φ(u)du. The following is essentially due to T. T. Cheng (1949, Theorem I): Theorem 1 (T. T. Cheng). We have for any 0 < λ < and k = 0, 1, 2,...

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1 EXPOSITION OF T. T. CHENG S EDGEWORTH APPROXIMATION OF THE POISSON DISTRIBUTION, IMPROVED DRAFT, August 5, by R. M. Dudley For < λ < + and k =,,..., let P(k,λ) := k j= e λ λ j /j! and let x := x(k,λ) := (k + λ)/ λ. For u R let φ(u) := (π) / exp( u /) and Φ(x) := x φ(u)du. The following is essentially due to T. T. Cheng (949, Theorem I): Theorem (T. T. Cheng). We have for any < λ < and k =,,,... that () P(k,λ) = Φ(x) + x 6 λ φ(x) + R the remainder R = R(k,λ) satisfies for all λ > () R δ λ := c λ + c λ 3/ + c 3 λ with c =.75, c =. and c 3 =.. Remarks. Cheng (949) gave the statement with c =.76, c =.43 and c 3 =.3. His proof as it stands gives the slightly smaller values of c and c 3 stated in Theorem. An improvement in the proof here will give the smaller c. Johnson, Kotz and Kemp (99, 4.5, (4.4)) give further terms in the formal Edgeworth expansion, implying that as λ +, ( ) x 5 + 7x 3 3x R(k,λ) = φ(x) + O(λ 3/ ). 7λ Now, x 5 7x 3 + 3x φ(x) is maximized at x. = ±.535. It follows that necessarily in Theorem, c.79. So it s possible that the given c =.75 could be substantially improved. Proof. The proof will be given in detail, following Cheng s proof but filling in further steps. The Poisson characteristic function is, for all t R, e λ (λe it ) j /j! = exp(λ(e it )). j= The series is absolutely and uniformly convergent, so for r =,,,... (3) π e irt e λ(eit ) dt = πe λ λ r /r!. Next, for t not a multiple of π, k e irt = (e i(k+)t )/(e it ) = (e it/ e i(k+ )t )/(i sin(t/)). r=

2 Also, π exp(λ(e it ))e irt dt = exp(λ(e it ))e irt dt. Thus by (3) P(k,λ) = exp(λ(e it )) eit/ e i(k+ )t π i sin(t/) + exp(λ(e it )) ei(k+ )t e it/ dt. i sin(t/) Let z := z(t,k,λ) := e iλsin t (e it/ e i(k+/)t )/(i sin(t/)). Then πp(k,λ) = = I + e λ(cos t ) R(z)dt e λ(cos t ) sin (( k + ) ) t λ sin t /(sin(t/))dt I := [ e λ(cos t ) cos(λ sin t) + cos(t/) ] sin(λ sin t) dt. sin(t/) Claim. I does not depend on λ. To prove the claim let Then G(λ,t) ( t := e λ(cos t ) sin = e λ sin (t/) sin ) + λ sin t / sin(t/) ( ( ) t t + λ sin cos ( t G(λ,t)/ λ = e λ(cos t ) cos(t + λ sin t). )) / sin(t/). Subclaim. For n =,,..., n G(λ,t) λ n λ(cos t ) = e ( ) n ( ) n j cos(jt + λ sin t). j j=

3 3 Proof of subclaim. For n = the statement holds. If it holds for a given n, then n+ G(λ,t) ( λ(cos t ) n [ = e ( ) n j (cost ) cos(jt + λ sin t) λ j) n+ j= ] (sin t) sin(jt + λ sin t) ( n [ ] = e λ(cos t ) ( ) j) n j cos((j + )t + λ sin t) cos(jt + λ sin t) j= n+ [( ) n = e λ(cos t ) (cos(mt + λ sin t)( ) n+ m + m m= which gives the subclaim by the Pascal s triangle identity. ( )] n m Now to continue with the proof of the Claim, G is an entire analytic function of the variables λ and t. (The possible singularities when t is a multiple of π are removable.) Thus, I is an entire analytic function of λ and we can differentiate under the integral sign. We have by the subclaim for all n n G(λ,t) λ n = λ= ( n ( ) n j j j= ) cos(jt), and cos(jt)dt = for all j =,,... Thus by analyticity di /dλ and the claim follows. So I I () = dt = π. It follows that (4) P(k,λ) = I := I(λ,k) (5) I := π Let (6) I := π (( e λ(cos t ) sin k + ) ) t λ sin t /(sin(t/))dt. By definition of x, λx + λ = k + (/). Let (7) J := π J := π exp( λt /) t sin[ λxt + λ(t sin t)]dt. π π exp( λt /)(/t)dt, exp( λ sin (t/)) exp( λt /) sin(t/) t J > because < sin u < u for < u π/. Then for the same reason and because sin u for all u, (8) I I J + J. Now J = J + J 3 J := π dt, exp( λ sin (t/)) cot(t/) exp( λt /)(/t)dt

