − f ′ (0) k = 1 , . . . , d, (8)then G ( m + d ) m (0) = h f ′ (0) f (0) i d f m +1 (0) H m,m + d ( m + 1) if d ≥ − m ≤ d < . (9)The third lemma, extracted from the proof of Theorem 1 in Arnold and Vil- lasenor (2013), plays a central role in the proof of the Theorem. Lemma 3
Let X be a non-negative random variable with pdf f ( x ). Assumethat f ( x ) is complex analytic for every x and f (0) >
0. If f ( k ) (0) = (cid:20) f ′ (0) f (0) (cid:21) k − f ′ (0) , k = 1 , , . . . , (10)then X is exponential with some positive parameter.Note that the assumptions for analyticity of f ( x ) and f (0) > used in the proof of Lemma 3 given in Arnold and Villasenor (2013).
3. Outline of the Main Steps in the Proof of the Theorem
The proof of the Theorem can be divided into four steps as follows.
Step 1:
Define d j := n − j + 1 and y j := z − x s − P j − k =1 x k for 1 ≤ j ≤ s .Show that the equality in distribution (2) is equivalent to Z z G n − s − ( x s ) Z y . . . Z y s − s − Y j =1 f ( d j x j ) f ( d s y s ) dx s − . . . dx dx s = f ( z ) Z z Z x . . . Z x s − s − Y k =1 f ( x k ) ! G n − s − ( x s ) dx s . . . dx . (11) Step 2:
Denote r j ( t ) := n − s + t + 1 − j − X k =1 i k ≤ j ≤ s, t ≥ , i k are integers. We shall write r j instead of r j ( t ). Also, introduce a i ,...,i s := d sr s − i s − s − Y j =1 d ji j , b i ,...,i s := (cid:18) r s i s + 1 (cid:19) s − Y j =1 (cid:18) r j + s − ji j (cid:19) . Prove (by differentiating (11) ( n + t ) times with respect to z and setting z = 0)that (11) implies r X i =0 · · · r s − X i s − =0 r s − X i s =0 a i ,...,i s s − Y j =1 f ( i j ) (0) f ( r s − i s − (0) G ( i s ) n − s − (0) (12)= r X i =0 · · · r s − X i s − =0 r s − X i s =0 b i ,...,i s s − Y j =1 f ( i j ) (0) f ( r s − i s − (0) G ( i s ) n − s − (0) . Step 3:
Using Lemma 1, prove that r X i =0 · · · r s − X i s − =0 r s − X i s =0 a i ,...,i s H n − s − ,i s ( n − s ) (13)= r X i =0 · · · r s − X i s − =0 r s − X i s =0 b i ,...,i s H n − s − ,i s ( n − s ) . Step 4:
Prove (10) by induction using the results from Step 2 and Step 3.The statement of the Theorem follows by Step 4 and Lemma 3.
4. Proofs of the Steps in Section 3
Let F n ( x ) and f n ( x ) denote the cdf and pdf, respectively, of the maximum X n : n . Obviously, F n ( x ) = F n ( x ). Let f n − ,n ( x ) denote the density of X n − / ( n −
1) + X n /n . Setting s = 2 in(2), for the density f LHS ( z | s = 2), say, of the left-hand side of (2) we find f LHS ( z | s = 2) = Z z f n − ( x ) f n − ,n ( z − x ) dx = Z z ( n − G n − ( x ) n ( n − Z z − x f ( nx ) f (( n − z − x − x )) dx dx = ( n ) Z z G n − ( x ) Z z − x f ( nx ) f (( n − z − x − x )) dx dx , n ) := n ( n − n − s = 3 in (2), we have f LHS ( z | s = 3) = ( n ) Z z G n − ( x ) Z z − x Z z − x − x Z z − x − x − x × f ( nx ) f (( n − x ) f (( n − z − x − x − x )) dx dx dx . Repeating this argument, we obtain for any s such that 2 ≤ s ≤ n − f LHS ( z )( n ) s +1 (14)= Z z G n − s − ( x s ) Z y . . . Z y s − s − Y j =1 f ( d j x j ) f ( d s y s ) dx s − . . . dx dx s . For the density f RHS ( z ), say, of the right-hand side of (2), we have f RHS ( z ) = nf ( z ) F n − ( z )= nf ( z ) Z z f n − ( x ) dx = n ( n − f ( z ) Z z f ( x ) F n − ( x ) dx = n ( n − n − f ( z ) Z z Z x f ( x ) f ( x ) F n − ( x ) dx dx . Repeating this argument ( s −
