Maxima of Dirichlet and triangular arrays of gamma variables
aa r X i v : . [ m a t h . P R ] M a r Maxima of Dirichlet and triangular arrays of gamma variables
Arup Bose a , Amites Dasgupta a , Krishanu Maulik a , ∗ a Statistics and Mathematics Unit, Indian Statistical Institute, 203 B.T. Road, Kolkata 700108, India
Abstract
Consider a rowwise independent triangular array of gamma random variables with varying parameters. Under severaldifferent conditions on the shape parameter, we show that the sequence of row-maximums converges weakly afterlinear or power transformation. Depending on the parameter combinations, we obtain both Gumbel and non-Gumbellimits.The weak limits for maximum of the coordinates of certain Dirichlet vectors of increasing dimension are alsoobtained using the gamma representation.
Key words:
Random sequences, Triangular array, Maxima, Limit distribution, Gamma distribution, Dirichlet distribution,Gumbel distribution
Primary 60G70, 60F50, Secondary 60F10
1. Introduction
Suppose { Y n } is a sequence of i.i.d. random variables and M ∗ n = max { Y , . . . , Y n } . Necessary and sufficientconditions for the weak convergence of M ∗ n under linear normalisation are well known. See for example,Fisher and Tippett (1928), Gnedenko (1943), de Haan (1970). In particular, let Y n be i.i.d. standard normalvariables and let G denote the Gumbel distribution G ( x ) = exp( − e − x ) . Then (cf. Leadbetter et al., 1983, Theorem 1.5.3),lim n →∞ P [ M ∗ n ≤ c n x + d n ] = G ( x ) , where c n = 1 √ n and d n = p n − log log n + log(4 π )2 √ n . (1)Now let ( Y n , . . . , Y nn ) be a triangular sequence of random variables and let M n = max { Y n , . . . , Y nn } .The question of convergence of M n has been addressed under a variety of conditions. ∗ Corresponding author.
Email addresses: [email protected] (Arup Bose), [email protected] (Amites Dasgupta), [email protected] (Krishanu Maulik).
URLs: (Arup Bose), (Krishanu Maulik).
Preprint submitted to Elsevier 26 October 2018 or example, let Y in be i.i.d. with Y n = (cid:0) P ≤ j ≤ α n U j − α n µ (cid:1) / ( σα / n ), where U j are i.i.d. with mean µ and standard deviation σ ; α n is a sequence of integers going to ∞ . Assuming that U j has a finite momentgenerating function in an open interval containing the origin and α ( R +1) / ( R +3) n log n → ∞ , (2)for some integer R ≥
0, Anderson et al. (1997) showed thatlim n →∞ P [ M n ≤ c n x + d n ] = G ( x )for c n as in (1) and some suitable sequences d n .Nadarajah and Mitov (2002) considered the maximums of triangular array of binomial, negative binomialand discrete uniform variables. The case of binomial triangular array is discussed with increasing number oftrials m n and fixed probability of success, p .Bose et al. (2007) considered the row-maximum of a triangular array with dependent rows. More precisely,for n -dimensional multinomial random variable with equally likely cells, the maximum of the coordinatesconverges to Gumbel law if number of trials increases fast enough.We consider { Y in } to be a triangular sequence such that, for each n , Y in are i.i.d. random variables havingGamma ( α n , 1) distribution. Also let X n = ( X n , . . . , X nn ) be an n -dimensional Dirichlet distribution withparameters α n , . . . , α n , β n supported on the n -dimensional simplex { x : 0 ≤ P ni =1 x i ≤ } and with densityΓ( nα n + β n )Γ( α n ) n Γ( β n ) n Y i =1 x i ! α n − n X i =1 x i ! β n . We investigate the problem of existence of the weak limits of the maxima M n = max { Y n , Y n , . . . , Y nn } , f M n = max { X n , X n , . . . , X nn } under linear or power transformation.In Section 2, we study the behavior of M n . When nα n → ∞ , the limit of centered and scaled M n isGumbel, see Theorems 2.1–2.5. In particular, if α n / log n →
0, we can take the scaling to be 1. If nα n has apositive, finite limit, in Theorem 2.6, we show that M n itself has a non-Gumbel limit. Under the assumption nα n →
0, the linear transformation of M n does not converge. However, in Theorem 2.7, we show that apower transformation leads to uniform limit.In Section 3, f M n is taken up. When nα n + β n → ∞ and nα n converges to a positive limit, the limitof centered and scaled f M n is still Gumbel. When nα n and nα n + β n both have finite, positive limits, f M n itself converges, but to a non-standard limit, cf. Theorem 3.1. When nα n → nα n /β n converges in[0 , ∞ ], in Theorem 3.2, we show that a power transformation of scaled f M n converges to a mixture of uniformdistribution on (0 ,
1) and a point mass at 1.
