Limit Distribution of Two Skellam Distributions, Conditionally on Their Equality
aa r X i v : . [ m a t h . P R ] F e b Limit Distribution of Two Skellam Distributions,Conditionally on Their Equality ´Elie de Panafieu ∗ , Fran¸cois Durand † Nokia Bell Labs France
February 23, 2021
This note provides a proof of the following proposition.
Proposition 1.
Let A n , B n , C n and D n denote independent random Poisson variables of respective param-eters nτ A , nτ B , nτ C and nτ D . We denote X n = ( A n − B n | A n − B n = C n − D n ) .1. If τ A = τ B = 0 , or τ B = τ C = 0 , or τ C = τ D = 0 , or τ D = τ A = 0 , then the distribution of X n is aDirac measure in 0.2. Otherwise, E ( X n ) = nE + E ′ + o (1) , V ( X n ) = nV + V ′ + o (1) , where E = τ A τ C − τ B τ D p ( τ A + τ D )( τ B + τ C ) ,E ′ = − τ A τ C − τ B τ D τ A + τ D )( τ B + τ C ) ,V = ( τ A τ C + 2 τ A τ B + τ B τ D )( τ A τ C + 2 τ C τ D + τ B τ D )2 (cid:2) ( τ A + τ D )( τ B + τ C ) (cid:3) ,V ′ = − ( τ A τ C + τ B τ D )( τ A τ B + τ C τ D ) + 4 τ A τ B τ C τ D (cid:2) ( τ A + τ D )( τ B + τ C ) (cid:3) . And the distribution of ( X n − nE ) / √ nV is asymptotically Gaussian. Since case 1 is trivial, we only need to prove case 2 and thus assume that τ A + τ B > τ B + τ C > τ C + τ D > τ D + τ A > A n and B n , and between C n and D n , we also assume that τ A τ C ≥ τ B τ D .Intuitively, it means that in X n = ( A n − B n | A n − B n = C n − D n ), the “positive” forces A n and C n arestronger than the “negative” ones, B n and D n . It corresponds to the cases where E , the main term in theasymptotic development of the expectation, will be proved nonnegative. ∗ depanafi[email protected] † [email protected] Generic Case: All Coefficients Are Positive
For this section, we add the assumption that τ A , τ B , τ C and τ D are all positive. This assumption will beremoved in the second section.Let P n ( u ) denote the probability generating function P n ( u ) = X m ∈ Z P ( A n − B n = m | A n − B n = C n − D n ) u m . Introducing the function F n ( u ) = X a,b,c,d ≥ a − b = c − d P ( A n = a ) P ( B n = b ) P ( C n = c ) P ( D n = d ) u a − b , we obtain P n ( u ) = F n ( u ) F n (1) . Expressing the probabilities explicitly, the expression becomes F n ( u ) = e − n ( τ A + τ B + τ C + τ D ) X a,b,c,d ≥ a − b = c − d ( nτ A ) a a ! ( nτ B ) b b ! ( nτ C ) c c ! ( nτ D ) d d ! u a − b . Introducing the variable g = a − b = c − d , we obtain F n ( u ) = e − n ( τ A + τ B + τ C + τ D ) G n ( u ) , where G n ( u ) = X b,d ≥ g ≥ max ( − b, − d ) ( nτ A ) ( b + g ) ( b + g )! ( nτ B ) b b ! ( nτ C ) ( d + g ) ( d + g )! ( nτ D ) d d ! u g . The Stirling approximation is introduced and we define the values x = bn , y = dn , z = gn . Then: G n ( u ) = X b,d ≥ g ≥ max ( − b, − d ) ψ n ( x, y, z ) e − nφ u ( x,y,z ) , where ψ n ( x, y, z ) = ( n ( x + z )) n ( x + z ) e − n ( x + z ) ( n ( x + z ))! ( nx ) nx e − nx ( nx )! ( n ( y + z )) n ( y + z ) e − n ( y + z ) ( n ( y + z ))! ( ny ) ny e − ny ( ny )! ,φ u ( x, y, z ) = x log( x ) − x (1 + log( τ B )) + ( x + z ) log( x + z ) − ( x + z )(1 + log( τ A ))+ y log( y ) − y (1 + log( τ D )) + ( y + z ) log( y + z ) − ( y + z )(1 + log( τ C )) − z log( u ) . The above expression is well defined because we assumed that τ A , τ B , τ C and τ D are all positive.We will soon see that the main contributions to G n ( u ) come from the vicinity of the minimum of φ u . Wewill first compute this main contribution, then prove that the rest of the sum is negligible.The function φ u is convex with a unique minimum. Therefore, there is a small enough vicinity V ofthis minimum such that φ u is larger anywhere outside this vicinity than anywhere inside. Without loss ofgenerality, we assume that V does not contain the origin. Thus, uniformly on V , we have ψ n ( x, y, z ) ∼ ψ ( x, y, z )(2 πn ) where ψ ( x, y, z ) := 1 p x ( x + z ) y ( y + z ) . We apply the following classical lemma (Laplace method) to extract the asymptotics.2 emma 1 (Laplace Method) . Consider a compact set C of R d and the series I n = X k ∈ Z d ≥ k /n ∈C ψ ( k /n ) e − nφ ( k /n ) where ψ, φ are differentiable functions from C to R . Assume furthermore that φ has a unique global minimum θ ∗ which is located in the interior of C , φ is three-times differentiable in a neighborhood of θ ∗ , Hessian φ ( θ ∗ ) > and ψ ( θ ∗ ) = 0 . Then I n ∼ (2 πn ) d/ ψ ( θ ∗ ) e − nφ ( θ ∗ ) p Hessian φ ( θ ∗ ) . Proof.