4 4 and J 3 := exp( λ sin (t/)) cos(t/) dt. π sin(t/) It s easily shown that cotu = u + u 3 O(u3 ) as u, from which J <. In J we can thus replace by lim δ Now < δ δ sin(δ/) δ exp( λ sin (t/)) cot(t/)dt =. We have setting v := sin(t/) that sin(δ/) exp( λv /)dv/v. δ exp( λv /)dv/v < dv/v [ln(δ) ln( sin(δ/))] sin(δ/) as δ. Thus J = exp( λv /)dv/v <, so (9) J < J 3. We have () J < 3π π t 3 exp( λt /π )dt because exp( λt /)(/t) < 6 t3 exp( λt /π ) for all t > π. Next, () J 3 < { ( ( )) ( )} exp( λ sin (t/)) sin (t/) cos(t/) t t + cos sin dt π sin(t/) beause cos u < (sin u) ( ) cos u for < u < π/, as follows from < v( v), < v = cos u <. It will be shown that () J 3 < 8π v exp( λv /)dv + 3π t 3 exp( λt /π )dt. For this, the first summand on the right in (), via the subsitution v = sin(t/), gives the first term on the right in (). For the other terms it will be shown for u := t/ that ( cos u)(sinu) exp( λ(sin u 4u /π ) < u 3 / for < u < π/. We have sinu > u/π on this range by concavity, < sin u < u, and cos u < u / by derivatives, so indeed () holds. By (8), (), (9), and () we get (3) 3π = 64π I I < t 3 exp( λt /π )dt + v 8π exp( λv /)dv u exp( λu/π )du + d 8πλ exp( λv /)dv ( ) = π 64π λ + ( 8πλ e λ ) < For I itself we have by (6) I = π + π π3 + < λ 8πλ 8πλ λ t exp( λt /) sin( λxt) cos(λ(t sin t))dt t exp( λt /) cos( λxt) sin(λ(t sin t))dt

5 5 (4) J 4 := π = J 4 + π + J 5 + π t exp( λt /) sin( λxt)dt t exp( λt /) cos( λxt)λt 3 dt/6 t exp( λt /) sin( λxt)[cos(λ(t sin t)) ]dt, J 5 := π t exp( λt /) cos( λxt)[sin(λ(t sin t)) λt 3 /6]dt. Since sin u for all real u and cosv v / for all real v by derivatives, J 4 π t exp( λt /)λ (t sin t) dt/, and since t sin t t 3 /6 for all t by derivatives, a substitution using the gamma function gives (5) J 4 λ 7π t 5 exp( λt /)dt = 9πλ. For J 5 we have sin t < t for all t >, which applied twice gives sin(λ(t sin t)) λt 3 /6 λ(t sin t t 3 /6). We have moreover sint t + t 3 /6 t 5 / for all t by derivatives. It follows by another substitution and gamma function that (6) J 5 < λ π = λ 4π = λ 3/ π t 4 exp( λt /)dt u 3/ exp( λu/)du 8 <./λ3/. By (4) we have (7) I = J 4 + J 5 + U + U (8) U := π (9) U := λ 6π exp( λt /) sin( λxt)dt/t, exp( λt /)t cos( λxt)dt.

6 6 Setting u := λt gives U = π = 4π = 4π exp( u /) sin(ux)du/u x exp( u /)[e iux e iux ]du/(iu) exp( u /) x x e iut dt du = ) (u it) exp ( t dudt 4π x = x π e t / dt = x e t / dt = Φ(x) 4π π, x and by an integration by parts we get U = 6π e u / [cos(ux) xu sin(ux)]du, λ e u / cos(ux)du = Another integration by parts for the other term in U gives exp( (u /) iux)du = π/ exp( x /). U = x 6 πλ exp( x /). gives the first two terms of the expansion in the theorem. Using We see that U +U + I + = P(k,λ) from (4), we get by (8), (3), (7), (5) and (6), ( R(k,λ) < I I + J 4 + J 5 < 8 + ) 9 which finishes the proof of the theorem. REFERENCES πλ +.. +, λ3/ λ Cheng, Tseng Tung (949). The normal approximation to the Poisson distribution and a proof of a conjecture of Ramanujan. Bull. Amer. Math. Soc. 55, Johnson, N. L., Kotz, S., and Kemp, A. W. (99). Univariate Discrete Distributions, d ed. Wiley, New York.