2. Maximum of triangular array of Gamma random variables
The centering and scaling depends on the nature of the sequence α n . The first case is similar to Proposition2 of Anderson et al. (1997). Throughout the article, we use D → to denote convergence in distribution. Theorem 2.1
Assume that α n / log n → ∞ . Then r nα n ( M n − α n − b n √ α n ) D → G, where b n is the unique solution, in the region b n ∼ √ n , of log z + 12 log(2 π ) + z √ α n − α n log (cid:18) z √ α n (cid:19) = log n. (3)2bserve that in this case, Y n can be considered to be the “sum” of α n many i.i.d. random variables, eachof which is distributed as unit Exponential random variable. This set up is similar to that of Proposition 2of Anderson et al. (1997) mentioned earlier but we have the added advantage that the random variables aregamma distributed. It may also be noted that the condition α n / log n → ∞ , is the limiting form ( R = ∞ )of (2). Almost verbatim repetition of their argument yields the proof of Theorem 2.1. We omit the detailsbut point out that their Lemma 2 continues to hold if we replace the degree R polynomial in that lemma withthe corresponding power series ( R = ∞ ). Using the moment generating function of the gamma distribution,the j -th coefficient of the power series simplifies to ( − j +1 / ( j + 2), for j ≥
1. This yields the definingequation for b n given in (3) above.Let the centering and scaling required in general be d n and c n respectively and let x n = c n x + d n . (4)Suppose x n is such that P [ Y n > x n ] →
0. Then − log P [ M n ≤ x n ] = − log P [ Y n < x n ] n ∼ nP [ Y n > x n ] . (5)Motivated by the above, and noting that Y n has the Gamma( α n ,
1) distribution, define A n = n Γ( α n ) Z ∞ x n e − u u α n − du = nP [ Y n > x n ] . Integrating by parts, we immediately have A n = B n + ( α n − C n , (6)where B n = n Γ( α n ) e − x n x α n − n (7)and C n = n Γ( α n ) Z ∞ x n e − u u α n − du ≤ A n x n , (8)which provides us with an upper bound for A n : A n ≤ − | α n − | x n B n . (9)For fixed k , if α n > k , we also obtain a lower bound via integration by parts k times repeatedly: A n = B n k − X j =1 j Y i =1 α n − ix n + k Y i =1 ( α n − i ) n Γ( α n ) Z ∞ x n e − u u α n − k − du ≥ B n k − X j =1 (cid:18) α n − kx n (cid:19) j = B n − α n − kx n " − (cid:18) α n − kx n (cid:19) k ≥ B n − α n − kx n " − (cid:18) α n x n (cid:19) k . (10)When α n / log n remains bounded away from both 0 and ∞ , we have α n → ∞ . The proof of the followingTheorem requires careful use of both the bounds (9) and (10). Theorem 2.2
Assume that α n / log n remains bounded away from and ∞ . Then (cid:18) − α n ζ n (cid:19) ( M n − ζ n ) + log (cid:18) − α n ζ n (cid:19) D → G, where ζ n /α n is the unique solution bigger than of z = 1 + log nα n − log √ π + log α n α n + (cid:18) − α n (cid:19) log z. (11)3 roof. We start with the solution of (11). Observe that (log √ π + log α n ) /α n →