There are many variants of this classic result. The one-dimensional case ( d = 1) is treated by Masoero(2015). There, the approximation of the sum by a Riemann integral is justified. The same transformationapplies to the multivariate case. The asymptotics of the triple integral is then obtained by a multivariateLaplace method, see e.g. Pemantle and Wilson (2013).Denoting x ∗ , y ∗ , z ∗ the minimal point of φ u , we conclude that the contribution from V to the sum G n ( u )has an asymptotics of the form e − nφ ( x ⋆ ,y ⋆ ,z ⋆ ) multiplied by a polynomial term. Because φ u outside of V islarger than φ u ( x ⋆ , y ⋆ , z ⋆ ) + ǫ for some positive ǫ , we conclude that the contribution of the rest of the sum isexponentially small and therefore negligible in the asymptotics. (Also: outside of V , we use Stirling boundsto bound ψ n ( x, y, z )).In order to get the asymptotics of G n ( u ), what remains to do is to evaluate ψ ( x ∗ , y ∗ , z ∗ ), φ u ( x ∗ , y ∗ , z ∗ )and Hessian φ u ( x ∗ , y ∗ , z ∗ ). The minimal point of φ u is characterized by the system x ∗ ( x ∗ + z ∗ ) = τ A τ B ,y ∗ ( y ∗ + z ∗ ) = τ C τ D , ( x ∗ + z ∗ )( y ∗ + z ∗ ) = τ A τ C u. The solution is given by: x ∗ = τ B √ u vuut τ A √ u + τ D √ u τ C √ u + τ B √ u ,y ∗ = τ D √ u vuut τ C √ u + τ B √ u τ A √ u + τ D √ u ,z ∗ = τ A τ C u − τ B τ D u r(cid:16) τ A √ u + τ D √ u (cid:17) (cid:16) τ C √ u + τ B √ u (cid:17) . We have x ∗ > y ∗ >
0. Moreover, the assumption τ A τ C ≥ τ B τ D ensures that for u close enough to 1, z ∗ > max( − x ∗ , − y ∗ ). As a consequence, ( x ∗ , y ∗ , z ∗ ) is on the interior of the integration zone defining G n ( u ),which validates the approximation by a Riemann integral mentioned above.Simple algebra leads to: ψ ( x ∗ , y ∗ , z ∗ ) = 1 √ τ A τ B τ C τ D ,φ u ( x ∗ , y ∗ , z ∗ ) = − s(cid:18) τ A √ u + τ D √ u (cid:19) (cid:18) τ C √ u + τ B √ u (cid:19) , Hessian φ u ( x ∗ , y ∗ , z ∗ ) = 2 τ A τ B τ C τ D s(cid:18) τ A √ u + τ D √ u (cid:19) (cid:18) τ C √ u + τ B √ u (cid:19) . G n ( u ) ∼ √ πn γ ( u ) − exp (cid:16) n p γ ( u ) (cid:17) , where γ ( u ) = (cid:18) τ A √ u + τ D √ u (cid:19) (cid:18) τ C √ u + τ B √ u (cid:19) . To obtain the convergence in distribution to a Gaussian law, we will apply the
Quasi-powers Theorem ,due to Hwang (1998), which proof is also given by Flajolet and Sedgewick (2009) (Lemma IX.1) (we use aslightly weaker version because we are not interested into the speed of convergence).