2) more times we obtain f RHS ( z )( n ) s +1 = f ( z ) Z z Z x . . . Z x s − s − Y k =1 f ( x k ) ! G n − s − ( x s ) dx s . . . dx . (15)Combining (14) and (15) we obtain (11). Define K n,s − ( y ) := Z y . . . Z y s − s − Y j =1 f ( d j x j ) f ( d s y s ) dx s − . . . dx . Observing that K ( i ) n,s − (0) = 0 when i < s − G ( i ) d s +2 (0) = 0 for i < d s +2 ,for the ( n + t )th derivative of the left-hand side of (11) at 0, we obtain ddz n + t (cid:26)Z z G d s +2 ( x s ) K n,s − ( z − x s ) dx s (cid:27) | z =0 (16)6 ddz n + t − (cid:26) G d s +2 ( z ) K n, s − (0) + Z z G d s +2 ( x s ) K ′ n,s − ( z − x s ) dx s (cid:27) | z =0 = ddz n + t − s (cid:26) G d s +2 ( z ) K ( s − n,s − (0) + Z z G d s +2 ( x s ) K ( s ) n,s − ( z − x s ) dx s (cid:27) | z =0 = n − s + t X i = n − s − G ( i ) d s +2 (0) K ( n + t − − i ) n,s − (0) . Using the recursive relation K n,s − ( u ) = Z u f ( nx ) K n − ,s − ( u − x ) dx ≤ s ≤ n − , one can show by induction that the m th derivative of K n,s − ( u ) at 0 for any m ≥ s − n ≥ K ( m ) n,s − (0) = l X i =0 · · · l s − X i s − =0 s − Y j =1 d i j j f ( i j ) (0) d l s s f ( l s ) (0) , (17)where l j = m − s + 1 − P j − l =1 i l for 1 ≤ j ≤ s . We omit the derivation of (17)here. Substituting (17) into (16) and changing the indexes of summation, onecan see that the last sum in (16) equals r X i =0 · · · r s − X i s − =0 s − Y j =1 d i j j f ( i j ) (0) r s − X i s =0 d r s − i s − s f ( r s − i s − (0) G ( i s ) d s +2 (0) . (18)where, as before, r j = n − s + t + 1 − P j − k =0 i k for 1 ≤ j ≤ s . Thus, we haveobtained the left-hand side of (12). We turn to the right-hand side of (12). Denote L ( x | x , . . . , x s ) := Z x . . . Z x s − s − Y k =1 f ( x k ) ! G d s +2 ( x s ) dx s . . . dx . With this notation for the ( n + t )th derivative of f RHS ( z ) / ( n ) s +1 at 0, we find ddz n + t (cid:26) f ( z ) Z z L ( x | x , . . . , x s ) dx (cid:27) | z =0 = n + t X i =0 (cid:18) n + ti (cid:19) f ( i ) (0) ddz n + t − i (cid:26)Z z f ( x ) L ( x | x , . . . , x s ) dx (cid:27) | z =0 = n + t − X i =0 (cid:18) n + ti (cid:19) f ( i ) (0) ddz n + t − − i (cid:26) f ( z ) Z z L ( x | x , . . . , x s ) dx (cid:27) | z =0 r j := n + t − s + 1 − P j − k =1 i k for j = 1 , . . . , s . Repeating the lastargument, it is not difficult to obtain ddz n + t (cid:26) f ( z ) Z z L ( x | x , . . . , x s ) dx (cid:27) | z =0 (19)= r X i =0 · · · r s − X i s − =0 s − Y j =1 (cid:18) r j + s − ji j (cid:19) f ( i j ) (0) ddz r s (cid:26) f ( z ) Z z G d s +2 ( x s ) dx s (cid:27) | z =0 = r X i =0 · · · r s − X i s − =0 s − Y j =1 (cid:18) r j + s − ji j (cid:19) f ( i j ) (0) r s − X i s =0 (cid:18) r s i s + 1 (cid:19) f ( r s − i s − (0) G ( i s ) d s +2 (0) . Combining (18) and (19) we prove Step 2.