0, since α n → ∞ .Further, since log n/α n is bounded away from 0, for all large n , 1 + log n/α n − (log √ π + log α n ) /α n isbounded away from 1. Thus, (11) has unique solution ζ n /α n bigger than 1 for all large n and hence ζ n → ∞ .We have, ζ n /α n >
1, for all large n . If for some subsequence { n k } , ζ n k /α n k →
1, then using (11), we alsohave log n k /α n k →
0, which is a contradiction. If for some subsequence { n k } , ζ n k /α n k → ∞ , then using (11),we have 1 = 1 ζ n k /α n k + log n k /α n k ζ n k /α n k − log √ π + log α n k ( ζ n k /α n k ) · α n k + (cid:18) − α n k (cid:19) log( ζ n k /α n k ) ζ n k /α n k and the right side converges to 0, which is again a contradiction. Thus, ζ n /α n is bounded away from both1 and ∞ .Corresponding to the choice of centering and scaling, we have x n = x − log (cid:16) − α n ζ n (cid:17) − α n ζ n + ζ n . (12)First observe that, since ζ n /α n is bounded away from both 1 and ∞ , we have x n ∼ ζ n . Hence,log x n = log ζ n + x − log (cid:16) − α n ζ n (cid:17) ζ n (cid:16) − α n ζ n (cid:17) + O (cid:18) ζ n (cid:19) . (13)Now, using Stirling’s approximation, (7), (12), (13) and the fact α n − O ( ζ n ) = o ( ζ n ), we have, aftercollecting terms, − log B n = x " − α n ζ n − α n − ζ n − α n ζ n + ζ n − log (cid:16) − α n ζ n (cid:17) − α n ζ n − log n − ( α n −
1) log ζ n + α n − ζ n log (cid:16) − α n ζ n (cid:17) − α n ζ n + log √ π − α n + (cid:18) α n − (cid:19) log α n + o (1)= α n " ζ n α n − − log nα n + log √ π + log α n α n − (cid:18) − α n (cid:19) log ζ n α n + x ζ n (cid:16) − α n ζ n (cid:17) − log (cid:16) − α n ζ n (cid:17) − α n ζ n (cid:20) − α n ζ n + 1 ζ n (cid:21) + o (1) (14)= x (1 + o (1)) − log (cid:18) − α n ζ n (cid:19) + o (1) , where the first term of (14) vanishes since ζ n /α n satisfies (11). Thus, we have (cid:18) − α n ζ n (cid:19) − B n ∼ e − x . (15)Hence, using the upper bound (9) and the facts α n → ∞ and x n ∼ ζ n , we havelim sup A n ≤ lim B n − α n − x n = lim B n − α n ζ n = e − x . (16)Also, since α n /ζ n is bounded away from 1 and x n ∼ ζ n , given any ε >
0, we can fix a positive integer K ,such that ( α n /x n ) K < ε , for all large n . Hence, using the lower bound (10), since for all large n , α n > K and ( α n /x n ) K < ε hold, we havelim inf A n ≥ lim inf B n − α n − Kx n (1 − ε ) = (1 − ε ) e − x . ε > A n ≥ e − x . Combining with (16), we get A n → e − x , which completes the proof. ✷ When nα n → ∞ , but α n = o (log n ), it turns out that c n = 1 and the limiting distribution is G . However,the choices of d n vary according to the specific limiting behavior of α n . In general, we have the followinglemma, which is used repeatedly in the subsequent developments. Lemma 2.1
Suppose c n and d n are such that for all x ∈ R , B n → − log F ( x ) , (17) and ( α n −
1) = o ( x n ) . (18) Then A n → − log F ( x ) and hence M n − d n c n D → F. Proof.