Lemma 2 (Quasi-powers) . Assume that the Laplace transform E ( e sX n ) of a sequence of random variables X n is analytic in a neighborhood of , and has an asymptotics of the form E ( e sX n ) ∼ n → + ∞ e β n f ( s )+ g ( s ) , with β n → + ∞ as n → + ∞ , and f ( s ) , g ( s ) analytic on a neighborhood of . Assume also the condition f ′′ (0) = 0 . Under these assumptions, the mean and variance of X n satisfy E ( X n ) = β n f ′ (0) + g ′ (0) + o (1) , V ( X n ) = β n f ′′ (0) + g ′′ (0) + o (1) , and the distribution of ( X n − β n f ′ (0)) / p β n f ′′ (0) is asymptotically Gaussian. We apply Lemma 2 to X n = ( A n − B n | A n − B n = C n − D n ). Using the asymptotics of G n , we have: E ( e sX n ) = F n ( e s ) F n (1) ∼ n → + ∞ exp (cid:20) n (cid:16)p γ ( e s ) − p γ (1) (cid:17) −
14 (log( γ ( e s )) − log( γ (1))) (cid:21) . The result of Proposition 1 is then obtained by application of Lemma 2 with β n = n,f ( s ) = 2 (cid:16)p γ ( e s ) − p γ (1) (cid:17) ,g ( s ) = −
14 (log( γ ( e s )) − log( γ (1))) . The assumptions τ A + τ B > τ B + τ C > τ C + τ D > τ D + τ A > f ′′ (0) = ( τ A τ C + 2 τ A τ B + τ B τ D )( τ A τ C + 2 τ C τ D + τ B τ D )2 (cid:2) ( τ A + τ D )( τ B + τ C ) (cid:3) is positive. We now consider the case where one or several coefficients τ vanish. Considering our assumptions τ A + τ B > τ B + τ C > τ C + τ D > τ D + τ A > τ A τ C ≥ τ B τ D , there are only two cases, up tosymmetries: • τ B = 0 and the other coefficients are positive, • τ B = τ D = 0 and the other coefficients are positive.In both cases, the proof is based on the same principle as in the first section. The main difference is thatthe triple sum is replaced by a double sum in the first case, and by a simple sum in the second case.4 .1 τ B = 0 and the other coefficients are positive The probability generating function becomes P n ( u ) = X m ∈ Z P ( A n = m | A n = C n − D n ) u m . Introducing the function F n ( u ) = X a,c,d ≥ a = c − d P ( A n = a ) P ( C n = c ) P ( D n = d ) u a , we obtain P n ( u ) = F n ( u ) F n (1) . Expressing the probabilities explicitly, the expression becomes F n ( u ) = e − n ( τ A + τ C + τ D ) X a,c,d ≥ a = c − d ( nτ A ) a a ! ( nτ C ) c c ! ( nτ D ) d d ! u a and we obtain F n ( u ) = e − n ( τ A + τ C + τ D ) G n ( u ) , where G n ( u ) = X a,d ≥ ( nτ A ) a a ! ( nτ C ) a + d ( a + d )! ( nτ D ) d d ! u a . The Stirling approximation is introduced and we define the values x = an and y = dn . Then: G n ( u ) = X a,d ≥ ψ n ( x, y ) e − nφ u ( x,y ) , where ψ n ( x, y ) = ( nx ) nx e − nx ( nx )! ( n ( x + y )) n ( x + y ) e − n ( x + y ) ( n ( x + y ))! ( ny ) ny e − ny ( ny )! ,φ u ( x, y ) = x log( x ) − x (1 + log( τ A )) + ( x + y ) log( x + y ) − ( x + y )(1 + log( τ C ))+ y log( y ) − y (1 + log( τ D )) − x log( u ) . The minimum of φ u is obtained for: x ∗ = τ A τ C u √ τ A τ C u + τ C τ D ,y ∗ = τ C τ D √ τ A τ C u + τ C τ D . This leads to: ψ n ( x ∗ , y ∗ ) ∼ πn ) ( τ A τ C u + τ C τ D ) √ τ A τ C τ D u ,φ u ( x ∗ , y ∗ ) = − √ τ A τ C u + τ C τ D , Hessian φ u ( x ∗ , y ∗ ) = 2 τ A τ C u + τ C τ D τ A τ C τ D u . G n ( u ) ∼ √ πn γ ( u ) − exp (cid:16) n p γ ( u ) (cid:17) , where γ ( u ) has the same expression as in Section 1, applied to the particular case τ B = 0. From this point,the end of the proof is the same as in Section 1. τ B = τ D = 0 and the other coefficients are positive In this case, we have F n ( u ) = e − n ( τ A + τ C ) X a,c ≥ a = c ( nτ A ) a a ! ( nτ C ) c c ! u a , which leads to a simple sum (instead of a double or triple sum): G n ( u ) = X a ≥ ( nτ A ) a a ! ( nτ C ) a a ! u a . As usual, we define the value x = an and obtain G n ( u ) = X a ≥ ψ n ( x ) e − nφ u ( x ) , where ψ n ( x, y ) = (cid:18) ( nx ) nx e − nx ( nx )! (cid:19) φ u ( x, y ) = x (cid:0) x ) − − log( τ A τ C u ) (cid:1) . The minimum of φ u is obtained for x ∗ = √ τ A τ C u , which leads to: ψ n ( x ∗ ) ∼ πn √ τ A τ C u ,φ u ( x ∗ ) = − √ τ A τ C u, Hessian φ u ( x ∗ ) = 2 √ τ A τ C u . Applying Lemma 1, we obtain G n ( u ) ∼ √ πn γ ( u ) − exp (cid:16) n p γ ( u ) (cid:17) , where γ ( u ) has the same expression as in Section 1, applied to the particular case τ B = τ D = 0. We thenconclude like in Section 1. References
P. Flajolet and R. Sedgewick.
Analytic Combinatorics . Cambridge University Press, 2009.H.-K. Hwang. On convergence rates in the central limit theorems for combinatorial structures.
EuropeanJournal of Combinatorics , 19(3):329–343, 1998. 6. Masoero. A laplace’s method for series and the semiclassical analysis of epidemiological models. arXiv ,2015.R. Pemantle and M. C. Wilson.