We shall simplify the right-hand side of (13), working on the most inner sumfirst and moving to the outer ones later. Applying (6), we see that r s − X i s − =0 (cid:18) r s + 1 i s − (cid:19) r s − X i s =0 (cid:18) r s i s + 1 (cid:19) H n − s − ,i s ( n − s )= 1 n − s r s − X i s − =0 (cid:18) r s + 1 i s − (cid:19) H n − s,r s − − i s ( n − s + 1)= 1( n − s )( n − s + 1) H n − s +1 ,r s − +1 − i s − ( n − s + 2) . Furthermore, since H n − s +1 , ( n − s + 1) = 0, applying (6) again, we have r s − X i s − =0 (cid:18) r s − + 2 i s − (cid:19) H n − s +1 ,r s − +1 − i s − ( n − s + 2)= r s − +1 X i s − =0 (cid:18) r s − + 2 i s − (cid:19) H n − s +1 ,r s − +1 − i s − ( n − s + 2)= 1( n − s + 2) H n − s +2 ,r s − +2 − i s − ( n − s + 3) . Repeating this argument for the rest of the sums on the right-hand side of (13), we find r X i =0 · · · r s − X i s − =0 s − Y j =1 (cid:18) r j + s − ji j (cid:19) r s − X i s =0 (cid:18) r s i s + 1 (cid:19) H n − s − ,i s ( n − s ) (20)= H n − ,n + t ( n )( n − s , d s +2 = n − s −
1. Let us turn to the left-hand side of (13). Recall that H t,i s ( · ) = 0 for 0 ≤ i s ≤ t −
1. Similarly to the arguments in the simplificationof the right-hand side above, applying (7) instead of (6), we obtain r X i =0 · · · r s − X i s − =0 s − Y j =1 d i j j r s − X i s =0 d r s − i s − s H n − s − ,i s ( n − s ) = H n − ,n + t ( n )( n − s . (21)Equations (20) and (21) imply (13), which completes the proof of Step 3. Denote c i ,...,i s := a i ,...,i s − b i ,...,i s . With this notation and taking intoaccount that G ( i s ) n − s − (0) = 0 when i s < n − s −
1, we write (12) as r X i =0 · · · r s − X i s − =0 r s − X i s = n − s − c i ,...,i s s − Y j =1 f ( i j ) (0) f ( r s − i s − (0) G ( i s ) n − s − (0) = 0 . (22)We shall prove (10) by (strong) induction on k . The base case k = 1 is trivial.Assuming (10) for k ≤ t , we shall prove it for k = t + 1, where t stands for any positive integer. First, observe that since the order of the derivative of f ( x )in (12) must be nonnegative, we have r s − i s − ≥
0. Combining this with i s ≥ n − s −
1, we see that s − X k =1 i k ≤ t + 1 . (23)To extract the terms with a factor f ( t +1) (0), we shall split the sum in (22) intotwo as follows X I\I c i ,...,i s s − Y j =1 f ( i j ) (0) f ( r s − i s − (0) G ( i s ) n − s − (0) (24)+ f s − (0) f ( t +1) (0) G ( n − s − n − s − (0) X I c i ,...,i s = 0 , where I = { ( i , . . . , i s ) : 0 ≤ i j ≤ r j , ≤ j ≤ s − , ≤ i s ≤ r s − } and I isthe set of vectors ( i , . . . , i s ) such that i s = n − s − s − t + 1 and the others are zeros.Notice that by Lemma 2 G ( n − s − n − s − (0) = f n − s (0) H n − s − ,n − s − ( n − s ) . (25)9onsider the first sum in (24) (the one over I \ I ). Inequality (23) along with the definition of the index set I \ I implies that all derivatives of f ( x ) includedin the product term have order less than or equal to t . Therefore, applying theinduction assumption to f ( i j ) (0) for i j ≥