From the upper bound (9), we havelim sup A n ≤ lim n →∞ − | α n − | x n lim n →∞ B n = − log F ( x ) . Thus A n is bounded. Also, since ( α n − /x n →
0, we have, using (8), | α n − | C n ≤ | α n − | x n A n → . Then, (6) gives us lim A n = lim B n = − log F ( x ). ✷ As an illustration, suppose α n = α for all n . It is well-known that, in this case, the limiting distributionis Gumbel with the centering, d n = log n + ( α −
1) log log n − log Γ( α ), c n = 1 and M n − d n D → G . See forexample, Resnick (1987, pp. 72-73). This follows from the above Lemma since x n = x + d n ∼ log n and − log B n = − log n + log Γ( α ) + x n − ( α −
1) log x n = x + ( α −
1) log log nx n → x. We begin with the case where α n → ∞ , but α n = o (log n ). Theorem 2.3
Assume that α n → ∞ , such that α n = o (log n ) . Then M n − log n − ( α n −
1) log log n − ξ n + log Γ( α n ) D → G, where ξ n / log n is the unique positive solution of z = α n − n log (cid:20) − α n log n log α n log n + α n − log log n log n + 12 log α n log n + z (cid:21) . (19) Proof.
First we consider the solution of (19). Define ε n = α n − log log n log n + 12 log α n log n − α n log n log α n log n = α n log n − α n − n log α n log n −
12 log α n log n ∼ − α n log n log α n log n , since α n → ∞ , but α n = o (log n ). Thus we have ε n > ε n →
0. Also m n := log n/ ( α n − →∞ . With these notations, (19) becomes e m n z = 1 + ε n + z. ε n > z n . For positive z n , we have 1 + ε n + z n = e m n z n > m n z n , so that z n < ε n / ( m n − →
0. Hence, ξ n = o (log n ) . (20)Using Stirling’s formula, we write x n = x + d n = x + log n + ( α n −
1) log log n + ξ n − log Γ( α n ) (21)= x + log n + ( α n −
1) log log n + ξ n − log √ π + α n − ( α n − /
2) log α n + o (1)= log n " − α n log n log α n log n + ξ n log n + α n − log log n log n + 12 log α n log n + x − log √ π log n + o (cid:18) n (cid:19) = R n log n " x − log √ πR n log n + o (cid:18) R n log n (cid:19) , (22)where R n = 1 − α n log n log α n log n + α n − log log n log n + 12 log α n log n + ξ n log n → , (23)using (20). Hence x + d n = x n ∼ d n ∼ log n (24)and α n − ∼ α n = o ( x n ), which gives us (18). Since ξ n / log n is a solution of (19), we have, using (23), and α n / log n → ξ n = ( α n −
1) log R n . (25)Also, using (7), (21) and (22), we have − log B n = x + ξ n − ( α n −
1) log x n log n = x + ξ n − ( α n −
1) log R n − ( α n −
1) log " x − log √ πR n log n + o (cid:18) R n log n (cid:19) = x − ( α n −
1) log " x − log √ πR n log n + o (cid:18) R n log n (cid:19) ∼ x − α n − n ( x − log √ π ) → x, which gives us (17) and completes the proof using Lemma 2.1. ✷ Next we consider the cases where α n is bounded above, but nα n → ∞ . Here the centering d n depends onthe limiting behavior of log α n / log n . We separate out two cases, depending on whether the ratio log α n / log n converges to 0, or is bounded away from 0. We consider the former case first, which includes the case α n = α ,discussed earlier. Theorem 2.4
Suppose α n is bounded above, but log α n = o (log n ) . Then M n − log n − ( α n −
1) log log n + log Γ( α n ) D → G. Proof.
In this case, x n = x + d n = x + log n + ( α n −
1) log log n − log Γ( α n ) . (26)If α n →
0, then log Γ( α n ) = − log α n + o (1) = o (log n ). Otherwise, α n is bounded away from both 0 and ∞ .Hence log Γ( α n ) is bounded and is o (log n ). In either case, we have x n ∼ log n. (27)6lso note that d n ∼ x n ∼ log n ∼ log( nα n ) . (28)As α n is bounded, we have | α n − | = o ( x n ), which gives us (18). Using (7) and (26) and the fact α n isbounded and (27), we have − log B n = x − ( α n −
1) log x n log n → x. This shows (17) and completes the proof of the theorem using Lemma 2.1. ✷ Next we consider the case where α n is bounded above and log α n / log n is bounded away from 0. Theorem 2.5
Assume that α n is bounded above, nα n → ∞ and log α n / log n is bounded away from . Then M n − log( nα n ) − ( α n −
1) log log( nα n ) D → G. Proof.