1, we have s − Y j =1 f ( i j ) (0) = f (0)[ f ′ (0)] P s − k =1 i k if ( i , . . . , i s − ) = (0 , . . . , i , . . . , i s − ) = (0 , . . . , . (26)It is not difficult to see that over the index set I \ I we have r s − i s − ≤ t andtherefore, applying the induction assumption, we obtain for n − s − ≤ i s ≤ r s − f ( r s − i s − (0) = (cid:20) f ′ (0) f (0) (cid:21) r s − i s − f ′ (0) . (27)It remains to study the factor G ( i s ) n − s − (0). Since i s ≤ r s − ≤ n − s + t − P s − k =1 i k ,we have that i s − ( n − s − ≤ t + 1 − P s − k =1 i k . We consider two cases asfollows. (i) Let P s − k =1 i k ≥
1. Then i s − ( n − s − ≤ t and, under the inductionassumption, applying Lemma 2 with m = n − s − d = i s − ( n − s − ≤ t ,we have G ( i s ) n − s − (0) = (cid:20) f ′ (0) f (0) (cid:21) i s − n + s +1 f n − s (0) H n − s − ,i s ( n − s ) . (28)(ii) Let P s − k =1 i k = 0. If i s ≤ n − s + t −
1, then (28) holds. If i s = n − s + t ,then we see that c ,..., ,n − s + t = 0 . (29)Combining (25)-(29), it is not difficult to obtain that, under the induction as-sumption, (24) implies (cid:20) f ′ (0) f (0) (cid:21) t f ′ (0) X I\I c i ,...,i s H n − s − ,i s ( n − s )+ f ( t +1) (0) X I c i ,...,i s H n − s − ,n − s − ( n − s ) = 0 . Thus, to prove (10) for k = t + 1, it is sufficient to prove X I\I c i ,...,i s H n − s − ,i s ( n − s ) + X I c i ,...,i s H n − s − ,n − s − ( n − s ) = 010r, equivalently, X I a i ,...,i s H n − s − ,i s ( n − s ) = X I b i ,...,i s H n − s − ,i s ( n − s ) . This is equivalent to (13) proven to be true in Step 3. Therefore, the proof ofStep 4 is complete.
5. Proofs of Lemma 1 and Lemma 2
It is known (cf. Ruiz, 1996) that for any non-negative integer n and real x H n,i ( x ) = n ! if i = n ;0 if 0 ≤ i < n. This information will be useful in the proofs of the lemmas that follow.
Proof of Lemma 1. (i) By the definition of H n,i ( x ) in (4), we obtain r − X j =0 (cid:18) rj (cid:19) H s − ,j ( s ) = s − X i =0 ( − i (cid:18) s − i (cid:19) r − X j =0 (cid:18) rj (cid:19) ( s − i ) j = s − X i =0 ( − i (cid:18) s − i (cid:19) [( s + 1 − i ) r − ( s − i ) r ]= ( s + 1) r − (cid:20) s r + (cid:18) s − (cid:19) s r (cid:21) + . . . + ( − s − (cid:20)(cid:18) s − s − (cid:19) s + 2 s (cid:21) + ( − s = ( s + 1) r − (cid:18) s (cid:19) s r + . . . + ( − s − (cid:18) ss − (cid:19) r + ( − s = s X j =0 ( − j (cid:18) sj (cid:19) ( s + 1 − j ) r = H s,r ( s + 1) . (ii) Indeed, using the definition of H s,j ( x ) in (4), we have r − X j =0 (cid:18) rj + 1 (cid:19) H s,j ( s + 1) = r − X j =0 (cid:18) rj + 1 (cid:19) s X i =0 ( − i (cid:18) si (cid:19) ( s + 1 − i ) j = s X i =0 ( − i (cid:18) si (cid:19) r X k =1 (cid:18) rk (cid:19) ( s + 1 − i ) k − = s X i =0 ( − i (cid:18) si (cid:19) s + 1 − i " r X k =0 (cid:18) rk (cid:19) ( s + 1 − i ) k −
11 1 s + 1 s X i =0 ( − i (cid:18) s + 1 i (cid:19) [( s + 2 − i ) r − s + 1 s +1 X i =0 ( − i (cid:18) s + 1 i (cid:19) ( s + 2 − i ) r = 1 s + 1 H s +1 ,r ( s + 2) . (iii) We have r − X j =0 ( s + 2) r − − j H s,j ( s + 1) = r − X j =0 ( s + 2) r − − j s X i =0 ( − i (cid:18) si (cid:19) ( s + 1 − i ) j = s X i =0 ( − i (cid:18) si (cid:19) ( s + 2) r − r − X j =0 (cid:18) s + 1 − is + 2 (cid:19) j = s X i =0 ( − i (cid:18) si (cid:19) i + 1 [( s + 2) r − ( s + 1 − i ) r ]= 1 s + 1 s X i =0 ( − i (cid:18) s + 1 i + 1 (cid:19) [( s + 2) r − ( s + 1 − i ) r ]= 1 s + 1 s +1 X j =0 ( − j (cid:18) s + 1 j (cid:19) ( s + 2 − j ) r = 1 s + 1 H s +1 ,r ( s + 2) . Proof of Lemma 2. (i) If − m ≤ d <