From the given conditions, we have log α n → −∞ and hence α n →
0. Here x n = x + d n = x + log( nα n ) + ( α n −
1) log log( nα n ) . (29)Since nα n → ∞ and α n →
0, we have d n ∼ x n ∼ log( nα n ) → ∞ (30)and thus, α n − o ( x n ), which gives us (18).Also, using (7), (29), (30) and the fact α n →
0, we have − log B n = x + log Γ( α n + 1) − ( α n −
1) log x n log( nα n ) → x, Thus we have (17) and the proof is completed using Lemma 2.1. ✷ Next we consider α n , which goes to 0 at a faster rate. We first look at the case nα n → α ∈ (0 , ∞ ). In thiscase, the maximum M n itself converges to a non-degenerate limiting distribution, which is parametrized by α . This distribution is not one of the three standard classes of the extreme value distributions. Theorem 2.6
Assume nα n → α ∈ (0 , ∞ ) . Then, for all x > , we have P [ M n ≤ x ] → F α ( x ) := exp (cid:18) − α Z ∞ x e − u u du (cid:19) , x ≥ . (31) Proof.
The proof follows immediately from (5). Using the dominated convergence theorem, since P ( Y n >x ) →
0, we have for all x > − log P [ M n ≤ x ] ∼ nP [ Y n > x ] ∼ nα n Z ∞ x e − u u α n − du → α Z ∞ x e − u u du. ✷ When nα n →
0, there does not exist any non-degenerate limit distribution under linear transformations.However, a power transformation gives Uniform (0 ,
1) as the limiting distribution. The idea behind the powerscaling is contained in the following lemma. This is used later in Section 3 as well.
Lemma 2.2
Let V n be Gamma ( δ n , random variables, where δ n → . Then V δ n n D → U , where U is aUniform (0 , random variable. Also, for all k > , we have, E [ V kn ] → / (1 + k ) . Proof.
Observe that for any k >
0, we have E (cid:2) V kδ n n (cid:3) = Γ( δ n (1 + k )) / Γ( δ n ) ∼ / (1 + k ) = E [ U k ]. Then theresult follows easily. ✷ Thus, Y α n n is approximately distributed as Uniform (0 , n -thpower of the maximum of n i.i.d. Uniform (0 ,
1) random variable is again Uniform (0 , M nα n n to converge to Uniform (0 ,
1) distribution.
Theorem 2.7
Assume that nα n → . Then, for all < x < , P [ M nα n n ≤ x ] → x. roof. For any 0 < x <
1, we have P [ M nα n n ≤ x ] = ( α n ) Z x / ( nαn ) e − u u α n − du ) n = ( α n Γ( α n ) Z x /n e − u /αn du ) n = ( x /n Γ( α n + 1) − α n + 1) Z x /n (1 − e − u /αn ) du ) n = x { Γ( α n + 1) } n ( − n Z x /n n (1 − e − u /αn ) x /n du ) n . (32)Since Γ( x ) is continuously differentiable on (0 , ∞ ), we have { Γ( α n + 1) } n = (1 + O ( α n )) n = 1 + O ( nα n ) → . (33)We also have, Z x /n n (1 − e − u /αn ) x /n du ≤ nx /n Z x /n u /α n du = nα n x / ( nα n ) α n + 1 → , since x / ( nα n ) →
0, for all 0 < x <
1. Thus, we have, ( − n Z x /n n (1 − e − u /αn ) x /n du ) n → . The conclusion follows from (32) and (33). ✷
3. Maximum of coordinates of exchangeable Dirichlet vectors with increasing dimension
We now extend the results to the maximum of Dirichlet distributions. The discussion is closely re-lated to the Gamma representation of Dirichlet: Recall X n is an n -dimensional vector having Dirichlet( α n , . . . , α n ; β n ) distribution. Let { Y in : 1 ≤ i ≤ n } be i.i.d. Gamma ( α n , 1) random variables and Z n beanother independent Gamma ( β n , 1) random variable defined on the same probability space. Then X n D = (cid:18) Y n P ni =1 Y in + Z n , . . . , Y nn P ni =1 Y in + Z n (cid:19) . (34)This allows us to obtain the limiting distribution corresponding to each case of i.i.d. Gamma randomvariables. For further calculations, it helps to define T n = ( P ni =1 Y in + Z n ) / ( nα n + β n ). Further, if M n =max { Y n , Y n , . . . , Y nn } , as before, then ( nα n + β n ) f M n D = M n /T n . So for some centering d n and scaling c n ,we shall have ( nα n + β n ) f M n − d n c n D = M n /T n − d n c n = 1 T n (cid:18) M n − d n c n − d n c n ( T n − (cid:19) . Note that, if nα n + β n → ∞ , we have T n P →
1. Thus, we have the following result as a simple application ofSlutsky’s theorem, which we use repeatedly to obtain the results in Dirichlet case.
Proposition 3.1
Assume nα n + β n → ∞ . Further assume that ( M n − d n ) /c n D → F . Then [( nα n + β n ) f M n − d n ] /c n D → F , whenever d n c n ( T n − P → . (35) holds. We now obtain the results for f M n as a corollary to Proposition 3.1 above, whenever nα n → α ∈ (0 , ∞ ]and nα n + β n → ∞ . 8 orollary 3.1 Assume that nα n → α ∈ (0 , ∞ ] and nα n + β n → ∞ . Then ( nα n + β n ) f M n − d n c n D → G. with the centering d n = α n − b n √ α n , if log n = o ( α n ) ,ζ n − (cid:18) − α n ζ n (cid:19) log (cid:18) − α n ζ n (cid:19) , if α n log n is bounded away from both and ∞ , log n + ( α n −
1) log log n − log Γ( α n ) + ξ n , if α n log n → and α n → ∞ , log n + ( α n −
1) log log n − log Γ( α n ) , if log α n = o (log n ) and α n bounded , log( nα n ) + ( α n −
1) log log( nα n ) , if α n is bounded, nα n → ∞ and log α n log n is bounded away from , , if nα n → α ∈ (0 , ∞ ) and β n → ∞ , (36) and the scaling c n = r α n n , if log n = o ( α n ) , (cid:18) − α n ζ n (cid:19) − , if α n log n bounded from both and ∞ , , if α n log n → and nα n → ∞ ,β n nα n + β n , if nα n → α ∈ (0 , ∞ ) and β n → ∞ , (37) where, as in Theorem 2.1, b n is the unique root of (3) in the region b n ∼ √ n , as in Theorem 2.2, ζ n /α n is the root of (11) which is bigger than and, as in Theorem 2.3, ξ n / log n is the unique positive rootof (19) .When nα n → α ∈ (0 , ∞ ) and β n → ∞ , the statement simplifies to β n f M n D → F α . (38) Proof.
We already have ( M n − d n ) /c n converges weakly to the appropriate limit, G or F α from the corre-sponding theorems in Section 2. We only verify (35) one by one by showing V ar (cid:20) d n c n ( T n − (cid:21) = (cid:18) d n c n (cid:19) nα n + β n → . First consider log n = o ( α n ). From Theorem 2.1, we have d n ∼ α n . Hence d n /c n ∼ √ α n log n . Thus, (cid:18) d n c n (cid:19) nα n + β n ∼ α n log nnα n + β n ≤ nn → . Next we consider the cases where α n → ∞ and α n / log n is bounded above. We consider the cases wherethe ratio converges to 0 or stays bounded away from 0 together. In the latter case, we know from Theorem 2.2that α n /ζ n is bounded away from 1, and thus, in (36), we have d n ∼ ζ n = O ( α n ) = O (log n ). Also, in (37), c n stays bounded away from 0 and 1. Hence d n /c n = O (log n ). If α n = o (log n ), then, by (24), we have d n ∼ log n . Thus, in either case, d n /c n = O (log n ). Hence, (cid:18) d n c n (cid:19) nα n + β n ∼ log nnα n + β n ≤ log nn → . Next we consider the cases, where α n is bounded, nα n → ∞ and log α n / log n is bounded above. In thesecases, c n = 1. Using (28) and (30), we have d n ∼ log( nα n ). Then, (cid:18) d n c n (cid:19) nα n + β n ∼ log ( nα n ) nα n + β n ≤ log ( nα n ) nα n → . nα n → α ∈ (0 , ∞ ) and β n → ∞ is trivial since d n = 0. The simplified form (38)follows from Slutsky’s theorem, since nα n + β n ∼ β n in this case. ✷ Next we consider the case, ( nα n + β n ) remains bounded and hence Proposition 3.1 does not hold. Theorem 3.1
Assume that nα n → α ∈ (0 , ∞ ) and β n → β ∈ [0 , ∞ ) . Then f M n D → H, where H is a distribution supported on (0 , ∞ ) with k -th moment given by µ k /γ k where γ k is the k th momentof the Gamma ( α + β, distribution and µ k is the k -th moment of the distribution F α , given by µ k = α Z ∞ x k − exp (cid:18) − x − α Z ∞ x e − u u du (cid:19) dx. Proof.