0, then G ( m + d ) m (0) = 0 because all the terms in the expansion of G ( m + d ) m (0) have a factor F (0) = 0.(ii) Let d = 0. We shall prove (9) by induction on m . One can verify directlythe case m = 1. Assuming (9) for m = k , we shall prove it for m = k + 1. Since G k +1 ( x ) = F ( x ) G k ( x ), applying (i), we see that G ( k +1) k +1 (0) = k +1 X j =0 (cid:18) k + 1 j (cid:19) F ( j ) (0) G ( k +1 − j ) k (0)= F (0) G ( k +1) k (0) + ( k + 1) F ′ (0) G ( k ) k (0) + k +1 X j =2 (cid:18) k + 1 j (cid:19) F ( j ) (0) G ( k +1 − j ) k (0)= ( k + 1)! f k +2 (0) , which completes the proof of (ii). (iii) Let d > m be any positive integer. For simplicity, we will write f ( j ) := f ( j ) (0) below. 12a) Let m = 1. If d = 1, then we have G (2)1 (0) = 3 f ′ f = f ′ f H , (2) since H , (2) = 3. Thus, (9) is true for d = 1. Next, assuming (9) for G ( k )1 (0), weshall prove it for G ( k +1)1 (0). Since G ( x ) = F ( x ) f ( x ), using (8) we obtain G ( k +1)1 (0) = k +1 X j =1 (cid:18) k + 1 j (cid:19) f ( j − f ( k +1 − j ) = k +1 X j =1 (cid:18) k + 1 j (cid:19) (cid:18) f ′ f (cid:19) j − f ′ (cid:18) f ′ f (cid:19) k − j f ′ = (cid:18) f ′ f (cid:19) k − ( f ′ ) k +1 X j =1 (cid:18) k + 1 j (cid:19) = (cid:18) f ′ f (cid:19) k − ( f ′ ) (2 k +1 − (cid:18) f ′ f (cid:19) k f H , k (2) . This completes the proof for the case (a) m = 1 and any d > m = 1 , , . . . k and any d > m = k + 1 and any d >
0. Since G k +1 ( x ) = F ( x ) G k ( x ), by (8) and theinduction assumption, we obtain G ( k +1+ d ) k +1 (0) = k +1+ d X j =1 (cid:18) k + 1 + dj (cid:19) f ( j − G ( k +1+ d − j ) k (0)= d +1 X j =1 (cid:18) k + 1 + dj (cid:19) f ( j − G ( k +1+ d − j ) k (0)= d +1 X j =1 (cid:18) k + 1 + dj (cid:19) (cid:18) f ′ f (cid:19) j − f ′ (cid:18) f ′ f (cid:19) d − j f k +1 H k,k +1+ d − j ( m )= (cid:18) f ′ f (cid:19) d f k +2 k +1+ d X j =1 (cid:18) k + 1 + dj (cid:19) H k,k +1+ d − j ( k + 1)= (cid:18) f ′ f (cid:19) d f k +2 k + d X i =0 (cid:18) k + 1 + di (cid:19) H k,i ( k + 1)= (cid:18) f ′ f (cid:19) d f k +2 H k +1 ,k +1+ d ( k + 2) , where the last equality follows from (5) with s = k + 1 and r = k + 1 + d . This proves the induction step (b). Now (iii) follows from (a) and (b).13 . Concluding Remarks We study the distributional equation X + T d = Y , where the shift (translator) T is a sum of i.i.d. random variables without a specified distribution. The mainresult in this paper is a characterization of the exponential distribution via a relationship involving a pair of maxima of i.i.d. continuous random variables.As a corollary, we prove that the Sukhatme-R´enyi decomposition of maxima isalso a characterization property for the exponential distribution.The proof of the main theorem uses a new technique based on an argumentfrom Arnold and Villasenor (2013), which requires analyticity of the density function. It is an open question if this assumption can be weakened. Acknowledgements
We thank the reviewers and the associate editor for their constructive cri-tique and suggestions. The first author was partially supported by the NFSRat the MES of Bulgaria, Grant No DFNI-I02/17 while being on leave from the
Institute of Mathematics and Informatics at the Bulgarian Academy of Sciences.
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