From the representation (34) f M n D = M n /S n , where S n = P ni =1 Y in + Z n . Further, ( Y n /S n , . . . , Y nn /S n )is independent of S n and hence M n /S n is independent of S n .From Theorem 2.6, we have M n D → F α and S n converges weakly to Gamma ( α + β ) distribution. Furthersince { M n /S n } is bounded, it is tight. Hence for any subsequence n k there is a further subsequence n k ( l ) , suchthat { M n k ( l ) /S n k ( l ) , S n k ( l ) } converges weakly to say ( V, W ) where W has Gamma ( α + β ) distribution. Since M n /S n is independent of S n , V and W are independent. Hence V W has distribution F α . This implies that E ( V W ) k = µ k , that is E ( V k ) = µ k /γ k . Since V has support [0 , ✷ Now suppose nα n →
0. In this case, as for Gamma, no linear transformation of f M n will have a limitingdistribution, however a power transformation will converge. We use Lemma 2.2 to obtain the followingtheorem. Theorem 3.2
Assume that nα n → and nα n /β n → λ ∈ [0 , ∞ ] . Then, (cid:16) σ n f M n (cid:17) nα n D → U λ where U λ is the distribution of B λ U + (1 − B λ ) , U and B λ are independent, P ( B λ = 1) = λ = 1 − P ( B λ =0) , U is uniform (0 , , and σ n = ( β n , when β n → ∞ , , otherwise.When λ = ∞ , we interpret B ∞ as the random variable degenerate at . Proof.
We first consider the case λ = 0. Define S n = P ni =1 Y in + Z n as before.When β n → ∞ , clearly S n / ( nα n + β n ) P → nα n + β n ∼ β n . Thus, S n /β n P →
1, and hence ( S n /β n ) nα n P →
1. If β n is bounded away from 0 and ∞ , S n , which has Gamma ( nα n + β n ) distribution, is a tight sequenceon (0 , ∞ ) and hence S nα n n P → β n →
0, by Lemma 2.2, S nα n + β n n converges to Uniform (0 ,
1) distribution weakly. Hence S nα n n P → λ = 0, S nα n n P →
1. Also, from Theorem 2.7, we have M n D → U . Thus by Slutsky’s theoremand the gamma representation (34), we have the required result when λ = 0.When λ ∈ (0 , ∞ ], the proof is very similar to that of Theorem 3.1. When λ = ∞ , we shall interpret λ/ (1 + λ ) = 1 and 1 / (1 + λ ) = 0. From Theorem 2.7, we have that M nα n n → U . Also, from Lemma 2.2,we have that S nα n + β n n D → U , and hence S nα n n D → U λ/ (1+ λ ) . Then arguing as in Theorem 3.1, { M nα n n /S nα n n } converges weakly and the k -th moment of the limit is given by E [ U k ] E (cid:2) U λk/ (1+ λ ) (cid:3) = 1 + λ + k (1 + k )(1 + λ ) = λ λ + 11 + λ
11 + k , which is the k -th moment of the required limiting distribution. ✷ Acknowledgement.
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